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

Optimal. Leaf size=25 \[ \frac{1}{2} \sinh (a) \text{Chi}\left (b x^2\right )+\frac{1}{2} \cosh (a) \text{Shi}\left (b x^2\right ) \]

[Out]

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

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Rubi [A]  time = 0.0353859, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5318, 5317, 5316} \[ \frac{1}{2} \sinh (a) \text{Chi}\left (b x^2\right )+\frac{1}{2} \cosh (a) \text{Shi}\left (b x^2\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b*x^2]/x,x]

[Out]

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

Rule 5318

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 5317

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

Rule 5316

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

Rubi steps

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

Mathematica [A]  time = 0.0135722, size = 23, normalized size = 0.92 \[ \frac{1}{2} \left (\sinh (a) \text{Chi}\left (b x^2\right )+\cosh (a) \text{Shi}\left (b x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b*x^2]/x,x]

[Out]

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

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Maple [A]  time = 0.012, size = 27, normalized size = 1.1 \begin{align*}{\frac{{{\rm e}^{-a}}{\it Ei} \left ( 1,b{x}^{2} \right ) }{4}}-{\frac{{{\rm e}^{a}}{\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,x)

[Out]

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

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Maxima [A]  time = 1.25246, size = 32, normalized size = 1.28 \begin{align*} -\frac{1}{4} \,{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + \frac{1}{4} \,{\rm Ei}\left (b x^{2}\right ) e^{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.58646, size = 104, normalized size = 4.16 \begin{align*} \frac{1}{4} \,{\left ({\rm Ei}\left (b x^{2}\right ) -{\rm Ei}\left (-b x^{2}\right )\right )} \cosh \left (a\right ) + \frac{1}{4} \,{\left ({\rm Ei}\left (b x^{2}\right ) +{\rm Ei}\left (-b x^{2}\right )\right )} \sinh \left (a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b*x**2)/x, x)

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Giac [A]  time = 1.11746, size = 32, normalized size = 1.28 \begin{align*} -\frac{1}{4} \,{\rm Ei}\left (-b x^{2}\right ) e^{\left (-a\right )} + \frac{1}{4} \,{\rm Ei}\left (b x^{2}\right ) e^{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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