Integrand size = 32, antiderivative size = 100 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=-\frac {15 a^3 \text {arctanh}(\sin (e+f x))}{2 c f}-\frac {10 a^3 \tan (e+f x)}{c f}-\frac {5 a^3 \sec (e+f x) \tan (e+f x)}{2 c f}-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{f (c-c \sec (e+f x))} \] Output:
-15/2*a^3*arctanh(sin(f*x+e))/c/f-10*a^3*tan(f*x+e)/c/f-5/2*a^3*sec(f*x+e) *tan(f*x+e)/c/f-2*a*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.52 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=-\frac {a^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^3 \tan (e+f x)}{7 c f \sqrt {2-2 \sec (e+f x)}} \] Input:
Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x]),x]
Output:
-1/7*(a^3*Hypergeometric2F1[3/2, 7/2, 9/2, (1 + Sec[e + f*x])/2]*(1 + Sec[ e + f*x])^3*Tan[e + f*x])/(c*f*Sqrt[2 - 2*Sec[e + f*x]])
Time = 0.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4445, 3042, 4275, 3042, 4254, 24, 4534, 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.
| \(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^3}{c-c \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {5 a \int \sec (e+f x) (\sec (e+f x) a+a)^2dx}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2dx}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle -\frac {5 a \left (2 a^2 \int \sec ^2(e+f x)dx+\int \sec (e+f x) \left (\sec ^2(e+f x) a^2+a^2\right )dx\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (2 a^2 \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx+\int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {5 a \left (\int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx-\frac {2 a^2 \int 1d(-\tan (e+f x))}{f}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {5 a \left (\int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+\frac {2 a^2 \tan (e+f x)}{f}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 4534 |
| \(\displaystyle -\frac {5 a \left (\frac {3}{2} a^2 \int \sec (e+f x)dx+\frac {2 a^2 \tan (e+f x)}{f}+\frac {a^2 \tan (e+f x) \sec (e+f x)}{2 f}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (\frac {3}{2} a^2 \int \csc \left (e+f x+\frac {\pi }{2}\right )dx+\frac {2 a^2 \tan (e+f x)}{f}+\frac {a^2 \tan (e+f x) \sec (e+f x)}{2 f}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {5 a \left (\frac {3 a^2 \text {arctanh}(\sin (e+f x))}{2 f}+\frac {2 a^2 \tan (e+f x)}{f}+\frac {a^2 \tan (e+f x) \sec (e+f x)}{2 f}\right )}{c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{f (c-c \sec (e+f x))}\) |
Input:
Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x]),x]
Output:
(-2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])) - (5*a *((3*a^2*ArcTanh[Sin[e + f*x]])/(2*f) + (2*a^2*Tan[e + f*x])/f + (a^2*Sec[ e + f*x]*Tan[e + f*x])/(2*f)))/c
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_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), x] - Simp[d*((2*n - 1)/(b*(2*m + 1))) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && LtQ[m, -2^ (-1)] && IntegerQ[2*m]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Time = 0.26 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.06
| method | result | size |
| parallelrisch | \(\frac {15 a^{3} \left (\left (1+\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )+\left (-1-\cos \left (2 f x +2 e \right )\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )-\frac {14 \cot \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (f x +e \right )-\frac {12 \cos \left (2 f x +2 e \right )}{7}-\frac {11}{7}\right )}{15}\right )}{2 c f \left (1+\cos \left (2 f x +2 e \right )\right )}\) | \(106\) |
| derivativedivides | \(\frac {8 a^{3} \left (\frac {1}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {7}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16}-\frac {1}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {7}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f c}\) | \(112\) |
| default | \(\frac {8 a^{3} \left (\frac {1}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {7}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{16}-\frac {1}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}+\frac {7}{16 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {15 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{16}+\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f c}\) | \(112\) |
| risch | \(\frac {i a^{3} \left (17 \,{\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{3 i \left (f x +e \right )}+39 \,{\mathrm e}^{2 i \left (f x +e \right )}-7 \,{\mathrm e}^{i \left (f x +e \right )}+24\right )}{f c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 c f}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 c f}\) | \(134\) |
| norman | \(\frac {-\frac {8 a^{3}}{c f}+\frac {33 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {40 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}+\frac {15 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}+\frac {15 a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 c f}-\frac {15 a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 c f}\) | \(153\) |
Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x,method=_RETURNVERBOSE )
Output:
15/2*a^3*((1+cos(2*f*x+2*e))*ln(tan(1/2*f*x+1/2*e)-1)+(-1-cos(2*f*x+2*e))* ln(tan(1/2*f*x+1/2*e)+1)-14/15*cot(1/2*f*x+1/2*e)*(cos(f*x+e)-12/7*cos(2*f *x+2*e)-11/7))/c/f/(1+cos(2*f*x+2*e))
Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.25 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=-\frac {15 \, a^{3} \cos \left (f x + e\right )^{2} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 15 \, a^{3} \cos \left (f x + e\right )^{2} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) - 48 \, a^{3} \cos \left (f x + e\right )^{3} - 34 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} \cos \left (f x + e\right ) + 2 \, a^{3}}{4 \, c f \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="fri cas")
Output:
-1/4*(15*a^3*cos(f*x + e)^2*log(sin(f*x + e) + 1)*sin(f*x + e) - 15*a^3*co s(f*x + e)^2*log(-sin(f*x + e) + 1)*sin(f*x + e) - 48*a^3*cos(f*x + e)^3 - 34*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 2*a^3)/(c*f*cos(f*x + e)^2* sin(f*x + e))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=- \frac {a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx\right )}{c} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e)),x)
Output:
-a**3*(Integral(sec(e + f*x)/(sec(e + f*x) - 1), x) + Integral(3*sec(e + f *x)**2/(sec(e + f*x) - 1), x) + Integral(3*sec(e + f*x)**3/(sec(e + f*x) - 1), x) + Integral(sec(e + f*x)**4/(sec(e + f*x) - 1), x))/c
Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (97) = 194\).
Time = 0.04 (sec) , antiderivative size = 387, normalized size of antiderivative = 3.87 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=-\frac {a^{3} {\left (\frac {2 \, {\left (\frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 1\right )}}{\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {3 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 6 \, a^{3} {\left (\frac {\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1}{\frac {c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c}\right )} + 6 \, a^{3} {\left (\frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c} - \frac {\log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c} - \frac {\cos \left (f x + e\right ) + 1}{c \sin \left (f x + e\right )}\right )} - \frac {2 \, a^{3} {\left (\cos \left (f x + e\right ) + 1\right )}}{c \sin \left (f x + e\right )}}{2 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="max ima")
Output:
-1/2*(a^3*(2*(5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*sin(f*x + e)^4/(co s(f*x + e) + 1)^4 - 1)/(c*sin(f*x + e)/(cos(f*x + e) + 1) - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*log (sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c - 3*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c) + 6*a^3*((3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c*sin( f*x + e)/(cos(f*x + e) + 1) - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + log (sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c - log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c) + 6*a^3*(log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c - log(sin( f*x + e)/(cos(f*x + e) + 1) - 1)/c - (cos(f*x + e) + 1)/(c*sin(f*x + e))) - 2*a^3*(cos(f*x + e) + 1)/(c*sin(f*x + e)))/f
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=-\frac {\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c} - \frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c} - \frac {16 \, a^{3}}{c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - \frac {2 \, {\left (7 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} c}}{2 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x, algorithm="gia c")
Output:
-1/2*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c - 15*a^3*log(abs(tan(1/2 *f*x + 1/2*e) - 1))/c - 16*a^3/(c*tan(1/2*f*x + 1/2*e)) - 2*(7*a^3*tan(1/2 *f*x + 1/2*e)^3 - 9*a^3*tan(1/2*f*x + 1/2*e))/((tan(1/2*f*x + 1/2*e)^2 - 1 )^2*c))/f
Time = 12.42 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=\frac {15\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-25\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+8\,a^3}{f\,\left (c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5-2\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}-\frac {15\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c\,f} \] Input:
int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))),x)
Output:
(15*a^3*tan(e/2 + (f*x)/2)^4 - 25*a^3*tan(e/2 + (f*x)/2)^2 + 8*a^3)/(f*(c* tan(e/2 + (f*x)/2) - 2*c*tan(e/2 + (f*x)/2)^3 + c*tan(e/2 + (f*x)/2)^5)) - (15*a^3*atanh(tan(e/2 + (f*x)/2)))/(c*f)
Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 2.24 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{c-c \sec (e+f x)} \, dx=\frac {a^{3} \left (15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-30 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+30 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-50 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+16\right )}{2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right )} \] Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e)),x)
Output:
(a**3*(15*log(tan((e + f*x)/2) - 1)*tan((e + f*x)/2)**5 - 30*log(tan((e + f*x)/2) - 1)*tan((e + f*x)/2)**3 + 15*log(tan((e + f*x)/2) - 1)*tan((e + f *x)/2) - 15*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2)**5 + 30*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2)**3 - 15*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2) + 30*tan((e + f*x)/2)**4 - 50*tan((e + f*x)/2)**2 + 16))/(2*tan(( e + f*x)/2)*c*f*(tan((e + f*x)/2)**4 - 2*tan((e + f*x)/2)**2 + 1))