\(\int x^3 \cosh (a+b x^2) \, dx\) [1]

Optimal result

Integrand size = 12, antiderivative size = 34 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=-\frac {\cosh \left (a+b x^2\right )}{2 b^2}+\frac {x^2 \sinh \left (a+b x^2\right )}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5429, 3377, 2718} \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\frac {x^2 \sinh \left (a+b x^2\right )}{2 b}-\frac {\cosh \left (a+b x^2\right )}{2 b^2} \]

[In]

Int[x^3*Cosh[a + b*x^2],x]

[Out]

-1/2*Cosh[a + b*x^2]/b^2 + (x^2*Sinh[a + b*x^2])/(2*b)

Rule 2718

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

Rule 3377

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 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 x \cosh (a+b x) \, dx,x,x^2\right ) \\ & = \frac {x^2 \sinh \left (a+b x^2\right )}{2 b}-\frac {\text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\frac {-\cosh \left (a+b x^2\right )+b x^2 \sinh \left (a+b x^2\right )}{2 b^2} \]

[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)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(\frac {1-\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right ) x^{2} b}{b^{2} \left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )^{2}-1\right )}\)	\(41\)
risch	\(\frac {\left (b \,x^{2}-1\right ) {\mathrm e}^{b \,x^{2}+a}}{4 b^{2}}-\frac {\left (b \,x^{2}+1\right ) {\mathrm e}^{-b \,x^{2}-a}}{4 b^{2}}\)	\(45\)
meijerg	\(-\frac {\cosh \left (a \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (b \,x^{2}\right )}{2 \sqrt {\pi }}-\frac {b \,x^{2} \sinh \left (b \,x^{2}\right )}{2 \sqrt {\pi }}\right )}{b^{2}}+\frac {\sinh \left (a \right ) \left (\cosh \left (b \,x^{2}\right ) b \,x^{2}-\sinh \left (b \,x^{2}\right )\right )}{2 b^{2}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\frac {b x^{2} \sinh \left (b x^{2} + a\right ) - \cosh \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\begin {cases} \frac {x^{2} \sinh {\left (a + b x^{2} \right )}}{2 b} - \frac {\cosh {\left (a + b x^{2} \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \cosh {\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (30) = 60\).

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.35 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\frac {1}{4} \, x^{4} \cosh \left (b x^{2} + a\right ) - \frac {1}{8} \, b {\left (\frac {{\left (b^{2} x^{4} e^{a} - 2 \, b x^{2} e^{a} + 2 \, e^{a}\right )} e^{\left (b x^{2}\right )}}{b^{3}} + \frac {{\left (b^{2} x^{4} + 2 \, b x^{2} + 2\right )} e^{\left (-b x^{2} - a\right )}}{b^{3}}\right )} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (30) = 60\).

Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.21 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=\frac {{\left (b x^{2} + a - 1\right )} e^{\left (b x^{2} + a\right )} - {\left (b x^{2} + a + 1\right )} e^{\left (-b x^{2} - a\right )}}{4 \, b^{2}} - \frac {a e^{\left (b x^{2} + a\right )} - a e^{\left (-b x^{2} - a\right )}}{4 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int x^3 \cosh \left (a+b x^2\right ) \, dx=-\frac {\mathrm {cosh}\left (b\,x^2+a\right )-b\,x^2\,\mathrm {sinh}\left (b\,x^2+a\right )}{2\,b^2} \]

[In]

int(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)