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

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

[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)

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Rubi [A]
time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 = {5448, 5426, 5425, 5424} \begin {gather*} -\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 ) \end {gather*}

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 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*} \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*}

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Mathematica [A]
time = 0.03, size = 49, normalized size = 0.89 \begin {gather*} \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 ) \end {gather*}

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 = 1.05, size = 55, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{-3 a} \expIntegral \left (1, 3 x^{2} b \right )}{16}-\frac {3 \,{\mathrm e}^{-a} \expIntegral \left (1, x^{2} b \right )}{16}+\frac {3 \,{\mathrm e}^{a} \expIntegral \left (1, -x^{2} b \right )}{16}-\frac {{\mathrm e}^{3 a} \expIntegral \left (1, -3 x^{2} b \right )}{16}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 50, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.38, size = 83, normalized size = 1.51 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh ^{3}{\left (a + b x^{2} \right )}}{x}\, dx \end {gather*}

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 = 0.43, size = 50, normalized size = 0.91 \begin {gather*} \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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {sinh}\left (b\,x^2+a\right )}^3}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)^3/x,x)

[Out]

int(sinh(a + b*x^2)^3/x, x)

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