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]

1/2*sinh(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 = {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 2637

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

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]))

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*}

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Mathematica [B]  time = 0.01, 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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fricas [A]  time = 0.45, size = 13, normalized size = 0.87 \[ \frac {\sinh \left (b x^{2} + a\right )}{2 \, b} \]

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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giac [B]  time = 0.11, size = 27, normalized size = 1.80 \[ \frac {e^{\left (b x^{2} + a\right )} - e^{\left (-b x^{2} - a\right )}}{4 \, b} \]

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

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maple [A]  time = 0.04, size = 14, normalized size = 0.93 \[ \frac {\sinh \left (b \,x^{2}+a \right )}{2 b} \]

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 = 0.32, size = 13, normalized size = 0.87 \[ \frac {\sinh \left (b x^{2} + a\right )}{2 \, b} \]

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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mupad [B]  time = 0.05, size = 13, normalized size = 0.87 \[ \frac {\mathrm {sinh}\left (b\,x^2+a\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(a + b*x^2),x)

[Out]

sinh(a + b*x^2)/(2*b)

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sympy [A]  time = 0.21, size = 19, normalized size = 1.27 \[ \begin {cases} \frac {\sinh {\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \cosh {\relax (a )}}{2} & \text {otherwise} \end {cases} \]

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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