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

Optimal result

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

[Out]

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

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 = {5428, 3377, 2717} \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=\frac {x^2 \cosh \left (a+b x^2\right )}{2 b}-\frac {\sinh \left (a+b x^2\right )}{2 b^2} \]

[In]

Int[x^3*Sinh[a + b*x^2],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 5428

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[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 \sinh (a+b x) \, dx,x,x^2\right ) \\ & = \frac {x^2 \cosh \left (a+b x^2\right )}{2 b}-\frac {\text {Subst}\left (\int \cosh (a+b x) \, dx,x,x^2\right )}{2 b} \\ & = \frac {x^2 \cosh \left (a+b x^2\right )}{2 b}-\frac {\sinh \left (a+b x^2\right )}{2 b^2} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

Integrate[x^3*Sinh[a + b*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {\cosh \left (x^{2} b +a \right ) b \,x^{2}-\sinh \left (x^{2} b +a \right )}{2 b^{2}}\)	\(30\)
risch	\(\frac {\left (x^{2} b -1\right ) {\mathrm e}^{x^{2} b +a}}{4 b^{2}}+\frac {\left (x^{2} b +1\right ) {\mathrm e}^{-x^{2} b -a}}{4 b^{2}}\)	\(45\)
meijerg	\(-\frac {\sinh \left (a \right ) \sqrt {\pi }\, \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (x^{2} b \right )}{2 \sqrt {\pi }}-\frac {x^{2} b \sinh \left (x^{2} b \right )}{2 \sqrt {\pi }}\right )}{b^{2}}+\frac {\cosh \left (a \right ) \left (\cosh \left (x^{2} b \right ) x^{2} b -\sinh \left (x^{2} b \right )\right )}{2 b^{2}}\)	\(71\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=\frac {b x^{2} \cosh \left (b x^{2} + a\right ) - \sinh \left (b x^{2} + a\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=\begin {cases} \frac {x^{2} \cosh {\left (a + b x^{2} \right )}}{2 b} - \frac {\sinh {\left (a + b x^{2} \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh {\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (30) = 60\).

Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.38 \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=\frac {1}{4} \, x^{4} \sinh \left (b x^{2} + a\right ) - \frac {1}{8} \, b {\left (\frac {{\left (b^{2} x^{4} e^{a} - 2 \, b x^{2} e^{a} + 2 \, e^{a}\right )} e^{\left (b x^{2}\right )}}{b^{3}} - \frac {{\left (b^{2} x^{4} + 2 \, b x^{2} + 2\right )} e^{\left (-b x^{2} - a\right )}}{b^{3}}\right )} \]

[In]

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

[Out]

1/4*x^4*sinh(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. 73 vs. \(2 (30) = 60\).

Time = 0.36 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.15 \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=\frac {{\left (b x^{2} + a - 1\right )} e^{\left (b x^{2} + a\right )} + {\left (b x^{2} + a + 1\right )} e^{\left (-b x^{2} - a\right )}}{4 \, b^{2}} - \frac {a e^{\left (b x^{2} + a\right )} + a e^{\left (-b x^{2} - a\right )}}{4 \, b^{2}} \]

[In]

integrate(x^3*sinh(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 = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int x^3 \sinh \left (a+b x^2\right ) \, dx=-\frac {\mathrm {sinh}\left (b\,x^2+a\right )-b\,x^2\,\mathrm {cosh}\left (b\,x^2+a\right )}{2\,b^2} \]

[In]

int(x^3*sinh(a + b*x^2),x)

[Out]

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