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

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

[Out]

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

________________________________________________________________________________________

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, 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 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 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))^{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 = 0.60, size = 92, normalized size = 0.89 \[ \frac {2}{3} \sqrt {a \cosh (x)+b \sinh (x)} \left (-b \sqrt {1-\frac {a^2}{b^2}} \sqrt {\cosh ^2\left (\tanh ^{-1}\left (\frac {a}{b}\right )+x\right )} \text {sech}\left (\tanh ^{-1}\left (\frac {a}{b}\right )+x\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\sinh ^2\left (x+\tanh ^{-1}\left (\frac {a}{b}\right )\right )\right )+a \sinh (x)+b \cosh (x)\right ) \]

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

________________________________________________________________________________________

fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}^{\frac {3}{2}}, x\right ) \]

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)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}^{\frac {3}{2}}\,{d x} \]

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)

________________________________________________________________________________________

maple [A]  time = 0.60, size = 171, normalized size = 1.66 \[ -\frac {\sqrt {-\sqrt {a^{2}-b^{2}}\, \left (\sinh ^{3}\relax (x )\right )}\, \left (\cosh \relax (x ) \sqrt {-\sqrt {a^{2}-b^{2}}\, \left (\sinh ^{3}\relax (x )\right )}\, \sqrt {\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}\, \left (a^{2}-b^{2}\right )+\sinh \relax (x ) \left (a^{2}-b^{2}\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}\, \cosh \relax (x )}{\sqrt {-\sqrt {a^{2}-b^{2}}\, \left (\sinh ^{3}\relax (x )\right )}}\right )\right )}{2 \sinh \relax (x )^{2} \sqrt {a^{2}-b^{2}}\, \sqrt {\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}\, \sqrt {-\sinh \relax (x ) \sqrt {a^{2}-b^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \cosh \relax (x) + b \sinh \relax (x)\right )}^{\frac {3}{2}}\,{d x} \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a\,\mathrm {cosh}\relax (x)+b\,\mathrm {sinh}\relax (x)\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x) + b*sinh(x))^(3/2),x)

[Out]

int((a*cosh(x) + b*sinh(x))^(3/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \cosh {\relax (x )} + b \sinh {\relax (x )}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________