Integrand size = 41, antiderivative size = 194 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {(4 A-7 B+10 C) \text {arctanh}(\sin (c+d x))}{2 a^2 d}+\frac {(5 A-8 B+12 C) \tan (c+d x)}{a^2 d}-\frac {(4 A-7 B+10 C) \sec (c+d x) \tan (c+d x)}{2 a^2 d}-\frac {(4 A-7 B+10 C) \sec ^3(c+d x) \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^4(c+d x) \tan (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {(5 A-8 B+12 C) \tan ^3(c+d x)}{3 a^2 d} \] Output:
-1/2*(4*A-7*B+10*C)*arctanh(sin(d*x+c))/a^2/d+(5*A-8*B+12*C)*tan(d*x+c)/a^ 2/d-1/2*(4*A-7*B+10*C)*sec(d*x+c)*tan(d*x+c)/a^2/d-1/3*(4*A-7*B+10*C)*sec( d*x+c)^3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+C)*sec(d*x+c)^4*tan(d*x+ c)/d/(a+a*sec(d*x+c))^2+1/3*(5*A-8*B+12*C)*tan(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(1069\) vs. \(2(194)=388\).
Time = 7.56 (sec) , antiderivative size = 1069, normalized size of antiderivative = 5.51 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a* Sec[c + d*x])^2,x]
Output:
(4*(4*A - 7*B + 10*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] - Sin[c/ 2 + (d*x)/2]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*C os[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (4*(4*A - 7*B + 10*C)*Cos[c/2 + (d*x)/2]^4*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]* (A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (Cos[c/2 + (d*x)/2]*Sec[c/2] *Sec[c]*Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-48*A*Sin[ (d*x)/2] + 45*B*Sin[(d*x)/2] - 6*C*Sin[(d*x)/2] + 132*A*Sin[(3*d*x)/2] - 2 01*B*Sin[(3*d*x)/2] + 310*C*Sin[(3*d*x)/2] - 120*A*Sin[c - (d*x)/2] + 195* B*Sin[c - (d*x)/2] - 306*C*Sin[c - (d*x)/2] + 48*A*Sin[c + (d*x)/2] - 51*B *Sin[c + (d*x)/2] + 42*C*Sin[c + (d*x)/2] - 120*A*Sin[2*c + (d*x)/2] + 189 *B*Sin[2*c + (d*x)/2] - 270*C*Sin[2*c + (d*x)/2] - 8*A*Sin[c + (3*d*x)/2] - B*Sin[c + (3*d*x)/2] + 50*C*Sin[c + (3*d*x)/2] + 72*A*Sin[2*c + (3*d*x)/ 2] - 81*B*Sin[2*c + (3*d*x)/2] + 90*C*Sin[2*c + (3*d*x)/2] - 68*A*Sin[3*c + (3*d*x)/2] + 119*B*Sin[3*c + (3*d*x)/2] - 170*C*Sin[3*c + (3*d*x)/2] + 8 4*A*Sin[c + (5*d*x)/2] - 129*B*Sin[c + (5*d*x)/2] + 198*C*Sin[c + (5*d*x)/ 2] - 9*B*Sin[2*c + (5*d*x)/2] + 42*C*Sin[2*c + (5*d*x)/2] + 48*A*Sin[3*c + (5*d*x)/2] - 57*B*Sin[3*c + (5*d*x)/2] + 66*C*Sin[3*c + (5*d*x)/2] - 36*A *Sin[4*c + (5*d*x)/2] + 63*B*Sin[4*c + (5*d*x)/2] - 90*C*Sin[4*c + (5*d*x) /2] + 48*A*Sin[2*c + (7*d*x)/2] - 75*B*Sin[2*c + (7*d*x)/2] + 114*C*Sin...
Time = 1.10 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 4572, 25, 3042, 4507, 27, 3042, 4274, 3042, 4254, 2009, 4255, 3042, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int -\frac {\sec ^4(c+d x) (a (A-4 B+4 C)-3 a (A-B+2 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\sec ^4(c+d x) (a (A-4 B+4 C)-3 a (A-B+2 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a (A-4 B+4 C)-3 a (A-B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle -\frac {\frac {\int 3 \sec ^3(c+d x) \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \sec (c+d x)\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {3 \int \sec ^3(c+d x) \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \sec (c+d x)\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a^2 (4 A-7 B+10 C)-a^2 (5 A-8 B+12 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \sec ^3(c+d x)dx-a^2 (5 A-8 B+12 C) \int \sec ^4(c+d x)dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-a^2 (5 A-8 B+12 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {\frac {3 \left (\frac {a^2 (5 A-8 B+12 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}+a^2 (4 A-7 B+10 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a^2 (5 A-8 B+12 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 (5 A-8 B+12 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 (5 A-8 B+12 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {3 \left (a^2 (4 A-7 B+10 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a^2 (5 A-8 B+12 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )}{a^2}+\frac {(4 A-7 B+10 C) \tan (c+d x) \sec ^3(c+d x)}{d (\sec (c+d x)+1)}}{3 a^2}-\frac {(A-B+C) \tan (c+d x) \sec ^4(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
Output:
-1/3*((A - B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) - (((4*A - 7*B + 10*C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(1 + Sec[c + d*x])) + (3*(a^2*(4*A - 7*B + 10*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x] *Tan[c + d*x])/(2*d)) + (a^2*(5*A - 8*B + 12*C)*(-Tan[c + d*x] - Tan[c + d *x]^3/3))/d))/a^2)/(3*a^2)
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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.68 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.13
| method | result | size |
| parallelrisch | \(\frac {6 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (A -\frac {7 B}{4}+\frac {5 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-6 \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (A -\frac {7 B}{4}+\frac {5 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+7 \left (\left (A -\frac {43 B}{28}+\frac {33 C}{14}\right ) \cos \left (3 d x +3 c \right )+\frac {\left (13 A -19 B +30 C \right ) \cos \left (2 d x +2 c \right )}{7}+\frac {\left (\frac {5 A}{2}-4 B +6 C \right ) \cos \left (4 d x +4 c \right )}{7}+\left (3 A -\frac {117 B}{28}+\frac {95 C}{14}\right ) \cos \left (d x +c \right )+\frac {3 A}{2}-\frac {15 B}{7}+\frac {26 C}{7}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \,a^{2} \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(219\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-6 C +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A -5 B +10 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-10 C +7 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {6 C -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A -5 B +10 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 A -7 B +10 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{2 d \,a^{2}}\) | \(260\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {-6 C +2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 A -5 B +10 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (-10 C +7 B -4 A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {6 C -2 B}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {2 A -5 B +10 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (4 A -7 B +10 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {2 C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}}{2 d \,a^{2}}\) | \(260\) |
| norman | \(\frac {\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 a d}+\frac {\left (5 A -8 B +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a d}-\frac {\left (9 A -13 B +21 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {2 \left (47 A -77 B +115 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {\left (61 A -94 B +143 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}-\frac {\left (77 A -125 B +185 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\left (217 A -349 B +521 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a}+\frac {\left (4 A -7 B +10 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{2} d}-\frac {\left (4 A -7 B +10 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{2} d}\) | \(279\) |
| risch | \(\frac {i \left (12 A \,{\mathrm e}^{8 i \left (d x +c \right )}+170 C \,{\mathrm e}^{6 i \left (d x +c \right )}+48 C +20 A -32 B -21 B \,{\mathrm e}^{8 i \left (d x +c \right )}-119 B \,{\mathrm e}^{6 i \left (d x +c \right )}+68 A \,{\mathrm e}^{6 i \left (d x +c \right )}+120 A \,{\mathrm e}^{4 i \left (d x +c \right )}+306 C \,{\mathrm e}^{4 i \left (d x +c \right )}+84 A \,{\mathrm e}^{2 i \left (d x +c \right )}+198 C \,{\mathrm e}^{2 i \left (d x +c \right )}+36 A \,{\mathrm e}^{7 i \left (d x +c \right )}+120 A \,{\mathrm e}^{5 i \left (d x +c \right )}+132 A \,{\mathrm e}^{3 i \left (d x +c \right )}+48 A \,{\mathrm e}^{i \left (d x +c \right )}+90 C \,{\mathrm e}^{7 i \left (d x +c \right )}+270 C \,{\mathrm e}^{5 i \left (d x +c \right )}+310 C \,{\mathrm e}^{3 i \left (d x +c \right )}+114 C \,{\mathrm e}^{i \left (d x +c \right )}-75 B \,{\mathrm e}^{i \left (d x +c \right )}+30 C \,{\mathrm e}^{8 i \left (d x +c \right )}-129 B \,{\mathrm e}^{2 i \left (d x +c \right )}-189 B \,{\mathrm e}^{5 i \left (d x +c \right )}-195 B \,{\mathrm e}^{4 i \left (d x +c \right )}-201 B \,{\mathrm e}^{3 i \left (d x +c \right )}-63 B \,{\mathrm e}^{7 i \left (d x +c \right )}\right )}{3 a^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a^{2} d}-\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 a^{2} d}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{2} d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a^{2} d}+\frac {7 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 a^{2} d}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{2} d}\) | \(467\) |
Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x,meth od=_RETURNVERBOSE)
Output:
1/3*(6*(cos(3*d*x+3*c)+3*cos(d*x+c))*(A-7/4*B+5/2*C)*ln(tan(1/2*d*x+1/2*c) -1)-6*(cos(3*d*x+3*c)+3*cos(d*x+c))*(A-7/4*B+5/2*C)*ln(tan(1/2*d*x+1/2*c)+ 1)+7*((A-43/28*B+33/14*C)*cos(3*d*x+3*c)+1/7*(13*A-19*B+30*C)*cos(2*d*x+2* c)+1/7*(5/2*A-4*B+6*C)*cos(4*d*x+4*c)+(3*A-117/28*B+95/14*C)*cos(d*x+c)+3/ 2*A-15/7*B+26/7*C)*sec(1/2*d*x+1/2*c)^2*tan(1/2*d*x+1/2*c))/d/a^2/(cos(3*d *x+3*c)+3*cos(d*x+c))
Time = 0.08 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.40 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {3 \, {\left ({\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{5} + 2 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (4 \, A - 7 \, B + 10 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (5 \, A - 8 \, B + 12 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (28 \, A - 43 \, B + 66 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (A - B + 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} + 2 \, a^{2} d \cos \left (d x + c\right )^{4} + a^{2} d \cos \left (d x + c\right )^{3}\right )}} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="fricas")
Output:
-1/12*(3*((4*A - 7*B + 10*C)*cos(d*x + c)^5 + 2*(4*A - 7*B + 10*C)*cos(d*x + c)^4 + (4*A - 7*B + 10*C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 3*((4 *A - 7*B + 10*C)*cos(d*x + c)^5 + 2*(4*A - 7*B + 10*C)*cos(d*x + c)^4 + (4 *A - 7*B + 10*C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(5*A - 8*B + 12*C)*cos(d*x + c)^4 + (28*A - 43*B + 66*C)*cos(d*x + c)^3 + 6*(A - B + 2*C)*cos(d*x + c)^2 + (3*B - 2*C)*cos(d*x + c) + 2*C)*sin(d*x + c))/(a^2*d *cos(d*x + c)^5 + 2*a^2*d*cos(d*x + c)^4 + a^2*d*cos(d*x + c)^3)
\[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{4}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))* *2,x)
Output:
(Integral(A*sec(c + d*x)**4/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + I ntegral(B*sec(c + d*x)**5/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Int egral(C*sec(c + d*x)**6/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (184) = 368\).
Time = 0.05 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.92 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="maxima")
Output:
1/6*(C*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^2 - 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 30*log(sin(d*x + c)/(co s(d*x + c) + 1) + 1)/a^2 + 30*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2 ) - B*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) - 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 - 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d *x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 21*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 21*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) + A*((15*si n(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 12*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 12*log(sin(d*x + c)/(co s(d*x + c) + 1) - 1)/a^2 + 12*sin(d*x + c)/((a^2 - a^2*sin(d*x + c)^2/(cos (d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d
Time = 0.26 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.56 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {3 \, {\left (4 \, A - 7 \, B + 10 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 21 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 27 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2, x, algorithm="giac")
Output:
-1/6*(3*(4*A - 7*B + 10*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 3*(4*A - 7*B + 10*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 - 15*B*tan(1/2*d*x + 1/2*c)^5 + 30*C*tan(1/2*d*x + 1/2*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 + 24*B*tan(1/2*d*x + 1/2*c)^3 - 40*C*tan(1/2*d *x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) - 9*B*tan(1/2*d*x + 1/2*c) + 18*C *tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*a^2) - (A*a^4*tan(1 /2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2 *c)^3 + 15*A*a^4*tan(1/2*d*x + 1/2*c) - 21*B*a^4*tan(1/2*d*x + 1/2*c) + 27 *C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d
Time = 12.98 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.12 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A-3\,B+5\,C}{2\,a^2}+\frac {2\,\left (A-B+C\right )}{a^2}\right )}{d}-\frac {\left (2\,A-5\,B+10\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,B-4\,A-\frac {40\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A-3\,B+6\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-a^2\right )}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A-7\,B+10\,C\right )}{a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B+C\right )}{6\,a^2\,d} \] Input:
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a/cos(c + d*x))^2),x)
Output:
(tan(c/2 + (d*x)/2)*((A - 3*B + 5*C)/(2*a^2) + (2*(A - B + C))/a^2))/d - ( tan(c/2 + (d*x)/2)*(2*A - 3*B + 6*C) + tan(c/2 + (d*x)/2)^5*(2*A - 5*B + 1 0*C) - tan(c/2 + (d*x)/2)^3*(4*A - 8*B + (40*C)/3))/(d*(3*a^2*tan(c/2 + (d *x)/2)^2 - 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 - a^2)) - (atanh(tan(c/2 + (d*x)/2))*(4*A - 7*B + 10*C))/(a^2*d) + (tan(c/2 + (d*x) /2)^3*(A - B + C))/(6*a^2*d)
Time = 0.18 (sec) , antiderivative size = 810, normalized size of antiderivative = 4.18 \[ \int \frac {\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
Output:
(12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**6*a - 21*log(tan((c + d*x) /2) - 1)*tan((c + d*x)/2)**6*b + 30*log(tan((c + d*x)/2) - 1)*tan((c + d*x )/2)**6*c - 36*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 63*log(ta n((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b - 90*log(tan((c + d*x)/2) - 1)*t an((c + d*x)/2)**4*c + 36*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 63*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b + 90*log(tan((c + d*x )/2) - 1)*tan((c + d*x)/2)**2*c - 12*log(tan((c + d*x)/2) - 1)*a + 21*log( tan((c + d*x)/2) - 1)*b - 30*log(tan((c + d*x)/2) - 1)*c - 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*a + 21*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*b - 30*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**6*c + 36*l og(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 63*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*b + 90*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)* *4*c - 36*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*a + 63*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b - 90*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 12*log(tan((c + d*x)/2) + 1)*a - 21*log(tan((c + d*x)/2) + 1)*b + 30*log(tan((c + d*x)/2) + 1)*c + tan((c + d*x)/2)**9*a - tan((c + d*x)/2)**9*b + tan((c + d*x)/2)**9*c + 12*tan((c + d*x)/2)**7*a - 18*tan ((c + d*x)/2)**7*b + 24*tan((c + d*x)/2)**7*c - 54*tan((c + d*x)/2)**5*a + 90*tan((c + d*x)/2)**5*b - 138*tan((c + d*x)/2)**5*c + 68*tan((c + d*x)/2 )**3*a - 110*tan((c + d*x)/2)**3*b + 160*tan((c + d*x)/2)**3*c - 27*tan...