### 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*(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]]

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Rubi [A]  time = 0.0501036, antiderivative size = 103, 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 = {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 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])

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]

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*}

Mathematica [C]  time = 0.813822, size = 193, normalized size = 1.87 $\frac{(a \cosh (x)+b \sinh (x)) \left (b \left (a^2+b^2\right ) \sinh (2 x)+6 a \left (a^2-b^2\right )+2 a b^2 \cosh (2 x)\right )-\frac{3 (a-b)^2 (a+b)^2 \left (b \sinh \left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right ) \text{HypergeometricPFQ}\left (\left \{-\frac{1}{2},-\frac{1}{4}\right \},\left \{\frac{3}{4}\right \},\cosh ^2\left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right )\right )+\sqrt{-\sinh ^2\left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right )} \left (2 a \cosh \left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right )-b \sinh \left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right )\right )\right )}{a \sqrt{1-\frac{b^2}{a^2}} \sqrt{-\sinh ^2\left (\tanh ^{-1}\left (\frac{b}{a}\right )+x\right )}}}{5 b \sqrt{a \cosh (x)+b \sinh (x)}}$

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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Maple [A]  time = 0.152, size = 51, normalized size = 0.5 \begin{align*}{ \left ( -{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{3}}{3} \left ({a}^{2}-{b}^{2} \right ) ^{{\frac{3}{2}}}}+ \left ({a}^{2}-{b}^{2} \right ) ^{{\frac{3}{2}}}\cosh \left ( x \right ) \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((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., size = 0, normalized size = 0. \begin{align*} \int{\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((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 ({\left (a^{2} \cosh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2}\right )} \sqrt{a \cosh \left (x\right ) + b \sinh \left (x\right )}, x\right ) \end{align*}

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

[Out]

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