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

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

[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 = 0.05, 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 = {3073, 3078, 2639} \[ \frac {2}{5} (a \sinh (x)+b \cosh (x)) (a \cosh (x)+b \sinh (x))^{3/2}-\frac {6 i \left (a^2-b^2\right ) \sqrt {a \cosh (x)+b \sinh (x)} E\left (\left .\frac {1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{5 \sqrt {\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}}} \]

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 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]

Rule 3073

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 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 (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]  time = 0.86, size = 193, normalized size = 1.87 \[ \frac {(a \cosh (x)+b \sinh (x)) \left (b \left (a^2+b^2\right ) \sinh (2 x)+6 a \left (a^2-b^2\right )+2 a b^2 \cosh (2 x)\right )-\frac {3 (a-b)^2 (a+b)^2 \left (b \sinh \left (\tanh ^{-1}\left (\frac {b}{a}\right )+x\right ) \, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cosh ^2\left (x+\tanh ^{-1}\left (\frac {b}{a}\right )\right )\right )+\sqrt {-\sinh ^2\left (\tanh ^{-1}\left (\frac {b}{a}\right )+x\right )} \left (2 a \cosh \left (\tanh ^{-1}\left (\frac {b}{a}\right )+x\right )-b \sinh \left (\tanh ^{-1}\left (\frac {b}{a}\right )+x\right )\right )\right )}{a \sqrt {1-\frac {b^2}{a^2}} \sqrt {-\sinh ^2\left (\tanh ^{-1}\left (\frac {b}{a}\right )+x\right )}}}{5 b \sqrt {a \cosh (x)+b \sinh (x)}} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} \cosh \relax (x)^{2} + 2 \, a b \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2}\right )} \sqrt {a \cosh \relax (x) + b \sinh \relax (x)}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.41, size = 51, normalized size = 0.50 \[ \frac {-\frac {\left (a^{2}-b^{2}\right )^{\frac {3}{2}} \left (\cosh ^{3}\relax (x )\right )}{3}+\left (a^{2}-b^{2}\right )^{\frac {3}{2}} \cosh \relax (x )}{\sqrt {-\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 \[ \int {\left (a\,\mathrm {cosh}\relax (x)+b\,\mathrm {sinh}\relax (x)\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x) + b*sinh(x))^(5/2),x)

[Out]

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

[Out]

Timed out

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