3.1.24 \(\int \frac {\text {Shi}(a+b x)}{x^3} \, dx\) [24]

Optimal. Leaf size=111 \[ \frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}-\frac {b^2 \text {Chi}(b x) \sinh (a)}{2 a^2}-\frac {b \sinh (a+b x)}{2 a x}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {\text {Shi}(a+b x)}{2 x^2} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6667, 6874, 3378, 3384, 3379, 3382} \begin {gather*} -\frac {b^2 \sinh (a) \text {Chi}(b x)}{2 a^2}+\frac {b^2 \text {Shi}(a+b x)}{2 a^2}-\frac {b^2 \cosh (a) \text {Shi}(b x)}{2 a^2}+\frac {b^2 \cosh (a) \text {Chi}(b x)}{2 a}+\frac {b^2 \sinh (a) \text {Shi}(b x)}{2 a}-\frac {\text {Shi}(a+b x)}{2 x^2}-\frac {b \sinh (a+b x)}{2 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[SinhIntegral[a + b*x]/x^3,x]

[Out]

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

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 3382

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 3384

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 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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 86, normalized size = 0.77 \begin {gather*} \frac {b^2 x^2 \text {Chi}(b x) (a \cosh (a)-\sinh (a))-a b x \sinh (a+b x)+b^2 x^2 (-\cosh (a)+a \sinh (a)) \text {Shi}(b x)-a^2 \text {Shi}(a+b x)+b^2 x^2 \text {Shi}(a+b x)}{2 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.18, size = 0, normalized size = 0.00 \[\int \frac {\hyperbolicSineIntegral \left (b x +a \right )}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(Shi(b*x+a)/x^3,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+a)/x^3,x, algorithm="maxima")

[Out]

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

[Out]

integral(sinh_integral(b*x + a)/x^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {Shi}{\left (a + b x \right )}}{x^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Shi(a + b*x)/x**3, x)

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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+a)/x^3,x, algorithm="giac")

[Out]

integrate(Shi(b*x + a)/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {sinhint}\left (a+b\,x\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinhint(a + b*x)/x^3,x)

[Out]

int(sinhint(a + b*x)/x^3, x)

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