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

Optimal. Leaf size=15 \[ \frac {\cosh \left (a+b x^2\right )}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5428, 2718} \begin {gather*} \frac {\cosh \left (a+b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^2],x]

[Out]

Cosh[a + b*x^2]/(2*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 \sinh \left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int \sinh (a+b x) \, dx,x,x^2\right )\\ &=\frac {\cosh \left (a+b x^2\right )}{2 b}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(31\) vs. \(2(15)=30\).
time = 0.01, size = 31, normalized size = 2.07 \begin {gather*} \frac {\cosh (a) \cosh \left (b x^2\right )}{2 b}+\frac {\sinh (a) \sinh \left (b x^2\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x^2],x]

[Out]

(Cosh[a]*Cosh[b*x^2])/(2*b) + (Sinh[a]*Sinh[b*x^2])/(2*b)

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Maple [A]
time = 0.23, size = 14, normalized size = 0.93

method	result	size
derivativedivides	\(\frac {\cosh \left (x^{2} b +a \right )}{2 b}\)	\(14\)
default	\(\frac {\cosh \left (x^{2} b +a \right )}{2 b}\)	\(14\)
risch	\(\frac {{\mathrm e}^{x^{2} b +a}}{4 b}+\frac {{\mathrm e}^{-x^{2} b -a}}{4 b}\)	\(31\)
meijerg	\(\frac {\sinh \left (a \right ) \sinh \left (x^{2} b \right )}{2 b}-\frac {\cosh \left (a \right ) \sqrt {\pi }\, \left (\frac {1}{\sqrt {\pi }}-\frac {\cosh \left (x^{2} b \right )}{\sqrt {\pi }}\right )}{2 b}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cosh \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cosh(b*x^2 + a)/b

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Fricas [A]
time = 0.39, size = 13, normalized size = 0.87 \begin {gather*} \frac {\cosh \left (b x^{2} + a\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cosh(b*x^2 + a)/b

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Sympy [A]
time = 0.07, size = 19, normalized size = 1.27 \begin {gather*} \begin {cases} \frac {\cosh {\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh {\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),x)

[Out]

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

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Giac [A]
time = 0.48, size = 25, normalized size = 1.67 \begin {gather*} \frac {e^{\left (b x^{2} + a\right )} + e^{\left (-b x^{2} - a\right )}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(e^(b*x^2 + a) + e^(-b*x^2 - a))/b

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Mupad [B]
time = 0.37, size = 13, normalized size = 0.87 \begin {gather*} \frac {\mathrm {cosh}\left (b\,x^2+a\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(a + b*x^2),x)

[Out]

cosh(a + b*x^2)/(2*b)

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