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

Optimal. Leaf size=31 \[ \frac{\sinh \left (a+b x^2\right ) \cosh \left (a+b x^2\right )}{4 b}+\frac{x^2}{4} \]

[Out]

x^2/4 + (Cosh[a + b*x^2]*Sinh[a + b*x^2])/(4*b)

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Rubi [A]  time = 0.0282266, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5321, 2635, 8} \[ \frac{\sinh \left (a+b x^2\right ) \cosh \left (a+b x^2\right )}{4 b}+\frac{x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x^2]^2,x]

[Out]

x^2/4 + (Cosh[a + b*x^2]*Sinh[a + b*x^2])/(4*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x \cosh ^2\left (a+b x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \cosh ^2(a+b x) \, dx,x,x^2\right )\\ &=\frac{\cosh \left (a+b x^2\right ) \sinh \left (a+b x^2\right )}{4 b}+\frac{1}{4} \operatorname{Subst}\left (\int 1 \, dx,x,x^2\right )\\ &=\frac{x^2}{4}+\frac{\cosh \left (a+b x^2\right ) \sinh \left (a+b x^2\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0221609, size = 27, normalized size = 0.87 \[ \frac{2 \left (a+b x^2\right )+\sinh \left (2 \left (a+b x^2\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x^2]^2,x]

[Out]

(2*(a + b*x^2) + Sinh[2*(a + b*x^2)])/(8*b)

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Maple [A]  time = 0.009, size = 34, normalized size = 1.1 \begin{align*}{\frac{1}{2\,b} \left ({\frac{\cosh \left ( b{x}^{2}+a \right ) \sinh \left ( b{x}^{2}+a \right ) }{2}}+{\frac{b{x}^{2}}{2}}+{\frac{a}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03834, size = 51, normalized size = 1.65 \begin{align*} \frac{1}{4} \, x^{2} + \frac{e^{\left (2 \, b x^{2} + 2 \, a\right )}}{16 \, b} - \frac{e^{\left (-2 \, b x^{2} - 2 \, a\right )}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 + 1/16*e^(2*b*x^2 + 2*a)/b - 1/16*e^(-2*b*x^2 - 2*a)/b

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Fricas [A]  time = 1.81328, size = 66, normalized size = 2.13 \begin{align*} \frac{b x^{2} + \cosh \left (b x^{2} + a\right ) \sinh \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)^2,x, algorithm="fricas")

[Out]

1/4*(b*x^2 + cosh(b*x^2 + a)*sinh(b*x^2 + a))/b

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Sympy [A]  time = 0.546164, size = 60, normalized size = 1.94 \begin{align*} \begin{cases} - \frac{x^{2} \sinh ^{2}{\left (a + b x^{2} \right )}}{4} + \frac{x^{2} \cosh ^{2}{\left (a + b x^{2} \right )}}{4} + \frac{\sinh{\left (a + b x^{2} \right )} \cosh{\left (a + b x^{2} \right )}}{4 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \cosh ^{2}{\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)**2,x)

[Out]

Piecewise((-x**2*sinh(a + b*x**2)**2/4 + x**2*cosh(a + b*x**2)**2/4 + sinh(a + b*x**2)*cosh(a + b*x**2)/(4*b),
Ne(b, 0)), (x**2*cosh(a)**2/2, True))

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Giac [B]  time = 1.30069, size = 73, normalized size = 2.35 \begin{align*} \frac{4 \, b x^{2} -{\left (2 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} + 1\right )} e^{\left (-2 \, b x^{2} - 2 \, a\right )} + 4 \, a + e^{\left (2 \, b x^{2} + 2 \, a\right )}}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*b*x^2 - (2*e^(2*b*x^2 + 2*a) + 1)*e^(-2*b*x^2 - 2*a) + 4*a + e^(2*b*x^2 + 2*a))/b