3.1.82 \(\int \frac {(e x)^{-1+2 n}}{(a+b \sec (c+d x^n))^2} \, dx\) [82]

3.1.82.1 Optimal result

Integrand size = 24, antiderivative size = 757 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\frac {(e x)^{2 n}}{2 a^2 e n}-\frac {i b^3 x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d e n}+\frac {2 i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d e n}+\frac {i b^3 x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d e n}-\frac {2 i b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d e n}+\frac {b^2 x^{-2 n} (e x)^{2 n} \log \left (b+a \cos \left (c+d x^n\right )\right )}{a^2 \left (a^2-b^2\right ) d^2 e n}-\frac {b^3 x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2 e n}+\frac {2 b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2 e n}+\frac {b^3 x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \left (-a^2+b^2\right )^{3/2} d^2 e n}-\frac {2 b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (c+d x^n\right )}}{b+\sqrt {-a^2+b^2}}\right )}{a^2 \sqrt {-a^2+b^2} d^2 e n}+\frac {b^2 x^{-n} (e x)^{2 n} \sin \left (c+d x^n\right )}{a \left (a^2-b^2\right ) d e n \left (b+a \cos \left (c+d x^n\right )\right )} \]

output

1/2*(e*x)^(2*n)/a^2/e/n+b^2*(e*x)^(2*n)*ln(b+a*cos(c+d*x^n))/a^2/(a^2-b^2) 
/d^2/e/n/(x^(2*n))-I*b^3*(e*x)^(2*n)*ln(1+a*exp(I*(c+d*x^n))/(b-(-a^2+b^2) 
^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d/e/n/(x^n)+I*b^3*(e*x)^(2*n)*ln(1+a*exp(I*( 
c+d*x^n))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d/e/n/(x^n)-b^3*(e*x) 
^(2*n)*polylog(2,-a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^ 
(3/2)/d^2/e/n/(x^(2*n))+b^3*(e*x)^(2*n)*polylog(2,-a*exp(I*(c+d*x^n))/(b+( 
-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/d^2/e/n/(x^(2*n))+b^2*(e*x)^(2*n)*s 
in(c+d*x^n)/a/(a^2-b^2)/d/e/n/(x^n)/(b+a*cos(c+d*x^n))+2*I*b*(e*x)^(2*n)*l 
n(1+a*exp(I*(c+d*x^n))/(b-(-a^2+b^2)^(1/2)))/a^2/d/e/n/(x^n)/(-a^2+b^2)^(1 
/2)-2*I*b*(e*x)^(2*n)*ln(1+a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^(1/2)))/a^2/d/ 
e/n/(x^n)/(-a^2+b^2)^(1/2)+2*b*(e*x)^(2*n)*polylog(2,-a*exp(I*(c+d*x^n))/( 
b-(-a^2+b^2)^(1/2)))/a^2/d^2/e/n/(x^(2*n))/(-a^2+b^2)^(1/2)-2*b*(e*x)^(2*n 
)*polylog(2,-a*exp(I*(c+d*x^n))/(b+(-a^2+b^2)^(1/2)))/a^2/d^2/e/n/(x^(2*n) 
)/(-a^2+b^2)^(1/2)
 

3.1.82.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2450\) vs. \(2(757)=1514\).

Time = 10.69 (sec) , antiderivative size = 2450, normalized size of antiderivative = 3.24 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\text {Result too large to show} \]

input

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

output

(-2*b*x^(1 - 2*n)*(e*x)^(-1 + 2*n)*(b + a*Cos[c + d*x^n])^2*(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)*(ArcTanh[((a + b)*Cot[(c + d*x^n)/2])/Sqrt[a^2 - b^2]] - ArcTa 
nh[((a - b)*Tan[(c + d*x^n)/2])/Sqrt[a^2 - b^2]]))*Log[Sqrt[a^2 - b^2]/(Sq 
rt[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[1 - ((b - I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*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[1 
- ((b + I*Sqrt[a^2 - b^2])*(a + b - Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))/( 
a*(a + b + Sqrt[a^2 - b^2]*Tan[(c + d*x^n)/2]))] + I*(PolyLog[2, ((b - I*S 
qrt[a^2 - b^2])*(a + b - Sqrt[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]*Tan[(c + d*x^n)/2]))/(a*(a + b + Sqrt[a^2 - b^2 
]*Tan[(c + d*x^n)/2]))]))*Sec[c + d*x^n]^2)/((a^2 - b^2)^(3/2)*d^2*n*(a + 
b*Sec[c + d*x^n])^2) + (b^3*x^(1 - 2*n)*(e*x)^(-1 + 2*n)*(b + a*Cos[c +...
 

3.1.82.3 Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 603, normalized size of antiderivative = 0.80, 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.

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

\(\Big \downarrow \) 4696

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^{2 n-1}}{\left (a+b \sec \left (d x^n+c\right )\right )^2}dx}{e}\)

\(\Big \downarrow \) 4692

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{\left (a+b \sec \left (d x^n+c\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \frac {x^n}{\left (a+b \csc \left (d x^n+c+\frac {\pi }{2}\right )\right )^2}dx^n}{e n}\)

\(\Big \downarrow \) 4679

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \int \left (-\frac {2 b x^n}{a^2 \left (b+a \cos \left (d x^n+c\right )\right )}+\frac {x^n}{a^2}+\frac {b^2 x^n}{a^2 \left (b+a \cos \left (d x^n+c\right )\right )^2}\right )dx^n}{e n}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^{-2 n} (e x)^{2 n} \left (\frac {2 b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}-\frac {2 b \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \sqrt {b^2-a^2}}+\frac {b^2 \log \left (a \cos \left (c+d x^n\right )+b\right )}{a^2 d^2 \left (a^2-b^2\right )}+\frac {2 i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d \sqrt {b^2-a^2}}-\frac {2 i b x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a^2 d \sqrt {b^2-a^2}}+\frac {b^2 x^n \sin \left (c+d x^n\right )}{a d \left (a^2-b^2\right ) \left (a \cos \left (c+d x^n\right )+b\right )}-\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}+\frac {b^3 \operatorname {PolyLog}\left (2,-\frac {a e^{i \left (d x^n+c\right )}}{b+\sqrt {b^2-a^2}}\right )}{a^2 d^2 \left (b^2-a^2\right )^{3/2}}-\frac {i b^3 x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{b-\sqrt {b^2-a^2}}\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}+\frac {i b^3 x^n \log \left (1+\frac {a e^{i \left (c+d x^n\right )}}{\sqrt {b^2-a^2}+b}\right )}{a^2 d \left (b^2-a^2\right )^{3/2}}+\frac {x^{2 n}}{2 a^2}\right )}{e n}\)

input

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

output

((e*x)^(2*n)*(x^(2*n)/(2*a^2) - (I*b^3*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/( 
b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*d) + ((2*I)*b*x^n*Log[1 + 
(a*E^(I*(c + d*x^n)))/(b - Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) + 
(I*b^3*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a 
^2 + b^2)^(3/2)*d) - ((2*I)*b*x^n*Log[1 + (a*E^(I*(c + d*x^n)))/(b + Sqrt[ 
-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]*d) + (b^2*Log[b + a*Cos[c + d*x^n]])/ 
(a^2*(a^2 - b^2)*d^2) - (b^3*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(b - Sqrt[ 
-a^2 + b^2]))])/(a^2*(-a^2 + b^2)^(3/2)*d^2) + (2*b*PolyLog[2, -((a*E^(I*( 
c + d*x^n)))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (b^3*P 
olyLog[2, -((a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 + b^2]))])/(a^2*(-a^2 + b 
^2)^(3/2)*d^2) - (2*b*PolyLog[2, -((a*E^(I*(c + d*x^n)))/(b + Sqrt[-a^2 + 
b^2]))])/(a^2*Sqrt[-a^2 + b^2]*d^2) + (b^2*x^n*Sin[c + d*x^n])/(a*(a^2 - b 
^2)*d*(b + a*Cos[c + d*x^n]))))/(e*n*x^(2*n))
 

3.1.82.3.1 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]
 

3.1.82.4 Maple [C] (warning: unable to verify)

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

Time = 1.74 (sec) , antiderivative size = 3010, normalized size of antiderivative = 3.98

method	result	size
risch	\(\text {Expression too large to display}\)	\(3010\)

input

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

output

1/2/a^2/n*x*exp(1/2*(2*n-1)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*Pi*cs 
gn(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)))+2*I*b^2/a^2/(a^2-b^2)/d/n*x^n/(a*exp(2*I*(c+d*x^n))+2*b*exp( 
I*(c+d*x^n))+a)*(-1)^(1/2*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*(e^n)^2*(b*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*csgn(I*e)*csgn( 
I*e*x)^2-Pi*csgn(I*x)*csgn(I*e*x)^2+Pi*csgn(I*e*x)^3+2*d*x^n+2*c))+exp(1/2 
*I*Pi*csgn(I*e*x)*(-2*csgn(I*e)*csgn(I*x)*n+2*csgn(I*e)*csgn(I*e*x)*n+2*cs 
gn(I*x)*csgn(I*e*x)*n-2*csgn(I*e*x)^2*n-csgn(I*e)*csgn(I*e*x)-csgn(I*x)*cs 
gn(I*e*x)+csgn(I*e*x)^2))*a)/e-2*I*b/(a^2-b^2)^2/d*(exp(2*I*c)*b^2-a^2*exp 
(2*I*c))^(1/2)/n/e*(e^n)^2*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*c 
sgn(I*e*x)^3-Pi*csgn(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))*x^n*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*b^3/a^2/(a^2-b^2)^2/d*(exp(2*I*c)*b^ 
2-a^2*exp(2*I*c))^(1/2)/n/e*(e^n)^2*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*csgn(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))*x^n*ln((-a*...
 

3.1.82.5 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. 2503 vs. \(2 (705) = 1410\).

Time = 0.57 (sec) , antiderivative size = 2503, normalized size of antiderivative = 3.31 \[ \int \frac {(e x)^{-1+2 n}}{\left (a+b \sec \left (c+d x^n\right )\right )^2} \, dx=\text {Too large to display} \]

input

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

output

1/2*((a^5 - 2*a^3*b^2 + a*b^4)*d^2*e^(2*n - 1)*x^(2*n)*cos(d*x^n + c) + (a 
^4*b - 2*a^2*b^3 + b^5)*d^2*e^(2*n - 1)*x^(2*n) + 2*(a^3*b^2 - a*b^4)*d*e^ 
(2*n - 1)*x^n*sin(d*x^n + c) - ((2*a^4*b - a^2*b^3)*e^(2*n - 1)*sqrt(-(a^2 
 - b^2)/a^2)*cos(d*x^n + c) + (2*a^3*b^2 - a*b^4)*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*s 
qrt(-(a^2 - b^2)/a^2) + I*b)*sin(d*x^n + c) + a)/a + 1) - ((2*a^4*b - a^2* 
b^3)*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*cos(d*x^n + c) + (2*a^3*b^2 - a*b^ 
4)*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) + ((2*a^4*b - a^2*b^3)*e^(2*n - 1)*sqrt(-(a^2 - b^2)/a^2)*cos(d* 
x^n + c) + (2*a^3*b^2 - a*b^4)*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) + ((2*a^4*b - a^2*b^3)*e^(2*n - 1)*sqr 
t(-(a^2 - b^2)/a^2)*cos(d*x^n + c) + (2*a^3*b^2 - a*b^4)*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^3*b^2 
 - a*b^4 - I*(2*a^4*b - a^2*b^3)*c*sqrt(-(a^2 - b^2)/a^2))*e^(2*n - 1)*cos 
(d*x^n + c) + (a^2*b^3 - b^5 - I*(2*a^3*b^2 - a*b^4)*c*sqrt(-(a^2 - b^2)/a 
^2))*e^(2*n - 1))*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^3*b^2 - a*b^4 + I*(2*a^4*b - a^2*b^3)*c...
 

3.1.82.6 Sympy [F]

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

input

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

output

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

3.1.82.7 Maxima [F]

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

input

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

output

1/2*(4*a*b^3*e^(2*n)*x^n*sin(d*x^n + c) + (a^4 - a^2*b^2)*d*e^(2*n)*x^(2*n 
)*cos(2*d*x^n + 2*c)^2 + 4*(a^2*b^2 - b^4)*d*e^(2*n)*x^(2*n)*cos(d*x^n + c 
)^2 + (a^4 - a^2*b^2)*d*e^(2*n)*x^(2*n)*sin(2*d*x^n + 2*c)^2 + 4*(a^2*b^2 
- b^4)*d*e^(2*n)*x^(2*n)*sin(d*x^n + c)^2 + 4*(a^3*b - a*b^3)*d*e^(2*n)*x^ 
(2*n)*cos(d*x^n + c) + (a^4 - a^2*b^2)*d*e^(2*n)*x^(2*n) - 2*(2*a*b^3*e^(2 
*n)*x^n*sin(d*x^n + c) - 2*(a^3*b - a*b^3)*d*e^(2*n)*x^(2*n)*cos(d*x^n + c 
) - (a^4 - a^2*b^2)*d*e^(2*n)*x^(2*n))*cos(2*d*x^n + 2*c) + 2*((a^6 - a^4* 
b^2)*d*e*n*cos(2*d*x^n + 2*c)^2 + 4*(a^4*b^2 - a^2*b^4)*d*e*n*cos(d*x^n + 
c)^2 + (a^6 - a^4*b^2)*d*e*n*sin(2*d*x^n + 2*c)^2 + 4*(a^5*b - a^3*b^3)*d* 
e*n*sin(2*d*x^n + 2*c)*sin(d*x^n + c) + 4*(a^4*b^2 - a^2*b^4)*d*e*n*sin(d* 
x^n + c)^2 + 4*(a^5*b - a^3*b^3)*d*e*n*cos(d*x^n + c) + (a^6 - a^4*b^2)*d* 
e*n + 2*(2*(a^5*b - a^3*b^3)*d*e*n*cos(d*x^n + c) + (a^6 - a^4*b^2)*d*e*n) 
*cos(2*d*x^n + 2*c))*integrate(2*(a^2*b^4*e^(2*n)*x^n*cos(2*c)*sin(2*d*x^n 
) + a^2*b^4*e^(2*n)*x^n*cos(2*d*x^n)*sin(2*c) - 2*(a^3*b^3 - a*b^5)*e^(2*n 
)*x^n*cos(c)*sin(d*x^n) - 2*(a^3*b^3 - a*b^5)*e^(2*n)*x^n*cos(d*x^n)*sin(c 
) + (a^3*b^3*e^(2*n)*x^n*sin(d*x^n + c) - (2*a^5*b - a^3*b^3)*d*e^(2*n)*x^ 
(2*n)*cos(d*x^n + c))*cos(2*d*x^n + 2*c) - ((2*a^5*b - 3*a^3*b^3 + a*b^5)* 
d*e^(2*n)*x^(2*n) - (a*b^5*e^(2*n)*x^n*sin(2*c) + (2*a^3*b^3 - a*b^5)*d*e^ 
(2*n)*x^(2*n)*cos(2*c))*cos(2*d*x^n) + 2*((2*a^4*b^2 - 3*a^2*b^4 + b^6)*d* 
e^(2*n)*x^(2*n)*cos(c) + (a^2*b^4 - b^6)*e^(2*n)*x^n*sin(c))*cos(d*x^n)...
 

3.1.82.8 Giac [F]

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

input

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

output

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

3.1.82.9 Mupad [F(-1)]

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

input

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

output

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