Integrand size = 40, antiderivative size = 202 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {(6 B-13 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}+\frac {8 (9 B-19 C) \tan (c+d x)}{15 a^3 d}-\frac {(6 B-13 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}+\frac {(B-C) \sec ^4(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 B-11 C) \sec ^3(c+d x) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {4 (9 B-19 C) \sec ^2(c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
-1/2*(6*B-13*C)*arctanh(sin(d*x+c))/a^3/d+8/15*(9*B-19*C)*tan(d*x+c)/a^3/d -1/2*(6*B-13*C)*sec(d*x+c)*tan(d*x+c)/a^3/d+1/5*(B-C)*sec(d*x+c)^4*tan(d*x +c)/d/(a+a*sec(d*x+c))^3+1/15*(6*B-11*C)*sec(d*x+c)^3*tan(d*x+c)/a/d/(a+a* sec(d*x+c))^2+4/15*(9*B-19*C)*sec(d*x+c)^2*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c ))
Time = 1.89 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {-\frac {3 (B-C) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (480 \text {arctanh}(\sin (c+d x)) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x)-2 \left (20 \sin \left (\frac {1}{2} (c+d x)\right )+57 \sin \left (\frac {3}{2} (c+d x)\right )+45 \sin \left (\frac {5}{2} (c+d x)\right )+12 \sin \left (\frac {7}{2} (c+d x)\right )\right )\right )}{(1+\sec (c+d x))^3}+5 C \left (42 \text {arctanh}(\sin (c+d x))-\frac {(37+60 \cos (c+d x)+43 \cos (2 (c+d x))+16 \cos (3 (c+d x))) \sec (c+d x) \tan (c+d x)}{(1+\cos (c+d x))^2}\right )}{60 a^3 d} \]
((-3*(B - C)*Cos[(c + d*x)/2]*Sec[c + d*x]^4*(480*ArcTanh[Sin[c + d*x]]*Co s[(c + d*x)/2]^5*Cos[c + d*x] - 2*(20*Sin[(c + d*x)/2] + 57*Sin[(3*(c + d* x))/2] + 45*Sin[(5*(c + d*x))/2] + 12*Sin[(7*(c + d*x))/2])))/(1 + Sec[c + d*x])^3 + 5*C*(42*ArcTanh[Sin[c + d*x]] - ((37 + 60*Cos[c + d*x] + 43*Cos [2*(c + d*x)] + 16*Cos[3*(c + d*x)])*Sec[c + d*x]*Tan[c + d*x])/(1 + Cos[c + d*x])^2))/(60*a^3*d)
Time = 1.46 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4560, 3042, 4507, 3042, 4507, 3042, 4507, 3042, 4274, 3042, 4254, 24, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \frac {\sec ^5(c+d x) (B+C \sec (c+d x))}{(a \sec (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^5 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\int \frac {\sec ^4(c+d x) (4 a (B-C)-a (2 B-7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (4 a (B-C)-a (2 B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \frac {\sec ^3(c+d x) \left (3 a^2 (6 B-11 C)-a^2 (18 B-43 C) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (3 a^2 (6 B-11 C)-a^2 (18 B-43 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\frac {\int \sec ^2(c+d x) \left (8 a^3 (9 B-19 C)-15 a^3 (6 B-13 C) \sec (c+d x)\right )dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (8 a^3 (9 B-19 C)-15 a^3 (6 B-13 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {8 a^3 (9 B-19 C) \int \sec ^2(c+d x)dx-15 a^3 (6 B-13 C) \int \sec ^3(c+d x)dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {8 a^3 (9 B-19 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx-15 a^3 (6 B-13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {-\frac {8 a^3 (9 B-19 C) \int 1d(-\tan (c+d x))}{d}-15 a^3 (6 B-13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 B-19 C) \tan (c+d x)}{d}-15 a^3 (6 B-13 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 B-19 C) \tan (c+d x)}{d}-15 a^3 (6 B-13 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (9 B-19 C) \tan (c+d x)}{d}-15 a^3 (6 B-13 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}+\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {4 a^2 (9 B-19 C) \tan (c+d x) \sec ^2(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {8 a^3 (9 B-19 C) \tan (c+d x)}{d}-15 a^3 (6 B-13 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )}{a^2}}{3 a^2}+\frac {a (6 B-11 C) \tan (c+d x) \sec ^3(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}+\frac {(B-C) \tan (c+d x) \sec ^4(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
((B - C)*Sec[c + d*x]^4*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((a*( 6*B - 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + (( 4*a^2*(9*B - 19*C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((8*a^3*(9*B - 19*C)*Tan[c + d*x])/d - 15*a^3*(6*B - 13*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*a^2)
3.4.48.3.1 Defintions of rubi rules used
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_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.35 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.87
| method | result | size |
| parallelrisch | \(\frac {720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B -\frac {13 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-720 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (B -\frac {13 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+72 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \left (\left (\frac {21 B}{2}-\frac {87 C}{4}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {19 B}{4}-\frac {239 C}{24}\right ) \cos \left (3 d x +3 c \right )+\left (B -\frac {19 C}{9}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {191 B}{12}-\frac {259 C}{8}\right ) \cos \left (d x +c \right )+\frac {19 B}{2}-\frac {677 C}{36}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(175\) |
| derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 C -12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +\left (-26 C +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-14 C +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 C -12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| norman | \(\frac {\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}+\frac {\left (3 B -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 a d}-\frac {25 \left (9 B -19 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 a d}-\frac {\left (25 B -51 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (27 B -59 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{12 a d}+\frac {\left (345 B -721 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (549 B -1165 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{12 a d}-\frac {\left (3123 B -6613 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5} a^{2}}+\frac {\left (6 B -13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}-\frac {\left (6 B -13 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(280\) |
| risch | \(\frac {i \left (90 B \,{\mathrm e}^{8 i \left (d x +c \right )}-195 C \,{\mathrm e}^{8 i \left (d x +c \right )}+450 B \,{\mathrm e}^{7 i \left (d x +c \right )}-975 C \,{\mathrm e}^{7 i \left (d x +c \right )}+1050 B \,{\mathrm e}^{6 i \left (d x +c \right )}-2275 C \,{\mathrm e}^{6 i \left (d x +c \right )}+1650 B \,{\mathrm e}^{5 i \left (d x +c \right )}-3575 C \,{\mathrm e}^{5 i \left (d x +c \right )}+2094 B \,{\mathrm e}^{4 i \left (d x +c \right )}-4329 C \,{\mathrm e}^{4 i \left (d x +c \right )}+1830 B \,{\mathrm e}^{3 i \left (d x +c \right )}-3805 C \,{\mathrm e}^{3 i \left (d x +c \right )}+1278 B \,{\mathrm e}^{2 i \left (d x +c \right )}-2673 C \,{\mathrm e}^{2 i \left (d x +c \right )}+630 B \,{\mathrm e}^{i \left (d x +c \right )}-1325 C \,{\mathrm e}^{i \left (d x +c \right )}+144 B -304 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}+\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}-\frac {13 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a^{3} d}\) | \(324\) |
1/240*(720*(1+cos(2*d*x+2*c))*(B-13/6*C)*ln(tan(1/2*d*x+1/2*c)-1)-720*(1+c os(2*d*x+2*c))*(B-13/6*C)*ln(tan(1/2*d*x+1/2*c)+1)+72*sec(1/2*d*x+1/2*c)^4 *((21/2*B-87/4*C)*cos(2*d*x+2*c)+(19/4*B-239/24*C)*cos(3*d*x+3*c)+(B-19/9* C)*cos(4*d*x+4*c)+(191/12*B-259/8*C)*cos(d*x+c)+19/2*B-677/36*C)*tan(1/2*d *x+1/2*c))/d/a^3/(1+cos(2*d*x+2*c))
Time = 0.27 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.46 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left ({\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (9 \, B - 19 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (114 \, B - 239 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (234 \, B - 479 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 15 \, C\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
-1/60*(15*((6*B - 13*C)*cos(d*x + c)^5 + 3*(6*B - 13*C)*cos(d*x + c)^4 + 3 *(6*B - 13*C)*cos(d*x + c)^3 + (6*B - 13*C)*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 15*((6*B - 13*C)*cos(d*x + c)^5 + 3*(6*B - 13*C)*cos(d*x + c)^4 + 3*(6*B - 13*C)*cos(d*x + c)^3 + (6*B - 13*C)*cos(d*x + c)^2)*log(-sin(d* x + c) + 1) - 2*(16*(9*B - 19*C)*cos(d*x + c)^4 + 3*(114*B - 239*C)*cos(d* x + c)^3 + (234*B - 479*C)*cos(d*x + c)^2 + 15*(2*B - 3*C)*cos(d*x + c) + 15*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3 *d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
\[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {B \sec ^{5}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{6}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
(Integral(B*sec(c + d*x)**5/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**6/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3
Time = 0.22 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.87 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \]
integrate(sec(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
-1/60*(C*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) - 7*sin(d*x + c)^3/(cos(d* x + c) + 1)^3)/(a^3 - 2*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^3*sin( d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) + 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 - a^3* sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c )/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d
Time = 0.34 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (6 \, B - 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (6 \, B - 13 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
-1/60*(30*(6*B - 13*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(6*B - 13*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(2*B*tan(1/2*d*x + 1/2*c )^3 - 7*C*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c) + 5*C*tan(1/2* d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 30*B*a^12*tan(1/2*d*x + 1/ 2*c)^3 - 40*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 255*B*a^12*tan(1/2*d*x + 1/2*c ) - 465*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 15.73 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^3}+\frac {3\,\left (3\,B-5\,C\right )}{4\,a^3}+\frac {2\,B-10\,C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B-7\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,B-5\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{4\,a^3}+\frac {3\,B-5\,C}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}-\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,B-13\,C\right )}{a^3\,d} \]
(tan(c/2 + (d*x)/2)*((3*(B - C))/(2*a^3) + (3*(3*B - 5*C))/(4*a^3) + (2*B - 10*C)/(4*a^3)))/d - (tan(c/2 + (d*x)/2)^3*(2*B - 7*C) - tan(c/2 + (d*x)/ 2)*(2*B - 5*C))/(d*(a^3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) + (tan(c/2 + (d*x)/2)^3*((B - C)/(4*a^3) + (3*B - 5*C)/(12*a^3)))/ d + (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) - (atanh(tan(c/2 + (d*x)/2)) *(6*B - 13*C))/(a^3*d)