\(\int \frac {x^2 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx\) [412]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {x^2 \left (1+c^2 x^2\right )}{b c (a+b \text {arcsinh}(c x))}-\frac {\text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right ) \sinh \left (\frac {4 a}{b}\right )}{2 b^2 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c^3} \]

[Out]

-x^2*(c^2*x^2+1)/b/c/(a+b*arcsinh(c*x))+1/2*cosh(4*a/b)*Shi(4*(a+b*arcsinh(c*x))/b)/b^2/c^3-1/2*Chi(4*(a+b*arc
sinh(c*x))/b)*sinh(4*a/b)/b^2/c^3

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {5814, 5780, 5556, 12, 3384, 3379, 3382} \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=-\frac {\sinh \left (\frac {4 a}{b}\right ) \text {Chi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c^3}+\frac {\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (\frac {4 (a+b \text {arcsinh}(c x))}{b}\right )}{2 b^2 c^3}-\frac {x^2 \left (c^2 x^2+1\right )}{b c (a+b \text {arcsinh}(c x))} \]

[In]

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

[Out]

-((x^2*(1 + c^2*x^2))/(b*c*(a + b*ArcSinh[c*x]))) - (CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b]*Sinh[(4*a)/b])/(
2*b^2*c^3) + (Cosh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(2*b^2*c^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 5780

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

Rule 5814

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[(f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*((a + b*ArcSinh[c*x])^(n + 1)/(b*c*(n + 1))), x] + (-Dist[f*(m/(b*c*(
n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(
n + 1), x], x] - Dist[c*((m + 2*p + 1)/(b*f*(n + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(
1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d]
&& LtQ[n, -1] && IGtQ[2*p, 0] && NeQ[m + 2*p + 1, 0] && IGtQ[m, -3]

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \sqrt {1+c^2 x^2}}{(a+b \text {arcsinh}(c x))^2} \, dx=\frac {-\frac {2 b c^2 x^2 \left (1+c^2 x^2\right )}{a+b \text {arcsinh}(c x)}-\text {Chi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right ) \sinh \left (\frac {4 a}{b}\right )+\cosh \left (\frac {4 a}{b}\right ) \text {Shi}\left (4 \left (\frac {a}{b}+\text {arcsinh}(c x)\right )\right )}{2 b^2 c^3} \]

[In]

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

[Out]

((-2*b*c^2*x^2*(1 + c^2*x^2))/(a + b*ArcSinh[c*x]) - CoshIntegral[4*(a/b + ArcSinh[c*x])]*Sinh[(4*a)/b] + Cosh
[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])])/(2*b^2*c^3)

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.55

method	result	size
default	\(-\frac {4 b \,c^{4} x^{4}+4 b \,c^{2} x^{2}+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )-{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) b \,\operatorname {arcsinh}\left (c x \right )+{\mathrm e}^{-\frac {4 a}{b}} \operatorname {Ei}_{1}\left (-4 \,\operatorname {arcsinh}\left (c x \right )-\frac {4 a}{b}\right ) a -{\mathrm e}^{\frac {4 a}{b}} \operatorname {Ei}_{1}\left (4 \,\operatorname {arcsinh}\left (c x \right )+\frac {4 a}{b}\right ) a}{4 c^{3} b^{2} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}\)	\(144\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-((c^2*x^4 + x^2)*(c^2*x^2 + 1) + (c^3*x^5 + c*x^3)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sqrt(c^2*x^2 + 1)*a*b*c^
2*x + a*b*c + (b^2*c^3*x^2 + sqrt(c^2*x^2 + 1)*b^2*c^2*x + b^2*c)*log(c*x + sqrt(c^2*x^2 + 1))) + integrate(((
4*c^3*x^4 + c*x^2)*(c^2*x^2 + 1)^(3/2) + 2*(4*c^4*x^5 + 4*c^2*x^3 + x)*(c^2*x^2 + 1) + (4*c^5*x^6 + 7*c^3*x^4
+ 3*c*x^2)*sqrt(c^2*x^2 + 1))/(a*b*c^5*x^4 + (c^2*x^2 + 1)*a*b*c^3*x^2 + 2*a*b*c^3*x^2 + a*b*c + (b^2*c^5*x^4
+ (c^2*x^2 + 1)*b^2*c^3*x^2 + 2*b^2*c^3*x^2 + b^2*c + 2*(b^2*c^4*x^3 + b^2*c^2*x)*sqrt(c^2*x^2 + 1))*log(c*x +
sqrt(c^2*x^2 + 1)) + 2*(a*b*c^4*x^3 + a*b*c^2*x)*sqrt(c^2*x^2 + 1)), x)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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