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

Optimal. Leaf size=82 \[ \frac{2 b x^{-n} (e x)^n \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2+b^2}}\right )}{a d e n \sqrt{a^2+b^2}}+\frac{(e x)^n}{a e n} \]

[Out]

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

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Rubi [A]  time = 0.143895, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5441, 5437, 3783, 2660, 618, 204} \[ \frac{2 b x^{-n} (e x)^n \tanh ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{a^2+b^2}}\right )}{a d e n \sqrt{a^2+b^2}}+\frac{(e x)^n}{a e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5441

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*((e_)*(x_))^(m_.), x_Symbol] :> Dist[(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]

Rule 5437

Int[((a_.) + Csch[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(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[Simplif
y[(m + 1)/n], 0] && IntegerQ[p]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{(e x)^{-1+n}}{a+b \text{csch}\left (c+d x^n\right )} \, dx &=\frac{\left (x^{-n} (e x)^n\right ) \int \frac{x^{-1+n}}{a+b \text{csch}\left (c+d x^n\right )} \, dx}{e}\\ &=\frac{\left (x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{a+b \text{csch}(c+d x)} \, dx,x,x^n\right )}{e n}\\ &=\frac{(e x)^n}{a e n}-\frac{\left (x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a \sinh (c+d x)}{b}} \, dx,x,x^n\right )}{a e n}\\ &=\frac{(e x)^n}{a e n}+\frac{\left (2 i x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{2 i a x}{b}+x^2} \, dx,x,i \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )\right )}{a d e n}\\ &=\frac{(e x)^n}{a e n}-\frac{\left (4 i x^{-n} (e x)^n\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1+\frac{a^2}{b^2}\right )-x^2} \, dx,x,-\frac{2 i a}{b}+2 i \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )\right )}{a d e n}\\ &=\frac{(e x)^n}{a e n}+\frac{2 b x^{-n} (e x)^n \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}-\tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )\right )}{\sqrt{a^2+b^2}}\right )}{a \sqrt{a^2+b^2} d e n}\\ \end{align*}

Mathematica [A]  time = 0.168064, size = 84, normalized size = 1.02 \[ \frac{(e x)^n \left (-\frac{2 b x^{-n} \tan ^{-1}\left (\frac{a-b \tanh \left (\frac{1}{2} \left (c+d x^n\right )\right )}{\sqrt{-a^2-b^2}}\right )}{\sqrt{-a^2-b^2}}+c x^{-n}+d\right )}{a d e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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)

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Maple [C]  time = 0.135, size = 319, normalized size = 3.9 \begin{align*}{\frac{x}{an}{{\rm e}^{{\frac{ \left ( -1+n \right ) \left ( -i\pi \,{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) +i\pi \,{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}+i\pi \,{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}-i \left ({\it csgn} \left ( iex \right ) \right ) ^{3}\pi +2\,\ln \left ( x \right ) +2\,\ln \left ( e \right ) \right ) }{2}}}}}-2\,{\frac{b{{\rm e}^{-i/2\pi \,n{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) }}{{\rm e}^{i/2\pi \,n{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{i/2\pi \,n{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{-i/2\pi \,n \left ({\it csgn} \left ( iex \right ) \right ) ^{3}}}{{\rm e}^{i/2\pi \,{\it csgn} \left ( ie \right ){\it csgn} \left ( ix \right ){\it csgn} \left ( iex \right ) }}{{\rm e}^{-i/2\pi \,{\it csgn} \left ( ie \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{-i/2\pi \,{\it csgn} \left ( ix \right ) \left ({\it csgn} \left ( iex \right ) \right ) ^{2}}}{{\rm e}^{i/2\pi \, \left ({\it csgn} \left ( iex \right ) \right ) ^{3}}}{e}^{n}{{\rm e}^{c}}}{aned\sqrt{-{a}^{2}{{\rm e}^{2\,c}}-{{\rm e}^{2\,c}}{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a{{\rm e}^{2\,c+d{x}^{n}}}+2\,{{\rm e}^{c}}b}{\sqrt{-{a}^{2}{{\rm e}^{2\,c}}-{{\rm e}^{2\,c}}{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*b/a/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)*exp(-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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -2 \, b e^{n} \int \frac{e^{\left (d x^{n} + n \log \left (x\right ) + c\right )}}{a^{2} e x e^{\left (2 \, d x^{n} + 2 \, c\right )} + 2 \, a b e x e^{\left (d x^{n} + c\right )} - a^{2} e x}\,{d x} + \frac{e^{n - 1} x^{n}}{a n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Fricas [B]  time = 2.23637, size = 728, normalized size = 8.88 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((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)) + sqrt(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*cosh((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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e x\right )^{n - 1}}{a + b \operatorname{csch}{\left (c + d x^{n} \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e x\right )^{n - 1}}{b \operatorname{csch}\left (d x^{n} + c\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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