\(\int x \cosh ^3(a+b x^2) \, dx\) [17]

Optimal result

Integrand size = 12, antiderivative size = 33 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {\sinh \left (a+b x^2\right )}{2 b}+\frac {\sinh ^3\left (a+b x^2\right )}{6 b} \]

[Out]

1/2*sinh(b*x^2+a)/b+1/6*sinh(b*x^2+a)^3/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5429, 2713} \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {\sinh ^3\left (a+b x^2\right )}{6 b}+\frac {\sinh \left (a+b x^2\right )}{2 b} \]

[In]

Int[x*Cosh[a + b*x^2]^3,x]

[Out]

Sinh[a + b*x^2]/(2*b) + Sinh[a + b*x^2]^3/(6*b)

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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}{2} \text {Subst}\left (\int \cosh ^3(a+b x) \, dx,x,x^2\right ) \\ & = \frac {i \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh \left (a+b x^2\right )\right )}{2 b} \\ & = \frac {\sinh \left (a+b x^2\right )}{2 b}+\frac {\sinh ^3\left (a+b x^2\right )}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {\sinh \left (a+b x^2\right )}{2 b}+\frac {\sinh ^3\left (a+b x^2\right )}{6 b} \]

[In]

Integrate[x*Cosh[a + b*x^2]^3,x]

[Out]

Sinh[a + b*x^2]/(2*b) + Sinh[a + b*x^2]^3/(6*b)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {\left (\frac {2}{3}+\frac {\cosh \left (b \,x^{2}+a \right )^{2}}{3}\right ) \sinh \left (b \,x^{2}+a \right )}{2 b}\)	\(28\)
default	\(\frac {\left (\frac {2}{3}+\frac {\cosh \left (b \,x^{2}+a \right )^{2}}{3}\right ) \sinh \left (b \,x^{2}+a \right )}{2 b}\)	\(28\)
parallelrisch	\(\frac {\sinh \left (3 b \,x^{2}+3 a \right )+9 \sinh \left (b \,x^{2}+a \right )}{24 b}\)	\(28\)
risch	\(\frac {{\mathrm e}^{3 b \,x^{2}+3 a}}{48 b}+\frac {3 \,{\mathrm e}^{b \,x^{2}+a}}{16 b}-\frac {3 \,{\mathrm e}^{-b \,x^{2}-a}}{16 b}-\frac {{\mathrm e}^{-3 b \,x^{2}-3 a}}{48 b}\)	\(63\)

[In]

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

[Out]

1/2/b*(2/3+1/3*cosh(b*x^2+a)^2)*sinh(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {\sinh \left (b x^{2} + a\right )^{3} + 3 \, {\left (\cosh \left (b x^{2} + a\right )^{2} + 3\right )} \sinh \left (b x^{2} + a\right )}{24 \, b} \]

[In]

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

[Out]

1/24*(sinh(b*x^2 + a)^3 + 3*(cosh(b*x^2 + a)^2 + 3)*sinh(b*x^2 + a))/b

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\begin {cases} - \frac {\sinh ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {\sinh {\left (a + b x^{2} \right )} \cosh ^{2}{\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \cosh ^{3}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-sinh(a + b*x**2)**3/(3*b) + sinh(a + b*x**2)*cosh(a + b*x**2)**2/(2*b), Ne(b, 0)), (x**2*cosh(a)**
3/2, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (29) = 58\).

Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.88 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {e^{\left (3 \, b x^{2} + 3 \, a\right )}}{48 \, b} + \frac {3 \, e^{\left (b x^{2} + a\right )}}{16 \, b} - \frac {3 \, e^{\left (-b x^{2} - a\right )}}{16 \, b} - \frac {e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{48 \, b} \]

[In]

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

[Out]

1/48*e^(3*b*x^2 + 3*a)/b + 3/16*e^(b*x^2 + a)/b - 3/16*e^(-b*x^2 - a)/b - 1/48*e^(-3*b*x^2 - 3*a)/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=-\frac {{\left (9 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )} - e^{\left (3 \, b x^{2} + 3 \, a\right )} - 9 \, e^{\left (b x^{2} + a\right )}}{48 \, b} \]

[In]

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

[Out]

-1/48*((9*e^(2*b*x^2 + 2*a) + 1)*e^(-3*b*x^2 - 3*a) - e^(3*b*x^2 + 3*a) - 9*e^(b*x^2 + a))/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int x \cosh ^3\left (a+b x^2\right ) \, dx=\frac {{\mathrm {sinh}\left (b\,x^2+a\right )}^3+3\,\mathrm {sinh}\left (b\,x^2+a\right )}{6\,b} \]

[In]

int(x*cosh(a + b*x^2)^3,x)

[Out]

(3*sinh(a + b*x^2) + sinh(a + b*x^2)^3)/(6*b)