3.1.9 \(\int x^2 \cosh ^2(a+b x^2) \, dx\) [9]

Optimal. Leaf size=99 \[ \frac {x^3}{6}+\frac {e^{-2 a} \sqrt {\frac {\pi }{2}} \text {Erf}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}-\frac {e^{2 a} \sqrt {\frac {\pi }{2}} \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}+\frac {x \sinh \left (2 a+2 b x^2\right )}{8 b} \]

[Out]

1/6*x^3+1/8*x*sinh(2*b*x^2+2*a)/b+1/64*erf(x*2^(1/2)*b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/exp(2*a)-1/64*exp(2*a)*
erfi(x*2^(1/2)*b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5449, 5433, 5406, 2235, 2236} \begin {gather*} \frac {\sqrt {\frac {\pi }{2}} e^{-2 a} \text {Erf}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}-\frac {\sqrt {\frac {\pi }{2}} e^{2 a} \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}+\frac {x \sinh \left (2 a+2 b x^2\right )}{8 b}+\frac {x^3}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Cosh[a + b*x^2]^2,x]

[Out]

x^3/6 + (Sqrt[Pi/2]*Erf[Sqrt[2]*Sqrt[b]*x])/(32*b^(3/2)*E^(2*a)) - (E^(2*a)*Sqrt[Pi/2]*Erfi[Sqrt[2]*Sqrt[b]*x]
)/(32*b^(3/2)) + (x*Sinh[2*a + 2*b*x^2])/(8*b)

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2236

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erf[(c + d*x)*Rt[(-b)*Log[F],
 2]]/(2*d*Rt[(-b)*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5406

Int[Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] - Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 5433

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sinh[c +
d*x^n]/(d*n)), x] - Dist[e^n*((m - n + 1)/(d*n)), Int[(e*x)^(m - n)*Sinh[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[0, n, m + 1]

Rule 5449

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(
e*x)^m, (a + b*Cosh[c + d*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[p, 1] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^2 \cosh ^2\left (a+b x^2\right ) \, dx &=\int \left (\frac {x^2}{2}+\frac {1}{2} x^2 \cosh \left (2 a+2 b x^2\right )\right ) \, dx\\ &=\frac {x^3}{6}+\frac {1}{2} \int x^2 \cosh \left (2 a+2 b x^2\right ) \, dx\\ &=\frac {x^3}{6}+\frac {x \sinh \left (2 a+2 b x^2\right )}{8 b}-\frac {\int \sinh \left (2 a+2 b x^2\right ) \, dx}{8 b}\\ &=\frac {x^3}{6}+\frac {x \sinh \left (2 a+2 b x^2\right )}{8 b}+\frac {\int e^{-2 a-2 b x^2} \, dx}{16 b}-\frac {\int e^{2 a+2 b x^2} \, dx}{16 b}\\ &=\frac {x^3}{6}+\frac {e^{-2 a} \sqrt {\frac {\pi }{2}} \text {erf}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}-\frac {e^{2 a} \sqrt {\frac {\pi }{2}} \text {erfi}\left (\sqrt {2} \sqrt {b} x\right )}{32 b^{3/2}}+\frac {x \sinh \left (2 a+2 b x^2\right )}{8 b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.16, size = 101, normalized size = 1.02 \begin {gather*} \frac {3 \sqrt {2 \pi } \text {Erf}\left (\sqrt {2} \sqrt {b} x\right ) (\cosh (2 a)-\sinh (2 a))-3 \sqrt {2 \pi } \text {Erfi}\left (\sqrt {2} \sqrt {b} x\right ) (\cosh (2 a)+\sinh (2 a))+8 \sqrt {b} x \left (4 b x^2+3 \sinh \left (2 \left (a+b x^2\right )\right )\right )}{192 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cosh[a + b*x^2]^2,x]

[Out]

(3*Sqrt[2*Pi]*Erf[Sqrt[2]*Sqrt[b]*x]*(Cosh[2*a] - Sinh[2*a]) - 3*Sqrt[2*Pi]*Erfi[Sqrt[2]*Sqrt[b]*x]*(Cosh[2*a]
 + Sinh[2*a]) + 8*Sqrt[b]*x*(4*b*x^2 + 3*Sinh[2*(a + b*x^2)]))/(192*b^(3/2))

________________________________________________________________________________________

Maple [A]
time = 1.52, size = 90, normalized size = 0.91

method result size
risch \(\frac {x^{3}}{6}-\frac {{\mathrm e}^{-2 a} x \,{\mathrm e}^{-2 x^{2} b}}{16 b}+\frac {{\mathrm e}^{-2 a} \sqrt {\pi }\, \sqrt {2}\, \erf \left (x \sqrt {2}\, \sqrt {b}\right )}{64 b^{\frac {3}{2}}}+\frac {{\mathrm e}^{2 a} x \,{\mathrm e}^{2 x^{2} b}}{16 b}-\frac {{\mathrm e}^{2 a} \sqrt {\pi }\, \erf \left (\sqrt {-2 b}\, x \right )}{32 b \sqrt {-2 b}}\) \(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3-1/16*exp(-2*a)/b*x*exp(-2*x^2*b)+1/64*exp(-2*a)/b^(3/2)*Pi^(1/2)*2^(1/2)*erf(x*2^(1/2)*b^(1/2))+1/16*e
xp(2*a)/b*x*exp(2*x^2*b)-1/32*exp(2*a)/b*Pi^(1/2)/(-2*b)^(1/2)*erf((-2*b)^(1/2)*x)

________________________________________________________________________________________

Maxima [A]
time = 0.48, size = 95, normalized size = 0.96 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {-b} x\right ) e^{\left (2 \, a\right )}}{64 \, \sqrt {-b} b} + \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\sqrt {2} \sqrt {b} x\right ) e^{\left (-2 \, a\right )}}{64 \, b^{\frac {3}{2}}} + \frac {x e^{\left (2 \, b x^{2} + 2 \, a\right )}}{16 \, b} - \frac {x e^{\left (-2 \, b x^{2} - 2 \, a\right )}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/64*sqrt(2)*sqrt(pi)*erf(sqrt(2)*sqrt(-b)*x)*e^(2*a)/(sqrt(-b)*b) + 1/64*sqrt(2)*sqrt(pi)*erf(sqrt(
2)*sqrt(b)*x)*e^(-2*a)/b^(3/2) + 1/16*x*e^(2*b*x^2 + 2*a)/b - 1/16*x*e^(-2*b*x^2 - 2*a)/b

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 427 vs. \(2 (71) = 142\).
time = 0.39, size = 427, normalized size = 4.31 \begin {gather*} \frac {32 \, b^{2} x^{3} \cosh \left (b x^{2} + a\right )^{2} + 12 \, b x \cosh \left (b x^{2} + a\right )^{4} + 48 \, b x \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{3} + 12 \, b x \sinh \left (b x^{2} + a\right )^{4} + 3 \, \sqrt {2} \sqrt {\pi } {\left (\cosh \left (b x^{2} + a\right )^{2} \cosh \left (2 \, a\right ) + {\left (\cosh \left (2 \, a\right ) + \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} + \cosh \left (b x^{2} + a\right )^{2} \sinh \left (2 \, a\right ) + 2 \, {\left (\cosh \left (b x^{2} + a\right ) \cosh \left (2 \, a\right ) + \cosh \left (b x^{2} + a\right ) \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )\right )} \sqrt {-b} \operatorname {erf}\left (\sqrt {2} \sqrt {-b} x\right ) + 3 \, \sqrt {2} \sqrt {\pi } {\left (\cosh \left (b x^{2} + a\right )^{2} \cosh \left (2 \, a\right ) + {\left (\cosh \left (2 \, a\right ) - \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} - \cosh \left (b x^{2} + a\right )^{2} \sinh \left (2 \, a\right ) + 2 \, {\left (\cosh \left (b x^{2} + a\right ) \cosh \left (2 \, a\right ) - \cosh \left (b x^{2} + a\right ) \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )\right )} \sqrt {b} \operatorname {erf}\left (\sqrt {2} \sqrt {b} x\right ) + 8 \, {\left (4 \, b^{2} x^{3} + 9 \, b x \cosh \left (b x^{2} + a\right )^{2}\right )} \sinh \left (b x^{2} + a\right )^{2} - 12 \, b x + 16 \, {\left (4 \, b^{2} x^{3} \cosh \left (b x^{2} + a\right ) + 3 \, b x \cosh \left (b x^{2} + a\right )^{3}\right )} \sinh \left (b x^{2} + a\right )}{192 \, {\left (b^{2} \cosh \left (b x^{2} + a\right )^{2} + 2 \, b^{2} \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right ) + b^{2} \sinh \left (b x^{2} + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(32*b^2*x^3*cosh(b*x^2 + a)^2 + 12*b*x*cosh(b*x^2 + a)^4 + 48*b*x*cosh(b*x^2 + a)*sinh(b*x^2 + a)^3 + 12
*b*x*sinh(b*x^2 + a)^4 + 3*sqrt(2)*sqrt(pi)*(cosh(b*x^2 + a)^2*cosh(2*a) + (cosh(2*a) + sinh(2*a))*sinh(b*x^2
+ a)^2 + cosh(b*x^2 + a)^2*sinh(2*a) + 2*(cosh(b*x^2 + a)*cosh(2*a) + cosh(b*x^2 + a)*sinh(2*a))*sinh(b*x^2 +
a))*sqrt(-b)*erf(sqrt(2)*sqrt(-b)*x) + 3*sqrt(2)*sqrt(pi)*(cosh(b*x^2 + a)^2*cosh(2*a) + (cosh(2*a) - sinh(2*a
))*sinh(b*x^2 + a)^2 - cosh(b*x^2 + a)^2*sinh(2*a) + 2*(cosh(b*x^2 + a)*cosh(2*a) - cosh(b*x^2 + a)*sinh(2*a))
*sinh(b*x^2 + a))*sqrt(b)*erf(sqrt(2)*sqrt(b)*x) + 8*(4*b^2*x^3 + 9*b*x*cosh(b*x^2 + a)^2)*sinh(b*x^2 + a)^2 -
 12*b*x + 16*(4*b^2*x^3*cosh(b*x^2 + a) + 3*b*x*cosh(b*x^2 + a)^3)*sinh(b*x^2 + a))/(b^2*cosh(b*x^2 + a)^2 + 2
*b^2*cosh(b*x^2 + a)*sinh(b*x^2 + a) + b^2*sinh(b*x^2 + a)^2)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \cosh ^{2}{\left (a + b x^{2} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*cosh(a + b*x**2)**2, x)

________________________________________________________________________________________

Giac [A]
time = 0.41, size = 97, normalized size = 0.98 \begin {gather*} \frac {1}{6} \, x^{3} + \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\sqrt {2} \sqrt {-b} x\right ) e^{\left (2 \, a\right )}}{64 \, \sqrt {-b} b} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\sqrt {2} \sqrt {b} x\right ) e^{\left (-2 \, a\right )}}{64 \, b^{\frac {3}{2}}} + \frac {x e^{\left (2 \, b x^{2} + 2 \, a\right )}}{16 \, b} - \frac {x e^{\left (-2 \, b x^{2} - 2 \, a\right )}}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 + 1/64*sqrt(2)*sqrt(pi)*erf(-sqrt(2)*sqrt(-b)*x)*e^(2*a)/(sqrt(-b)*b) - 1/64*sqrt(2)*sqrt(pi)*erf(-sqr
t(2)*sqrt(b)*x)*e^(-2*a)/b^(3/2) + 1/16*x*e^(2*b*x^2 + 2*a)/b - 1/16*x*e^(-2*b*x^2 - 2*a)/b

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\mathrm {cosh}\left (b\,x^2+a\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cosh(a + b*x^2)^2,x)

[Out]

int(x^2*cosh(a + b*x^2)^2, x)

________________________________________________________________________________________