3.1 \(\int x^3 \cosh (a+b x^2) \, dx\)

Optimal. Leaf size=34 \[ \frac{x^2 \sinh \left (a+b x^2\right )}{2 b}-\frac{\cosh \left (a+b x^2\right )}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.036081, antiderivative size = 34, 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, 3296, 2638} \[ \frac{x^2 \sinh \left (a+b x^2\right )}{2 b}-\frac{\cosh \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Cosh[a + b*x^2],x]

[Out]

-Cosh[a + b*x^2]/(2*b^2) + (x^2*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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

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

Rubi steps

\begin{align*} \int x^3 \cosh \left (a+b x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \cosh (a+b x) \, dx,x,x^2\right )\\ &=\frac{x^2 \sinh \left (a+b x^2\right )}{2 b}-\frac{\operatorname{Subst}\left (\int \sinh (a+b x) \, dx,x,x^2\right )}{2 b}\\ &=-\frac{\cosh \left (a+b x^2\right )}{2 b^2}+\frac{x^2 \sinh \left (a+b x^2\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0330678, size = 31, normalized size = 0.91 \[ \frac{b x^2 \sinh \left (a+b x^2\right )-\cosh \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Cosh[a + b*x^2],x]

[Out]

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

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Maple [A]  time = 0.013, size = 45, normalized size = 1.3 \begin{align*}{\frac{ \left ( b{x}^{2}-1 \right ){{\rm e}^{b{x}^{2}+a}}}{4\,{b}^{2}}}-{\frac{ \left ( b{x}^{2}+1 \right ){{\rm e}^{-b{x}^{2}-a}}}{4\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cosh(b*x^2+a),x)

[Out]

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

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Maxima [B]  time = 0.985214, size = 108, normalized size = 3.18 \begin{align*} \frac{1}{4} \, x^{4} \cosh \left (b x^{2} + a\right ) - \frac{1}{8} \, b{\left (\frac{{\left (b^{2} x^{4} e^{a} - 2 \, b x^{2} e^{a} + 2 \, e^{a}\right )} e^{\left (b x^{2}\right )}}{b^{3}} + \frac{{\left (b^{2} x^{4} + 2 \, b x^{2} + 2\right )} e^{\left (-b x^{2} - a\right )}}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x^2+a),x, algorithm="maxima")

[Out]

1/4*x^4*cosh(b*x^2 + a) - 1/8*b*((b^2*x^4*e^a - 2*b*x^2*e^a + 2*e^a)*e^(b*x^2)/b^3 + (b^2*x^4 + 2*b*x^2 + 2)*e
^(-b*x^2 - a)/b^3)

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Fricas [A]  time = 1.79538, size = 69, normalized size = 2.03 \begin{align*} \frac{b x^{2} \sinh \left (b x^{2} + a\right ) - \cosh \left (b x^{2} + a\right )}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x^2+a),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.03247, size = 36, normalized size = 1.06 \begin{align*} \begin{cases} \frac{x^{2} \sinh{\left (a + b x^{2} \right )}}{2 b} - \frac{\cosh{\left (a + b x^{2} \right )}}{2 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{4} \cosh{\left (a \right )}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*cosh(b*x**2+a),x)

[Out]

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

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Giac [A]  time = 1.29896, size = 65, normalized size = 1.91 \begin{align*} \frac{\frac{{\left (b x^{2} - 1\right )} e^{\left (b x^{2} + a\right )}}{b} - \frac{{\left (b x^{2} + 1\right )} e^{\left (-b x^{2} - a\right )}}{b}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x^2+a),x, algorithm="giac")

[Out]

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