\(\int \frac {\sinh ^3(a+b x^2)}{x} \, dx\) [19]

Optimal result

Integrand size = 14, antiderivative size = 55 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=-\frac {3}{8} \text {Chi}\left (b x^2\right ) \sinh (a)+\frac {1}{8} \text {Chi}\left (3 b x^2\right ) \sinh (3 a)-\frac {3}{8} \cosh (a) \text {Shi}\left (b x^2\right )+\frac {1}{8} \cosh (3 a) \text {Shi}\left (3 b x^2\right ) \]

[Out]

-3/8*cosh(a)*Shi(b*x^2)+1/8*cosh(3*a)*Shi(3*b*x^2)-3/8*Chi(b*x^2)*sinh(a)+1/8*Chi(3*b*x^2)*sinh(3*a)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5448, 5426, 5425, 5424} \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=-\frac {3}{8} \sinh (a) \text {Chi}\left (b x^2\right )+\frac {1}{8} \sinh (3 a) \text {Chi}\left (3 b x^2\right )-\frac {3}{8} \cosh (a) \text {Shi}\left (b x^2\right )+\frac {1}{8} \cosh (3 a) \text {Shi}\left (3 b x^2\right ) \]

[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 5424

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

Rule 5425

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

Rule 5426

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 5448

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{8} \left (-3 \text {Chi}\left (b x^2\right ) \sinh (a)+\text {Chi}\left (3 b x^2\right ) \sinh (3 a)-3 \cosh (a) \text {Shi}\left (b x^2\right )+\cosh (3 a) \text {Shi}\left (3 b x^2\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.25

method	result	size
risch	\(-\frac {{\mathrm e}^{6 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (-3 x^{2} b \right )}{16}+\frac {3 \,{\mathrm e}^{4 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (-x^{2} b \right )}{16}-\frac {3 \,{\mathrm e}^{2 a} {\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (x^{2} b \right )}{16}+\frac {{\mathrm e}^{-3 a} \operatorname {Ei}_{1}\left (3 x^{2} b \right )}{16}\)	\(69\)

[In]

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

[Out]

-1/16*exp(6*a)*exp(-3*a)*Ei(1,-3*x^2*b)+3/16*exp(4*a)*exp(-3*a)*Ei(1,-x^2*b)-3/16*exp(2*a)*exp(-3*a)*Ei(1,x^2*
b)+1/16*exp(-3*a)*Ei(1,3*x^2*b)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.51 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\left ({\rm Ei}\left (3 \, b x^{2}\right ) - {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \cosh \left (3 \, a\right ) - \frac {3}{16} \, {\left ({\rm Ei}\left (b x^{2}\right ) - {\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + \frac {1}{16} \, {\left ({\rm Ei}\left (3 \, b x^{2}\right ) + {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \sinh \left (3 \, a\right ) - \frac {3}{16} \, {\left ({\rm Ei}\left (b x^{2}\right ) + {\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right ) \]

[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)

Sympy [F]

\[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\int \frac {\sinh ^{3}{\left (a + b x^{2} \right )}}{x}\, dx \]

[In]

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

[Out]

Integral(sinh(a + b*x**2)**3/x, x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - \frac {1}{16} \, {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - \frac {3}{16} \, {\rm Ei}\left (b x^{2}\right ) e^{a} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\frac {1}{16} \, {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} + \frac {3}{16} \, {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - \frac {1}{16} \, {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - \frac {3}{16} \, {\rm Ei}\left (b x^{2}\right ) e^{a} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sinh ^3\left (a+b x^2\right )}{x} \, dx=\int \frac {{\mathrm {sinh}\left (b\,x^2+a\right )}^3}{x} \,d x \]

[In]

int(sinh(a + b*x^2)^3/x,x)

[Out]

int(sinh(a + b*x^2)^3/x, x)