### 3.591 $$\int (a \cosh (x)+b \sinh (x))^{3/2} \, dx$$

Optimal. Leaf size=103 $\frac{2}{3} (a \sinh (x)+b \cosh (x)) \sqrt{a \cosh (x)+b \sinh (x)}-\frac{2 i \left (a^2-b^2\right ) \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} \text{EllipticF}\left (\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right ),2\right )}{3 \sqrt{a \cosh (x)+b \sinh (x)}}$

[Out]

(2*(b*Cosh[x] + a*Sinh[x])*Sqrt[a*Cosh[x] + b*Sinh[x]])/3 - (((2*I)/3)*(a^2 - b^2)*EllipticF[(I*x - ArcTan[a,
(-I)*b])/2, 2]*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/Sqrt[a*Cosh[x] + b*Sinh[x]]

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Rubi [A]  time = 0.0532752, 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, 2641} $\frac{2}{3} (a \sinh (x)+b \cosh (x)) \sqrt{a \cosh (x)+b \sinh (x)}-\frac{2 i \left (a^2-b^2\right ) \sqrt{\frac{a \cosh (x)+b \sinh (x)}{\sqrt{a^2-b^2}}} F\left (\left .\frac{1}{2} \left (i x-\tan ^{-1}(a,-i b)\right )\right |2\right )}{3 \sqrt{a \cosh (x)+b \sinh (x)}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^(3/2),x]

[Out]

(2*(b*Cosh[x] + a*Sinh[x])*Sqrt[a*Cosh[x] + b*Sinh[x]])/3 - (((2*I)/3)*(a^2 - b^2)*EllipticF[(I*x - ArcTan[a,
(-I)*b])/2, 2]*Sqrt[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/Sqrt[a*Cosh[x] + b*Sinh[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])

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]

Rubi steps

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

Mathematica [C]  time = 0.573857, size = 92, normalized size = 0.89 $\frac{2}{3} \sqrt{a \cosh (x)+b \sinh (x)} \left (-b \sqrt{1-\frac{a^2}{b^2}} \sqrt{\cosh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )} \text{sech}\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right ) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{2}\right \},\left \{\frac{5}{4}\right \},-\sinh ^2\left (\tanh ^{-1}\left (\frac{a}{b}\right )+x\right )\right )+a \sinh (x)+b \cosh (x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^(3/2),x]

[Out]

(2*(b*Cosh[x] - Sqrt[1 - a^2/b^2]*b*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*Sinh[x])*Sqrt[a*Cosh[x] + b*Sinh[x]])/3

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Maple [A]  time = 0.238, size = 171, normalized size = 1.7 \begin{align*} -{\frac{1}{2\, \left ( \sinh \left ( x \right ) \right ) ^{2}}\sqrt{-\sqrt{{a}^{2}-{b}^{2}} \left ( \sinh \left ( x \right ) \right ) ^{3}} \left ( \cosh \left ( x \right ) \sqrt{\sinh \left ( x \right ) \sqrt{{a}^{2}-{b}^{2}}}\sqrt{-\sqrt{{a}^{2}-{b}^{2}} \left ( \sinh \left ( x \right ) \right ) ^{3}} \left ({a}^{2}-{b}^{2} \right ) +\sinh \left ( x \right ) \arctan \left ({\cosh \left ( x \right ) \sqrt{\sinh \left ( x \right ) \sqrt{{a}^{2}-{b}^{2}}}{\frac{1}{\sqrt{-\sqrt{{a}^{2}-{b}^{2}} \left ( \sinh \left ( x \right ) \right ) ^{3}}}}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}{\frac{1}{\sqrt{\sinh \left ( x \right ) \sqrt{{a}^{2}-{b}^{2}}}}}{\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))^(3/2),x)

[Out]

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

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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{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*cosh(x) + b*sinh(x))^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}^{\frac{3}{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="fricas")

[Out]

integral((a*cosh(x) + b*sinh(x))^(3/2), 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))**(3/2),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{3}{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^(3/2),x, algorithm="giac")

[Out]

integrate((a*cosh(x) + b*sinh(x))^(3/2), x)