3.13 \(\int \frac{\sinh ^2(a+b x^2)}{x^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-2 a} \sqrt{b} \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )+\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{2 a} \sqrt{b} \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )-\frac{\sinh ^2\left (a+b x^2\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.0676429, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {5330, 5617, 5314, 5298, 2204, 2205} \[ -\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-2 a} \sqrt{b} \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )+\frac{1}{2} \sqrt{\frac{\pi }{2}} e^{2 a} \sqrt{b} \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )-\frac{\sinh ^2\left (a+b x^2\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]^2/x^2,x]

[Out]

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

Rule 5330

Int[(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_)]^(p_), x_Symbol] :> -Simp[Sinh[a + b*x^n]^p/((n - 1)*x^(n - 1)), x
] + Dist[(b*n*p)/(n - 1), Int[Sinh[a + b*x^n]^(p - 1)*Cosh[a + b*x^n], x], x] /; FreeQ[{a, b}, x] && IntegersQ
[n, p] && EqQ[m + n, 0] && GtQ[p, 1] && NeQ[n, 1]

Rule 5617

Int[Cosh[w_]^(p_.)*(u_.)*Sinh[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sinh[2*v]^p, x], x] /; EqQ[w, v] && In
tegerQ[p]

Rule 5314

Int[((a_.) + (b_.)*Sinh[u_])^(p_.), x_Symbol] :> Int[(a + b*Sinh[ExpandToSum[u, x]])^p, x] /; FreeQ[{a, b, p},
x] && BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rule 5298

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 2204

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 2205

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]

Rubi steps

\begin{align*} \int \frac{\sinh ^2\left (a+b x^2\right )}{x^2} \, dx &=-\frac{\sinh ^2\left (a+b x^2\right )}{x}+(4 b) \int \cosh \left (a+b x^2\right ) \sinh \left (a+b x^2\right ) \, dx\\ &=-\frac{\sinh ^2\left (a+b x^2\right )}{x}+(2 b) \int \sinh \left (2 \left (a+b x^2\right )\right ) \, dx\\ &=-\frac{\sinh ^2\left (a+b x^2\right )}{x}+(2 b) \int \sinh \left (2 a+2 b x^2\right ) \, dx\\ &=-\frac{\sinh ^2\left (a+b x^2\right )}{x}-b \int e^{-2 a-2 b x^2} \, dx+b \int e^{2 a+2 b x^2} \, dx\\ &=-\frac{1}{2} \sqrt{b} e^{-2 a} \sqrt{\frac{\pi }{2}} \text{erf}\left (\sqrt{2} \sqrt{b} x\right )+\frac{1}{2} \sqrt{b} e^{2 a} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\sqrt{2} \sqrt{b} x\right )-\frac{\sinh ^2\left (a+b x^2\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.225957, size = 94, normalized size = 1.07 \[ \frac{\sqrt{2 \pi } \sqrt{b} x (\sinh (2 a)-\cosh (2 a)) \text{Erf}\left (\sqrt{2} \sqrt{b} x\right )+\sqrt{2 \pi } \sqrt{b} x (\sinh (2 a)+\cosh (2 a)) \text{Erfi}\left (\sqrt{2} \sqrt{b} x\right )-4 \sinh ^2\left (a+b x^2\right )}{4 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]^2/x^2,x]

[Out]

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

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Maple [A]  time = 0.038, size = 86, normalized size = 1. \begin{align*}{\frac{1}{2\,x}}-{\frac{{{\rm e}^{-2\,a}}{{\rm e}^{-2\,b{x}^{2}}}}{4\,x}}-{\frac{{{\rm e}^{-2\,a}}\sqrt{\pi }\sqrt{2}}{4}\sqrt{b}{\it Erf} \left ( x\sqrt{2}\sqrt{b} \right ) }-{\frac{{{\rm e}^{2\,a}}{{\rm e}^{2\,b{x}^{2}}}}{4\,x}}+{\frac{{{\rm e}^{2\,a}}b\sqrt{\pi }}{2}{\it Erf} \left ( \sqrt{-2\,b}x \right ){\frac{1}{\sqrt{-2\,b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x^2+a)^2/x^2,x)

[Out]

1/2/x-1/4*exp(-2*a)/x*exp(-2*b*x^2)-1/4*exp(-2*a)*b^(1/2)*Pi^(1/2)*2^(1/2)*erf(x*2^(1/2)*b^(1/2))-1/4*exp(2*a)
/x*exp(2*b*x^2)+1/2*exp(2*a)*b*Pi^(1/2)/(-2*b)^(1/2)*erf((-2*b)^(1/2)*x)

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Maxima [A]  time = 1.17014, size = 82, normalized size = 0.93 \begin{align*} -\frac{\sqrt{2} \sqrt{b x^{2}} e^{\left (-2 \, a\right )} \Gamma \left (-\frac{1}{2}, 2 \, b x^{2}\right )}{8 \, x} - \frac{\sqrt{2} \sqrt{-b x^{2}} e^{\left (2 \, a\right )} \Gamma \left (-\frac{1}{2}, -2 \, b x^{2}\right )}{8 \, x} + \frac{1}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(b*x^2)*e^(-2*a)*gamma(-1/2, 2*b*x^2)/x - 1/8*sqrt(2)*sqrt(-b*x^2)*e^(2*a)*gamma(-1/2, -2*b*x
^2)/x + 1/2/x

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Fricas [B]  time = 1.86221, size = 1058, normalized size = 12.02 \begin{align*} -\frac{\cosh \left (b x^{2} + a\right )^{4} + 4 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{3} + \sinh \left (b x^{2} + a\right )^{4} + \sqrt{2} \sqrt{\pi }{\left (x \cosh \left (b x^{2} + a\right )^{2} \cosh \left (2 \, a\right ) + x \cosh \left (b x^{2} + a\right )^{2} \sinh \left (2 \, a\right ) +{\left (x \cosh \left (2 \, a\right ) + x \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} + 2 \,{\left (x \cosh \left (b x^{2} + a\right ) \cosh \left (2 \, a\right ) + x \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 ) + \sqrt{2} \sqrt{\pi }{\left (x \cosh \left (b x^{2} + a\right )^{2} \cosh \left (2 \, a\right ) - x \cosh \left (b x^{2} + a\right )^{2} \sinh \left (2 \, a\right ) +{\left (x \cosh \left (2 \, a\right ) - x \sinh \left (2 \, a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} + 2 \,{\left (x \cosh \left (b x^{2} + a\right ) \cosh \left (2 \, a\right ) - x \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 ) + 2 \,{\left (3 \, \cosh \left (b x^{2} + a\right )^{2} - 1\right )} \sinh \left (b x^{2} + a\right )^{2} - 2 \, \cosh \left (b x^{2} + a\right )^{2} + 4 \,{\left (\cosh \left (b x^{2} + a\right )^{3} - \cosh \left (b x^{2} + a\right )\right )} \sinh \left (b x^{2} + a\right ) + 1}{4 \,{\left (x \cosh \left (b x^{2} + a\right )^{2} + 2 \, x \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right ) + x \sinh \left (b x^{2} + a\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(cosh(b*x^2 + a)^4 + 4*cosh(b*x^2 + a)*sinh(b*x^2 + a)^3 + sinh(b*x^2 + a)^4 + sqrt(2)*sqrt(pi)*(x*cosh(b
*x^2 + a)^2*cosh(2*a) + x*cosh(b*x^2 + a)^2*sinh(2*a) + (x*cosh(2*a) + x*sinh(2*a))*sinh(b*x^2 + a)^2 + 2*(x*c
osh(b*x^2 + a)*cosh(2*a) + x*cosh(b*x^2 + a)*sinh(2*a))*sinh(b*x^2 + a))*sqrt(-b)*erf(sqrt(2)*sqrt(-b)*x) + sq
rt(2)*sqrt(pi)*(x*cosh(b*x^2 + a)^2*cosh(2*a) - x*cosh(b*x^2 + a)^2*sinh(2*a) + (x*cosh(2*a) - x*sinh(2*a))*si
nh(b*x^2 + a)^2 + 2*(x*cosh(b*x^2 + a)*cosh(2*a) - x*cosh(b*x^2 + a)*sinh(2*a))*sinh(b*x^2 + a))*sqrt(b)*erf(s
qrt(2)*sqrt(b)*x) + 2*(3*cosh(b*x^2 + a)^2 - 1)*sinh(b*x^2 + a)^2 - 2*cosh(b*x^2 + a)^2 + 4*(cosh(b*x^2 + a)^3
- cosh(b*x^2 + a))*sinh(b*x^2 + a) + 1)/(x*cosh(b*x^2 + a)^2 + 2*x*cosh(b*x^2 + a)*sinh(b*x^2 + a) + x*sinh(b
*x^2 + a)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)**2/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (b x^{2} + a\right )^{2}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(b*x^2 + a)^2/x^2, x)