3.1.78 \(\int \frac {(e x)^{-1+n}}{a+b \text {csch}(c+d x^n)} \, dx\) [78]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 = {5549, 5545, 3868, 2739, 632, 210} \begin {gather*} \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} \end {gather*}

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 210

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 632

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 2739

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 3868

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[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 5545

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 5549

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]

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.13, size = 84, normalized size = 1.02 \begin {gather*} \frac {(e x)^n \left (d+c x^{-n}-\frac {2 b x^{-n} \text {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} \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 4.32, size = 319, normalized size = 3.89

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/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)*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)-b^2*exp(2*c))^(1/2)*arctan(1/2*(2*a*exp(2*c
+d*x^n)+2*exp(c)*b)/(-a^2*exp(2*c)-b^2*exp(2*c))^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (79) = 158\).
time = 0.51, size = 218, normalized size = 2.66 \begin {gather*} \frac {\sqrt {a^{2} + b^{2}} {\left (b \cosh \left (n - 1\right ) + b \sinh \left (n - 1\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 (n - 1\right ) + {\left (a^{2} + b^{2}\right )} d \sinh \left (n - 1\right )\right )} \cosh \left (n \log \left (x\right )\right ) + {\left ({\left (a^{2} + b^{2}\right )} d \cosh \left (n - 1\right ) + {\left (a^{2} + b^{2}\right )} d \sinh \left (n - 1\right )\right )} \sinh \left (n \log \left (x\right )\right )}{{\left (a^{3} + a b^{2}\right )} d n} \end {gather*}

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]

(sqrt(a^2 + b^2)*(b*cosh(n - 1) + b*sinh(n - 1))*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) + (a^2 + b^2)
*d*sinh(n - 1))*cosh(n*log(x)) + ((a^2 + b^2)*d*cosh(n - 1) + (a^2 + b^2)*d*sinh(n - 1))*sinh(n*log(x)))/((a^3
+ a*b^2)*d*n)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (e x\right )^{n - 1}}{a + b \operatorname {csch}{\left (c + d x^{n} \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 8.66, size = 410, normalized size = 5.00 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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)

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