\(\int \frac {(e x)^{-1+2 n}}{a+b \sec (c+d x^n)} \, dx\) [79]

Optimal result

Integrand size = 24, antiderivative size = 328 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n}}{2 a e n}+\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}-\frac {i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d e n}+\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n}-\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a \sqrt {-a^2+b^2} d^2 e n} \] Output:

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/(-a^2+b^2)^(1/2)/d/e/n/(x^n)-I*b*(e*x)^(2*n)*ln(1+a*exp(I*(c+d 
*x^n))/(b+(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d/e/n/(x^n)+b*(e*x)^(2*n)* 
polylog(2,-a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a/(-a^2+b^2)^(1/2)/d^2 
/e/n/(x^(2*n))-b*(e*x)^(2*n)*polylog(2,-a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^( 
1/2)))/a/(-a^2+b^2)^(1/2)/d^2/e/n/(x^(2*n))
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(861\) vs. \(2(328)=656\).

Time = 1.60 (sec) , antiderivative size = 861, normalized size of antiderivative = 2.62 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx =\text {Too large to display} \] Input:

Integrate[(e*x)^(-1 + 2*n)/(a + b*Sec[c + d*x^n]),x]
 

Output:

((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]))
 

Rubi [A] (verified)

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

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}-\frac {b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a d^2 \sqrt {b^2-a^2}}+\frac {i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a d \sqrt {b^2-a^2}}-\frac {i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a d \sqrt {b^2-a^2}}+\frac {x^{2 n}}{2 a}\right )}{e n}\)

Input:

Int[(e*x)^(-1 + 2*n)/(a + b*Sec[c + d*x^n]),x]
 

Output:

((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))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4679

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]
 

rule 4692

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]
 

rule 4696

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.40 (sec) , antiderivative size = 753, normalized size of antiderivative = 2.30

method	result	size
risch	\(\frac {x \,{\mathrm e}^{\frac {\left (-1+2 n \right ) \left (-i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right ) \pi +i \operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi +i \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2} \pi -i \operatorname {csgn}\left (i e x \right )^{3} \pi +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}\)	\(753\)

Input:

int((e*x)^(-1+2*n)/(a+b*sec(c+d*x^n)),x,method=_RETURNVERBOSE)
 

Output:

1/2/a/n*x*exp(1/2*(-1+2*n)*(-I*csgn(I*e)*csgn(I*x)*csgn(I*e*x)*Pi+I*csgn(I 
*e)*csgn(I*e*x)^2*Pi+I*csgn(I*x)*csgn(I*e*x)^2*Pi-I*csgn(I*e*x)^3*Pi+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)))+ 
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))-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*cs 
gn(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))
 

Fricas [B] (verification not implemented)

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 = 0.28 (sec) , antiderivative size = 1268, normalized size of antiderivative = 3.87 \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\text {Too large to display} \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sec(c+d*x^n)),x, algorithm="fricas")
 

Output:

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^...
 

Sympy [F]

\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int \frac {\left (e x\right )^{2 n - 1}}{a + b \sec {\left (c + d x^{n} \right )}}\, dx \] Input:

integrate((e*x)**(-1+2*n)/(a+b*sec(c+d*x**n)),x)
 

Output:

Integral((e*x)**(2*n - 1)/(a + b*sec(c + d*x**n)), x)
 

Maxima [F]

\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sec(c+d*x^n)),x, algorithm="maxima")
 

Output:

-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)
 

Giac [F]

\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int { \frac {\left (e x\right )^{2 \, n - 1}}{b \sec \left (d x^{n} + c\right ) + a} \,d x } \] Input:

integrate((e*x)^(-1+2*n)/(a+b*sec(c+d*x^n)),x, algorithm="giac")
 

Output:

integrate((e*x)^(2*n - 1)/(b*sec(d*x^n + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\int \frac {{\left (e\,x\right )}^{2\,n-1}}{a+\frac {b}{\cos \left (c+d\,x^n\right )}} \,d x \] Input:

int((e*x)^(2*n - 1)/(a + b/cos(c + d*x^n)),x)
 

Output:

int((e*x)^(2*n - 1)/(a + b/cos(c + d*x^n)), x)
 

Reduce [F]

\[ \int \frac {(e x)^{-1+2 n}}{a+b \sec \left (c+d x^n\right )} \, dx=\frac {e^{2 n} \left (x^{2 n}+4 \left (\int \frac {x^{2 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} a^{2} x -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} b^{2} x -a^{2} x -2 a b x -b^{2} x}d x \right ) a b n +4 \left (\int \frac {x^{2 n} \tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2}}{\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} a^{2} x -\tan \left (\frac {x^{n} d}{2}+\frac {c}{2}\right )^{2} b^{2} x -a^{2} x -2 a b x -b^{2} x}d x \right ) b^{2} n \right )}{2 e n \left (a +b \right )} \] Input:

int((e*x)^(-1+2*n)/(a+b*sec(c+d*x^n)),x)
 

Output:

(e**(2*n)*(x**(2*n) + 4*int((x**(2*n)*tan((x**n*d + c)/2)**2)/(tan((x**n*d 
 + c)/2)**2*a**2*x - tan((x**n*d + c)/2)**2*b**2*x - a**2*x - 2*a*b*x - b* 
*2*x),x)*a*b*n + 4*int((x**(2*n)*tan((x**n*d + c)/2)**2)/(tan((x**n*d + c) 
/2)**2*a**2*x - tan((x**n*d + c)/2)**2*b**2*x - a**2*x - 2*a*b*x - b**2*x) 
,x)*b**2*n))/(2*e*n*(a + b))