3.1.47 \(\int \frac {\cosh (b x) \text {Shi}(b x)}{x^2} \, dx\) [47]

Optimal. Leaf size=44 \[ b \text {Chi}(2 b x)-\frac {\sinh (2 b x)}{2 x}-\frac {\cosh (b x) \text {Shi}(b x)}{x}+\frac {1}{2} b \text {Shi}(b x)^2 \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6685, 6818, 12, 5556, 3378, 3382} \begin {gather*} b \text {Chi}(2 b x)+\frac {1}{2} b \text {Shi}(b x)^2-\frac {\text {Shi}(b x) \cosh (b x)}{x}-\frac {\sinh (2 b x)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Cosh[b*x]*SinhIntegral[b*x])/x^2,x]

[Out]

b*CoshIntegral[2*b*x] - Sinh[2*b*x]/(2*x) - (Cosh[b*x]*SinhIntegral[b*x])/x + (b*SinhIntegral[b*x]^2)/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 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 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 6685

Int[Cosh[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(e
+ f*x)^(m + 1)*Cosh[a + b*x]*(SinhIntegral[c + d*x]/(f*(m + 1))), x] + (-Dist[b/(f*(m + 1)), Int[(e + f*x)^(m
+ 1)*Sinh[a + b*x]*SinhIntegral[c + d*x], x], x] - Dist[d/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Cosh[a + b*x]*(S
inh[c + d*x]/(c + d*x)), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[m, -1]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 44, normalized size = 1.00 \begin {gather*} b \text {Chi}(2 b x)-\frac {\sinh (2 b x)}{2 x}-\frac {\cosh (b x) \text {Shi}(b x)}{x}+\frac {1}{2} b \text {Shi}(b x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Cosh[b*x]*SinhIntegral[b*x])/x^2,x]

[Out]

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

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Maple [F]
time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {\cosh \left (b x \right ) \hyperbolicSineIntegral \left (b x \right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x)*Shi(b*x)/x^2,x)

[Out]

int(cosh(b*x)*Shi(b*x)/x^2,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(cosh(b*x)*Shi(b*x)/x^2,x, algorithm="maxima")

[Out]

integrate(Shi(b*x)*cosh(b*x)/x^2, 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(cosh(b*x)*Shi(b*x)/x^2,x, algorithm="fricas")

[Out]

integral(cosh(b*x)*sinh_integral(b*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x)*Shi(b*x)/x**2,x)

[Out]

Integral(cosh(b*x)*Shi(b*x)/x**2, 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(cosh(b*x)*Shi(b*x)/x^2,x, algorithm="giac")

[Out]

integrate(Shi(b*x)*cosh(b*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sinhint(b*x)*cosh(b*x))/x^2,x)

[Out]

int((sinhint(b*x)*cosh(b*x))/x^2, x)

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