3.19 \(\int \frac{\sinh ^3(a+b x^2)}{x} \, dx\)

Optimal. Leaf size=55 \[ -\frac{3}{8} \sinh (a) \text{Chi}\left (b x^2\right )+\frac{1}{8} \sinh (3 a) \text{Chi}\left (3 b x^2\right )-\frac{3}{8} \cosh (a) \text{Shi}\left (b x^2\right )+\frac{1}{8} \cosh (3 a) \text{Shi}\left (3 b x^2\right ) \]

[Out]

(-3*CoshIntegral[b*x^2]*Sinh[a])/8 + (CoshIntegral[3*b*x^2]*Sinh[3*a])/8 - (3*Cosh[a]*SinhIntegral[b*x^2])/8 +
(Cosh[3*a]*SinhIntegral[3*b*x^2])/8

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Rubi [A]  time = 0.0939522, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5340, 5318, 5317, 5316} \[ -\frac{3}{8} \sinh (a) \text{Chi}\left (b x^2\right )+\frac{1}{8} \sinh (3 a) \text{Chi}\left (3 b x^2\right )-\frac{3}{8} \cosh (a) \text{Shi}\left (b x^2\right )+\frac{1}{8} \cosh (3 a) \text{Shi}\left (3 b x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]^3/x,x]

[Out]

(-3*CoshIntegral[b*x^2]*Sinh[a])/8 + (CoshIntegral[3*b*x^2]*Sinh[3*a])/8 - (3*Cosh[a]*SinhIntegral[b*x^2])/8 +
(Cosh[3*a]*SinhIntegral[3*b*x^2])/8

Rule 5340

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

Rule 5318

Int[Sinh[(c_) + (d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Dist[Sinh[c], Int[Cosh[d*x^n]/x, x], x] + Dist[Cosh[c], In
t[Sinh[d*x^n]/x, x], x] /; FreeQ[{c, d, n}, x]

Rule 5317

Int[Cosh[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CoshIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rule 5316

Int[Sinh[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinhIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

Rubi steps

\begin{align*} \int \frac{\sinh ^3\left (a+b x^2\right )}{x} \, dx &=\int \left (-\frac{3 \sinh \left (a+b x^2\right )}{4 x}+\frac{\sinh \left (3 a+3 b x^2\right )}{4 x}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\sinh \left (3 a+3 b x^2\right )}{x} \, dx-\frac{3}{4} \int \frac{\sinh \left (a+b x^2\right )}{x} \, dx\\ &=-\left (\frac{1}{4} (3 \cosh (a)) \int \frac{\sinh \left (b x^2\right )}{x} \, dx\right )+\frac{1}{4} \cosh (3 a) \int \frac{\sinh \left (3 b x^2\right )}{x} \, dx-\frac{1}{4} (3 \sinh (a)) \int \frac{\cosh \left (b x^2\right )}{x} \, dx+\frac{1}{4} \sinh (3 a) \int \frac{\cosh \left (3 b x^2\right )}{x} \, dx\\ &=-\frac{3}{8} \text{Chi}\left (b x^2\right ) \sinh (a)+\frac{1}{8} \text{Chi}\left (3 b x^2\right ) \sinh (3 a)-\frac{3}{8} \cosh (a) \text{Shi}\left (b x^2\right )+\frac{1}{8} \cosh (3 a) \text{Shi}\left (3 b x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0336521, size = 49, normalized size = 0.89 \[ \frac{1}{8} \left (-3 \sinh (a) \text{Chi}\left (b x^2\right )+\sinh (3 a) \text{Chi}\left (3 b x^2\right )-3 \cosh (a) \text{Shi}\left (b x^2\right )+\cosh (3 a) \text{Shi}\left (3 b x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]^3/x,x]

[Out]

(-3*CoshIntegral[b*x^2]*Sinh[a] + CoshIntegral[3*b*x^2]*Sinh[3*a] - 3*Cosh[a]*SinhIntegral[b*x^2] + Cosh[3*a]*
SinhIntegral[3*b*x^2])/8

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Maple [A]  time = 0.038, size = 55, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{-3\,a}}{\it Ei} \left ( 1,3\,b{x}^{2} \right ) }{16}}-{\frac{3\,{{\rm e}^{-a}}{\it Ei} \left ( 1,b{x}^{2} \right ) }{16}}-{\frac{{{\rm e}^{3\,a}}{\it Ei} \left ( 1,-3\,b{x}^{2} \right ) }{16}}+{\frac{3\,{{\rm e}^{a}}{\it Ei} \left ( 1,-b{x}^{2} \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(b*x^2+a)^3/x,x)

[Out]

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

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Maxima [A]  time = 1.32201, size = 68, normalized size = 1.24 \begin{align*} \frac{1}{16} \,{\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac{3}{16} \,{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - \frac{1}{16} \,{\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - \frac{3}{16} \,{\rm Ei}\left (b x^{2}\right ) e^{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.64296, size = 231, normalized size = 4.2 \begin{align*} \frac{1}{16} \,{\left ({\rm Ei}\left (3 \, b x^{2}\right ) -{\rm Ei}\left (-3 \, b x^{2}\right )\right )} \cosh \left (3 \, a\right ) - \frac{3}{16} \,{\left ({\rm Ei}\left (b x^{2}\right ) -{\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + \frac{1}{16} \,{\left ({\rm Ei}\left (3 \, b x^{2}\right ) +{\rm Ei}\left (-3 \, b x^{2}\right )\right )} \sinh \left (3 \, a\right ) - \frac{3}{16} \,{\left ({\rm Ei}\left (b x^{2}\right ) +{\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(Ei(3*b*x^2) - Ei(-3*b*x^2))*cosh(3*a) - 3/16*(Ei(b*x^2) - Ei(-b*x^2))*cosh(a) + 1/16*(Ei(3*b*x^2) + Ei(-
3*b*x^2))*sinh(3*a) - 3/16*(Ei(b*x^2) + Ei(-b*x^2))*sinh(a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh ^{3}{\left (a + b x^{2} \right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)**3/x, x)

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Giac [A]  time = 1.27941, size = 68, normalized size = 1.24 \begin{align*} \frac{1}{16} \,{\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac{3}{16} \,{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - \frac{1}{16} \,{\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - \frac{3}{16} \,{\rm Ei}\left (b x^{2}\right ) e^{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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