3.6.92 \(\int (a \cosh (x)+b \sinh (x))^{5/2} \, dx\) [592]

Optimal. Leaf size=103 \[ \frac {2}{5} (b \cosh (x)+a \sinh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \]

[Out]

2/5*(b*cosh(x)+a*sinh(x))*(a*cosh(x)+b*sinh(x))^(3/2)-6/5*I*(a^2-b^2)*(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))*EllipticE(sin(1/2*I*x-1/2*arctan(a,-I*b)),2^(1/2))*(a*cosh(x)+b*sinh(x))^(1/
2)/((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 103, 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 = {3152, 3157, 2719} \begin {gather*} \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)} E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \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])*(a*Cosh[x] + b*Sinh[x])^(3/2))/5 - (((6*I)/5)*(a^2 - b^2)*EllipticE[(I*x - ArcTan[a
, (-I)*b])/2, 2]*Sqrt[a*Cosh[x] + b*Sinh[x]])/Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]]

Rule 2719

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

Rule 3152

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)), x] + Dist[(n - 1)*((a^2 + b^2)/n), 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] &&  !IntegerQ[
(n - 1)/2] && GtQ[n, 1]

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 (a \cosh (x)+b \sinh (x))^{5/2} \, dx &=\frac {2}{5} (b \cosh (x)+a \sinh (x)) (a \cosh (x)+b \sinh (x))^{3/2}+\frac {1}{5} \left (3 \left (a^2-b^2\right )\right ) \int \sqrt {a \cosh (x)+b \sinh (x)} \, dx\\ &=\frac {2}{5} (b \cosh (x)+a \sinh (x)) (a \cosh (x)+b \sinh (x))^{3/2}+\frac {\left (3 \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)}\right ) \int \sqrt {\cosh \left (x+i \tan ^{-1}(a,-i b)\right )} \, dx}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\\ &=\frac {2}{5} (b \cosh (x)+a \sinh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right ) \sqrt {a \cosh (x)+b \sinh (x)}}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.57, size = 193, normalized size = 1.87 \begin {gather*} \frac {(a \cosh (x)+b \sinh (x)) \left (6 a \left (a^2-b^2\right )+2 a b^2 \cosh (2 x)+b \left (a^2+b^2\right ) \sinh (2 x)\right )-\frac {3 (a-b)^2 (a+b)^2 \left (b \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cosh ^2\left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )\right ) \sinh \left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )+\sqrt {-\sinh ^2\left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )} \left (2 a \cosh \left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )-b \sinh \left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )\right )\right )}{a \sqrt {1-\frac {b^2}{a^2}} \sqrt {-\sinh ^2\left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )}}}{5 b \sqrt {a \cosh (x)+b \sinh (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a*Cosh[x] + b*Sinh[x])*(6*a*(a^2 - b^2) + 2*a*b^2*Cosh[2*x] + b*(a^2 + b^2)*Sinh[2*x]) - (3*(a - b)^2*(a + b
)^2*(b*HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cosh[x + ArcTanh[b/a]]^2]*Sinh[x + ArcTanh[b/a]] + Sqrt[-Sinh[x
+ ArcTanh[b/a]]^2]*(2*a*Cosh[x + ArcTanh[b/a]] - b*Sinh[x + ArcTanh[b/a]])))/(a*Sqrt[1 - b^2/a^2]*Sqrt[-Sinh[x
 + ArcTanh[b/a]]^2]))/(5*b*Sqrt[a*Cosh[x] + b*Sinh[x]])

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Maple [A]
time = 2.22, size = 42, normalized size = 0.41

method result size
default \(-\frac {\left (\left (a +b \right ) \left (a -b \right )\right )^{\frac {3}{2}} \left (\frac {\left (\cosh ^{3}\left (x \right )\right )}{3}-\cosh \left (x \right )\right )}{\sqrt {-\sinh \left (x \right ) \sqrt {a^{2}-b^{2}}}}\) \(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(-sinh(x)*(a^2-b^2)^(1/2))^(1/2)*((a+b)*(a-b))^(3/2)*(1/3*cosh(x)^3-cosh(x))

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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((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 = 0.15, size = 274, normalized size = 2.66 \begin {gather*} -\frac {12 \, {\left (\sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 2 \, \sqrt {2} {\left (a^{2} - b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt {2} {\left (a^{2} - b^{2}\right )} \sinh \left (x\right )^{2}\right )} \sqrt {a + b} {\rm weierstrassZeta}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (a - b\right )}}{a + b}, 0, \cosh \left (x\right ) + \sinh \left (x\right )\right )\right ) - {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (x\right )^{4} - 12 \, {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )^{2} + 6 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{2} - 2 \, a^{2} + 2 \, b^{2}\right )} \sinh \left (x\right )^{2} - a^{2} + 2 \, a b - b^{2} + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (x\right )^{3} - 6 \, {\left (a^{2} - b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \sqrt {a \cosh \left (x\right ) + b \sinh \left (x\right )}}{10 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/10*(12*(sqrt(2)*(a^2 - b^2)*cosh(x)^2 + 2*sqrt(2)*(a^2 - b^2)*cosh(x)*sinh(x) + sqrt(2)*(a^2 - b^2)*sinh(x)
^2)*sqrt(a + b)*weierstrassZeta(-4*(a - b)/(a + b), 0, weierstrassPInverse(-4*(a - b)/(a + b), 0, cosh(x) + si
nh(x))) - ((a^2 + 2*a*b + b^2)*cosh(x)^4 + 4*(a^2 + 2*a*b + b^2)*cosh(x)*sinh(x)^3 + (a^2 + 2*a*b + b^2)*sinh(
x)^4 - 12*(a^2 - b^2)*cosh(x)^2 + 6*((a^2 + 2*a*b + b^2)*cosh(x)^2 - 2*a^2 + 2*b^2)*sinh(x)^2 - a^2 + 2*a*b -
b^2 + 4*((a^2 + 2*a*b + b^2)*cosh(x)^3 - 6*(a^2 - b^2)*cosh(x))*sinh(x))*sqrt(a*cosh(x) + b*sinh(x)))/(cosh(x)
^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)

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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((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((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 {\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((a*cosh(x) + b*sinh(x))^(5/2),x)

[Out]

int((a*cosh(x) + b*sinh(x))^(5/2), x)

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