Integrand size = 32, antiderivative size = 119 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=\frac {5 a^3 \text {arctanh}(\sin (e+f x))}{c^2 f}+\frac {5 a^3 \tan (e+f x)}{c^2 f}-\frac {2 a (a+a \sec (e+f x))^2 \tan (e+f x)}{3 f (c-c \sec (e+f x))^2}+\frac {10 \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (c^2-c^2 \sec (e+f x)\right )} \] Output:
5*a^3*arctanh(sin(f*x+e))/c^2/f+5*a^3*tan(f*x+e)/c^2/f-2/3*a*(a+a*sec(f*x+ e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^2+10/3*(a^3+a^3*sec(f*x+e))*tan(f*x+e) /f/(c^2-c^2*sec(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.55 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.55 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=-\frac {a^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},\frac {1}{2} (1+\sec (e+f x))\right ) (1+\sec (e+f x))^3 \tan (e+f x)}{14 c^2 f \sqrt {2-2 \sec (e+f x)}} \] Input:
Integrate[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^2,x]
Output:
-1/14*(a^3*Hypergeometric2F1[5/2, 7/2, 9/2, (1 + Sec[e + f*x])/2]*(1 + Sec [e + f*x])^3*Tan[e + f*x])/(c^2*f*Sqrt[2 - 2*Sec[e + f*x]])
Time = 0.71 (sec) , antiderivative size = 117, 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, 4445, 3042, 4274, 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.
| \(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^3}{(c-c \sec (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {5 a \int \frac {\sec (e+f x) (\sec (e+f x) a+a)^2}{c-c \sec (e+f x)}dx}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}{c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \int \sec (e+f x) (\sec (e+f x) a+a)dx}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )dx}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (a \int \sec ^2(e+f x)dx+a \int \sec (e+f x)dx\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (a \int \csc \left (e+f x+\frac {\pi }{2}\right )dx+a \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (a \int \csc \left (e+f x+\frac {\pi }{2}\right )dx-\frac {a \int 1d(-\tan (e+f x))}{f}\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle -\frac {5 a \left (-\frac {3 a \left (a \int \csc \left (e+f x+\frac {\pi }{2}\right )dx+\frac {a \tan (e+f x)}{f}\right )}{c}-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {5 a \left (-\frac {2 \tan (e+f x) \left (a^2 \sec (e+f x)+a^2\right )}{f (c-c \sec (e+f x))}-\frac {3 a \left (\frac {a \text {arctanh}(\sin (e+f x))}{f}+\frac {a \tan (e+f x)}{f}\right )}{c}\right )}{3 c}-\frac {2 a \tan (e+f x) (a \sec (e+f x)+a)^2}{3 f (c-c \sec (e+f x))^2}\) |
Input:
Int[(Sec[e + f*x]*(a + a*Sec[e + f*x])^3)/(c - c*Sec[e + f*x])^2,x]
Output:
(-2*a*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(3*f*(c - c*Sec[e + f*x])^2) - (5*a*((-2*(a^2 + a^2*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])) - (3*a*((a*ArcTanh[Sin[e + f*x]])/f + (a*Tan[e + f*x])/f))/c))/(3*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_)), 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_)]*(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]
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {4 a^{3} \left (-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{2}}\) | \(97\) |
| default | \(\frac {4 a^{3} \left (-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}-\frac {1}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}-\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{f \,c^{2}}\) | \(97\) |
| parallelrisch | \(-\frac {5 a^{3} \left (\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (f x +e \right )-\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (f x +e \right )+\frac {17 \cot \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (\cos \left (f x +e \right )-\frac {23 \cos \left (2 f x +2 e \right )}{68}-\frac {29}{68}\right )}{15}\right )}{\cos \left (f x +e \right ) c^{2} f}\) | \(101\) |
| risch | \(-\frac {2 i a^{3} \left (12 \,{\mathrm e}^{4 i \left (f x +e \right )}-51 \,{\mathrm e}^{3 i \left (f x +e \right )}+41 \,{\mathrm e}^{2 i \left (f x +e \right )}-57 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{3 f \,c^{2} \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3}}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{c^{2} f}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{c^{2} f}\) | \(134\) |
| norman | \(\frac {\frac {4 a^{3}}{3 c f}+\frac {4 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c f}-\frac {22 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{c f}+\frac {80 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{3 c f}-\frac {10 a^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}-\frac {5 a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{c^{2} f}+\frac {5 a^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{c^{2} f}\) | \(178\) |
Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x,method=_RETURNVERBO SE)
Output:
4/f*a^3/c^2*(-1/4/(tan(1/2*f*x+1/2*e)+1)+5/4*ln(tan(1/2*f*x+1/2*e)+1)-1/4/ (tan(1/2*f*x+1/2*e)-1)-5/4*ln(tan(1/2*f*x+1/2*e)-1)-1/3/tan(1/2*f*x+1/2*e) ^3-2/tan(1/2*f*x+1/2*e))
Time = 0.12 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=-\frac {46 \, a^{3} \cos \left (f x + e\right )^{3} - 22 \, a^{3} \cos \left (f x + e\right )^{2} - 62 \, a^{3} \cos \left (f x + e\right ) + 6 \, a^{3} - 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right ) + 15 \, {\left (a^{3} \cos \left (f x + e\right )^{2} - a^{3} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) \sin \left (f x + e\right )}{6 \, {\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="f ricas")
Output:
-1/6*(46*a^3*cos(f*x + e)^3 - 22*a^3*cos(f*x + e)^2 - 62*a^3*cos(f*x + e) + 6*a^3 - 15*(a^3*cos(f*x + e)^2 - a^3*cos(f*x + e))*log(sin(f*x + e) + 1) *sin(f*x + e) + 15*(a^3*cos(f*x + e)^2 - a^3*cos(f*x + e))*log(-sin(f*x + e) + 1)*sin(f*x + e))/((c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e))*sin(f*x + e))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=\frac {a^{3} \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} - 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{c^{2}} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**2,x)
Output:
a**3*(Integral(sec(e + f*x)/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x) + I ntegral(3*sec(e + f*x)**2/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x) + Int egral(3*sec(e + f*x)**3/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x) + Integ ral(sec(e + f*x)**4/(sec(e + f*x)**2 - 2*sec(e + f*x) + 1), x))/c**2
Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (117) = 234\).
Time = 0.04 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.93 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=-\frac {a^{3} {\left (\frac {\frac {14 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {27 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 1}{\frac {c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {c^{2} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}}\right )} - 3 \, a^{3} {\left (\frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{2}} - \frac {{\left (\frac {9 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}\right )} + \frac {3 \, a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}} - \frac {a^{3} {\left (\frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{3}}{c^{2} \sin \left (f x + e\right )^{3}}}{6 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="m axima")
Output:
-1/6*(a^3*((14*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 27*sin(f*x + e)^4/(co s(f*x + e) + 1)^4 + 1)/(c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - c^2*sin( f*x + e)^5/(cos(f*x + e) + 1)^5) - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^2 + 12*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^2) - 3*a^3*(6*log (sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^2 - 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^2 - (9*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)*(cos(f*x + e) + 1)^3/(c^2*sin(f*x + e)^3)) + 3*a^3*(3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)*(cos(f*x + e) + 1)^3/(c^2*sin(f*x + e)^3) - a^3*(3*sin(f*x + e)^ 2/(cos(f*x + e) + 1)^2 - 1)*(cos(f*x + e) + 1)^3/(c^2*sin(f*x + e)^3))/f
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.97 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=\frac {\frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{c^{2}} - \frac {15 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{c^{2}} - \frac {6 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2}} - \frac {4 \, {\left (6 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a^{3}\right )}}{c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{3 \, f} \] Input:
integrate(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x, algorithm="g iac")
Output:
1/3*(15*a^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/c^2 - 15*a^3*log(abs(tan(1/ 2*f*x + 1/2*e) - 1))/c^2 - 6*a^3*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2* e)^2 - 1)*c^2) - 4*(6*a^3*tan(1/2*f*x + 1/2*e)^2 + a^3)/(c^2*tan(1/2*f*x + 1/2*e)^3))/f
Time = 11.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=\frac {10\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{c^2\,f}+\frac {-10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\frac {20\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{3}+\frac {4\,a^3}{3}}{c^2\,f\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )} \] Input:
int((a + a/cos(e + f*x))^3/(cos(e + f*x)*(c - c/cos(e + f*x))^2),x)
Output:
(10*a^3*atanh(tan(e/2 + (f*x)/2)))/(c^2*f) + ((20*a^3*tan(e/2 + (f*x)/2)^2 )/3 - 10*a^3*tan(e/2 + (f*x)/2)^4 + (4*a^3)/3)/(c^2*f*tan(e/2 + (f*x)/2)^3 *(tan(e/2 + (f*x)/2)^2 - 1))
Time = 0.17 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^3}{(c-c \sec (e+f x))^2} \, dx=\frac {a^{3} \left (-15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}-15 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+20 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+4\right )}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c^{2} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )} \] Input:
int(sec(f*x+e)*(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^2,x)
Output:
(a**3*( - 15*log(tan((e + f*x)/2) - 1)*tan((e + f*x)/2)**5 + 15*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)**5 - 15*log(tan((e + f*x)/2) + 1)*tan((e + f*x)/2)**3 - 30*tan(( e + f*x)/2)**4 + 20*tan((e + f*x)/2)**2 + 4))/(3*tan((e + f*x)/2)**3*c**2* f*(tan((e + f*x)/2)**2 - 1))