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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5320, 3296, 2637} \[ \frac {x^2 \cosh \left (a+b x^2\right )}{2 b}-\frac {\sinh \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[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 2637

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

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 5320

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 0.91 \[ \frac {b x^2 \cosh \left (a+b x^2\right )-\sinh \left (a+b x^2\right )}{2 b^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.50, size = 29, normalized size = 0.85 \[ \frac {b x^{2} \cosh \left (b x^{2} + a\right ) - \sinh \left (b x^{2} + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.26, size = 47, normalized size = 1.38 \[ \frac {\frac {{\left (b x^{2} - 1\right )} e^{\left (b x^{2} + a\right )}}{b} + \frac {{\left (b x^{2} + 1\right )} e^{\left (-b x^{2} - a\right )}}{b}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*sinh(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

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maple [A]  time = 0.02, size = 45, normalized size = 1.32 \[ \frac {\left (b \,x^{2}-1\right ) {\mathrm e}^{b \,x^{2}+a}}{4 b^{2}}+\frac {\left (b \,x^{2}+1\right ) {\mathrm e}^{-b \,x^{2}-a}}{4 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(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 = 0.31, size = 81, normalized size = 2.38 \[ \frac {1}{4} \, x^{4} \sinh \left (b x^{2} + a\right ) - \frac {1}{8} \, b {\left (\frac {{\left (b^{2} x^{4} e^{a} - 2 \, b x^{2} e^{a} + 2 \, e^{a}\right )} e^{\left (b x^{2}\right )}}{b^{3}} - \frac {{\left (b^{2} x^{4} + 2 \, b x^{2} + 2\right )} e^{\left (-b x^{2} - a\right )}}{b^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.09, size = 28, normalized size = 0.82 \[ -\frac {\mathrm {sinh}\left (b\,x^2+a\right )-b\,x^2\,\mathrm {cosh}\left (b\,x^2+a\right )}{2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.80, size = 36, normalized size = 1.06 \[ \begin {cases} \frac {x^{2} \cosh {\left (a + b x^{2} \right )}}{2 b} - \frac {\sinh {\left (a + b x^{2} \right )}}{2 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh {\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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