\(\int \frac {x^2 \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx\) [357]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3}-\frac {\log (a+b \text {arcsinh}(c x))}{8 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3} \]

[Out]

1/8*Chi(4*(a+b*arcsinh(c*x))/b)*cosh(4*a/b)/b/c^3-1/8*ln(a+b*arcsinh(c*x))/b/c^3-1/8*Shi(4*(a+b*arcsinh(c*x))/
b)*sinh(4*a/b)/b/c^3

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {5819, 5556, 3384, 3379, 3382} \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3}-\frac {\log (a+b \text {arcsinh}(c x))}{8 b c^3} \]

[In]

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

[Out]

(Cosh[(4*a)/b]*CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(8*b*c^3) - Log[a + b*ArcSinh[c*x]]/(8*b*c^3) - (Sinh
[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(8*b*c^3)

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 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 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 {\cosh ^2\left (\frac {a}{b}-\frac {x}{b}\right ) \sinh ^2\left (\frac {a}{b}-\frac {x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c^3} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{8 x}+\frac {\cosh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{8 x}\right ) \, dx,x,a+b \text {arcsinh}(c x)\right )}{b c^3} \\ & = -\frac {\log (a+b \text {arcsinh}(c x))}{8 b c^3}+\frac {\text {Subst}\left (\int \frac {\cosh \left (\frac {4 a}{b}-\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^3} \\ & = -\frac {\log (a+b \text {arcsinh}(c x))}{8 b c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\cosh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Subst}\left (\int \frac {\sinh \left (\frac {4 x}{b}\right )}{x} \, dx,x,a+b \text {arcsinh}(c x)\right )}{8 b c^3} \\ & = \frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3}-\frac {\log (a+b \text {arcsinh}(c x))}{8 b c^3}-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{8 b c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.79 \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{a+b \text {arcsinh}(c x)} \, dx=\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )-\log (a+b \text {arcsinh}(c x))-\sinh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{8 b c^3} \]

[In]

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

[Out]

(Cosh[(4*a)/b]*CoshIntegral[4*(a/b + ArcSinh[c*x])] - Log[a + b*ArcSinh[c*x]] - Sinh[(4*a)/b]*SinhIntegral[4*(
a/b + ArcSinh[c*x])])/(8*b*c^3)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right )+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right )+2 \ln \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{3} b}\)	\(67\)

[In]

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

[Out]

-1/16*(exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)+exp(-4*a/b)*Ei(1,-4*arcsinh(c*x)-4*a/b)+2*ln(a+b*arcsinh(c*x)))/c
^3/b

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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