Integrand size = 33, antiderivative size = 195 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^4 A x+\frac {a^4 (48 A+35 B+28 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 (40 A+35 B+28 C) \tan (c+d x)}{8 d}+\frac {a (5 B+4 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 d}+\frac {(20 A+35 B+28 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{60 d}+\frac {(32 A+35 B+28 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \] Output:
a^4*A*x+1/8*a^4*(48*A+35*B+28*C)*arctanh(sin(d*x+c))/d+1/8*a^4*(40*A+35*B+ 28*C)*tan(d*x+c)/d+1/20*a*(5*B+4*C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/5*C* (a+a*sec(d*x+c))^4*tan(d*x+c)/d+1/60*(20*A+35*B+28*C)*(a^2+a^2*sec(d*x+c)) ^2*tan(d*x+c)/d+1/24*(32*A+35*B+28*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d
Time = 5.62 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.02 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 \left (120 A d x+120 (4 A+B) \coth ^{-1}(\sin (c+d x))+15 (16 A+27 B+28 C) \text {arctanh}(\sin (c+d x))+840 A \tan (c+d x)+960 B \tan (c+d x)+960 C \tan (c+d x)+240 A \sec (c+d x) \tan (c+d x)+405 B \sec (c+d x) \tan (c+d x)+420 C \sec (c+d x) \tan (c+d x)+30 B \sec ^3(c+d x) \tan (c+d x)+120 C \sec ^3(c+d x) \tan (c+d x)+40 A \tan ^3(c+d x)+160 B \tan ^3(c+d x)+320 C \tan ^3(c+d x)+24 C \tan ^5(c+d x)\right )}{120 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x ]
Output:
(a^4*(120*A*d*x + 120*(4*A + B)*ArcCoth[Sin[c + d*x]] + 15*(16*A + 27*B + 28*C)*ArcTanh[Sin[c + d*x]] + 840*A*Tan[c + d*x] + 960*B*Tan[c + d*x] + 96 0*C*Tan[c + d*x] + 240*A*Sec[c + d*x]*Tan[c + d*x] + 405*B*Sec[c + d*x]*Ta n[c + d*x] + 420*C*Sec[c + d*x]*Tan[c + d*x] + 30*B*Sec[c + d*x]^3*Tan[c + d*x] + 120*C*Sec[c + d*x]^3*Tan[c + d*x] + 40*A*Tan[c + d*x]^3 + 160*B*Ta n[c + d*x]^3 + 320*C*Tan[c + d*x]^3 + 24*C*Tan[c + d*x]^5))/(120*d)
Time = 1.23 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.485, Rules used = {3042, 4542, 3042, 4405, 3042, 4405, 27, 3042, 4405, 27, 3042, 4402, 3042, 4254, 24, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (a \sec (c+d x)+a)^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 4542 |
\(\displaystyle \frac {\int (\sec (c+d x) a+a)^4 (5 a A+a (5 B+4 C) \sec (c+d x))dx}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (5 a A+a (5 B+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4405 |
\(\displaystyle \frac {\frac {1}{4} \int (\sec (c+d x) a+a)^3 \left (20 A a^2+(20 A+35 B+28 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (20 A a^2+(20 A+35 B+28 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4405 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int 5 (\sec (c+d x) a+a)^2 \left (12 A a^3+(32 A+35 B+28 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \int (\sec (c+d x) a+a)^2 \left (12 A a^3+(32 A+35 B+28 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (12 A a^3+(32 A+35 B+28 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4405 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {1}{2} \int 3 (\sec (c+d x) a+a) \left (8 A a^4+(40 A+35 B+28 C) \sec (c+d x) a^4\right )dx+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int (\sec (c+d x) a+a) \left (8 A a^4+(40 A+35 B+28 C) \sec (c+d x) a^4\right )dx+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (8 A a^4+(40 A+35 B+28 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )dx+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4402 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (a^5 (40 A+35 B+28 C) \int \sec ^2(c+d x)dx+a^5 (48 A+35 B+28 C) \int \sec (c+d x)dx+8 a^5 A x\right )+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (a^5 (48 A+35 B+28 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a^5 (40 A+35 B+28 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+8 a^5 A x\right )+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (-\frac {a^5 (40 A+35 B+28 C) \int 1d(-\tan (c+d x))}{d}+a^5 (48 A+35 B+28 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a^5 A x\right )+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (a^5 (48 A+35 B+28 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^5 (40 A+35 B+28 C) \tan (c+d x)}{d}+8 a^5 A x\right )+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )+\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {a^2 (5 B+4 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac {1}{4} \left (\frac {5}{3} \left (\frac {3}{2} \left (\frac {a^5 (48 A+35 B+28 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^5 (40 A+35 B+28 C) \tan (c+d x)}{d}+8 a^5 A x\right )+\frac {(32 A+35 B+28 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )+\frac {a^3 (20 A+35 B+28 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )}{5 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}\) |
Input:
Int[(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(C*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(5*d) + ((a^2*(5*B + 4*C)*(a + a*S ec[c + d*x])^3*Tan[c + d*x])/(4*d) + ((a^3*(20*A + 35*B + 28*C)*(a + a*Sec [c + d*x])^2*Tan[c + d*x])/(3*d) + (5*(((32*A + 35*B + 28*C)*(a^5 + a^5*Se c[c + d*x])*Tan[c + d*x])/(2*d) + (3*(8*a^5*A*x + (a^5*(48*A + 35*B + 28*C )*ArcTanh[Sin[c + d*x]])/d + (a^5*(40*A + 35*B + 28*C)*Tan[c + d*x])/d))/2 ))/3)/4)/(5*a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d Int[Csc[e + f*x]^2, x], x ] + Simp[(b*c + a*d) Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f }, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 *m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(b*(m + 1)) I nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 3.80 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.32
method | result | size |
parts | \(a^{4} A x +\frac {\left (4 a^{4} A +B \,a^{4}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (B \,a^{4}+4 a^{4} C \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (a^{4} A +4 B \,a^{4}+6 a^{4} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a^{4} A +6 B \,a^{4}+4 a^{4} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (6 a^{4} A +4 B \,a^{4}+a^{4} C \right ) \tan \left (d x +c \right )}{d}-\frac {a^{4} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(258\) |
parallelrisch | \(-\frac {6 a^{4} \left (\left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {35 B}{48}+\frac {7 C}{12}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {35 B}{48}+\frac {7 C}{12}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {5 d x A \cos \left (3 d x +3 c \right )}{6}-\frac {d x A \cos \left (5 d x +5 c \right )}{6}+\frac {\left (-\frac {31 B}{4}-4 A -11 C \right ) \sin \left (2 d x +2 c \right )}{3}+\frac {\left (-32 A -38 B -\frac {77 C}{2}\right ) \sin \left (3 d x +3 c \right )}{9}+\left (-\frac {7 C}{6}-\frac {2 A}{3}-\frac {9 B}{8}\right ) \sin \left (4 d x +4 c \right )+\frac {\left (-10 B -\frac {83 C}{10}-10 A \right ) \sin \left (5 d x +5 c \right )}{9}-\frac {5 d x A \cos \left (d x +c \right )}{3}-\frac {22 \left (\frac {14 B}{11}+A +\frac {35 C}{22}\right ) \sin \left (d x +c \right )}{9}\right )}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(280\) |
norman | \(\frac {a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-a^{4} A x +5 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {a^{4} \left (40 A +35 B +28 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{4} \left (72 A +93 B +100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{4} \left (272 A +245 B +196 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {4 a^{4} \left (295 A +280 B +224 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a^{4} \left (368 A +395 B +316 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{4} \left (48 A +35 B +28 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{4} \left (48 A +35 B +28 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(318\) |
derivativedivides | \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \tan \left (d x +c \right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \tan \left (d x +c \right )+6 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(406\) |
default | \(\frac {a^{4} A \left (d x +c \right )+B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{4} C \tan \left (d x +c \right )+4 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 B \,a^{4} \tan \left (d x +c \right )+4 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 a^{4} A \tan \left (d x +c \right )+6 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-6 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{4} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(406\) |
risch | \(a^{4} A x -\frac {i a^{4} \left (-720 A \,{\mathrm e}^{8 i \left (d x +c \right )}-1920 C \,{\mathrm e}^{6 i \left (d x +c \right )}-664 C -800 A -800 B -480 B \,{\mathrm e}^{8 i \left (d x +c \right )}-2880 B \,{\mathrm e}^{6 i \left (d x +c \right )}-3120 A \,{\mathrm e}^{6 i \left (d x +c \right )}-4880 A \,{\mathrm e}^{4 i \left (d x +c \right )}-4720 C \,{\mathrm e}^{4 i \left (d x +c \right )}-3280 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3200 C \,{\mathrm e}^{2 i \left (d x +c \right )}+480 A \,{\mathrm e}^{7 i \left (d x +c \right )}-480 A \,{\mathrm e}^{3 i \left (d x +c \right )}-240 A \,{\mathrm e}^{i \left (d x +c \right )}+1320 C \,{\mathrm e}^{7 i \left (d x +c \right )}-1320 C \,{\mathrm e}^{3 i \left (d x +c \right )}-420 C \,{\mathrm e}^{i \left (d x +c \right )}-405 B \,{\mathrm e}^{i \left (d x +c \right )}-120 C \,{\mathrm e}^{8 i \left (d x +c \right )}-3520 B \,{\mathrm e}^{2 i \left (d x +c \right )}-5120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-930 B \,{\mathrm e}^{3 i \left (d x +c \right )}+240 A \,{\mathrm e}^{9 i \left (d x +c \right )}+405 B \,{\mathrm e}^{9 i \left (d x +c \right )}+420 C \,{\mathrm e}^{9 i \left (d x +c \right )}+930 B \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {7 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(460\) |
Input:
int((a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
a^4*A*x+(4*A*a^4+B*a^4)/d*ln(sec(d*x+c)+tan(d*x+c))+(B*a^4+4*C*a^4)/d*(-(- 1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))) -(A*a^4+4*B*a^4+6*C*a^4)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(4*A*a^4+6*B *a^4+4*C*a^4)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+ (6*A*a^4+4*B*a^4+C*a^4)/d*tan(d*x+c)-a^4*C/d*(-8/15-1/5*sec(d*x+c)^4-4/15* sec(d*x+c)^2)*tan(d*x+c)
Time = 0.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.01 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {240 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (48 \, A + 35 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (48 \, A + 35 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (100 \, A + 100 \, B + 83 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 15 \, {\left (16 \, A + 27 \, B + 28 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 20 \, B + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 24 \, C a^{4}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= "fricas")
Output:
1/240*(240*A*a^4*d*x*cos(d*x + c)^5 + 15*(48*A + 35*B + 28*C)*a^4*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(48*A + 35*B + 28*C)*a^4*cos(d*x + c)^5* log(-sin(d*x + c) + 1) + 2*(8*(100*A + 100*B + 83*C)*a^4*cos(d*x + c)^4 + 15*(16*A + 27*B + 28*C)*a^4*cos(d*x + c)^3 + 8*(5*A + 20*B + 34*C)*a^4*cos (d*x + c)^2 + 30*(B + 4*C)*a^4*cos(d*x + c) + 24*C*a^4)*sin(d*x + c))/(d*c os(d*x + c)^5)
\[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{4} \left (\int A\, dx + \int 4 A \sec {\left (c + d x \right )}\, dx + \int 6 A \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec {\left (c + d x \right )}\, dx + \int 4 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 B \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
Output:
a**4*(Integral(A, x) + Integral(4*A*sec(c + d*x), x) + Integral(6*A*sec(c + d*x)**2, x) + Integral(4*A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x) **4, x) + Integral(B*sec(c + d*x), x) + Integral(4*B*sec(c + d*x)**2, x) + Integral(6*B*sec(c + d*x)**3, x) + Integral(4*B*sec(c + d*x)**4, x) + Int egral(B*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**2, x) + Integral(4* C*sec(c + d*x)**3, x) + Integral(6*C*sec(c + d*x)**4, x) + Integral(4*C*se c(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))
Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (183) = 366\).
Time = 0.05 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.47 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 240 \, {\left (d x + c\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{4} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 15 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 960 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 240 \, B a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 1440 \, A a^{4} \tan \left (d x + c\right ) + 960 \, B a^{4} \tan \left (d x + c\right ) + 240 \, C a^{4} \tan \left (d x + c\right )}{240 \, d} \] Input:
integrate((a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= "maxima")
Output:
1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 240*(d*x + c)*A*a^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 16*(3*tan(d*x + c)^5 + 10*ta n(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 15*B*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^ 4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*C*a^4*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2 *sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 240*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 360*B*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 240*C*a^4*(2*sin(d*x + c )/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 960*A*a^4*log(sec(d*x + c) + tan(d*x + c)) + 240*B*a^4*log(sec(d*x + c) + tan(d*x + c)) + 1440*A*a^4*tan(d*x + c) + 960*B*a^4*tan(d*x + c) + 240*C*a ^4*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.81 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {120 \, {\left (d x + c\right )} A a^{4} + 15 \, {\left (48 \, A a^{4} + 35 \, B a^{4} + 28 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (48 \, A a^{4} + 35 \, B a^{4} + 28 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (600 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 525 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2720 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2450 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 1960 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4720 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4480 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3584 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3680 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3950 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3160 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1080 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1395 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1500 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \] Input:
integrate((a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= "giac")
Output:
1/120*(120*(d*x + c)*A*a^4 + 15*(48*A*a^4 + 35*B*a^4 + 28*C*a^4)*log(abs(t an(1/2*d*x + 1/2*c) + 1)) - 15*(48*A*a^4 + 35*B*a^4 + 28*C*a^4)*log(abs(ta n(1/2*d*x + 1/2*c) - 1)) - 2*(600*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 525*B*a^4 *tan(1/2*d*x + 1/2*c)^9 + 420*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 2720*A*a^4*ta n(1/2*d*x + 1/2*c)^7 - 2450*B*a^4*tan(1/2*d*x + 1/2*c)^7 - 1960*C*a^4*tan( 1/2*d*x + 1/2*c)^7 + 4720*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 4480*B*a^4*tan(1/ 2*d*x + 1/2*c)^5 + 3584*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 3680*A*a^4*tan(1/2* d*x + 1/2*c)^3 - 3950*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 3160*C*a^4*tan(1/2*d* x + 1/2*c)^3 + 1080*A*a^4*tan(1/2*d*x + 1/2*c) + 1395*B*a^4*tan(1/2*d*x + 1/2*c) + 1500*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/ d
Time = 13.54 (sec) , antiderivative size = 996, normalized size of antiderivative = 5.11 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
int((a + a/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(30*A*a^4*sin(2*c + 2*d*x) + 80*A*a^4*sin(3*c + 3*d*x) + 15*A*a^4*sin(4*c + 4*d*x) + 25*A*a^4*sin(5*c + 5*d*x) + (465*B*a^4*sin(2*c + 2*d*x))/8 + 95 *B*a^4*sin(3*c + 3*d*x) + (405*B*a^4*sin(4*c + 4*d*x))/16 + 25*B*a^4*sin(5 *c + 5*d*x) + (165*C*a^4*sin(2*c + 2*d*x))/2 + (385*C*a^4*sin(3*c + 3*d*x) )/4 + (105*C*a^4*sin(4*c + 4*d*x))/4 + (83*C*a^4*sin(5*c + 5*d*x))/4 + 55* A*a^4*sin(c + d*x) + 70*B*a^4*sin(c + d*x) + (175*C*a^4*sin(c + d*x))/2 + (75*A*a^4*atan((2368*A^2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d*x)/2) + 784*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 2688*A*C*sin( c/2 + (d*x)/2) + 1960*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(2368*A^ 2 + 1225*B^2 + 784*C^2 + 3360*A*B + 2688*A*C + 1960*B*C)))*cos(3*c + 3*d*x ))/2 + (15*A*a^4*atan((2368*A^2*sin(c/2 + (d*x)/2) + 1225*B^2*sin(c/2 + (d *x)/2) + 784*C^2*sin(c/2 + (d*x)/2) + 3360*A*B*sin(c/2 + (d*x)/2) + 2688*A *C*sin(c/2 + (d*x)/2) + 1960*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*( 2368*A^2 + 1225*B^2 + 784*C^2 + 3360*A*B + 2688*A*C + 1960*B*C)))*cos(5*c + 5*d*x))/2 + 450*A*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d *x)/2)) + (2625*B*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2)))/8 + (525*C*a^4*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x )/2)))/2 + 225*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x) + 45*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(5*c + 5*d*x) + (2625*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(...
Time = 0.17 (sec) , antiderivative size = 751, normalized size of antiderivative = 3.85 \[ \int (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
int((a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**4*( - 720*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 5 25*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b - 420*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c + 1440*cos(c + d*x)*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 1050*cos(c + d*x)*log(tan((c + d* x)/2) - 1)*sin(c + d*x)**2*b + 840*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* sin(c + d*x)**2*c - 720*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a - 525*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*b - 420*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*c + 720*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a + 525*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b + 420*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 1440*cos(c + d*x)*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 1050*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 840*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*c + 720*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a + 525* cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 420*cos(c + d*x)*log(tan((c + d *x)/2) + 1)*c + 120*cos(c + d*x)*sin(c + d*x)**4*a*d*x - 240*cos(c + d*x)* sin(c + d*x)**3*a - 405*cos(c + d*x)*sin(c + d*x)**3*b - 420*cos(c + d*x)* sin(c + d*x)**3*c - 240*cos(c + d*x)*sin(c + d*x)**2*a*d*x + 240*cos(c + d *x)*sin(c + d*x)*a + 435*cos(c + d*x)*sin(c + d*x)*b + 540*cos(c + d*x)*si n(c + d*x)*c + 120*cos(c + d*x)*a*d*x + 800*sin(c + d*x)**5*a + 800*sin(c + d*x)**5*b + 664*sin(c + d*x)**5*c - 1640*sin(c + d*x)**3*a - 1760*sin...