\(\int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx\) [115]

Optimal result

Integrand size = 12, antiderivative size = 44 \[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\frac {1}{2} b \text {Chi}(b x)^2+b \text {Chi}(2 b x)-\frac {\text {Chi}(b x) \sinh (b x)}{x}-\frac {\sinh (2 b x)}{2 x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6686, 6818, 12, 5556, 3378, 3382} \[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\frac {1}{2} b \text {Chi}(b x)^2+b \text {Chi}(2 b x)-\frac {\text {Chi}(b x) \sinh (b x)}{x}-\frac {\sinh (2 b x)}{2 x} \]

[In]

Int[(CoshIntegral[b*x]*Sinh[b*x])/x^2,x]

[Out]

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

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 6686

Int[CoshIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_)*Sinh[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(e
+ f*x)^(m + 1)*Sinh[a + b*x]*(CoshIntegral[c + d*x]/(f*(m + 1))), x] + (-Dist[b/(f*(m + 1)), Int[(e + f*x)^(m
+ 1)*Cosh[a + b*x]*CoshIntegral[c + d*x], x], x] - Dist[d/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sinh[a + b*x]*(Co
sh[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*} \text {integral}& = -\frac {\text {Chi}(b x) \sinh (b x)}{x}+b \int \frac {\cosh (b x) \text {Chi}(b x)}{x} \, dx+b \int \frac {\cosh (b x) \sinh (b x)}{b x^2} \, dx \\ & = \frac {1}{2} b \text {Chi}(b x)^2-\frac {\text {Chi}(b x) \sinh (b x)}{x}+\int \frac {\cosh (b x) \sinh (b x)}{x^2} \, dx \\ & = \frac {1}{2} b \text {Chi}(b x)^2-\frac {\text {Chi}(b x) \sinh (b x)}{x}+\int \frac {\sinh (2 b x)}{2 x^2} \, dx \\ & = \frac {1}{2} b \text {Chi}(b x)^2-\frac {\text {Chi}(b x) \sinh (b x)}{x}+\frac {1}{2} \int \frac {\sinh (2 b x)}{x^2} \, dx \\ & = \frac {1}{2} b \text {Chi}(b x)^2-\frac {\text {Chi}(b x) \sinh (b x)}{x}-\frac {\sinh (2 b x)}{2 x}+b \int \frac {\cosh (2 b x)}{x} \, dx \\ & = \frac {1}{2} b \text {Chi}(b x)^2+b \text {Chi}(2 b x)-\frac {\text {Chi}(b x) \sinh (b x)}{x}-\frac {\sinh (2 b x)}{2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\frac {1}{2} b \text {Chi}(b x)^2+b \text {Chi}(2 b x)-\frac {\text {Chi}(b x) \sinh (b x)}{x}-\frac {\sinh (2 b x)}{2 x} \]

[In]

Integrate[(CoshIntegral[b*x]*Sinh[b*x])/x^2,x]

[Out]

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

Maple [F]

\[\int \frac {\operatorname {Chi}\left (b x \right ) \sinh \left (b x \right )}{x^{2}}d x\]

[In]

int(Chi(b*x)*sinh(b*x)/x^2,x)

[Out]

int(Chi(b*x)*sinh(b*x)/x^2,x)

Fricas [F]

\[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\int { \frac {{\rm Chi}\left (b x\right ) \sinh \left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Chi(b*x)*sinh(b*x)/x^2,x, algorithm="fricas")

[Out]

integral(cosh_integral(b*x)*sinh(b*x)/x^2, x)

Sympy [F]

\[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\int \frac {\sinh {\left (b x \right )} \operatorname {Chi}\left (b x\right )}{x^{2}}\, dx \]

[In]

integrate(Chi(b*x)*sinh(b*x)/x**2,x)

[Out]

Integral(sinh(b*x)*Chi(b*x)/x**2, x)

Maxima [F]

\[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\int { \frac {{\rm Chi}\left (b x\right ) \sinh \left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Chi(b*x)*sinh(b*x)/x^2,x, algorithm="maxima")

[Out]

integrate(Chi(b*x)*sinh(b*x)/x^2, x)

Giac [F]

\[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\int { \frac {{\rm Chi}\left (b x\right ) \sinh \left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(Chi(b*x)*sinh(b*x)/x^2,x, algorithm="giac")

[Out]

integrate(Chi(b*x)*sinh(b*x)/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {Chi}(b x) \sinh (b x)}{x^2} \, dx=\int \frac {\mathrm {coshint}\left (b\,x\right )\,\mathrm {sinh}\left (b\,x\right )}{x^2} \,d x \]

[In]

int((coshint(b*x)*sinh(b*x))/x^2,x)

[Out]

int((coshint(b*x)*sinh(b*x))/x^2, x)