3.223 \(\int \cosh (x) \sinh (4 x) \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{6} \cosh (3 x)+\frac {1}{10} \cosh (5 x) \]

[Out]

1/6*cosh(3*x)+1/10*cosh(5*x)

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Rubi [A]  time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4284} \[ \frac {1}{6} \cosh (3 x)+\frac {1}{10} \cosh (5 x) \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]*Sinh[4*x],x]

[Out]

Cosh[3*x]/6 + Cosh[5*x]/10

Rule 4284

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

Rubi steps

\begin {align*} \int \cosh (x) \sinh (4 x) \, dx &=\frac {1}{6} \cosh (3 x)+\frac {1}{10} \cosh (5 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ \frac {1}{6} \cosh (3 x)+\frac {1}{10} \cosh (5 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]*Sinh[4*x],x]

[Out]

Cosh[3*x]/6 + Cosh[5*x]/10

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fricas [B]  time = 0.42, size = 36, normalized size = 2.12 \[ \frac {1}{10} \, \cosh \relax (x)^{5} + \frac {1}{2} \, \cosh \relax (x) \sinh \relax (x)^{4} + \frac {1}{6} \, \cosh \relax (x)^{3} + \frac {1}{2} \, {\left (2 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x)^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(4*x),x, algorithm="fricas")

[Out]

1/10*cosh(x)^5 + 1/2*cosh(x)*sinh(x)^4 + 1/6*cosh(x)^3 + 1/2*(2*cosh(x)^3 + cosh(x))*sinh(x)^2

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giac [B]  time = 0.11, size = 27, normalized size = 1.59 \[ \frac {1}{60} \, {\left (5 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-5 \, x\right )} + \frac {1}{20} \, e^{\left (5 \, x\right )} + \frac {1}{12} \, e^{\left (3 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(4*x),x, algorithm="giac")

[Out]

1/60*(5*e^(2*x) + 3)*e^(-5*x) + 1/20*e^(5*x) + 1/12*e^(3*x)

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maple [A]  time = 0.10, size = 14, normalized size = 0.82 \[ \frac {8 \left (\cosh ^{5}\relax (x )\right )}{5}-\frac {4 \left (\cosh ^{3}\relax (x )\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*sinh(4*x),x)

[Out]

8/5*cosh(x)^5-4/3*cosh(x)^3

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maxima [B]  time = 0.34, size = 27, normalized size = 1.59 \[ \frac {1}{60} \, {\left (5 \, e^{\left (-2 \, x\right )} + 3\right )} e^{\left (5 \, x\right )} + \frac {1}{12} \, e^{\left (-3 \, x\right )} + \frac {1}{20} \, e^{\left (-5 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(4*x),x, algorithm="maxima")

[Out]

1/60*(5*e^(-2*x) + 3)*e^(5*x) + 1/12*e^(-3*x) + 1/20*e^(-5*x)

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mupad [B]  time = 1.44, size = 14, normalized size = 0.82 \[ \frac {4\,{\mathrm {cosh}\relax (x)}^3\,\left (6\,{\mathrm {cosh}\relax (x)}^2-5\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(4*x)*cosh(x),x)

[Out]

(4*cosh(x)^3*(6*cosh(x)^2 - 5))/15

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sympy [A]  time = 0.41, size = 20, normalized size = 1.18 \[ - \frac {\sinh {\relax (x )} \sinh {\left (4 x \right )}}{15} + \frac {4 \cosh {\relax (x )} \cosh {\left (4 x \right )}}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(4*x),x)

[Out]

-sinh(x)*sinh(4*x)/15 + 4*cosh(x)*cosh(4*x)/15

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