\(\int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx\) [392]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F]
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 25, antiderivative size = 54 \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right ) \sinh \left (\frac {a}{b}\right )}{b c^2}+\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{b c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {5819, 3384, 3379, 3382} \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\frac {\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{b c^2}-\frac {\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \text {arcsinh}(c x)}{b}\right )}{b c^2} \]

[In]

Int[x/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])),x]

[Out]

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

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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
 0]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=-\frac {\text {Chi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {arcsinh}(c x)\right )}{b c^2} \]

[In]

Integrate[x/(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])),x]

[Out]

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98

method result size
default \(\frac {{\mathrm e}^{\frac {a}{b}} \operatorname {Ei}_{1}\left (\operatorname {arcsinh}\left (c x \right )+\frac {a}{b}\right )-{\mathrm e}^{-\frac {a}{b}} \operatorname {Ei}_{1}\left (-\operatorname {arcsinh}\left (c x \right )-\frac {a}{b}\right )}{2 c^{2} b}\) \(53\)

[In]

int(x/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]

[In]

integrate(x/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*x^2 + 1)*x/(a*c^2*x^2 + (b*c^2*x^2 + b)*arcsinh(c*x) + a), x)

Sympy [F]

\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \sqrt {c^{2} x^{2} + 1}}\, dx \]

[In]

integrate(x/(a+b*asinh(c*x))/(c**2*x**2+1)**(1/2),x)

[Out]

Integral(x/((a + b*asinh(c*x))*sqrt(c**2*x**2 + 1)), x)

Maxima [F]

\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]

[In]

integrate(x/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*x) + a)), x)

Giac [F]

\[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int { \frac {x}{\sqrt {c^{2} x^{2} + 1} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}} \,d x } \]

[In]

integrate(x/(a+b*arcsinh(c*x))/(c^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(x/(sqrt(c^2*x^2 + 1)*(b*arcsinh(c*x) + a)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))} \, dx=\int \frac {x}{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {c^2\,x^2+1}} \,d x \]

[In]

int(x/((a + b*asinh(c*x))*(c^2*x^2 + 1)^(1/2)),x)

[Out]

int(x/((a + b*asinh(c*x))*(c^2*x^2 + 1)^(1/2)), x)