3.31 \(\int \sinh (a+\frac {b}{x}) \, dx\)

Optimal. Leaf size=33 \[ -b \cosh (a) \text {Chi}\left (\frac {b}{x}\right )-b \sinh (a) \text {Shi}\left (\frac {b}{x}\right )+x \sinh \left (a+\frac {b}{x}\right ) \]

[Out]

-b*Chi(b/x)*cosh(a)-b*Shi(b/x)*sinh(a)+x*sinh(a+b/x)

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Rubi [A]  time = 0.08, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5302, 3297, 3303, 3298, 3301} \[ -b \cosh (a) \text {Chi}\left (\frac {b}{x}\right )-b \sinh (a) \text {Shi}\left (\frac {b}{x}\right )+x \sinh \left (a+\frac {b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x],x]

[Out]

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

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 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]

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 5302

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^2
, x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && ILtQ[n, 0] && IntegerQ[p]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 33, normalized size = 1.00 \[ -b \cosh (a) \text {Chi}\left (\frac {b}{x}\right )-b \sinh (a) \text {Shi}\left (\frac {b}{x}\right )+x \sinh \left (a+\frac {b}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x],x]

[Out]

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

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fricas [A]  time = 0.44, size = 58, normalized size = 1.76 \[ -\frac {1}{2} \, {\left (b {\rm Ei}\left (\frac {b}{x}\right ) + b {\rm Ei}\left (-\frac {b}{x}\right )\right )} \cosh \relax (a) - \frac {1}{2} \, {\left (b {\rm Ei}\left (\frac {b}{x}\right ) - b {\rm Ei}\left (-\frac {b}{x}\right )\right )} \sinh \relax (a) + x \sinh \left (\frac {a x + b}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.14, size = 173, normalized size = 5.24 \[ -\frac {a b^{2} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )} - \frac {{\left (a x + b\right )} b^{2} {\rm Ei}\left (a - \frac {a x + b}{x}\right ) e^{\left (-a\right )}}{x} - b^{2} e^{\left (-\frac {a x + b}{x}\right )}}{2 \, {\left (a - \frac {a x + b}{x}\right )} b} - \frac {a b^{2} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a} - \frac {{\left (a x + b\right )} b^{2} {\rm Ei}\left (-a + \frac {a x + b}{x}\right ) e^{a}}{x} + b^{2} e^{\left (\frac {a x + b}{x}\right )}}{2 \, {\left (a - \frac {a x + b}{x}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 56, normalized size = 1.70 \[ \frac {b \,{\mathrm e}^{-a} \Ei \left (1, \frac {b}{x}\right )}{2}-\frac {{\mathrm e}^{-\frac {a x +b}{x}} x}{2}+\frac {b \,{\mathrm e}^{a} \Ei \left (1, -\frac {b}{x}\right )}{2}+\frac {{\mathrm e}^{\frac {a x +b}{x}} x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x),x)

[Out]

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

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maxima [A]  time = 0.36, size = 36, normalized size = 1.09 \[ -\frac {1}{2} \, {\left ({\rm Ei}\left (-\frac {b}{x}\right ) e^{\left (-a\right )} + {\rm Ei}\left (\frac {b}{x}\right ) e^{a}\right )} b + x \sinh \left (a + \frac {b}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {sinh}\left (a+\frac {b}{x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x),x)

[Out]

int(sinh(a + b/x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\left (a + \frac {b}{x} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b/x), x)

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