3.1.28 \(\int x^3 \sinh (a+b x^4) \, dx\) [28]

Optimal. Leaf size=15 \[ \frac {\cosh \left (a+b x^4\right )}{4 b} \]

[Out]

1/4*cosh(b*x^4+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5428, 2718} \begin {gather*} \frac {\cosh \left (a+b x^4\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sinh[a + b*x^4],x]

[Out]

Cosh[a + b*x^4]/(4*b)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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^3 \sinh \left (a+b x^4\right ) \, dx &=\frac {1}{4} \text {Subst}\left (\int \sinh (a+b x) \, dx,x,x^4\right )\\ &=\frac {\cosh \left (a+b x^4\right )}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} \frac {\cosh \left (a+b x^4\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sinh[a + b*x^4],x]

[Out]

Cosh[a + b*x^4]/(4*b)

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Maple [A]
time = 0.28, size = 14, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {\cosh \left (b \,x^{4}+a \right )}{4 b}\)	\(14\)
default	\(\frac {\cosh \left (b \,x^{4}+a \right )}{4 b}\)	\(14\)
risch	\(\frac {{\mathrm e}^{b \,x^{4}+a}}{8 b}+\frac {{\mathrm e}^{-b \,x^{4}-a}}{8 b}\)	\(31\)
meijerg	\(\frac {\sinh \left (a \right ) \sinh \left (b \,x^{4}\right )}{4 b}-\frac {\cosh \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (b \,x^{4}\right )}{\sqrt {\pi }}\right )}{4 b}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

1/4*cosh(b*x^4+a)/b

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Maxima [A]
time = 0.26, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cosh \left (b x^{4} + a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(b*x^4+a),x, algorithm="maxima")

[Out]

1/4*cosh(b*x^4 + a)/b

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Fricas [A]
time = 0.44, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cosh \left (b x^{4} + a\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*cosh(b*x^4 + a)/b

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Sympy [A]
time = 0.17, size = 19, normalized size = 1.27 \begin {gather*} \begin {cases} \frac {\cosh {\left (a + b x^{4} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh {\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*sinh(b*x**4+a),x)

[Out]

Piecewise((cosh(a + b*x**4)/(4*b), Ne(b, 0)), (x**4*sinh(a)/4, True))

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Giac [A]
time = 0.43, size = 25, normalized size = 1.67 \begin {gather*} \frac {e^{\left (b x^{4} + a\right )} + e^{\left (-b x^{4} - a\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(b*x^4+a),x, algorithm="giac")

[Out]

1/8*(e^(b*x^4 + a) + e^(-b*x^4 - a))/b

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Mupad [B]
time = 0.38, size = 13, normalized size = 0.87 \begin {gather*} \frac {\mathrm {cosh}\left (b\,x^4+a\right )}{4\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(a + b*x^4),x)

[Out]

cosh(a + b*x^4)/(4*b)

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