\(\int x^2 \cosh (x^3) \, dx\) [23]

Optimal result

Integrand size = 8, antiderivative size = 8 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {\sinh \left (x^3\right )}{3} \]

[Out]

1/3*sinh(x^3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5429, 2717} \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {\sinh \left (x^3\right )}{3} \]

[In]

Int[x^2*Cosh[x^3],x]

[Out]

Sinh[x^3]/3

Rule 2717

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

Rule 5429

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*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \cosh (x) \, dx,x,x^3\right ) \\ & = \frac {\sinh \left (x^3\right )}{3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.00 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {\sinh \left (x^3\right )}{3} \]

[In]

Integrate[x^2*Cosh[x^3],x]

[Out]

Sinh[x^3]/3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\sinh \left (x^{3}\right )}{3}\)	\(7\)
default	\(\frac {\sinh \left (x^{3}\right )}{3}\)	\(7\)
meijerg	\(\frac {\sinh \left (x^{3}\right )}{3}\)	\(7\)
parallelrisch	\(\frac {\sinh \left (x^{3}\right )}{3}\)	\(7\)
risch	\(\frac {{\mathrm e}^{x^{3}}}{6}-\frac {{\mathrm e}^{-x^{3}}}{6}\)	\(16\)

[In]

int(x^2*cosh(x^3),x,method=_RETURNVERBOSE)

[Out]

1/3*sinh(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {1}{3} \, \sinh \left (x^{3}\right ) \]

[In]

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

[Out]

1/3*sinh(x^3)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 5, normalized size of antiderivative = 0.62 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {\sinh {\left (x^{3} \right )}}{3} \]

[In]

integrate(x**2*cosh(x**3),x)

[Out]

sinh(x**3)/3

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {1}{3} \, \sinh \left (x^{3}\right ) \]

[In]

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

[Out]

1/3*sinh(x^3)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 15 vs. \(2 (6) = 12\).

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.88 \[ \int x^2 \cosh \left (x^3\right ) \, dx=-\frac {1}{6} \, e^{\left (-x^{3}\right )} + \frac {1}{6} \, e^{\left (x^{3}\right )} \]

[In]

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

[Out]

-1/6*e^(-x^3) + 1/6*e^(x^3)

Mupad [B] (verification not implemented)

Time = 1.51 (sec) , antiderivative size = 6, normalized size of antiderivative = 0.75 \[ \int x^2 \cosh \left (x^3\right ) \, dx=\frac {\mathrm {sinh}\left (x^3\right )}{3} \]

[In]

int(x^2*cosh(x^3),x)

[Out]

sinh(x^3)/3