3.201 \(\int \frac{1}{x \text{sech}^{\frac{5}{2}}(a+b \log (c x^n))} \, dx\)

Optimal. Leaf size=97 \[ \frac{2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac{6 i \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{5 b n} \]

[Out]

(((-6*I)/5)*Sqrt[Cosh[a + b*Log[c*x^n]]]*EllipticE[(I/2)*(a + b*Log[c*x^n]), 2]*Sqrt[Sech[a + b*Log[c*x^n]]])/
(b*n) + (2*Sinh[a + b*Log[c*x^n]])/(5*b*n*Sech[a + b*Log[c*x^n]]^(3/2))

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Rubi [A]  time = 0.0702866, antiderivative size = 97, normalized size of antiderivative = 1., 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 = {3769, 3771, 2639} \[ \frac{2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac{6 i \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )} \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )}{5 b n} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sech[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(((-6*I)/5)*Sqrt[Cosh[a + b*Log[c*x^n]]]*EllipticE[(I/2)*(a + b*Log[c*x^n]), 2]*Sqrt[Sech[a + b*Log[c*x^n]]])/
(b*n) + (2*Sinh[a + b*Log[c*x^n]])/(5*b*n*Sech[a + b*Log[c*x^n]]^(3/2))

Rule 3769

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

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

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]

Rubi steps

\begin{align*} \int \frac{1}{x \text{sech}^{\frac{5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\text{sech}^{\frac{5}{2}}(a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{n}\\ &=\frac{2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{\text{sech}(a+b x)}} \, dx,x,\log \left (c x^n\right )\right )}{5 n}\\ &=\frac{2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac{\left (3 \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )}\right ) \operatorname{Subst}\left (\int \sqrt{\cosh (a+b x)} \, dx,x,\log \left (c x^n\right )\right )}{5 n}\\ &=-\frac{6 i \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right ) \sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )}}{5 b n}+\frac{2 \sinh \left (a+b \log \left (c x^n\right )\right )}{5 b n \text{sech}^{\frac{3}{2}}\left (a+b \log \left (c x^n\right )\right )}\\ \end{align*}

Mathematica [A]  time = 0.128879, size = 87, normalized size = 0.9 \[ \frac{\sqrt{\text{sech}\left (a+b \log \left (c x^n\right )\right )} \left (\sinh \left (a+b \log \left (c x^n\right )\right )+\sinh \left (3 \left (a+b \log \left (c x^n\right )\right )\right )-12 i \sqrt{\cosh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac{1}{2} i \left (a+b \log \left (c x^n\right )\right )\right |2\right )\right )}{10 b n} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sech[a + b*Log[c*x^n]]^(5/2)),x]

[Out]

(Sqrt[Sech[a + b*Log[c*x^n]]]*((-12*I)*Sqrt[Cosh[a + b*Log[c*x^n]]]*EllipticE[(I/2)*(a + b*Log[c*x^n]), 2] + S
inh[a + b*Log[c*x^n]] + Sinh[3*(a + b*Log[c*x^n])]))/(10*b*n)

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Maple [B]  time = 0.345, size = 256, normalized size = 2.6 \begin{align*}{\frac{2}{5\,bn}\sqrt{ \left ( 2\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1 \right ) \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{2}} \left ( 8\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{7}-16\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{5}+10\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{3}-3\,\sqrt{- \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}+1}{\it EllipticE} \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) ,\sqrt{2} \right ) -2\,\cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ){\frac{1}{\sqrt{2\, \left ( \sinh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{4}+ \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{2}}}} \left ( \sinh \left ({\frac{a}{2}}+{\frac{b\ln \left ( c{x}^{n} \right ) }{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cosh \left ( a/2+1/2\,b\ln \left ( c{x}^{n} \right ) \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/sech(a+b*ln(c*x^n))^(5/2),x)

[Out]

2/5/n*((2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)*sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(8*cosh(1/2*a+1/2*b*ln(c*x^n))
^7-16*cosh(1/2*a+1/2*b*ln(c*x^n))^5+10*cosh(1/2*a+1/2*b*ln(c*x^n))^3-3*(-sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*
(-2*cosh(1/2*a+1/2*b*ln(c*x^n))^2+1)^(1/2)*EllipticE(cosh(1/2*a+1/2*b*ln(c*x^n)),2^(1/2))-2*cosh(1/2*a+1/2*b*l
n(c*x^n)))/(2*sinh(1/2*a+1/2*b*ln(c*x^n))^4+sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/sinh(1/2*a+1/2*b*ln(c*x^n))/(
2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)^(1/2)/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sech(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/(x*sech(b*log(c*x^n) + a)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{x \operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sech(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")

[Out]

integral(1/(x*sech(b*log(c*x^n) + a)^(5/2)), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sech(a+b*ln(c*x**n))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \operatorname{sech}\left (b \log \left (c x^{n}\right ) + a\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/sech(a+b*log(c*x^n))^(5/2),x, algorithm="giac")

[Out]

integrate(1/(x*sech(b*log(c*x^n) + a)^(5/2)), x)