\(\int \frac {(e x)^{-1+n}}{a+b \text {csch}(c+d x^n)} \, dx\) [85]

Optimal result

Integrand size = 22, antiderivative size = 82 \[ \int \frac {(e x)^{-1+n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {(e x)^n}{a e n}+\frac {2 b x^{-n} (e x)^n \text {arctanh}\left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {a^2+b^2}}\right )}{a \sqrt {a^2+b^2} d e n} \] Output:

(e*x)^n/a/e/n+2*b*(e*x)^n*arctanh((a-b*tanh(1/2*c+1/2*d*x^n))/(a^2+b^2)^(1 
/2))/a/(a^2+b^2)^(1/2)/d/e/n/(x^n)
 

Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \frac {(e x)^{-1+n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {(e x)^n \left (d+c x^{-n}-\frac {2 b x^{-n} \arctan \left (\frac {a-b \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}\right )}{a d e n} \] Input:

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

Output:

((e*x)^n*(d + c/x^n - (2*b*ArcTan[(a - b*Tanh[(c + d*x^n)/2])/Sqrt[-a^2 - 
b^2]])/(Sqrt[-a^2 - b^2]*x^n)))/(a*d*e*n)
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {5964, 5960, 3042, 4270, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 5964

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

\(\Big \downarrow \) 5960

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4270

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {4 i \int \frac {1}{-x^{2 n}-4 \left (\frac {a^2}{b^2}+1\right )}d\left (2 i \tanh \left (\frac {1}{2} \left (d x^n+c\right )\right )-\frac {2 i a}{b}\right )}{a d}\right )}{e n}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {x^{-n} (e x)^n \left (\frac {x^n}{a}-\frac {2 b \text {arctanh}\left (\frac {b \tanh \left (\frac {1}{2} \left (c+d x^n\right )\right )}{2 \sqrt {a^2+b^2}}\right )}{a d \sqrt {a^2+b^2}}\right )}{e n}\)

Input:

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

Output:

((e*x)^n*(x^n/a - (2*b*ArcTanh[(b*Tanh[(c + d*x^n)/2])/(2*Sqrt[a^2 + b^2]) 
])/(a*Sqrt[a^2 + b^2]*d)))/(e*n*x^n)
 

Defintions of rubi rules used

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 4270

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] 
- Simp[1/a   Int[1/(1 + (a/b)*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, 
x] && NeQ[a^2 - b^2, 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 3.

Time = 0.78 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.89

method	result	size
risch	\(\frac {x \,{\mathrm e}^{\frac {\left (-1+n \right ) \left (-i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )+i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}-i \pi \operatorname {csgn}\left (i e x \right )^{3}+2 \ln \left (x \right )+2 \ln \left (e \right )\right )}{2}}}{a n}-\frac {2 b \,{\mathrm e}^{-\frac {i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )}{2}} {\mathrm e}^{\frac {i \pi n \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{\frac {i \pi n \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi n \operatorname {csgn}\left (i e x \right )^{3}}{2}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i e \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i e x \right )^{2}}{2}} {\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (i e x \right )^{3}}{2}} e^{n} {\mathrm e}^{c} \arctan \left (\frac {2 a \,{\mathrm e}^{2 c +d \,x^{n}}+2 \,{\mathrm e}^{c} b}{2 \sqrt {-a^{2} {\mathrm e}^{2 c}-{\mathrm e}^{2 c} b^{2}}}\right )}{a n e d \sqrt {-a^{2} {\mathrm e}^{2 c}-{\mathrm e}^{2 c} b^{2}}}\)	\(319\)

Input:

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

Output:

1/a/n*x*exp(1/2*(-1+n)*(-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)))-2/a*b/n*exp(-1/2*I*Pi*n*csgn(I*e)*csgn(I*x)*csgn(I*e*x))*exp(1/2* 
I*Pi*n*csgn(I*e)*csgn(I*e*x)^2)*exp(1/2*I*Pi*n*csgn(I*x)*csgn(I*e*x)^2)*ex 
p(-1/2*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/e*exp(c)/d/(-a^2*exp(2*c)-exp(2*c)*b^ 
2)^(1/2)*arctan(1/2*(2*a*exp(2*c+d*x^n)+2*exp(c)*b)/(-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. 248 vs. \(2 (79) = 158\).

Time = 0.10 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.02 \[ \int \frac {(e x)^{-1+n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {{\left (a^{2} + b^{2}\right )} d \cosh \left ({\left (n - 1\right )} \log \left (e\right )\right ) \cosh \left (n \log \left (x\right )\right ) + {\left (a^{2} + b^{2}\right )} d \cosh \left (n \log \left (x\right )\right ) \sinh \left ({\left (n - 1\right )} \log \left (e\right )\right ) + {\left (\sqrt {a^{2} + b^{2}} b \cosh \left ({\left (n - 1\right )} \log \left (e\right )\right ) + \sqrt {a^{2} + b^{2}} b \sinh \left ({\left (n - 1\right )} \log \left (e\right )\right )\right )} \log \left (\frac {a b + {\left (a^{2} + b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \cosh \left (d \cosh \left (n \log \left (x\right )\right ) + d \sinh \left (n \log \left (x\right )\right ) + c\right ) - {\left (b^{2} + \sqrt {a^{2} + b^{2}} b\right )} \sinh \left (d \cosh \left (n \log \left (x\right )\right ) + d \sinh \left (n \log \left (x\right )\right ) + c\right ) + \sqrt {a^{2} + b^{2}} a}{a \sinh \left (d \cosh \left (n \log \left (x\right )\right ) + d \sinh \left (n \log \left (x\right )\right ) + c\right ) + b}\right ) + {\left ({\left (a^{2} + b^{2}\right )} d \cosh \left ({\left (n - 1\right )} \log \left (e\right )\right ) + {\left (a^{2} + b^{2}\right )} d \sinh \left ({\left (n - 1\right )} \log \left (e\right )\right )\right )} \sinh \left (n \log \left (x\right )\right )}{{\left (a^{3} + a b^{2}\right )} d n} \] Input:

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

Output:

((a^2 + b^2)*d*cosh((n - 1)*log(e))*cosh(n*log(x)) + (a^2 + b^2)*d*cosh(n* 
log(x))*sinh((n - 1)*log(e)) + (sqrt(a^2 + b^2)*b*cosh((n - 1)*log(e)) + s 
qrt(a^2 + b^2)*b*sinh((n - 1)*log(e)))*log((a*b + (a^2 + b^2 + sqrt(a^2 + 
b^2)*b)*cosh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) - (b^2 + sqrt(a^2 + 
b^2)*b)*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + sqrt(a^2 + b^2)*a) 
/(a*sinh(d*cosh(n*log(x)) + d*sinh(n*log(x)) + c) + b)) + ((a^2 + b^2)*d*c 
osh((n - 1)*log(e)) + (a^2 + b^2)*d*sinh((n - 1)*log(e)))*sinh(n*log(x)))/ 
((a^3 + a*b^2)*d*n)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 8.70 (sec) , antiderivative size = 410, normalized size of antiderivative = 5.00 \[ \int \frac {(e x)^{-1+n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {x\,{\left (e\,x\right )}^{n-1}}{a\,n}-\frac {\left (2\,\mathrm {atan}\left (\frac {x\,{\left (e\,x\right )}^{n-1}\,\sqrt {-a^2\,d^2\,n^2\,x^{2\,n}\,\left (a^2+b^2\right )}}{a\,d\,n\,x^n\,\sqrt {b^2\,x^2\,{\left (e\,x\right )}^{2\,n-2}}}\right )-2\,\mathrm {atan}\left (\frac {a^2\,{\mathrm {e}}^{d\,x^n}\,{\mathrm {e}}^c\,\left (\frac {2\,b\,x\,{\left (e\,x\right )}^{n-1}}{a^4\,d\,n\,x^n\,\sqrt {b^2\,x^2\,{\left (e\,x\right )}^{2\,n-2}}}+\frac {2\,b\,d\,n\,x^n\,{\left (e\,x\right )}^{1-n}\,\sqrt {b^2\,x^2\,{\left (e\,x\right )}^{2\,n-2}}}{a^2\,x\,\sqrt {-a^4\,d^2\,n^2\,x^{2\,n}-a^2\,b^2\,d^2\,n^2\,x^{2\,n}}\,\sqrt {-a^2\,d^2\,n^2\,x^{2\,n}\,\left (a^2+b^2\right )}}\right )\,\sqrt {-a^4\,d^2\,n^2\,x^{2\,n}-a^2\,b^2\,d^2\,n^2\,x^{2\,n}}}{2}-\frac {a\,d\,n\,x^n\,{\left (e\,x\right )}^{1-n}\,\sqrt {b^2\,x^2\,{\left (e\,x\right )}^{2\,n-2}}}{x\,\sqrt {-a^2\,d^2\,n^2\,x^{2\,n}\,\left (a^2+b^2\right )}}\right )\right )\,\sqrt {b^2\,x^2\,{\left (e\,x\right )}^{2\,n-2}}}{\sqrt {-a^4\,d^2\,n^2\,x^{2\,n}-a^2\,b^2\,d^2\,n^2\,x^{2\,n}}} \] Input:

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

Output:

(x*(e*x)^(n - 1))/(a*n) - ((2*atan((x*(e*x)^(n - 1)*(-a^2*d^2*n^2*x^(2*n)* 
(a^2 + b^2))^(1/2))/(a*d*n*x^n*(b^2*x^2*(e*x)^(2*n - 2))^(1/2))) - 2*atan( 
(a^2*exp(d*x^n)*exp(c)*((2*b*x*(e*x)^(n - 1))/(a^4*d*n*x^n*(b^2*x^2*(e*x)^ 
(2*n - 2))^(1/2)) + (2*b*d*n*x^n*(e*x)^(1 - n)*(b^2*x^2*(e*x)^(2*n - 2))^( 
1/2))/(a^2*x*(- a^4*d^2*n^2*x^(2*n) - a^2*b^2*d^2*n^2*x^(2*n))^(1/2)*(-a^2 
*d^2*n^2*x^(2*n)*(a^2 + b^2))^(1/2)))*(- a^4*d^2*n^2*x^(2*n) - a^2*b^2*d^2 
*n^2*x^(2*n))^(1/2))/2 - (a*d*n*x^n*(e*x)^(1 - n)*(b^2*x^2*(e*x)^(2*n - 2) 
)^(1/2))/(x*(-a^2*d^2*n^2*x^(2*n)*(a^2 + b^2))^(1/2))))*(b^2*x^2*(e*x)^(2* 
n - 2))^(1/2))/(- a^4*d^2*n^2*x^(2*n) - a^2*b^2*d^2*n^2*x^(2*n))^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {(e x)^{-1+n}}{a+b \text {csch}\left (c+d x^n\right )} \, dx=\frac {e^{n} \left (-2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x^{n} d +c} a i +b i}{\sqrt {a^{2}+b^{2}}}\right ) b i +x^{n} a^{2} d +x^{n} b^{2} d \right )}{a d e n \left (a^{2}+b^{2}\right )} \] Input:

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

Output:

(e**n*( - 2*sqrt(a**2 + b**2)*atan((e**(x**n*d + c)*a*i + b*i)/sqrt(a**2 + 
 b**2))*b*i + x**n*a**2*d + x**n*b**2*d))/(a*d*e*n*(a**2 + b**2))