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

Optimal result

Integrand size = 14, antiderivative size = 79 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=-\frac {\cosh \left (a+b x^2\right )}{3 b^2}-\frac {\cosh ^3\left (a+b x^2\right )}{18 b^2}+\frac {x^2 \sinh \left (a+b x^2\right )}{3 b}+\frac {x^2 \cosh ^2\left (a+b x^2\right ) \sinh \left (a+b x^2\right )}{6 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5429, 3391, 3377, 2718} \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=-\frac {\cosh ^3\left (a+b x^2\right )}{18 b^2}-\frac {\cosh \left (a+b x^2\right )}{3 b^2}+\frac {x^2 \sinh \left (a+b x^2\right )}{3 b}+\frac {x^2 \sinh \left (a+b x^2\right ) \cosh ^2\left (a+b x^2\right )}{6 b} \]

[In]

Int[x^3*Cosh[a + b*x^2]^3,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.70 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=-\frac {27 \cosh \left (a+b x^2\right )+\cosh \left (3 \left (a+b x^2\right )\right )-3 b x^2 \left (9 \sinh \left (a+b x^2\right )+\sinh \left (3 \left (a+b x^2\right )\right )\right )}{72 b^2} \]

[In]

Integrate[x^3*Cosh[a + b*x^2]^3,x]

[Out]

-1/72*(27*Cosh[a + b*x^2] + Cosh[3*(a + b*x^2)] - 3*b*x^2*(9*Sinh[a + b*x^2] + Sinh[3*(a + b*x^2)]))/b^2

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {\left (3 b \,x^{2}-1\right ) {\mathrm e}^{3 b \,x^{2}+3 a}}{144 b^{2}}+\frac {3 \left (b \,x^{2}-1\right ) {\mathrm e}^{b \,x^{2}+a}}{16 b^{2}}-\frac {3 \left (b \,x^{2}+1\right ) {\mathrm e}^{-b \,x^{2}-a}}{16 b^{2}}-\frac {\left (3 b \,x^{2}+1\right ) {\mathrm e}^{-3 b \,x^{2}-3 a}}{144 b^{2}}\)	\(93\)
parallelrisch	\(\frac {-9 \tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )^{5} x^{2} b +6 \tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )^{3} x^{2} b +9 \tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )^{4}-9 \tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right ) x^{2} b -12 \tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )^{2}+7}{9 b^{2} {\left (1+\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )\right )}^{3} {\left (\tanh \left (\frac {b \,x^{2}}{2}+\frac {a}{2}\right )-1\right )}^{3}}\)	\(123\)

[In]

int(x^3*cosh(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)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.22 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=\frac {3 \, b x^{2} \sinh \left (b x^{2} + a\right )^{3} - \cosh \left (b x^{2} + a\right )^{3} - 3 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{2} + 9 \, {\left (b x^{2} \cosh \left (b x^{2} + a\right )^{2} + 3 \, b x^{2}\right )} \sinh \left (b x^{2} + a\right ) - 27 \, \cosh \left (b x^{2} + a\right )}{72 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=\begin {cases} - \frac {x^{2} \sinh ^{3}{\left (a + b x^{2} \right )}}{3 b} + \frac {x^{2} \sinh {\left (a + b x^{2} \right )} \cosh ^{2}{\left (a + b x^{2} \right )}}{2 b} + \frac {\sinh ^{2}{\left (a + b x^{2} \right )} \cosh {\left (a + b x^{2} \right )}}{3 b^{2}} - \frac {7 \cosh ^{3}{\left (a + b x^{2} \right )}}{18 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{4} \cosh ^{3}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.27 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=\frac {{\left (3 \, b x^{2} e^{\left (3 \, a\right )} - e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x^{2}\right )}}{144 \, b^{2}} + \frac {3 \, {\left (b x^{2} e^{a} - e^{a}\right )} e^{\left (b x^{2}\right )}}{16 \, b^{2}} - \frac {3 \, {\left (b x^{2} + 1\right )} e^{\left (-b x^{2} - a\right )}}{16 \, b^{2}} - \frac {{\left (3 \, b x^{2} + 1\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{144 \, b^{2}} \]

[In]

integrate(x^3*cosh(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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (71) = 142\).

Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.43 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=\frac {3 \, {\left (b x^{2} + a\right )} e^{\left (3 \, b x^{2} + 3 \, a\right )} + 27 \, {\left (b x^{2} + a\right )} e^{\left (b x^{2} + a\right )} - 27 \, {\left (b x^{2} + a\right )} e^{\left (-b x^{2} - a\right )} - 3 \, {\left (b x^{2} + a\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )} - e^{\left (3 \, b x^{2} + 3 \, a\right )} - 27 \, e^{\left (b x^{2} + a\right )} - 27 \, e^{\left (-b x^{2} - a\right )} - e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{144 \, b^{2}} - \frac {a e^{\left (3 \, b x^{2} + 3 \, a\right )} + 9 \, a e^{\left (b x^{2} + a\right )} - {\left (9 \, a e^{\left (2 \, b x^{2} + 2 \, a\right )} + a\right )} e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{48 \, b^{2}} \]

[In]

integrate(x^3*cosh(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

Mupad [B] (verification not implemented)

Time = 1.62 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.89 \[ \int x^3 \cosh ^3\left (a+b x^2\right ) \, dx=\frac {\frac {x^2\,\mathrm {sinh}\left (b\,x^2+a\right )}{3}+\frac {x^2\,{\mathrm {cosh}\left (b\,x^2+a\right )}^2\,\mathrm {sinh}\left (b\,x^2+a\right )}{6}}{b}-\frac {{\mathrm {cosh}\left (b\,x^2+a\right )}^3}{18\,b^2}-\frac {\mathrm {cosh}\left (b\,x^2+a\right )}{3\,b^2} \]

[In]

int(x^3*cosh(a + b*x^2)^3,x)

[Out]

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