3.1.17 \(\int x \sinh ^3(a+b x^2) \, dx\) [17]

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

[Out]

-1/2*cosh(b*x^2+a)/b+1/6*cosh(b*x^2+a)^3/b

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Rubi [A]
time = 0.02, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 = {5428, 2713} \begin {gather*} \frac {\cosh ^3\left (a+b x^2\right )}{6 b}-\frac {\cosh \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^2]^3,x]

[Out]

-1/2*Cosh[a + b*x^2]/b + Cosh[a + b*x^2]^3/(6*b)

Rule 2713

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]

Rule 5428

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]))

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} -\frac {3 \cosh \left (a+b x^2\right )}{8 b}+\frac {\cosh \left (3 \left (a+b x^2\right )\right )}{24 b} \end {gather*}

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 = 0.35, size = 31, normalized size = 0.94

method	result	size
default	\(-\frac {3 \cosh \left (x^{2} b +a \right )}{8 b}+\frac {\cosh \left (3 x^{2} b +3 a \right )}{24 b}\)	\(31\)
risch	\(\frac {{\mathrm e}^{3 x^{2} b +3 a}}{48 b}-\frac {3 \,{\mathrm e}^{x^{2} b +a}}{16 b}-\frac {3 \,{\mathrm e}^{-x^{2} b -a}}{16 b}+\frac {{\mathrm e}^{-3 x^{2} b -3 a}}{48 b}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (29) = 58\).
time = 0.27, size = 62, normalized size = 1.88 \begin {gather*} \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 {gather*}

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 = 0.43, size = 46, normalized size = 1.39 \begin {gather*} \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 {gather*}

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 = 0.17, size = 44, normalized size = 1.33 \begin {gather*} \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 {gather*}

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 = 0.41, size = 56, normalized size = 1.70 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.06, size = 28, normalized size = 0.85 \begin {gather*} -\frac {3\,\mathrm {cosh}\left (b\,x^2+a\right )-{\mathrm {cosh}\left (b\,x^2+a\right )}^3}{6\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a + b*x^2)^3,x)

[Out]

-(3*cosh(a + b*x^2) - cosh(a + b*x^2)^3)/(6*b)

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