3.1.15 \(\int x^3 \sinh ^3(a+b x^2) \, dx\) [15]

Optimal. Leaf size=79 \[ -\frac {x^2 \cosh \left (a+b x^2\right )}{3 b}+\frac {\sinh \left (a+b x^2\right )}{3 b^2}+\frac {x^2 \cosh \left (a+b x^2\right ) \sinh ^2\left (a+b x^2\right )}{6 b}-\frac {\sinh ^3\left (a+b x^2\right )}{18 b^2} \]

[Out]

-1/3*x^2*cosh(b*x^2+a)/b+1/3*sinh(b*x^2+a)/b^2+1/6*x^2*cosh(b*x^2+a)*sinh(b*x^2+a)^2/b-1/18*sinh(b*x^2+a)^3/b^
2

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Rubi [A]
time = 0.06, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5428, 3391, 3377, 2717} \begin {gather*} -\frac {\sinh ^3\left (a+b x^2\right )}{18 b^2}+\frac {\sinh \left (a+b x^2\right )}{3 b^2}-\frac {x^2 \cosh \left (a+b x^2\right )}{3 b}+\frac {x^2 \sinh ^2\left (a+b x^2\right ) \cosh \left (a+b x^2\right )}{6 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sinh[a + b*x^2]^3,x]

[Out]

-1/3*(x^2*Cosh[a + b*x^2])/b + Sinh[a + b*x^2]/(3*b^2) + (x^2*Cosh[a + b*x^2]*Sinh[a + b*x^2]^2)/(6*b) - Sinh[
a + b*x^2]^3/(18*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 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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*} \int x^3 \sinh ^3\left (a+b x^2\right ) \, dx &=\frac {1}{2} \text {Subst}\left (\int x \sinh ^3(a+b x) \, dx,x,x^2\right )\\ &=\frac {x^2 \cosh \left (a+b x^2\right ) \sinh ^2\left (a+b x^2\right )}{6 b}-\frac {\sinh ^3\left (a+b x^2\right )}{18 b^2}-\frac {1}{3} \text {Subst}\left (\int x \sinh (a+b x) \, dx,x,x^2\right )\\ &=-\frac {x^2 \cosh \left (a+b x^2\right )}{3 b}+\frac {x^2 \cosh \left (a+b x^2\right ) \sinh ^2\left (a+b x^2\right )}{6 b}-\frac {\sinh ^3\left (a+b x^2\right )}{18 b^2}+\frac {\text {Subst}\left (\int \cosh (a+b x) \, dx,x,x^2\right )}{3 b}\\ &=-\frac {x^2 \cosh \left (a+b x^2\right )}{3 b}+\frac {\sinh \left (a+b x^2\right )}{3 b^2}+\frac {x^2 \cosh \left (a+b x^2\right ) \sinh ^2\left (a+b x^2\right )}{6 b}-\frac {\sinh ^3\left (a+b x^2\right )}{18 b^2}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 58, normalized size = 0.73 \begin {gather*} -\frac {27 b x^2 \cosh \left (a+b x^2\right )-3 b x^2 \cosh \left (3 \left (a+b x^2\right )\right )-27 \sinh \left (a+b x^2\right )+\sinh \left (3 \left (a+b x^2\right )\right )}{72 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sinh[a + b*x^2]^3,x]

[Out]

-1/72*(27*b*x^2*Cosh[a + b*x^2] - 3*b*x^2*Cosh[3*(a + b*x^2)] - 27*Sinh[a + b*x^2] + Sinh[3*(a + b*x^2)])/b^2

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Maple [A]
time = 0.95, size = 93, normalized size = 1.18

method	result	size
risch	\(\frac {\left (3 x^{2} b -1\right ) {\mathrm e}^{3 x^{2} b +3 a}}{144 b^{2}}-\frac {3 \left (x^{2} b -1\right ) {\mathrm e}^{x^{2} b +a}}{16 b^{2}}-\frac {3 \left (x^{2} b +1\right ) {\mathrm e}^{-x^{2} b -a}}{16 b^{2}}+\frac {\left (3 x^{2} b +1\right ) {\mathrm e}^{-3 x^{2} b -3 a}}{144 b^{2}}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(3*b*x^2-1)/b^2*exp(3*b*x^2+3*a)-3/16*(b*x^2-1)/b^2*exp(b*x^2+a)-3/16*(b*x^2+1)/b^2*exp(-b*x^2-a)+1/144*
(3*b*x^2+1)/b^2*exp(-3*b*x^2-3*a)

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Maxima [A]
time = 0.27, size = 100, normalized size = 1.27 \begin {gather*} \frac {{\left (3 \, b x^{2} e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x^{2}\right )}}{144 \, b^{2}} - \frac {3 \, {\left (b x^{2} e^{a} - e^{a}\right )} e^{\left (b x^{2}\right )}}{16 \, b^{2}} - \frac {3 \, {\left (b x^{2} + 1\right )} e^{\left (-b x^{2} - a\right )}}{16 \, b^{2}} + \frac {{\left (3 \, b x^{2} + 1\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{144 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(3*b*x^2*e^(3*a) - e^(3*a))*e^(3*b*x^2)/b^2 - 3/16*(b*x^2*e^a - e^a)*e^(b*x^2)/b^2 - 3/16*(b*x^2 + 1)*e^
(-b*x^2 - a)/b^2 + 1/144*(3*b*x^2 + 1)*e^(-3*b*x^2 - 3*a)/b^2

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Fricas [A]
time = 0.41, size = 94, normalized size = 1.19 \begin {gather*} \frac {3 \, b x^{2} \cosh \left (b x^{2} + a\right )^{3} + 9 \, b x^{2} \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{2} - 27 \, b x^{2} \cosh \left (b x^{2} + a\right ) - \sinh \left (b x^{2} + a\right )^{3} - 3 \, {\left (\cosh \left (b x^{2} + a\right )^{2} - 9\right )} \sinh \left (b x^{2} + a\right )}{72 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(3*b*x^2*cosh(b*x^2 + a)^3 + 9*b*x^2*cosh(b*x^2 + a)*sinh(b*x^2 + a)^2 - 27*b*x^2*cosh(b*x^2 + a) - sinh(
b*x^2 + a)^3 - 3*(cosh(b*x^2 + a)^2 - 9)*sinh(b*x^2 + a))/b^2

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Sympy [A]
time = 0.40, size = 92, normalized size = 1.16 \begin {gather*} \begin {cases} \frac {x^{2} \sinh ^{2}{\left (a + b x^{2} \right )} \cosh {\left (a + b x^{2} \right )}}{2 b} - \frac {x^{2} \cosh ^{3}{\left (a + b x^{2} \right )}}{3 b} - \frac {7 \sinh ^{3}{\left (a + b x^{2} \right )}}{18 b^{2}} + \frac {\sinh {\left (a + b x^{2} \right )} \cosh ^{2}{\left (a + b x^{2} \right )}}{3 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \sinh ^{3}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**2*sinh(a + b*x**2)**2*cosh(a + b*x**2)/(2*b) - x**2*cosh(a + b*x**2)**3/(3*b) - 7*sinh(a + b*x**
2)**3/(18*b**2) + sinh(a + b*x**2)*cosh(a + b*x**2)**2/(3*b**2), Ne(b, 0)), (x**4*sinh(a)**3/4, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (71) = 142\).
time = 0.45, size = 192, normalized size = 2.43 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )} e^{\left (3 \, b x^{2} + 3 \, a\right )} - 27 \, {\left (b x^{2} + a\right )} e^{\left (b x^{2} + a\right )} - 27 \, {\left (b x^{2} + a\right )} e^{\left (-b x^{2} - a\right )} + 3 \, {\left (b x^{2} + a\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )} - e^{\left (3 \, b x^{2} + 3 \, a\right )} + 27 \, e^{\left (b x^{2} + a\right )} - 27 \, e^{\left (-b x^{2} - a\right )} + e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{144 \, b^{2}} - \frac {a e^{\left (3 \, b x^{2} + 3 \, a\right )} - 9 \, a e^{\left (b x^{2} + a\right )} - {\left (9 \, a e^{\left (2 \, b x^{2} + 2 \, a\right )} - a\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{48 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/144*(3*(b*x^2 + a)*e^(3*b*x^2 + 3*a) - 27*(b*x^2 + a)*e^(b*x^2 + a) - 27*(b*x^2 + a)*e^(-b*x^2 - a) + 3*(b*x
^2 + a)*e^(-3*b*x^2 - 3*a) - e^(3*b*x^2 + 3*a) + 27*e^(b*x^2 + a) - 27*e^(-b*x^2 - a) + e^(-3*b*x^2 - 3*a))/b^
2 - 1/48*(a*e^(3*b*x^2 + 3*a) - 9*a*e^(b*x^2 + a) - (9*a*e^(2*b*x^2 + 2*a) - a)*e^(-3*b*x^2 - 3*a))/b^2

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Mupad [B]
time = 0.13, size = 70, normalized size = 0.89 \begin {gather*} \frac {\frac {x^2\,{\mathrm {cosh}\left (b\,x^2+a\right )}^3}{6}-\frac {x^2\,\mathrm {cosh}\left (b\,x^2+a\right )}{2}}{b}+\frac {7\,\mathrm {sinh}\left (b\,x^2+a\right )}{18\,b^2}-\frac {{\mathrm {cosh}\left (b\,x^2+a\right )}^2\,\mathrm {sinh}\left (b\,x^2+a\right )}{18\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinh(a + b*x^2)^3,x)

[Out]

((x^2*cosh(a + b*x^2)^3)/6 - (x^2*cosh(a + b*x^2))/2)/b + (7*sinh(a + b*x^2))/(18*b^2) - (cosh(a + b*x^2)^2*si
nh(a + b*x^2))/(18*b^2)

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