Integrand size = 23, antiderivative size = 219 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=-\frac {\left (a^{2/3}+2 b^{2/3}\right ) \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{a} b^{4/3} f}-\frac {\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{b}+\sqrt [3]{a} \cos (e+f x)\right )}{3 \sqrt [3]{a} b^{4/3} f}+\frac {\left (a^{2/3}-2 b^{2/3}\right ) \log \left (b^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+a^{2/3} \cos ^2(e+f x)\right )}{6 \sqrt [3]{a} b^{4/3} f}-\frac {\log \left (b+a \cos ^3(e+f x)\right )}{3 a f}+\frac {\sec (e+f x)}{b f} \] Output:
-1/3*(a^(2/3)+2*b^(2/3))*arctan(1/3*(b^(1/3)-2*a^(1/3)*cos(f*x+e))*3^(1/2) /b^(1/3))*3^(1/2)/a^(1/3)/b^(4/3)/f-1/3*(a^(2/3)-2*b^(2/3))*ln(b^(1/3)+a^( 1/3)*cos(f*x+e))/a^(1/3)/b^(4/3)/f+1/6*(a^(2/3)-2*b^(2/3))*ln(b^(2/3)-a^(1 /3)*b^(1/3)*cos(f*x+e)+a^(2/3)*cos(f*x+e)^2)/a^(1/3)/b^(4/3)/f-1/3*ln(b+a* cos(f*x+e)^3)/a/f+sec(f*x+e)/b/f
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.54 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {3 b \log \left (\sec ^2\left (\frac {1}{2} (e+f x)\right )\right )-\text {RootSum}\left [-8 a+12 a \text {$\#$1}-6 a \text {$\#$1}^2+a \text {$\#$1}^3-b \text {$\#$1}^3\&,\frac {-4 a^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 a b \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+2 a^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}-8 a b \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}+a b \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2-b^2 \log \left (1-\text {$\#$1}+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \text {$\#$1}^2}{4 a-4 a \text {$\#$1}+a \text {$\#$1}^2-b \text {$\#$1}^2}\&\right ]+3 a \sec (e+f x)}{3 a b f} \] Input:
Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^3),x]
Output:
(3*b*Log[Sec[(e + f*x)/2]^2] - RootSum[-8*a + 12*a*#1 - 6*a*#1^2 + a*#1^3 - b*#1^3 & , (-4*a^2*Log[1 - #1 + Tan[(e + f*x)/2]^2] + 4*a*b*Log[1 - #1 + Tan[(e + f*x)/2]^2] + 2*a^2*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1 - 8*a*b*L og[1 - #1 + Tan[(e + f*x)/2]^2]*#1 + a*b*Log[1 - #1 + Tan[(e + f*x)/2]^2]* #1^2 - b^2*Log[1 - #1 + Tan[(e + f*x)/2]^2]*#1^2)/(4*a - 4*a*#1 + a*#1^2 - b*#1^2) & ] + 3*a*Sec[e + f*x])/(3*a*b*f)
Time = 0.54 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4626, 2373, 2009}
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.
| \(\displaystyle \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\tan (e+f x)^5}{a+b \sec (e+f x)^3}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(e+f x)\right )^2 \sec ^2(e+f x)}{a \cos ^3(e+f x)+b}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 2373 |
| \(\displaystyle -\frac {\int \left (\frac {\sec ^2(e+f x)}{b}+\frac {b \cos ^2(e+f x)-a \cos (e+f x)-2 b}{b \left (a \cos ^3(e+f x)+b\right )}\right )d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {\left (a^{2/3}+2 b^{2/3}\right ) \arctan \left (\frac {\sqrt [3]{b}-2 \sqrt [3]{a} \cos (e+f x)}{\sqrt {3} \sqrt [3]{b}}\right )}{\sqrt {3} \sqrt [3]{a} b^{4/3}}-\frac {\left (a^{2/3}-2 b^{2/3}\right ) \log \left (a^{2/3} \cos ^2(e+f x)-\sqrt [3]{a} \sqrt [3]{b} \cos (e+f x)+b^{2/3}\right )}{6 \sqrt [3]{a} b^{4/3}}+\frac {\left (a^{2/3}-2 b^{2/3}\right ) \log \left (\sqrt [3]{a} \cos (e+f x)+\sqrt [3]{b}\right )}{3 \sqrt [3]{a} b^{4/3}}+\frac {\log \left (a \cos ^3(e+f x)+b\right )}{3 a}-\frac {\sec (e+f x)}{b}}{f}\) |
Input:
Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^3),x]
Output:
-((((a^(2/3) + 2*b^(2/3))*ArcTan[(b^(1/3) - 2*a^(1/3)*Cos[e + f*x])/(Sqrt[ 3]*b^(1/3))])/(Sqrt[3]*a^(1/3)*b^(4/3)) + ((a^(2/3) - 2*b^(2/3))*Log[b^(1/ 3) + a^(1/3)*Cos[e + f*x]])/(3*a^(1/3)*b^(4/3)) - ((a^(2/3) - 2*b^(2/3))*L og[b^(2/3) - a^(1/3)*b^(1/3)*Cos[e + f*x] + a^(2/3)*Cos[e + f*x]^2])/(6*a^ (1/3)*b^(4/3)) + Log[b + a*Cos[e + f*x]^3]/(3*a) - Sec[e + f*x]/b)/f)
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[E xpandIntegrand[(c*x)^m*(Pq/(a + b*x^n)), x], x] /; FreeQ[{a, b, c, m}, x] & & PolyQ[Pq, x] && IntegerQ[n] && !IGtQ[m, 0]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(f *ff^(m + n*p - 1))^(-1) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff* x)^n)^p/x^(m + n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.29 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.06
| method | result | size |
| risch | \(\frac {i x}{a}+\frac {2 i e}{a f}+\frac {2 \,{\mathrm e}^{i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (27 a^{3} b^{4} f^{3} \textit {\_Z}^{3}+27 i a^{2} b^{4} f^{2} \textit {\_Z}^{2}+\left (-18 a^{3} b^{2} f -9 a \,b^{4} f \right ) \textit {\_Z} -i a^{4}+2 i a^{2} b^{2}-i b^{4}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\left (-\frac {18 a^{2} b^{3} f^{2} \textit {\_R}^{2}}{a^{3}+8 b^{2} a}-\frac {36 i b^{3} f a \textit {\_R}}{a^{3}+8 b^{2} a}+\frac {8 a^{2} b}{a^{3}+8 b^{2} a}+\frac {10 b^{3}}{a^{3}+8 b^{2} a}\right ) {\mathrm e}^{i \left (f x +e \right )}+1\right )\right )\) | \(233\) |
| derivativedivides | \(\frac {\frac {1}{b \cos \left (f x +e \right )}+\frac {2 b \left (\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )+a \left (-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )-\frac {b \ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}}{b}}{f}\) | \(267\) |
| default | \(\frac {\frac {1}{b \cos \left (f x +e \right )}+\frac {2 b \left (\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}-\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {2}{3}}}\right )+a \left (-\frac {\ln \left (\cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {1}{3}}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\ln \left (\cos \left (f x +e \right )^{2}-\left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x +e \right )+\left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \cos \left (f x +e \right )}{\left (\frac {b}{a}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 a \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )-\frac {b \ln \left (b +a \cos \left (f x +e \right )^{3}\right )}{3 a}}{b}}{f}\) | \(267\) |
Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)
Output:
I*x/a+2*I/a/f*e+2*exp(I*(f*x+e))/f/b/(exp(2*I*(f*x+e))+1)-I*sum(_R*ln(exp( 2*I*(f*x+e))+(-18/(a^3+8*a*b^2)*a^2*b^3*f^2*_R^2-36*I/(a^3+8*a*b^2)*b^3*f* a*_R+8/(a^3+8*a*b^2)*a^2*b+10/(a^3+8*a*b^2)*b^3)*exp(I*(f*x+e))+1),_R=Root Of(27*a^3*b^4*f^3*_Z^3+27*I*a^2*b^4*f^2*_Z^2+(-18*a^3*b^2*f-9*a*b^4*f)*_Z- I*a^4+2*I*a^2*b^2-I*b^4))
Result contains complex when optimal does not.
Time = 0.83 (sec) , antiderivative size = 4427, normalized size of antiderivative = 20.21 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{3}{\left (e + f x \right )}}\, dx \] Input:
integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**3),x)
Output:
Integral(tan(e + f*x)**5/(a + b*sec(e + f*x)**3), x)
Time = 0.11 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\frac {\frac {2 \, \sqrt {3} {\left (2 \, a b {\left (3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {b}{a}\right )} + 3 \, a^{2} \left (\frac {b}{a}\right )^{\frac {2}{3}} + 2 \, b^{2}\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {b}{a}\right )^{\frac {1}{3}} - 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{a b^{2}} - \frac {3 \, {\left (2 \, b {\left (\left (\frac {b}{a}\right )^{\frac {2}{3}} + 1\right )} - a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )} \log \left (\cos \left (f x + e\right )^{2} - \left (\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (\frac {b}{a}\right )^{\frac {2}{3}}\right )}{a b \left (\frac {b}{a}\right )^{\frac {2}{3}}} - \frac {6 \, {\left (b {\left (\left (\frac {b}{a}\right )^{\frac {2}{3}} - 2\right )} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right )} \log \left (\left (\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right )\right )}{a b \left (\frac {b}{a}\right )^{\frac {2}{3}}} + \frac {18}{b \cos \left (f x + e\right )}}{18 \, f} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="maxima")
Output:
1/18*(2*sqrt(3)*(2*a*b*(3*(b/a)^(1/3) - b/a) + 3*a^2*(b/a)^(2/3) + 2*b^2)* arctan(-1/3*sqrt(3)*((b/a)^(1/3) - 2*cos(f*x + e))/(b/a)^(1/3))/(a*b^2) - 3*(2*b*((b/a)^(2/3) + 1) - a*(b/a)^(1/3))*log(cos(f*x + e)^2 - (b/a)^(1/3) *cos(f*x + e) + (b/a)^(2/3))/(a*b*(b/a)^(2/3)) - 6*(b*((b/a)^(2/3) - 2) + a*(b/a)^(1/3))*log((b/a)^(1/3) + cos(f*x + e))/(a*b*(b/a)^(2/3)) + 18/(b*c os(f*x + e)))/f
Time = 0.22 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.05 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=-\frac {\log \left ({\left | a \cos \left (f x + e\right )^{3} + b \right |}\right )}{3 \, a f} + \frac {\sqrt {3} {\left (2 \, \left (-a^{2} b\right )^{\frac {1}{3}} b - \left (-a^{2} b\right )^{\frac {2}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {b}{a}\right )^{\frac {1}{3}} + 2 \, \cos \left (f x + e\right )\right )}}{3 \, \left (-\frac {b}{a}\right )^{\frac {1}{3}}}\right )}{3 \, a b^{2} f} + \frac {{\left (2 \, \left (-a^{2} b\right )^{\frac {1}{3}} b + \left (-a^{2} b\right )^{\frac {2}{3}}\right )} \log \left (\cos \left (f x + e\right )^{2} + \left (-\frac {b}{a}\right )^{\frac {1}{3}} \cos \left (f x + e\right ) + \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right )}{6 \, a b^{2} f} + \frac {1}{b f \cos \left (f x + e\right )} - \frac {{\left (a^{2} b f \left (-\frac {b}{a}\right )^{\frac {1}{3}} + 2 \, a b^{2} f\right )} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {b}{a}\right )^{\frac {1}{3}} + \cos \left (f x + e\right ) \right |}\right )}{3 \, a b^{3} f^{2}} \] Input:
integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x, algorithm="giac")
Output:
-1/3*log(abs(a*cos(f*x + e)^3 + b))/(a*f) + 1/3*sqrt(3)*(2*(-a^2*b)^(1/3)* b - (-a^2*b)^(2/3))*arctan(1/3*sqrt(3)*((-b/a)^(1/3) + 2*cos(f*x + e))/(-b /a)^(1/3))/(a*b^2*f) + 1/6*(2*(-a^2*b)^(1/3)*b + (-a^2*b)^(2/3))*log(cos(f *x + e)^2 + (-b/a)^(1/3)*cos(f*x + e) + (-b/a)^(2/3))/(a*b^2*f) + 1/(b*f*c os(f*x + e)) - 1/3*(a^2*b*f*(-b/a)^(1/3) + 2*a*b^2*f)*(-b/a)^(1/3)*log(abs (-(-b/a)^(1/3) + cos(f*x + e)))/(a*b^3*f^2)
Time = 17.54 (sec) , antiderivative size = 7402, normalized size of antiderivative = 33.80 \[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {Too large to display} \] Input:
int(tan(e + f*x)^5/(a + b/cos(e + f*x)^3),x)
Output:
symsum(log(-(262144*(148*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*b^17 - 1920*a*b^15 - 156*b^1 6*cos(e + f*x) + 300*b^16 + 16*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a* b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)^2*b^18 + 5232*a^2*b^14 - 7872*a^3*b^13 + 7080*a^4*b^12 - 3840*a^5*b^11 + 1200*a^6*b^10 - 192*a^7 *b^9 + 12*a^8*b^8 - 5916*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^2*b^15 + 4820*root(27*a^3* b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^ 4, z, k)*a^3*b^14 + 5933*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^4*b^13 - 12882*root(27*a^3 *b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a ^4, z, k)*a^5*b^12 + 8891*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^6*b^11 - 2872*root(27*a^3 *b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a ^4, z, k)*a^7*b^10 + 447*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^8*b^9 - 26*root(27*a^3*b^4 *z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^9*b^8 + root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3* b^2*z - 2*a^2*b^2 + b^4 + a^4, z, k)*a^10*b^7 + 1396*root(27*a^3*b^4*z^3 + 27*a^2*b^4*z^2 + 9*a*b^4*z + 18*a^3*b^2*z - 2*a^2*b^2 + b^4 + a^4, z, ...
\[ \int \frac {\tan ^5(e+f x)}{a+b \sec ^3(e+f x)} \, dx=\text {too large to display} \] Input:
int(tan(f*x+e)^5/(a+b*sec(f*x+e)^3),x)
Output:
( - 36*cos(e + f*x)**2*a**3*b - 72*cos(e + f*x)**2*a**2*b**2 - 30*cos(e + f*x)*int(sin(e + f*x)**5/(cos(e + f*x)*sin(e + f*x)**4*a**3 + 4*cos(e + f* x)*sin(e + f*x)**4*a**2*b + 4*cos(e + f*x)*sin(e + f*x)**4*a*b**2 - 2*cos( e + f*x)*sin(e + f*x)**2*a**3 - 8*cos(e + f*x)*sin(e + f*x)**2*a**2*b - 8* cos(e + f*x)*sin(e + f*x)**2*a*b**2 + cos(e + f*x)*a**3 + 4*cos(e + f*x)*a **2*b + 4*cos(e + f*x)*a*b**2 - sin(e + f*x)**2*a**2*b - 4*sin(e + f*x)**2 *a*b**2 - 4*sin(e + f*x)**2*b**3 + a**2*b + 4*a*b**2 + 4*b**3),x)*a**6*b*f - 111*cos(e + f*x)*int(sin(e + f*x)**5/(cos(e + f*x)*sin(e + f*x)**4*a**3 + 4*cos(e + f*x)*sin(e + f*x)**4*a**2*b + 4*cos(e + f*x)*sin(e + f*x)**4* a*b**2 - 2*cos(e + f*x)*sin(e + f*x)**2*a**3 - 8*cos(e + f*x)*sin(e + f*x) **2*a**2*b - 8*cos(e + f*x)*sin(e + f*x)**2*a*b**2 + cos(e + f*x)*a**3 + 4 *cos(e + f*x)*a**2*b + 4*cos(e + f*x)*a*b**2 - sin(e + f*x)**2*a**2*b - 4* sin(e + f*x)**2*a*b**2 - 4*sin(e + f*x)**2*b**3 + a**2*b + 4*a*b**2 + 4*b* *3),x)*a**5*b**2*f - 48*cos(e + f*x)*int(sin(e + f*x)**5/(cos(e + f*x)*sin (e + f*x)**4*a**3 + 4*cos(e + f*x)*sin(e + f*x)**4*a**2*b + 4*cos(e + f*x) *sin(e + f*x)**4*a*b**2 - 2*cos(e + f*x)*sin(e + f*x)**2*a**3 - 8*cos(e + f*x)*sin(e + f*x)**2*a**2*b - 8*cos(e + f*x)*sin(e + f*x)**2*a*b**2 + cos( e + f*x)*a**3 + 4*cos(e + f*x)*a**2*b + 4*cos(e + f*x)*a*b**2 - sin(e + f* x)**2*a**2*b - 4*sin(e + f*x)**2*a*b**2 - 4*sin(e + f*x)**2*b**3 + a**2*b + 4*a*b**2 + 4*b**3),x)*a**4*b**3*f + 192*cos(e + f*x)*int(sin(e + f*x)...