3.1.21 \(\int \frac {\sinh ^3(a+b x^2)}{x^3} \, dx\) [21]

Optimal. Leaf size=91 \[ -\frac {3}{8} b \cosh (a) \text {Chi}\left (b x^2\right )+\frac {3}{8} b \cosh (3 a) \text {Chi}\left (3 b x^2\right )+\frac {3 \sinh \left (a+b x^2\right )}{8 x^2}-\frac {\sinh \left (3 \left (a+b x^2\right )\right )}{8 x^2}-\frac {3}{8} b \sinh (a) \text {Shi}\left (b x^2\right )+\frac {3}{8} b \sinh (3 a) \text {Shi}\left (3 b x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5448, 5428, 3378, 3384, 3379, 3382} \begin {gather*} -\frac {3}{8} b \cosh (a) \text {Chi}\left (b x^2\right )+\frac {3}{8} b \cosh (3 a) \text {Chi}\left (3 b x^2\right )-\frac {3}{8} b \sinh (a) \text {Shi}\left (b x^2\right )+\frac {3}{8} b \sinh (3 a) \text {Shi}\left (3 b x^2\right )+\frac {3 \sinh \left (a+b x^2\right )}{8 x^2}-\frac {\sinh \left (3 \left (a+b x^2\right )\right )}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]^3/x^3,x]

[Out]

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

Rule 3378

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m
+ 1))), x] - Dist[f/(d*(m + 1)), Int[(c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[
m, -1]

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3384

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[c*(f/d) + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 5428

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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

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Mathematica [A]
time = 0.08, size = 90, normalized size = 0.99 \begin {gather*} -\frac {3 b x^2 \cosh (a) \text {Chi}\left (b x^2\right )-3 b x^2 \cosh (3 a) \text {Chi}\left (3 b x^2\right )-3 \sinh \left (a+b x^2\right )+\sinh \left (3 \left (a+b x^2\right )\right )+3 b x^2 \sinh (a) \text {Shi}\left (b x^2\right )-3 b x^2 \sinh (3 a) \text {Shi}\left (3 b x^2\right )}{8 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]^3/x^3,x]

[Out]

-1/8*(3*b*x^2*Cosh[a]*CoshIntegral[b*x^2] - 3*b*x^2*Cosh[3*a]*CoshIntegral[3*b*x^2] - 3*Sinh[a + b*x^2] + Sinh
[3*(a + b*x^2)] + 3*b*x^2*Sinh[a]*SinhIntegral[b*x^2] - 3*b*x^2*Sinh[3*a]*SinhIntegral[3*b*x^2])/x^2

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Maple [A]
time = 1.06, size = 120, normalized size = 1.32

method	result	size
risch	\(\frac {{\mathrm e}^{-3 a} {\mathrm e}^{-3 x^{2} b}}{16 x^{2}}-\frac {3 \,{\mathrm e}^{-3 a} b \expIntegral \left (1, 3 x^{2} b \right )}{16}-\frac {3 \,{\mathrm e}^{-a} {\mathrm e}^{-x^{2} b}}{16 x^{2}}+\frac {3 \,{\mathrm e}^{-a} b \expIntegral \left (1, x^{2} b \right )}{16}+\frac {3 \,{\mathrm e}^{a} {\mathrm e}^{x^{2} b}}{16 x^{2}}+\frac {3 \,{\mathrm e}^{a} b \expIntegral \left (1, -x^{2} b \right )}{16}-\frac {{\mathrm e}^{3 a} {\mathrm e}^{3 x^{2} b}}{16 x^{2}}-\frac {3 \,{\mathrm e}^{3 a} b \expIntegral \left (1, -3 x^{2} b \right )}{16}\)	\(120\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.34, size = 58, normalized size = 0.64 \begin {gather*} \frac {3}{16} \, b e^{\left (-3 \, a\right )} \Gamma \left (-1, 3 \, b x^{2}\right ) - \frac {3}{16} \, b e^{\left (-a\right )} \Gamma \left (-1, b x^{2}\right ) - \frac {3}{16} \, b e^{a} \Gamma \left (-1, -b x^{2}\right ) + \frac {3}{16} \, b e^{\left (3 \, a\right )} \Gamma \left (-1, -3 \, b x^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.50, size = 160, normalized size = 1.76 \begin {gather*} -\frac {2 \, \sinh \left (b x^{2} + a\right )^{3} - 3 \, {\left (b x^{2} {\rm Ei}\left (3 \, b x^{2}\right ) + b x^{2} {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \cosh \left (3 \, a\right ) + 3 \, {\left (b x^{2} {\rm Ei}\left (b x^{2}\right ) + b x^{2} {\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + 6 \, {\left (\cosh \left (b x^{2} + a\right )^{2} - 1\right )} \sinh \left (b x^{2} + a\right ) - 3 \, {\left (b x^{2} {\rm Ei}\left (3 \, b x^{2}\right ) - b x^{2} {\rm Ei}\left (-3 \, b x^{2}\right )\right )} \sinh \left (3 \, a\right ) + 3 \, {\left (b x^{2} {\rm Ei}\left (b x^{2}\right ) - b x^{2} {\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right )}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*sinh(b*x^2 + a)^3 - 3*(b*x^2*Ei(3*b*x^2) + b*x^2*Ei(-3*b*x^2))*cosh(3*a) + 3*(b*x^2*Ei(b*x^2) + b*x^2
*Ei(-b*x^2))*cosh(a) + 6*(cosh(b*x^2 + a)^2 - 1)*sinh(b*x^2 + a) - 3*(b*x^2*Ei(3*b*x^2) - b*x^2*Ei(-3*b*x^2))*
sinh(3*a) + 3*(b*x^2*Ei(b*x^2) - b*x^2*Ei(-b*x^2))*sinh(a))/x^2

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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^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)**3/x**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (80) = 160\).
time = 0.45, size = 223, normalized size = 2.45 \begin {gather*} \frac {3 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} - 3 \, a b^{2} {\rm Ei}\left (3 \, b x^{2}\right ) e^{\left (3 \, a\right )} - 3 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + 3 \, a b^{2} {\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + 3 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - 3 \, a b^{2} {\rm Ei}\left (-3 \, b x^{2}\right ) e^{\left (-3 \, a\right )} - 3 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (b x^{2}\right ) e^{a} + 3 \, a b^{2} {\rm Ei}\left (b x^{2}\right ) e^{a} - b^{2} e^{\left (3 \, b x^{2} + 3 \, a\right )} + 3 \, b^{2} e^{\left (b x^{2} + a\right )} - 3 \, b^{2} e^{\left (-b x^{2} - a\right )} + b^{2} e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{16 \, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {sinh}\left (b\,x^2+a\right )}^3}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)^3/x^3,x)

[Out]

int(sinh(a + b*x^2)^3/x^3, x)

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