3.194 \(\int \sinh (x) \sinh (4 x) \, dx\)

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

[Out]

-1/6*sinh(3*x)+1/10*sinh(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 = {4282} \[ \frac {1}{10} \sinh (5 x)-\frac {1}{6} \sinh (3 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sinh[3*x]/6 + Sinh[5*x]/10

Rule 4282

Int[sin[(a_.) + (b_.)*(x_)]*sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[a - c + (b - d)*x]/(2*(b - d)), x]
- Simp[Sin[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 \sinh (x) \sinh (4 x) \, dx &=-\frac {1}{6} \sinh (3 x)+\frac {1}{10} \sinh (5 x)\\ \end {align*}

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

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*Sinh[3*x] + Sinh[5*x]/10

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fricas [B]  time = 0.41, size = 36, normalized size = 2.12 \[ \frac {1}{10} \, \sinh \relax (x)^{5} + \frac {1}{6} \, {\left (6 \, \cosh \relax (x)^{2} - 1\right )} \sinh \relax (x)^{3} + \frac {1}{2} \, {\left (\cosh \relax (x)^{4} - \cosh \relax (x)^{2}\right )} \sinh \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.13, 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(sinh(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 = 0.16, size = 14, normalized size = 0.82 \[ -\frac {\sinh \left (3 x \right )}{6}+\frac {\sinh \left (5 x \right )}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sinh(3*x)+1/10*sinh(5*x)

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maxima [B]  time = 0.48, 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(sinh(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 = 0.06, size = 14, normalized size = 0.82 \[ \frac {4\,{\mathrm {sinh}\relax (x)}^3\,\left (6\,{\mathrm {sinh}\relax (x)}^2+5\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*sinh(x)^3*(6*sinh(x)^2 + 5))/15

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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