Integrand size = 24, antiderivative size = 328 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n}}{2 a e n}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}+\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n}-\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n} \]
1/2*(e*x)^(2*n)/a/e/n+I*b*(e*x)^(2*n)*ln(1+a*exp(I*(c+d*x^n))/(b-(-a^2+b^2 )^(1/2)))/a/d/e/n/(x^n)/(-a^2+b^2)^(1/2)-I*b*(e*x)^(2*n)*ln(1+a*exp(I*(c+d *x^n))/(b+(-a^2+b^2)^(1/2)))/a/d/e/n/(x^n)/(-a^2+b^2)^(1/2)+b*(e*x)^(2*n)* polylog(2,-a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a/d^2/e/n/(x^(2*n))/(- a^2+b^2)^(1/2)-b*(e*x)^(2*n)*polylog(2,-a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^( 1/2)))/a/d^2/e/n/(x^(2*n))/(-a^2+b^2)^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(861\) vs. \(2(328)=656\).
Time = 2.36 (sec) , antiderivative size = 861, normalized size of antiderivative = 2.62 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n} \left (b+a \cos \left (c+d x^n\right )\right ) \left (1-\frac {2 b x^{-2 n} \left (2 \left (c+d x^n\right ) \text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )-2 \left (c+\arccos \left (-\frac {b}{a}\right )\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+\left (\arccos \left (-\frac {b}{a}\right )-2 i \text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )+2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{-\frac {1}{2} i \left (c+d x^n\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (c+d x^n\right )}}\right )+\left (\arccos \left (-\frac {b}{a}\right )+2 i \left (\text {arctanh}\left (\frac {(a+b) \cot \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )-\text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right )\right ) \log \left (\frac {\sqrt {a^2-b^2} e^{\frac {1}{2} i \left (c+d x^n\right )}}{\sqrt {2} \sqrt {a} \sqrt {b+a \cos \left (c+d x^n\right )}}\right )-\left (\arccos \left (-\frac {b}{a}\right )-2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (a-b-i \sqrt {a^2-b^2}\right ) \left (1+i \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )-\left (\arccos \left (-\frac {b}{a}\right )+2 i \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \log \left (\frac {(a+b) \left (-i a+i b+\sqrt {a^2-b^2}\right ) \left (i+\tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )+i \left (\operatorname {PolyLog}\left (2,\frac {\left (b-i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )-\operatorname {PolyLog}\left (2,\frac {\left (b+i \sqrt {a^2-b^2}\right ) \left (a+b-\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}{a \left (a+b+\sqrt {a^2-b^2} \tan \left (\frac {1}{2} \left (c+d x^n\right )\right )\right )}\right )\right )\right )}{\sqrt {a^2-b^2} d^2}\right ) \sec \left (c+d x^n\right )}{2 a e n \left (a+b \sec \left (c+d x^n\right )\right )} \]
((e*x)^(2*n)*(b + a*Cos[c + d*x^n])*(1 - (2*b*(2*(c + d*x^n)*ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] - 2*(c + ArcCos[-(b/a)])*ArcTanh[ ((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + (ArcCos[-(b/a)] - (2*I)*Ar cTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] + (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[Sqrt[a^2 - b^2]/(Sqrt[2]*Sqrt [a]*E^((I/2)*(c + d*x^n))*Sqrt[b + a*Cos[c + d*x^n]])] + (ArcCos[-(b/a)] + (2*I)*(ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] - ArcTanh[(( a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]]))*Log[(Sqrt[a^2 - b^2]*E^((I/2 )*(c + d*x^n)))/(Sqrt[2]*Sqrt[a]*Sqrt[b + a*Cos[c + d*x^n]])] - (ArcCos[-( b/a)] - (2*I)*ArcTanh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[( (a + b)*(a - b - I*Sqrt[a^2 - b^2])*(1 + I*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))] - (ArcCos[-(b/a)] + (2*I)*ArcTanh[ ((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]])*Log[((a + b)*((-I)*a + I*b + Sqrt[a^2 - b^2])*(I + Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*T an[(c + d*x^n)/2]))] + I*(PolyLog[2, ((b - I*Sqrt[a^2 - b^2])*(a + b - Sqr t[a^2 - b^2]*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x ^n)/2]))] - PolyLog[2, ((b + I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*T an[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))])))/( Sqrt[a^2 - b^2]*d^2*x^(2*n)))*Sec[c + d*x^n])/(2*a*e*n*(a + b*Sec[c + d*x^ n]))
Time = 0.87 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4696, 4692, 3042, 4679, 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.
| \(\displaystyle \int \frac {(e x)^{2 n-1}}{a+b \sec \left (c+d x^n\right )} \, dx\) |
| \(\Big \downarrow \) 4696 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^{2 n-1}}{a+b \sec \left (d x^n+c\right )}dx}{e}\) |
| \(\Big \downarrow \) 4692 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{a+b \sec \left (d x^n+c\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{a+b \csc \left (d x^n+c+\frac {\pi }{2}\right )}dx^n}{e n}\) |
| \(\Big \downarrow \) 4679 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \left (\frac {x^n}{a}-\frac {b x^n}{a \left (b+a \cos \left (d x^n+c\right )\right )}\right )dx^n}{e n}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^{2 n}}{2 a}\right )}{e n}\) |
((e*x)^(2*n)*(x^(2*n)/(2*a) + (I*b*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) - (I*b*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2])])/(a*Sqrt[-a^2 + b^2]*d) + (b*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2]))])/(a*Sqrt[-a^2 + b^2]*d^2) - (b*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2]))])/(a*Sqrt [-a^2 + b^2]*d^2)))/(e*n*x^(2*n))
3.1.79.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt Q[m, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 1)/n], 0] && IntegerQ[p]
Int[((e_)*(x_))^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x _Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m]) Int[x^m*( a + b*Sec[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.66 (sec) , antiderivative size = 752, normalized size of antiderivative = 2.29
| method | result | size |
| risch | \(\frac {x \,{\mathrm e}^{\frac {\left (2 n -1\right ) \left (-i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-i \pi \operatorname {csgn}\left (i e x \right )^{3}+2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{2 a n}+\frac {\left (i x^{n} d \ln \left (\frac {a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}+{\mathrm e}^{i c} b +\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}{{\mathrm e}^{i c} b +\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}\right )-i x^{n} d \ln \left (\frac {a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}+{\mathrm e}^{i c} b -\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}{{\mathrm e}^{i c} b -\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}\right )-\operatorname {dilog}\left (\frac {a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}}{{\mathrm e}^{i c} b -\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}+\frac {{\mathrm e}^{i c} b}{{\mathrm e}^{i c} b -\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}-\frac {\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}{{\mathrm e}^{i c} b -\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}\right )+\operatorname {dilog}\left (\frac {a \,{\mathrm e}^{i \left (d \,x^{n}+2 c \right )}}{{\mathrm e}^{i c} b +\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}+\frac {{\mathrm e}^{i c} b}{{\mathrm e}^{i c} b +\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}+\frac {\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}{{\mathrm e}^{i c} b +\sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}}\right )\right ) \sqrt {{\mathrm e}^{2 i c} b^{2}-a^{2} {\mathrm e}^{2 i c}}\, e^{2 n} b \,{\mathrm e}^{-\frac {i \left (2 \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )-2 \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}-2 \pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}+2 \pi n \operatorname {csgn}\left (i e x \right )^{3}-\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+\pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-\pi \operatorname {csgn}\left (i e x \right )^{3}+2 c \right )}{2}}}{\left (a^{2}-b^{2}\right ) d^{2} n e a}\) | \(752\) |
1/2/a/n*x*exp(1/2*(2*n-1)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*Pi*csgn (I*e)*csgn(I*e*x)^2+I*Pi*csgn(I*x)*csgn(I*e*x)^2-I*Pi*csgn(I*e*x)^3+2*ln(x )+2*ln(e)))+1/(a^2-b^2)*(I*x^n*d*ln((a*exp(I*(d*x^n+2*c))+exp(I*c)*b+(exp( 2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))/(exp(I*c)*b+(exp(2*I*c)*b^2-a^2*exp(2*I* c))^(1/2)))-I*x^n*d*ln((a*exp(I*(d*x^n+2*c))+exp(I*c)*b-(exp(2*I*c)*b^2-a^ 2*exp(2*I*c))^(1/2))/(exp(I*c)*b-(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2)))-d ilog(a/(exp(I*c)*b-(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*exp(I*(d*x^n+2*c ))+1/(exp(I*c)*b-(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*exp(I*c)*b-1/(exp( I*c)*b-(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*(exp(2*I*c)*b^2-a^2*exp(2*I* c))^(1/2))+dilog(a/(exp(I*c)*b+(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*exp( I*(d*x^n+2*c))+1/(exp(I*c)*b+(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*exp(I* c)*b+1/(exp(I*c)*b+(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2))*(exp(2*I*c)*b^2- a^2*exp(2*I*c))^(1/2)))*(exp(2*I*c)*b^2-a^2*exp(2*I*c))^(1/2)/d^2/n/e*(e^n )^2*b/a*exp(-1/2*I*(2*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x)-2*Pi*n*csgn(I*e )*csgn(I*e*x)^2-2*Pi*n*csgn(I*x)*csgn(I*e*x)^2+2*Pi*n*csgn(I*e*x)^3-Pi*csg n(I*e)*csgn(I*x)*csgn(I*e*x)+Pi*csgn(I*e)*csgn(I*e*x)^2+Pi*csgn(I*x)*csgn( I*e*x)^2-Pi*csgn(I*e*x)^3+2*c))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1268 vs. \(2 (300) = 600\).
Time = 0.53 (sec) , antiderivative size = 1268, normalized size of antiderivative = 3.87 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\text {Too large to display} \]
1/2*(-I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^n + c) + 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) + I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(2*a*cos(d*x^n + c) - 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) + 2*b) - I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2 )/a^2)*log(-2*a*cos(d*x^n + c) + 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b ^2)/a^2) - 2*b) + I*a*b*c*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*log(-2*a*cos( d*x^n + c) - 2*I*a*sin(d*x^n + c) + 2*a*sqrt(-(a^2 - b^2)/a^2) - 2*b) + (a ^2 - b^2)*d^2*e^(2*n - 1)*x^(2*n) - a*b*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2) *dilog(-((a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) - (I*a*sqrt(-(a^2 - b^2)/a^2) + I*b)*sin(d*x^n + c) + a)/a + 1) - a*b*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*dilog(-((a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) - (-I*a* sqrt(-(a^2 - b^2)/a^2) - I*b)*sin(d*x^n + c) + a)/a + 1) + a*b*e^(2*n - 1) *sqrt(-(a^2 - b^2)/a^2)*dilog(((a*sqrt(-(a^2 - b^2)/a^2) - b)*cos(d*x^n + c) + (I*a*sqrt(-(a^2 - b^2)/a^2) - I*b)*sin(d*x^n + c) - a)/a + 1) + a*b*e ^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*dilog(((a*sqrt(-(a^2 - b^2)/a^2) - b)*co s(d*x^n + c) + (-I*a*sqrt(-(a^2 - b^2)/a^2) + I*b)*sin(d*x^n + c) - a)/a + 1) + (I*a*b*d*e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) + I*a*b*c*e^(2*n - 1 )*sqrt(-(a^2 - b^2)/a^2))*log(((a*sqrt(-(a^2 - b^2)/a^2) + b)*cos(d*x^n + c) - (I*a*sqrt(-(a^2 - b^2)/a^2) + I*b)*sin(d*x^n + c) + a)/a) + (-I*a*b*d *e^(2*n - 1)*x^n*sqrt(-(a^2 - b^2)/a^2) - I*a*b*c*e^(2*n - 1)*sqrt(-(a^...
\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int \frac {\left (e x\right )^{2 n - 1}}{a + b \sec {\left (c + d x^{n} \right )}}\, dx \]
\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \]
-1/2*(4*a*b*e^(2*n + 1)*n*integrate((a*x^(2*n)*cos(2*d*x^n + 2*c)*cos(d*x^ n + c) + 2*b*x^(2*n)*cos(d*x^n + c)^2 + a*x^(2*n)*sin(2*d*x^n + 2*c)*sin(d *x^n + c) + 2*b*x^(2*n)*sin(d*x^n + c)^2 + a*x^(2*n)*cos(d*x^n + c))/(a^3* e*x*cos(2*d*x^n + 2*c)^2 + 4*a*b^2*e*x*cos(d*x^n + c)^2 + a^3*e*x*sin(2*d* x^n + 2*c)^2 + 4*a^2*b*e*x*sin(2*d*x^n + 2*c)*sin(d*x^n + c) + 4*a*b^2*e*x *sin(d*x^n + c)^2 + 4*a^2*b*e*x*cos(d*x^n + c) + a^3*e*x + 2*(2*a^2*b*e*x* cos(d*x^n + c) + a^3*e*x)*cos(2*d*x^n + 2*c)), x) - e^(2*n)*x^(2*n))/(a*e* n)
\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int \frac {{\left (e\,x\right )}^{2\,n-1}}{a+\frac {b}{\cos \left (c+d\,x^n\right )}} \,d x \]