3.1.46 \(\int \frac {\sinh (a+\frac {b}{x^2})}{x} \, dx\) [46]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5426, 5425, 5424} \begin {gather*} -\frac {1}{2} \sinh (a) \text {Chi}\left (\frac {b}{x^2}\right )-\frac {1}{2} \cosh (a) \text {Shi}\left (\frac {b}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x,x]

[Out]

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 25, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (-\text {Chi}\left (\frac {b}{x^2}\right ) \sinh (a)-\cosh (a) \text {Shi}\left (\frac {b}{x^2}\right )\right ) \end {gather*}

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

________________________________________________________________________________________

Maple [A]
time = 0.30, size = 27, normalized size = 1.08

method	result	size
risch	\(-\frac {{\mathrm e}^{-a} \expIntegral \left (1, \frac {b}{x^{2}}\right )}{4}+\frac {{\mathrm e}^{a} \expIntegral \left (1, -\frac {b}{x^{2}}\right )}{4}\)	\(27\)
meijerg	\(-\frac {\cosh \left (a \right ) \hyperbolicSineIntegral \left (\frac {b}{x^{2}}\right )}{2}-\frac {\sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {2 \hyperbolicCosineIntegral \left (\frac {b}{x^{2}}\right )-2 \ln \left (\frac {b}{x^{2}}\right )-2 \gamma }{\sqrt {\pi }}+\frac {2 \gamma -4 \ln \left (x \right )+2 \ln \left (i b \right )}{\sqrt {\pi }}\right )}{4}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 24, normalized size = 0.96 \begin {gather*} \frac {1}{4} \, {\rm Ei}\left (-\frac {b}{x^{2}}\right ) e^{\left (-a\right )} - \frac {1}{4} \, {\rm Ei}\left (\frac {b}{x^{2}}\right ) e^{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.43, size = 39, normalized size = 1.56 \begin {gather*} -\frac {1}{4} \, {\left ({\rm Ei}\left (\frac {b}{x^{2}}\right ) - {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \cosh \left (a\right ) - \frac {1}{4} \, {\left ({\rm Ei}\left (\frac {b}{x^{2}}\right ) + {\rm Ei}\left (-\frac {b}{x^{2}}\right )\right )} \sinh \left (a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b/x**2)/x, x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(a + b/x^2)/x, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\frac {\mathrm {sinh}\left (a\right )\,\mathrm {coshint}\left (\frac {b}{x^2}\right )}{2}-\frac {\mathrm {cosh}\left (a\right )\,\mathrm {sinhint}\left (\frac {b}{x^2}\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)/x,x)

[Out]

- (sinh(a)*coshint(b/x^2))/2 - (cosh(a)*sinhint(b/x^2))/2

________________________________________________________________________________________