3.1.14 \(\int \frac {\sinh ^2(a+b x^2)}{x^3} \, dx\) [14]

Optimal. Leaf size=57 \[ \frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right ) \]

[Out]

1/4/x^2-1/4*cosh(2*b*x^2+2*a)/x^2+1/2*b*cosh(2*a)*Shi(2*b*x^2)+1/2*b*Chi(2*b*x^2)*sinh(2*a)

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Rubi [A]
time = 0.09, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {5448, 5429, 3378, 3384, 3379, 3382} \begin {gather*} \frac {1}{2} b \sinh (2 a) \text {Chi}\left (2 b x^2\right )+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right )-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]^2/x^3,x]

[Out]

1/(4*x^2) - Cosh[2*(a + b*x^2)]/(4*x^2) + (b*CoshIntegral[2*b*x^2]*Sinh[2*a])/2 + (b*Cosh[2*a]*SinhIntegral[2*
b*x^2])/2

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 5429

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[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 ^2\left (a+b x^2\right )}{x^3} \, dx &=\int \left (-\frac {1}{2 x^3}+\frac {\cosh \left (2 a+2 b x^2\right )}{2 x^3}\right ) \, dx\\ &=\frac {1}{4 x^2}+\frac {1}{2} \int \frac {\cosh \left (2 a+2 b x^2\right )}{x^3} \, dx\\ &=\frac {1}{4 x^2}+\frac {1}{4} \text {Subst}\left (\int \frac {\cosh (2 a+2 b x)}{x^2} \, dx,x,x^2\right )\\ &=\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Subst}\left (\int \frac {\sinh (2 a+2 b x)}{x} \, dx,x,x^2\right )\\ &=\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} (b \cosh (2 a)) \text {Subst}\left (\int \frac {\sinh (2 b x)}{x} \, dx,x,x^2\right )+\frac {1}{2} (b \sinh (2 a)) \text {Subst}\left (\int \frac {\cosh (2 b x)}{x} \, dx,x,x^2\right )\\ &=\frac {1}{4 x^2}-\frac {\cosh \left (2 \left (a+b x^2\right )\right )}{4 x^2}+\frac {1}{2} b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)+\frac {1}{2} b \cosh (2 a) \text {Shi}\left (2 b x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 46, normalized size = 0.81 \begin {gather*} \frac {1}{2} \left (b \text {Chi}\left (2 b x^2\right ) \sinh (2 a)-\frac {\sinh ^2\left (a+b x^2\right )}{x^2}+b \cosh (2 a) \text {Shi}\left (2 b x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]^2/x^3,x]

[Out]

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

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Maple [A]
time = 0.78, size = 69, normalized size = 1.21

method	result	size
risch	\(\frac {1}{4 x^{2}}-\frac {{\mathrm e}^{-2 a} {\mathrm e}^{-2 x^{2} b}}{8 x^{2}}+\frac {{\mathrm e}^{-2 a} b \expIntegral \left (1, 2 x^{2} b \right )}{4}-\frac {{\mathrm e}^{2 a} {\mathrm e}^{2 x^{2} b}}{8 x^{2}}-\frac {{\mathrm e}^{2 a} b \expIntegral \left (1, -2 x^{2} b \right )}{4}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4/x^2-1/8*exp(-2*a)/x^2*exp(-2*x^2*b)+1/4*exp(-2*a)*b*Ei(1,2*x^2*b)-1/8*exp(2*a)/x^2*exp(2*x^2*b)-1/4*exp(2*
a)*b*Ei(1,-2*x^2*b)

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Maxima [A]
time = 0.32, size = 36, normalized size = 0.63 \begin {gather*} -\frac {1}{4} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x^{2}\right ) + \frac {1}{4} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x^{2}\right ) + \frac {1}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*e^(-2*a)*gamma(-1, 2*b*x^2) + 1/4*b*e^(2*a)*gamma(-1, -2*b*x^2) + 1/4/x^2

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Fricas [A]
time = 0.36, size = 90, normalized size = 1.58 \begin {gather*} -\frac {\cosh \left (b x^{2} + a\right )^{2} - {\left (b x^{2} {\rm Ei}\left (2 \, b x^{2}\right ) - b x^{2} {\rm Ei}\left (-2 \, b x^{2}\right )\right )} \cosh \left (2 \, a\right ) + \sinh \left (b x^{2} + a\right )^{2} - {\left (b x^{2} {\rm Ei}\left (2 \, b x^{2}\right ) + b x^{2} {\rm Ei}\left (-2 \, b x^{2}\right )\right )} \sinh \left (2 \, a\right ) - 1}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(cosh(b*x^2 + a)^2 - (b*x^2*Ei(2*b*x^2) - b*x^2*Ei(-2*b*x^2))*cosh(2*a) + sinh(b*x^2 + a)^2 - (b*x^2*Ei(2
*b*x^2) + b*x^2*Ei(-2*b*x^2))*sinh(2*a) - 1)/x^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh ^{2}{\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)**2/x**3,x)

[Out]

Integral(sinh(a + b*x**2)**2/x**3, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (50) = 100\).
time = 0.44, size = 126, normalized size = 2.21 \begin {gather*} \frac {2 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (2 \, b x^{2}\right ) e^{\left (2 \, a\right )} - 2 \, a b^{2} {\rm Ei}\left (2 \, b x^{2}\right ) e^{\left (2 \, a\right )} - 2 \, {\left (b x^{2} + a\right )} b^{2} {\rm Ei}\left (-2 \, b x^{2}\right ) e^{\left (-2 \, a\right )} + 2 \, a b^{2} {\rm Ei}\left (-2 \, b x^{2}\right ) e^{\left (-2 \, a\right )} - b^{2} e^{\left (2 \, b x^{2} + 2 \, a\right )} - b^{2} e^{\left (-2 \, b x^{2} - 2 \, a\right )} + 2 \, b^{2}}{8 \, b^{2} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(b*x^2 + a)*b^2*Ei(2*b*x^2)*e^(2*a) - 2*a*b^2*Ei(2*b*x^2)*e^(2*a) - 2*(b*x^2 + a)*b^2*Ei(-2*b*x^2)*e^(-
2*a) + 2*a*b^2*Ei(-2*b*x^2)*e^(-2*a) - b^2*e^(2*b*x^2 + 2*a) - b^2*e^(-2*b*x^2 - 2*a) + 2*b^2)/(b^2*x^2)

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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 )}^2}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b*x^2)^2/x^3,x)

[Out]

int(sinh(a + b*x^2)^2/x^3, x)

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