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

Optimal. Leaf size=42 \[ \frac{1}{2} b \cosh (a) \text{Chi}\left (b x^2\right )+\frac{1}{2} b \sinh (a) \text{Shi}\left (b x^2\right )-\frac{\sinh \left (a+b x^2\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0922669, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5320, 3297, 3303, 3298, 3301} \[ \frac{1}{2} b \cosh (a) \text{Chi}\left (b x^2\right )+\frac{1}{2} b \sinh (a) \text{Shi}\left (b x^2\right )-\frac{\sinh \left (a+b x^2\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]/x^3,x]

[Out]

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

Rule 5320

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 3297

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 3303

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 3298

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 3301

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]

Rubi steps

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

Mathematica [A]  time = 0.0398667, size = 38, normalized size = 0.9 \[ \frac{1}{2} \left (b \cosh (a) \text{Chi}\left (b x^2\right )+b \sinh (a) \text{Shi}\left (b x^2\right )-\frac{\sinh \left (a+b x^2\right )}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]/x^3,x]

[Out]

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

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Maple [A]  time = 0.017, size = 58, normalized size = 1.4 \begin{align*}{\frac{{{\rm e}^{-a}}{{\rm e}^{-b{x}^{2}}}}{4\,{x}^{2}}}-{\frac{{{\rm e}^{-a}}b{\it Ei} \left ( 1,b{x}^{2} \right ) }{4}}-{\frac{{{\rm e}^{a}}{{\rm e}^{b{x}^{2}}}}{4\,{x}^{2}}}-{\frac{{{\rm e}^{a}}b{\it Ei} \left ( 1,-b{x}^{2} \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.27355, size = 53, normalized size = 1.26 \begin{align*} \frac{1}{4} \,{\left ({\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} +{\rm Ei}\left (b x^{2}\right ) e^{a}\right )} b - \frac{\sinh \left (b x^{2} + a\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.68495, size = 166, normalized size = 3.95 \begin{align*} \frac{{\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 ) +{\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 ) - 2 \, \sinh \left (b x^{2} + a\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)/x**3, x)

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Giac [B]  time = 1.15456, size = 147, normalized size = 3.5 \begin{align*} \frac{{\left (b x^{2} + a\right )} b^{2}{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} - a b^{2}{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} +{\left (b x^{2} + a\right )} b^{2}{\rm Ei}\left (b x^{2}\right ) e^{a} - a b^{2}{\rm Ei}\left (b x^{2}\right ) e^{a} - b^{2} e^{\left (b x^{2} + a\right )} + b^{2} e^{\left (-b x^{2} - a\right )}}{4 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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