3.283 \(\int \frac {1}{x \sinh ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=107 \[ -\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}-\frac {2 i \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (i a+i b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{b n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}} \]

[Out]

-2*cosh(a+b*ln(c*x^n))/b/n/sinh(a+b*ln(c*x^n))^(1/2)+2*I*(sin(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n))^2)^(1/2)/sin(1
/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n))*EllipticE(cos(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n)),2^(1/2))*sinh(a+b*ln(c*x^n))^
(1/2)/b/n/(I*sinh(a+b*ln(c*x^n)))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2636, 2640, 2639} \[ -\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}-\frac {2 i \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (i a+i b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{b n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sinh[a + b*Log[c*x^n]]^(3/2)),x]

[Out]

(-2*Cosh[a + b*Log[c*x^n]])/(b*n*Sqrt[Sinh[a + b*Log[c*x^n]]]) - ((2*I)*EllipticE[(I*a - Pi/2 + I*b*Log[c*x^n]
)/2, 2]*Sqrt[Sinh[a + b*Log[c*x^n]]])/(b*n*Sqrt[I*Sinh[a + b*Log[c*x^n]]])

Rule 2636

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1))/(b*d*(n +
1)), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{x \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sinh ^{\frac {3}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}+\frac {\operatorname {Subst}\left (\int \sqrt {\sinh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}+\frac {\sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )} \operatorname {Subst}\left (\int \sqrt {i \sinh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}\\ &=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}{b n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 80, normalized size = 0.75 \[ -\frac {2 \left (\cosh \left (a+b \log \left (c x^n\right )\right )-\sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{4} \left (-2 i a-2 i b \log \left (c x^n\right )+\pi \right )\right |2\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sinh[a + b*Log[c*x^n]]^(3/2)),x]

[Out]

(-2*(Cosh[a + b*Log[c*x^n]] - EllipticE[((-2*I)*a + Pi - (2*I)*b*Log[c*x^n])/4, 2]*Sqrt[I*Sinh[a + b*Log[c*x^n
]]]))/(b*n*Sqrt[Sinh[a + b*Log[c*x^n]]])

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")

[Out]

integral(1/(x*sinh(b*log(c*x^n) + a)^(3/2)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="giac")

[Out]

integrate(1/(x*sinh(b*log(c*x^n) + a)^(3/2)), x)

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maple [A]  time = 0.08, size = 212, normalized size = 1.98 \[ \frac {2 \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \EllipticE \left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+1}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \EllipticF \left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\cosh ^{2}\left (a +b \ln \left (c \,x^{n}\right )\right )\right )}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/sinh(a+b*ln(c*x^n))^(3/2),x)

[Out]

1/n*(2*(1-I*sinh(a+b*ln(c*x^n)))^(1/2)*2^(1/2)*(I*sinh(a+b*ln(c*x^n))+1)^(1/2)*(I*sinh(a+b*ln(c*x^n)))^(1/2)*E
llipticE((1-I*sinh(a+b*ln(c*x^n)))^(1/2),1/2*2^(1/2))-(1-I*sinh(a+b*ln(c*x^n)))^(1/2)*2^(1/2)*(I*sinh(a+b*ln(c
*x^n))+1)^(1/2)*(I*sinh(a+b*ln(c*x^n)))^(1/2)*EllipticF((1-I*sinh(a+b*ln(c*x^n)))^(1/2),1/2*2^(1/2))-2*cosh(a+
b*ln(c*x^n))^2)/cosh(a+b*ln(c*x^n))/sinh(a+b*ln(c*x^n))^(1/2)/b

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sinh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sinh(b*log(c*x^n) + a)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x\,{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*sinh(a + b*log(c*x^n))^(3/2)),x)

[Out]

int(1/(x*sinh(a + b*log(c*x^n))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sinh ^{\frac {3}{2}}{\left (a + b \log {\left (c x^{n} \right )} \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sinh(a+b*ln(c*x**n))**(3/2),x)

[Out]

Integral(1/(x*sinh(a + b*log(c*x**n))**(3/2)), x)

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