3.582 \(\int (a \cosh (x)+b \sinh (x))^3 \, dx\)

Optimal. Leaf size=35 \[ \left (a^2-b^2\right ) (a \sinh (x)+b \cosh (x))+\frac {1}{3} (a \sinh (x)+b \cosh (x))^3 \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3072} \[ \left (a^2-b^2\right ) (a \sinh (x)+b \cosh (x))+\frac {1}{3} (a \sinh (x)+b \cosh (x))^3 \]

Antiderivative was successfully verified.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^3,x]

[Out]

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

Rule 3072

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int
[(a^2 + b^2 - x^2)^((n - 1)/2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 + b^2, 0] && IGtQ[(n - 1)/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 63, normalized size = 1.80 \[ \frac {1}{12} \left (9 a \left (a^2-b^2\right ) \sinh (x)+a \left (a^2+3 b^2\right ) \sinh (3 x)+9 b \left (a^2-b^2\right ) \cosh (x)+b \left (3 a^2+b^2\right ) \cosh (3 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b*(a^2 - b^2)*Cosh[x] + b*(3*a^2 + b^2)*Cosh[3*x] + 9*a*(a^2 - b^2)*Sinh[x] + a*(a^2 + 3*b^2)*Sinh[3*x])/12

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fricas [B]  time = 0.41, size = 97, normalized size = 2.77 \[ \frac {1}{12} \, {\left (3 \, a^{2} b + b^{3}\right )} \cosh \relax (x)^{3} + \frac {1}{4} \, {\left (3 \, a^{2} b + b^{3}\right )} \cosh \relax (x) \sinh \relax (x)^{2} + \frac {1}{12} \, {\left (a^{3} + 3 \, a b^{2}\right )} \sinh \relax (x)^{3} + \frac {3}{4} \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) + \frac {1}{4} \, {\left (3 \, a^{3} - 3 \, a b^{2} + {\left (a^{3} + 3 \, a b^{2}\right )} \cosh \relax (x)^{2}\right )} \sinh \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a^2*b + b^3)*cosh(x)^3 + 1/4*(3*a^2*b + b^3)*cosh(x)*sinh(x)^2 + 1/12*(a^3 + 3*a*b^2)*sinh(x)^3 + 3/4*
(a^2*b - b^3)*cosh(x) + 1/4*(3*a^3 - 3*a*b^2 + (a^3 + 3*a*b^2)*cosh(x)^2)*sinh(x)

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giac [B]  time = 0.14, size = 134, normalized size = 3.83 \[ \frac {1}{24} \, a^{3} e^{\left (3 \, x\right )} + \frac {1}{8} \, a^{2} b e^{\left (3 \, x\right )} + \frac {1}{8} \, a b^{2} e^{\left (3 \, x\right )} + \frac {1}{24} \, b^{3} e^{\left (3 \, x\right )} + \frac {3}{8} \, a^{3} e^{x} + \frac {3}{8} \, a^{2} b e^{x} - \frac {3}{8} \, a b^{2} e^{x} - \frac {3}{8} \, b^{3} e^{x} - \frac {1}{24} \, {\left (9 \, a^{3} e^{\left (2 \, x\right )} - 9 \, a^{2} b e^{\left (2 \, x\right )} - 9 \, a b^{2} e^{\left (2 \, x\right )} + 9 \, b^{3} e^{\left (2 \, x\right )} + a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} e^{\left (-3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*a^3*e^(3*x) + 1/8*a^2*b*e^(3*x) + 1/8*a*b^2*e^(3*x) + 1/24*b^3*e^(3*x) + 3/8*a^3*e^x + 3/8*a^2*b*e^x - 3/
8*a*b^2*e^x - 3/8*b^3*e^x - 1/24*(9*a^3*e^(2*x) - 9*a^2*b*e^(2*x) - 9*a*b^2*e^(2*x) + 9*b^3*e^(2*x) + a^3 - 3*
a^2*b + 3*a*b^2 - b^3)*e^(-3*x)

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maple [A]  time = 0.46, size = 48, normalized size = 1.37 \[ b^{3} \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\relax (x )\right )}{3}\right ) \cosh \relax (x )+a \,b^{2} \left (\sinh ^{3}\relax (x )\right )+a^{2} b \left (\cosh ^{3}\relax (x )\right )+a^{3} \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\relax (x )\right )}{3}\right ) \sinh \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.39, size = 69, normalized size = 1.97 \[ a^{2} b \cosh \relax (x)^{3} + a b^{2} \sinh \relax (x)^{3} + \frac {1}{24} \, b^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} + e^{\left (-3 \, x\right )} - 9 \, e^{x}\right )} + \frac {1}{24} \, a^{3} {\left (e^{\left (3 \, x\right )} - 9 \, e^{\left (-x\right )} - e^{\left (-3 \, x\right )} + 9 \, e^{x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*b*cosh(x)^3 + a*b^2*sinh(x)^3 + 1/24*b^3*(e^(3*x) - 9*e^(-x) + e^(-3*x) - 9*e^x) + 1/24*a^3*(e^(3*x) - 9*e
^(-x) - e^(-3*x) + 9*e^x)

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mupad [B]  time = 0.10, size = 53, normalized size = 1.51 \[ {\mathrm {cosh}\relax (x)}^3\,\left (a^2\,b-\frac {2\,b^3}{3}\right )+{\mathrm {sinh}\relax (x)}^3\,\left (a\,b^2-\frac {2\,a^3}{3}\right )+a^3\,{\mathrm {cosh}\relax (x)}^2\,\mathrm {sinh}\relax (x)+b^3\,\mathrm {cosh}\relax (x)\,{\mathrm {sinh}\relax (x)}^2 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x) + b*sinh(x))^3,x)

[Out]

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

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sympy [B]  time = 0.33, size = 66, normalized size = 1.89 \[ - \frac {2 a^{3} \sinh ^{3}{\relax (x )}}{3} + a^{3} \sinh {\relax (x )} \cosh ^{2}{\relax (x )} + a^{2} b \cosh ^{3}{\relax (x )} + a b^{2} \sinh ^{3}{\relax (x )} + b^{3} \sinh ^{2}{\relax (x )} \cosh {\relax (x )} - \frac {2 b^{3} \cosh ^{3}{\relax (x )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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