3.3.55 \(\int \cosh ^2(x) \sinh ^3(x) \, dx\) [255]

Optimal. Leaf size=17 \[ -\frac {1}{3} \cosh ^3(x)+\frac {\cosh ^5(x)}{5} \]

[Out]

-1/3*cosh(x)^3+1/5*cosh(x)^5

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Rubi [A]
time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2645, 14} \begin {gather*} \frac {\cosh ^5(x)}{5}-\frac {\cosh ^3(x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2*Sinh[x]^3,x]

[Out]

-1/3*Cosh[x]^3 + Cosh[x]^5/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
 &&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {gather*} \begin {aligned} \text {Integral} &=-\text {Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cosh (x)\right )\\ &=-\text {Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cosh (x)\right )\\ &=-\frac {1}{3} \cosh ^3(x)+\frac {\cosh ^5(x)}{5}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 23, normalized size = 1.35 \begin {gather*} -\frac {\cosh (x)}{8}-\frac {1}{48} \cosh (3 x)+\frac {1}{80} \cosh (5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^2*Sinh[x]^3,x]

[Out]

-1/8*Cosh[x] - Cosh[3*x]/48 + Cosh[5*x]/80

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Maple [A]
time = 8.17, size = 14, normalized size = 0.82

method result size
derivativedivides \(-\frac {\left (\cosh ^{3}\left (x \right )\right )}{3}+\frac {\left (\cosh ^{5}\left (x \right )\right )}{5}\) \(14\)
default \(-\frac {\left (\cosh ^{3}\left (x \right )\right )}{3}+\frac {\left (\cosh ^{5}\left (x \right )\right )}{5}\) \(14\)
risch \(\frac {{\mathrm e}^{5 x}}{160}-\frac {{\mathrm e}^{3 x}}{96}-\frac {{\mathrm e}^{x}}{16}-\frac {{\mathrm e}^{-x}}{16}-\frac {{\mathrm e}^{-3 x}}{96}+\frac {{\mathrm e}^{-5 x}}{160}\) \(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)^3*cosh(x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*cosh(x)^3+1/5*cosh(x)^5

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (13) = 26\).
time = 0.37, size = 39, normalized size = 2.29 \begin {gather*} -\frac {1}{480} \, {\left (5 \, e^{\left (-2 \, x\right )} + 30 \, e^{\left (-4 \, x\right )} - 3\right )} e^{\left (5 \, x\right )} - \frac {1}{16} \, e^{\left (-x\right )} - \frac {1}{96} \, e^{\left (-3 \, x\right )} + \frac {1}{160} \, e^{\left (-5 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3*cosh(x)^2,x, algorithm="maxima")

[Out]

-1/480*(5*e^(-2*x) + 30*e^(-4*x) - 3)*e^(5*x) - 1/16*e^(-x) - 1/96*e^(-3*x) + 1/160*e^(-5*x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (13) = 26\).
time = 0.58, size = 42, normalized size = 2.47 \begin {gather*} \frac {1}{80} \, \cosh \left (x\right )^{5} + \frac {1}{16} \, \cosh \left (x\right ) \sinh \left (x\right )^{4} - \frac {1}{48} \, \cosh \left (x\right )^{3} + \frac {1}{16} \, {\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} - \frac {1}{8} \, \cosh \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3*cosh(x)^2,x, algorithm="fricas")

[Out]

1/80*cosh(x)^5 + 1/16*cosh(x)*sinh(x)^4 - 1/48*cosh(x)^3 + 1/16*(2*cosh(x)^3 - cosh(x))*sinh(x)^2 - 1/8*cosh(x
)

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Sympy [A]
time = 0.19, size = 19, normalized size = 1.12 \begin {gather*} \frac {\sinh ^{2}{\left (x \right )} \cosh ^{3}{\left (x \right )}}{3} - \frac {2 \cosh ^{5}{\left (x \right )}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)**3*cosh(x)**2,x)

[Out]

sinh(x)**2*cosh(x)**3/3 - 2*cosh(x)**5/15

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (13) = 26\).
time = 0.45, size = 37, normalized size = 2.18 \begin {gather*} -\frac {1}{480} \, {\left (30 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 3\right )} e^{\left (-5 \, x\right )} + \frac {1}{160} \, e^{\left (5 \, x\right )} - \frac {1}{96} \, e^{\left (3 \, x\right )} - \frac {1}{16} \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)^3*cosh(x)^2,x, algorithm="giac")

[Out]

-1/480*(30*e^(4*x) + 5*e^(2*x) - 3)*e^(-5*x) + 1/160*e^(5*x) - 1/96*e^(3*x) - 1/16*e^x

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Mupad [B]
time = 0.06, size = 14, normalized size = 0.82 \begin {gather*} \frac {{\mathrm {cosh}\left (x\right )}^3\,\left (3\,{\mathrm {cosh}\left (x\right )}^2-5\right )}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2*sinh(x)^3,x)

[Out]

(cosh(x)^3*(3*cosh(x)^2 - 5))/15

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Chatgpt [F] Failed to verify
time = 1.00, size = 22, normalized size = 1.29 \begin {gather*} \frac {\left (\cosh ^{4}\left (x \right )\right )}{4}-\frac {\left (\cosh ^{2}\left (x \right )\right )}{2}+\frac {3 \left (\sinh ^{4}\left (x \right )\right )}{8}+\frac {x}{2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

int(sinh(x)^3*cosh(x)^2,x)

[Out]

1/4*cosh(x)^4-1/2*cosh(x)^2+3/8*sinh(x)^4+1/2*x

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