3.3 \(\int x \cosh (a+b x^2) \, dx\)

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

[Out]

Sinh[a + b*x^2]/(2*b)

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Rubi [A]  time = 0.0149193, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5321, 2637} \[ \frac{\sinh \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x^2],x]

[Out]

Sinh[a + b*x^2]/(2*b)

Rule 5321

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.0097035, size = 31, normalized size = 2.07 \[ \frac{\sinh (a) \cosh \left (b x^2\right )}{2 b}+\frac{\cosh (a) \sinh \left (b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x^2],x]

[Out]

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

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Maple [A]  time = 0.005, size = 14, normalized size = 0.9 \begin{align*}{\frac{\sinh \left ( b{x}^{2}+a \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x^2+a),x)

[Out]

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

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Maxima [A]  time = 1.10586, size = 18, normalized size = 1.2 \begin{align*} \frac{\sinh \left (b x^{2} + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sinh(b*x^2 + a)/b

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Fricas [A]  time = 1.75891, size = 31, normalized size = 2.07 \begin{align*} \frac{\sinh \left (b x^{2} + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sinh(b*x^2 + a)/b

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Sympy [A]  time = 0.233205, size = 19, normalized size = 1.27 \begin{align*} \begin{cases} \frac{\sinh{\left (a + b x^{2} \right )}}{2 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \cosh{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x**2+a),x)

[Out]

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

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Giac [A]  time = 1.31239, size = 36, normalized size = 2.4 \begin{align*} \frac{e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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