3.1.8 \(\int \frac {\text {Shi}(b x)}{x^3} \, dx\) [8]

Optimal. Leaf size=46 \[ -\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6667, 12, 3378, 3379} \begin {gather*} \frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2}-\frac {\sinh (b x)}{4 x^2}-\frac {b \cosh (b x)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[SinhIntegral[b*x]/x^3,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 6667

Int[((c_.) + (d_.)*(x_))^(m_.)*SinhIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinhInte
gral[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sinh[a + b*x]/(a + b*x)), x], x] /
; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\text {Shi}(b x)}{x^3} \, dx &=-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{2} b \int \frac {\sinh (b x)}{b x^3} \, dx\\ &=-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{2} \int \frac {\sinh (b x)}{x^3} \, dx\\ &=-\frac {\sinh (b x)}{4 x^2}-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{4} b \int \frac {\cosh (b x)}{x^2} \, dx\\ &=-\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}-\frac {\text {Shi}(b x)}{2 x^2}+\frac {1}{4} b^2 \int \frac {\sinh (b x)}{x} \, dx\\ &=-\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 46, normalized size = 1.00 \begin {gather*} -\frac {b \cosh (b x)}{4 x}-\frac {\sinh (b x)}{4 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)-\frac {\text {Shi}(b x)}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[SinhIntegral[b*x]/x^3,x]

[Out]

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

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Maple [A]
time = 0.25, size = 48, normalized size = 1.04

method	result	size
derivativedivides	\(b^{2} \left (-\frac {\hyperbolicSineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b x}+\frac {\hyperbolicSineIntegral \left (b x \right )}{4}\right )\)	\(48\)
default	\(b^{2} \left (-\frac {\hyperbolicSineIntegral \left (b x \right )}{2 b^{2} x^{2}}-\frac {\sinh \left (b x \right )}{4 b^{2} x^{2}}-\frac {\cosh \left (b x \right )}{4 b x}+\frac {\hyperbolicSineIntegral \left (b x \right )}{4}\right )\)	\(48\)
meijerg	\(\frac {i \sqrt {\pi }\, b^{2} \left (\frac {4 i \cosh \left (b x \right )}{b x \sqrt {\pi }}+\frac {4 i \sinh \left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}+\frac {4 i \left (-b^{2} x^{2}+2\right ) \hyperbolicSineIntegral \left (b x \right )}{b^{2} x^{2} \sqrt {\pi }}\right )}{16}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(Shi(b*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Shi(b*x)/x^3,x, algorithm="maxima")

[Out]

integrate(Shi(b*x)/x^3, x)

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Fricas [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(Shi(b*x)/x^3,x, algorithm="fricas")

[Out]

integral(sinh_integral(b*x)/x^3, x)

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Sympy [A]
time = 0.48, size = 39, normalized size = 0.85 \begin {gather*} \frac {b^{2} \operatorname {Shi}{\left (b x \right )}}{4} - \frac {b \cosh {\left (b x \right )}}{4 x} - \frac {\sinh {\left (b x \right )}}{4 x^{2}} - \frac {\operatorname {Shi}{\left (b x \right )}}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Shi(b*x)/x**3,x)

[Out]

b**2*Shi(b*x)/4 - b*cosh(b*x)/(4*x) - sinh(b*x)/(4*x**2) - Shi(b*x)/(2*x**2)

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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(Shi(b*x)/x^3,x, algorithm="giac")

[Out]

integrate(Shi(b*x)/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \frac {b^2\,\mathrm {sinhint}\left (b\,x\right )}{4}-\frac {\frac {\mathrm {sinhint}\left (b\,x\right )}{2}+\frac {\mathrm {sinh}\left (b\,x\right )}{4}+\frac {b\,x\,\mathrm {cosh}\left (b\,x\right )}{4}}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinhint(b*x)/x^3,x)

[Out]

(b^2*sinhint(b*x))/4 - (sinhint(b*x)/2 + sinh(b*x)/4 + (b*x*cosh(b*x))/4)/x^2

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