3.595 \(\int \frac{1}{(a \cosh (x)+b \sinh (x))^{5/2}} \, dx\)

Optimal. Leaf size=116 \[ \frac{2 (a \sinh (x)+b \cosh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} \text{EllipticF}\left (\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right ),2\right )}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}} \]

[Out]

(2*(b*Cosh[x] + a*Sinh[x]))/(3*(a^2 - b^2)*(a*Cosh[x] + b*Sinh[x])^(3/2)) - (((2*I)/3)*EllipticF[(I*x - ArcTan
[a, (-I)*b])/2, 2]*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/((a^2 - b^2)*Sqrt[a*Cosh[x] + b*Sinh[x]])

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Rubi [A]  time = 0.0481128, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3076, 3078, 2641} \[ \frac{2 (a \sinh (x)+b \cosh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} F\left (\left .\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(-5/2),x]

[Out]

(2*(b*Cosh[x] + a*Sinh[x]))/(3*(a^2 - b^2)*(a*Cosh[x] + b*Sinh[x])^(3/2)) - (((2*I)/3)*EllipticF[(I*x - ArcTan
[a, (-I)*b])/2, 2]*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/((a^2 - b^2)*Sqrt[a*Cosh[x] + b*Sinh[x]])

Rule 3076

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

Rule 3078

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[c + d*x] +
b*Sin[c + d*x])^n/((a*Cos[c + d*x] + b*Sin[c + d*x])/Sqrt[a^2 + b^2])^n, Int[Cos[c + d*x - ArcTan[a, b]]^n, x]
, x] /; FreeQ[{a, b, c, d, n}, x] &&  !(GeQ[n, 1] || LeQ[n, -1]) &&  !(GtQ[a^2 + b^2, 0] || EqQ[a^2 + b^2, 0])

Rule 2641

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

Rubi steps

\begin{align*} \int \frac{1}{(a \cosh (x)+b \sinh (x))^{5/2}} \, dx &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}+\frac{\int \frac{1}{\sqrt{a \cosh (x)+b \sinh (x)}} \, dx}{3 \left (a^2-b^2\right )}\\ &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}+\frac{\sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} \int \frac{1}{\sqrt{\cosh \left (x+i \tan ^{-1}(a,-i b)\right )}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}}\\ &=\frac{2 (b \cosh (x)+a \sinh (x))}{3 \left (a^2-b^2\right ) (a \cosh (x)+b \sinh (x))^{3/2}}-\frac{2 i F\left (\left .\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}}}{3 \left (a^2-b^2\right ) \sqrt{a \cosh (x)+b \sinh (x)}}\\ \end{align*}

Mathematica [C]  time = 0.55822, size = 133, normalized size = 1.15 \[ -\frac{2 \left ((a \cosh (x)+b \sinh (x))^2 \sqrt{\cosh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )} \text{sech}\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},-\sinh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )\right )+b \sqrt{1-\frac{a^2}{b^2}} (a \sinh (x)+b \cosh (x))\right )}{3 b \sqrt{1-\frac{a^2}{b^2}} (b-a) (a+b) (a \cosh (x)+b \sinh (x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^(-5/2),x]

[Out]

(-2*(Sqrt[1 - a^2/b^2]*b*(b*Cosh[x] + a*Sinh[x]) + Sqrt[Cosh[x + ArcTanh[a/b]]^2]*HypergeometricPFQ[{1/4, 1/2}
, {5/4}, -Sinh[x + ArcTanh[a/b]]^2]*Sech[x + ArcTanh[a/b]]*(a*Cosh[x] + b*Sinh[x])^2))/(3*Sqrt[1 - a^2/b^2]*b*
(-a + b)*(a + b)*(a*Cosh[x] + b*Sinh[x])^(3/2))

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Maple [A]  time = 0.144, size = 37, normalized size = 0.3 \begin{align*} -{\frac{\cosh \left ( x \right ) }{ \left ({a}^{2}-{b}^{2} \right ) \sinh \left ( x \right ) }{\frac{1}{\sqrt{-\sinh \left ( x \right ) \sqrt{{a}^{2}-{b}^{2}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cosh(x)+b*sinh(x))^(5/2),x)

[Out]

-cosh(x)/(a^2-b^2)/sinh(x)/(-sinh(x)*(a^2-b^2)^(1/2))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*cosh(x) + b*sinh(x))^(-5/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \cosh \left (x\right ) + b \sinh \left (x\right )}}{a^{3} \cosh \left (x\right )^{3} + 3 \, a^{2} b \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, a b^{2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + b^{3} \sinh \left (x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*cosh(x) + b*sinh(x))/(a^3*cosh(x)^3 + 3*a^2*b*cosh(x)^2*sinh(x) + 3*a*b^2*cosh(x)*sinh(x)^2 +
b^3*sinh(x)^3), 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/(a*cosh(x)+b*sinh(x))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cosh(x)+b*sinh(x))^(5/2),x, algorithm="giac")

[Out]

integrate((a*cosh(x) + b*sinh(x))^(-5/2), x)