3.17 \(\int x \sinh ^3(a+b x^2) \, dx\)

Optimal. Leaf size=33 \[ \frac{\cosh ^3\left (a+b x^2\right )}{6 b}-\frac{\cosh \left (a+b x^2\right )}{2 b} \]

[Out]

-Cosh[a + b*x^2]/(2*b) + Cosh[a + b*x^2]^3/(6*b)

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Rubi [A]  time = 0.0330922, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5320, 2633} \[ \frac{\cosh ^3\left (a+b x^2\right )}{6 b}-\frac{\cosh \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^2]^3,x]

[Out]

-Cosh[a + b*x^2]/(2*b) + Cosh[a + b*x^2]^3/(6*b)

Rule 5320

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 2633

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

Rubi steps

\begin{align*} \int x \sinh ^3\left (a+b x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \sinh ^3(a+b x) \, dx,x,x^2\right )\\ &=-\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh \left (a+b x^2\right )\right )}{2 b}\\ &=-\frac{\cosh \left (a+b x^2\right )}{2 b}+\frac{\cosh ^3\left (a+b x^2\right )}{6 b}\\ \end{align*}

Mathematica [A]  time = 0.013943, size = 33, normalized size = 1. \[ \frac{\cosh \left (3 \left (a+b x^2\right )\right )}{24 b}-\frac{3 \cosh \left (a+b x^2\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x^2]^3,x]

[Out]

(-3*Cosh[a + b*x^2])/(8*b) + Cosh[3*(a + b*x^2)]/(24*b)

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Maple [A]  time = 0.006, size = 28, normalized size = 0.9 \begin{align*}{\frac{\cosh \left ( b{x}^{2}+a \right ) }{2\,b} \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( b{x}^{2}+a \right ) \right ) ^{2}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.05527, size = 84, normalized size = 2.55 \begin{align*} \frac{e^{\left (3 \, b x^{2} + 3 \, a\right )}}{48 \, b} - \frac{3 \, e^{\left (b x^{2} + a\right )}}{16 \, b} - \frac{3 \, e^{\left (-b x^{2} - a\right )}}{16 \, b} + \frac{e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/48*e^(3*b*x^2 + 3*a)/b - 3/16*e^(b*x^2 + a)/b - 3/16*e^(-b*x^2 - a)/b + 1/48*e^(-3*b*x^2 - 3*a)/b

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Fricas [A]  time = 1.78307, size = 116, normalized size = 3.52 \begin{align*} \frac{\cosh \left (b x^{2} + a\right )^{3} + 3 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{2} - 9 \, \cosh \left (b x^{2} + a\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.40858, size = 44, normalized size = 1.33 \begin{align*} \begin{cases} \frac{\sinh ^{2}{\left (a + b x^{2} \right )} \cosh{\left (a + b x^{2} \right )}}{2 b} - \frac{\cosh ^{3}{\left (a + b x^{2} \right )}}{3 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{3}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((sinh(a + b*x**2)**2*cosh(a + b*x**2)/(2*b) - cosh(a + b*x**2)**3/(3*b), Ne(b, 0)), (x**2*sinh(a)**3
/2, True))

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Giac [A]  time = 1.22626, size = 76, normalized size = 2.3 \begin{align*} -\frac{{\left (9 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )} - e^{\left (3 \, b x^{2} + 3 \, a\right )} + 9 \, e^{\left (b x^{2} + a\right )}}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^3,x, algorithm="giac")

[Out]

-1/48*((9*e^(2*b*x^2 + 2*a) - 1)*e^(-3*b*x^2 - 3*a) - e^(3*b*x^2 + 3*a) + 9*e^(b*x^2 + a))/b