\(\int \frac {(e x)^{-1+2 n}}{a+b \text {csch}(c+d x^n)} \, dx\) [86]

Optimal result

Integrand size = 24, antiderivative size = 291 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {(e x)^{2 n}}{2 a e n}-\frac {b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d e n}+\frac {b x^{-n} (e x)^{2 n} \log \left (1+\frac {a e^{c+d x^n}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d e n}-\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b-\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2 e n}+\frac {b x^{-2 n} (e x)^{2 n} \operatorname {PolyLog}\left (2,-\frac {a e^{c+d x^n}}{b+\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d^2 e n} \] Output:

1/2*(e*x)^(2*n)/a/e/n-b*(e*x)^(2*n)*ln(1+a*exp(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)*ln(1+a*exp(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 
(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(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 [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.04 (sec) , antiderivative size = 1181, normalized size of antiderivative = 4.06 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

((e*x)^(2*n)*Csch[c + d*x^n]*(1 - (2*b*(((-I)*Pi*ArcTanh[(-a + b*Tanh[(c + 
 d*x^n)/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] - (2*(c + I*ArcCos[((-I)*b)/ 
a])*ArcTan[((a - I*b)*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b^2 
]] + ((-2*I)*c + Pi - (2*I)*d*x^n)*ArcTanh[(((-I)*a + b)*Tan[((2*I)*c + Pi 
 + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b^2]] - (ArcCos[((-I)*b)/a] - 2*ArcTan[((a 
 - I*b)*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b^2]])*Log[((a + 
I*b)*(a - I*b + Sqrt[-a^2 - b^2])*(1 + I*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/ 
4]))/(a*(a + I*b + I*Sqrt[-a^2 - b^2]*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4]) 
)] - (ArcCos[((-I)*b)/a] + 2*ArcTan[((a - I*b)*Cot[((2*I)*c + Pi + (2*I)*d 
*x^n)/4])/Sqrt[-a^2 - b^2]])*Log[(I*(a + I*b)*(-a + I*b + Sqrt[-a^2 - b^2] 
)*(I + Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4]))/(a*(a + I*b + I*Sqrt[-a^2 - b 
^2]*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4]))] + (ArcCos[((-I)*b)/a] + 2*ArcTa 
n[((a - I*b)*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b^2]] - (2*I 
)*ArcTanh[(((-I)*a + b)*Tan[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b 
^2]])*Log[-(((-1)^(3/4)*Sqrt[-a^2 - b^2]*E^(-1/2*c - (d*x^n)/2))/(Sqrt[2]* 
Sqrt[(-I)*a]*Sqrt[b + a*Sinh[c + d*x^n]]))] + (ArcCos[((-I)*b)/a] - 2*ArcT 
an[((a - I*b)*Cot[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - b^2]] + (2* 
I)*ArcTanh[(((-I)*a + b)*Tan[((2*I)*c + Pi + (2*I)*d*x^n)/4])/Sqrt[-a^2 - 
b^2]])*Log[((-1)^(1/4)*Sqrt[-a^2 - b^2]*E^((c + d*x^n)/2))/(Sqrt[2]*Sqrt[( 
-I)*a]*Sqrt[b + a*Sinh[c + d*x^n]])] + I*(PolyLog[2, ((I*b + Sqrt[-a^2 ...
 

Rubi [A] (verified)

Time = 0.91 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.81, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {5964, 5960, 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.

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

\(\Big \downarrow \) 5964

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

\(\Big \downarrow \) 5960

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

((e*x)^(2*n)*(x^(2*n)/(2*a) - (b*x^n*Log[1 + (a*E^(c + d*x^n))/(b - Sqrt[a 
^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) + (b*x^n*Log[1 + (a*E^(c + d*x^n))/(b + 
 Sqrt[a^2 + b^2])])/(a*Sqrt[a^2 + b^2]*d) - (b*PolyLog[2, -((a*E^(c + d*x^ 
n))/(b - Sqrt[a^2 + b^2]))])/(a*Sqrt[a^2 + b^2]*d^2) + (b*PolyLog[2, -((a* 
E^(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 5960

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Csch[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 5964

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_)*(x_))^(m_.), 
x_Symbol] :> Simp[e^IntPart[m]*((e*x)^FracPart[m]/x^FracPart[m])   Int[x^m* 
(a + b*Csch[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 = 0.73 (sec) , antiderivative size = 577, normalized size of antiderivative = 1.98

method	result	size
risch	\(\frac {x \,{\mathrm e}^{\frac {\left (-1+2 n \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 {2 b \,{\mathrm e}^{-i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )} {\mathrm e}^{i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}} {\mathrm e}^{i \pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}} {\mathrm e}^{-i \pi n \operatorname {csgn}\left (i e x \right )^{3}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i e x \right )^{3}}{2}} e^{2 n} {\mathrm e}^{c} \left (\frac {x^{n} d \left (\ln \left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b -\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}{{\mathrm e}^{c} b -\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}\right )-\ln \left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b +\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}{{\mathrm e}^{c} b +\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}\right )\right )}{2 \sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}+\frac {\operatorname {dilog}\left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b -\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}{{\mathrm e}^{c} b -\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}\right )-\operatorname {dilog}\left (\frac {a \,{\mathrm e}^{2 c +d \,x^{n}}+{\mathrm e}^{c} b +\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}{{\mathrm e}^{c} b +\sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}\right )}{2 \sqrt {a^{2} {\mathrm e}^{2 c}+{\mathrm e}^{2 c} b^{2}}}\right )}{a e n \,d^{2}}\)	\(577\)

Input:

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

Output:

1/2/a/n*x*exp(1/2*(-1+2*n)*(-I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x)+I*Pi*csg 
n(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/a*b*exp(-I*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(I*Pi*n 
*csgn(I*e)*csgn(I*e*x)^2)*exp(I*Pi*n*csgn(I*x)*csgn(I*e*x)^2)*exp(-I*Pi*n* 
csgn(I*e*x)^3)*exp(1/2*I*Pi*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(-1/2*I*Pi 
*csgn(I*e)*csgn(I*e*x)^2)*exp(-1/2*I*Pi*csgn(I*x)*csgn(I*e*x)^2)*exp(1/2*I 
*Pi*csgn(I*e*x)^3)*(e^n)^2/e*exp(c)/n/d^2*(1/2*x^n*d*(ln((a*exp(2*c+d*x^n) 
+exp(c)*b-(a^2*exp(2*c)+exp(2*c)*b^2)^(1/2))/(exp(c)*b-(a^2*exp(2*c)+exp(2 
*c)*b^2)^(1/2)))-ln((a*exp(2*c+d*x^n)+exp(c)*b+(a^2*exp(2*c)+exp(2*c)*b^2) 
^(1/2))/(exp(c)*b+(a^2*exp(2*c)+exp(2*c)*b^2)^(1/2))))/(a^2*exp(2*c)+exp(2 
*c)*b^2)^(1/2)+1/2*(dilog((a*exp(2*c+d*x^n)+exp(c)*b-(a^2*exp(2*c)+exp(2*c 
)*b^2)^(1/2))/(exp(c)*b-(a^2*exp(2*c)+exp(2*c)*b^2)^(1/2)))-dilog((a*exp(2 
*c+d*x^n)+exp(c)*b+(a^2*exp(2*c)+exp(2*c)*b^2)^(1/2))/(exp(c)*b+(a^2*exp(2 
*c)+exp(2*c)*b^2)^(1/2))))/(a^2*exp(2*c)+exp(2*c)*b^2)^(1/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1183 vs. \(2 (271) = 542\).

Time = 0.13 (sec) , antiderivative size = 1183, normalized size of antiderivative = 4.07 \[ \int \frac {(e x)^{-1+2 n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((a^2 + b^2)*d^2*cosh((2*n - 1)*log(e))*cosh(n*log(x))^2 + (a^2 + b^2) 
*d^2*cosh(n*log(x))^2*sinh((2*n - 1)*log(e)) + ((a^2 + b^2)*d^2*cosh((2*n 
- 1)*log(e)) + (a^2 + b^2)*d^2*sinh((2*n - 1)*log(e)))*sinh(n*log(x))^2 - 
2*(a*b*sqrt((a^2 + b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*sqrt((a^2 + b^2) 
/a^2)*sinh((2*n - 1)*log(e)))*dilog(((a*sqrt((a^2 + b^2)/a^2) + b)*cosh(d* 
cosh(n*log(x)) + d*sinh(n*log(x)) + c) + (a*sqrt((a^2 + b^2)/a^2) + b)*sin 
h(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - a)/a + 1) + 2*(a*b*sqrt((a^2 
+ b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*sqrt((a^2 + b^2)/a^2)*sinh((2*n - 
 1)*log(e)))*dilog(-((a*sqrt((a^2 + b^2)/a^2) - b)*cosh(d*cosh(n*log(x)) + 
 d*sinh(n*log(x)) + c) + (a*sqrt((a^2 + b^2)/a^2) - b)*sinh(d*cosh(n*log(x 
)) + d*sinh(n*log(x)) + c) + a)/a + 1) - 2*(a*b*c*sqrt((a^2 + b^2)/a^2)*co 
sh((2*n - 1)*log(e)) + a*b*c*sqrt((a^2 + b^2)/a^2)*sinh((2*n - 1)*log(e))) 
*log(2*a*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + 2*a*sinh(d*cosh(n 
*log(x)) + d*sinh(n*log(x)) + c) + 2*a*sqrt((a^2 + b^2)/a^2) + 2*b) + 2*(a 
*b*c*sqrt((a^2 + b^2)/a^2)*cosh((2*n - 1)*log(e)) + a*b*c*sqrt((a^2 + b^2) 
/a^2)*sinh((2*n - 1)*log(e)))*log(2*a*cosh(d*cosh(n*log(x)) + d*sinh(n*log 
(x)) + c) + 2*a*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - 2*a*sqrt(( 
a^2 + b^2)/a^2) + 2*b) - 2*(a*b*d*sqrt((a^2 + b^2)/a^2)*cosh((2*n - 1)*log 
(e))*cosh(n*log(x)) + a*b*c*sqrt((a^2 + b^2)/a^2)*cosh((2*n - 1)*log(e)) + 
 (a*b*d*sqrt((a^2 + b^2)/a^2)*cosh(n*log(x)) + a*b*c*sqrt((a^2 + b^2)/a...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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