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

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

[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 = 0.05, 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 = {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 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]

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])

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]  time = 0.59, size = 133, normalized size = 1.15 \[ -\frac {2 \left ((a \cosh (x)+b \sinh (x))^2 \sqrt {\cosh ^2\left (\tanh ^{-1}\left (\frac {a}{b}\right )+x\right )} \text {sech}\left (\tanh ^{-1}\left (\frac {a}{b}\right )+x\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\sinh ^2\left (x+\tanh ^{-1}\left (\frac {a}{b}\right )\right )\right )+b \sqrt {1-\frac {a^2}{b^2}} (a \sinh (x)+b \cosh (x))\right )}{3 b \sqrt {1-\frac {a^2}{b^2}} (b-a) (a+b) (a \cosh (x)+b \sinh (x))^{3/2}} \]

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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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \cosh \relax (x) + b \sinh \relax (x)}}{a^{3} \cosh \relax (x)^{3} + 3 \, a^{2} b \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, a b^{2} \cosh \relax (x) \sinh \relax (x)^{2} + b^{3} \sinh \relax (x)^{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}^{\frac {5}{2}}}\,{d x} \]

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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maple [A]  time = 0.46, size = 37, normalized size = 0.32 \[ -\frac {\cosh \relax (x )}{\left (a^{2}-b^{2}\right ) \sinh \relax (x ) \sqrt {-\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}^{\frac {5}{2}}}\,{d x} \]

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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