3.6.95 \(\int \frac {1}{(a \cosh (x)+b \sinh (x))^{5/2}} \, dx\) [595]

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

[Out]

2/3*(b*cosh(x)+a*sinh(x))/(a^2-b^2)/(a*cosh(x)+b*sinh(x))^(3/2)-2/3*I*(cos(1/2*I*x-1/2*arctan(a,-I*b))^2)^(1/2
)/cos(1/2*I*x-1/2*arctan(a,-I*b))*EllipticF(sin(1/2*I*x-1/2*arctan(a,-I*b)),2^(1/2))*((a*cosh(x)+b*sinh(x))/(a
^2-b^2)^(1/2))^(1/2)/(a^2-b^2)/(a*cosh(x)+b*sinh(x))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 = {3155, 3157, 2720} \begin {gather*} \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)}} \end {gather*}

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 2720

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

Rule 3155

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 3157

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])

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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.42, size = 133, normalized size = 1.15 \begin {gather*} -\frac {2 \left (\sqrt {1-\frac {a^2}{b^2}} b (b \cosh (x)+a \sinh (x))+\sqrt {\cosh ^2\left (x+\tanh ^{-1}\left (\frac {a}{b}\right )\right )} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\sinh ^2\left (x+\tanh ^{-1}\left (\frac {a}{b}\right )\right )\right ) \text {sech}\left (x+\tanh ^{-1}\left (\frac {a}{b}\right )\right ) (a \cosh (x)+b \sinh (x))^2\right )}{3 \sqrt {1-\frac {a^2}{b^2}} b (-a+b) (a+b) (a \cosh (x)+b \sinh (x))^{3/2}} \end {gather*}

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 = 2.24, size = 37, normalized size = 0.32

method	result	size
default	\(-\frac {\cosh \left (x \right )}{\left (a^{2}-b^{2}\right ) \sinh \left (x \right ) \sqrt {-\sinh \left (x \right ) \sqrt {a^{2}-b^{2}}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.00, 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(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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 679, normalized size = 5.85 \begin {gather*} \frac {2 \, {\left ({\left (\sqrt {2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{4} + 4 \, \sqrt {2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sqrt {2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{4} + 2 \, \sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (3 \, \sqrt {2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} + \sqrt {2} {\left (a^{2} - b^{2}\right )}\right )} \sinh \left (x\right )^{2} + 4 \, {\left (\sqrt {2} {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{3} + \sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + \sqrt {2} {\left (a^{2} - 2 \, a b + b^{2}\right )}\right )} \sqrt {a + b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right ) + 2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{3} + 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{3} - {\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) + {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} - a^{2} + b^{2}\right )} \sinh \left (x\right )\right )} \sqrt {a \cosh \left (x\right ) + b \sinh \left (x\right )}\right )}}{3 \, {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5} + {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \sinh \left (x\right )^{4} + 2 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \left (x\right )^{2} + 2 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5} + 3 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{2} + 4 \, {\left ({\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} \cosh \left (x\right )^{3} + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \end {gather*}

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]

2/3*((sqrt(2)*(a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*sqrt(2)*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + sqrt(2)*(a^2 +
2*a*b + b^2)*sinh(x)^4 + 2*sqrt(2)*(a^2 - b^2)*cosh(x)^2 + 2*(3*sqrt(2)*(a^2 + 2*a*b + b^2)*cosh(x)^2 + sqrt(
2)*(a^2 - b^2))*sinh(x)^2 + 4*(sqrt(2)*(a^2 + 2*a*b + b^2)*cosh(x)^3 + sqrt(2)*(a^2 - b^2)*cosh(x))*sinh(x) +
sqrt(2)*(a^2 - 2*a*b + b^2))*sqrt(a + b)*weierstrassPInverse(-4*(a - b)/(a + b), 0, cosh(x) + sinh(x)) + 2*((a
^2 + 2*a*b + b^2)*cosh(x)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^2 + (a^2 + 2*a*b + b^2)*sinh(x)^3 - (a^2 -
b^2)*cosh(x) + (3*(a^2 + 2*a*b + b^2)*cosh(x)^2 - a^2 + b^2)*sinh(x))*sqrt(a*cosh(x) + b*sinh(x)))/(a^5 - a^4
*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)^4 +
4*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)*sinh(x)^3 + (a^5 + 3*a^4*b + 2*a^3*b^2 - 2*
a^2*b^3 - 3*a*b^4 - b^5)*sinh(x)^4 + 2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*cosh(x)^2 + 2*(a^5
+ a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5 + 3*(a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh
(x)^2)*sinh(x)^2 + 4*((a^5 + 3*a^4*b + 2*a^3*b^2 - 2*a^2*b^3 - 3*a*b^4 - b^5)*cosh(x)^3 + (a^5 + a^4*b - 2*a^3
*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*cosh(x))*sinh(x))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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(1/(a*cosh(x)+b*sinh(x))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a\,\mathrm {cosh}\left (x\right )+b\,\mathrm {sinh}\left (x\right )\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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