3.420 \(\int \frac{x^2 (1+c^2 x^2)^{3/2}}{(a+b \sinh ^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=219 \[ \frac{\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}-\frac{3 \sinh \left (\frac{6 a}{b}\right ) \text{Chi}\left (\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{4 b^2 c^3}+\frac{3 \cosh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^3}-\frac{x^2 \left (c^2 x^2+1\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-((x^2*(1 + c^2*x^2)^2)/(b*c*(a + b*ArcSinh[c*x]))) + (CoshIntegral[(2*(a + b*ArcSinh[c*x]))/b]*Sinh[(2*a)/b])
/(16*b^2*c^3) - (CoshIntegral[(4*(a + b*ArcSinh[c*x]))/b]*Sinh[(4*a)/b])/(4*b^2*c^3) - (3*CoshIntegral[(6*(a +
b*ArcSinh[c*x]))/b]*Sinh[(6*a)/b])/(16*b^2*c^3) - (Cosh[(2*a)/b]*SinhIntegral[(2*(a + b*ArcSinh[c*x]))/b])/(1
6*b^2*c^3) + (Cosh[(4*a)/b]*SinhIntegral[(4*(a + b*ArcSinh[c*x]))/b])/(4*b^2*c^3) + (3*Cosh[(6*a)/b]*SinhInteg
ral[(6*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^3)

________________________________________________________________________________________

Rubi [A]  time = 0.715753, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {5777, 5779, 5448, 3303, 3298, 3301} \[ \frac{\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}-\frac{3 \sinh \left (\frac{6 a}{b}\right ) \text{Chi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}-\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}+\frac{3 \cosh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}-\frac{x^2 \left (c^2 x^2+1\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(1 + c^2*x^2)^(3/2))/(a + b*ArcSinh[c*x])^2,x]

[Out]

-((x^2*(1 + c^2*x^2)^2)/(b*c*(a + b*ArcSinh[c*x]))) + (CoshIntegral[(2*a)/b + 2*ArcSinh[c*x]]*Sinh[(2*a)/b])/(
16*b^2*c^3) - (CoshIntegral[(4*a)/b + 4*ArcSinh[c*x]]*Sinh[(4*a)/b])/(4*b^2*c^3) - (3*CoshIntegral[(6*a)/b + 6
*ArcSinh[c*x]]*Sinh[(6*a)/b])/(16*b^2*c^3) - (Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcSinh[c*x]])/(16*b^2*c^
3) + (Cosh[(4*a)/b]*SinhIntegral[(4*a)/b + 4*ArcSinh[c*x]])/(4*b^2*c^3) + (3*Cosh[(6*a)/b]*SinhIntegral[(6*a)/
b + 6*ArcSinh[c*x]])/(16*b^2*c^3)

Rule 5777

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*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[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)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[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[m, -3] && IGtQ[2*p, 0]

Rule 5779

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

Rule 5448

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 3303

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 3298

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 3301

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]

Rubi steps

\begin{align*} \int \frac{x^2 \left (1+c^2 x^2\right )^{3/2}}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{2 \int \frac{x \left (1+c^2 x^2\right )}{a+b \sinh ^{-1}(c x)} \, dx}{b c}+\frac{(6 c) \int \frac{x^3 \left (1+c^2 x^2\right )}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{2 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac{6 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{2 \operatorname{Subst}\left (\int \left (\frac{\sinh (2 x)}{4 (a+b x)}+\frac{\sinh (4 x)}{8 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac{6 \operatorname{Subst}\left (\int \left (-\frac{3 \sinh (2 x)}{32 (a+b x)}+\frac{\sinh (6 x)}{32 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}\\ &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (6 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (4 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{9 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}\\ &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\cosh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac{\left (9 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}+\frac{\left (3 \cosh \left (\frac{6 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\sinh \left (\frac{2 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac{\left (9 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}-\frac{\sinh \left (\frac{4 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{4 a}{b}+4 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^3}-\frac{\left (3 \sinh \left (\frac{6 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{6 a}{b}+6 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^3}\\ &=-\frac{x^2 \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\text{Chi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{2 a}{b}\right )}{16 b^2 c^3}-\frac{\text{Chi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{4 a}{b}\right )}{4 b^2 c^3}-\frac{3 \text{Chi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right ) \sinh \left (\frac{6 a}{b}\right )}{16 b^2 c^3}-\frac{\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}+\frac{\cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (\frac{4 a}{b}+4 \sinh ^{-1}(c x)\right )}{4 b^2 c^3}+\frac{3 \cosh \left (\frac{6 a}{b}\right ) \text{Shi}\left (\frac{6 a}{b}+6 \sinh ^{-1}(c x)\right )}{16 b^2 c^3}\\ \end{align*}

Mathematica [A]  time = 0.730781, size = 306, normalized size = 1.4 \[ -\frac{-\sinh \left (\frac{2 a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+4 \sinh \left (\frac{4 a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text{Chi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+3 a \sinh \left (\frac{6 a}{b}\right ) \text{Chi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+3 b \sinh \left (\frac{6 a}{b}\right ) \sinh ^{-1}(c x) \text{Chi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+a \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+b \cosh \left (\frac{2 a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (2 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-4 a \cosh \left (\frac{4 a}{b}\right ) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-4 b \cosh \left (\frac{4 a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (4 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-3 a \cosh \left (\frac{6 a}{b}\right ) \text{Shi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-3 b \cosh \left (\frac{6 a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (6 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+16 b c^6 x^6+32 b c^4 x^4+16 b c^2 x^2}{16 b^2 c^3 \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(1 + c^2*x^2)^(3/2))/(a + b*ArcSinh[c*x])^2,x]

[Out]

-(16*b*c^2*x^2 + 32*b*c^4*x^4 + 16*b*c^6*x^6 - (a + b*ArcSinh[c*x])*CoshIntegral[2*(a/b + ArcSinh[c*x])]*Sinh[
(2*a)/b] + 4*(a + b*ArcSinh[c*x])*CoshIntegral[4*(a/b + ArcSinh[c*x])]*Sinh[(4*a)/b] + 3*a*CoshIntegral[6*(a/b
+ ArcSinh[c*x])]*Sinh[(6*a)/b] + 3*b*ArcSinh[c*x]*CoshIntegral[6*(a/b + ArcSinh[c*x])]*Sinh[(6*a)/b] + a*Cosh
[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*x])] + b*ArcSinh[c*x]*Cosh[(2*a)/b]*SinhIntegral[2*(a/b + ArcSinh[c*
x])] - 4*a*Cosh[(4*a)/b]*SinhIntegral[4*(a/b + ArcSinh[c*x])] - 4*b*ArcSinh[c*x]*Cosh[(4*a)/b]*SinhIntegral[4*
(a/b + ArcSinh[c*x])] - 3*a*Cosh[(6*a)/b]*SinhIntegral[6*(a/b + ArcSinh[c*x])] - 3*b*ArcSinh[c*x]*Cosh[(6*a)/b
]*SinhIntegral[6*(a/b + ArcSinh[c*x])])/(16*b^2*c^3*(a + b*ArcSinh[c*x]))

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Maple [B]  time = 0.298, size = 704, normalized size = 3.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x)

[Out]

1/16/c^3/(a+b*arcsinh(c*x))/b-1/64*(32*c^6*x^6-32*c^5*x^5*(c^2*x^2+1)^(1/2)+48*c^4*x^4-32*c^3*x^3*(c^2*x^2+1)^
(1/2)+18*c^2*x^2-6*c*x*(c^2*x^2+1)^(1/2)+1)/c^3/(a+b*arcsinh(c*x))/b+3/32/c^3/b^2*exp(6*a/b)*Ei(1,6*arcsinh(c*
x)+6*a/b)-1/32*(8*c^4*x^4-8*c^3*x^3*(c^2*x^2+1)^(1/2)+8*c^2*x^2-4*c*x*(c^2*x^2+1)^(1/2)+1)/c^3/(a+b*arcsinh(c*
x))/b+1/8/c^3/b^2*exp(4*a/b)*Ei(1,4*arcsinh(c*x)+4*a/b)+1/64*(2*c^2*x^2-2*c*x*(c^2*x^2+1)^(1/2)+1)/c^3/(a+b*ar
csinh(c*x))/b-1/32/c^3/b^2*exp(2*a/b)*Ei(1,2*arcsinh(c*x)+2*a/b)+1/64/c^3/b^2*(2*x^2*b*c^2+2*b*c*(c^2*x^2+1)^(
1/2)*x+2*arcsinh(c*x)*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*b+2*exp(-2*a/b)*Ei(1,-2*arcsinh(c*x)-2*a/b)*a+b)
/(a+b*arcsinh(c*x))-1/32/c^3/b^2*(8*x^4*b*c^4+8*(c^2*x^2+1)^(1/2)*x^3*b*c^3+8*x^2*b*c^2+4*b*c*(c^2*x^2+1)^(1/2
)*x+4*arcsinh(c*x)*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a/b)*b+4*Ei(1,-4*arcsinh(c*x)-4*a/b)*exp(-4*a/b)*a+b)/(a
+b*arcsinh(c*x))-1/64/c^3/b^2*(32*x^6*b*c^6+32*(c^2*x^2+1)^(1/2)*x^5*b*c^5+48*x^4*b*c^4+32*(c^2*x^2+1)^(1/2)*x
^3*b*c^3+18*x^2*b*c^2+6*b*c*(c^2*x^2+1)^(1/2)*x+6*arcsinh(c*x)*exp(-6*a/b)*Ei(1,-6*arcsinh(c*x)-6*a/b)*b+6*exp
(-6*a/b)*Ei(1,-6*arcsinh(c*x)-6*a/b)*a+b)/(a+b*arcsinh(c*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (c^{4} x^{6} + 2 \, c^{2} x^{4} + x^{2}\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (c^{5} x^{7} + 2 \, c^{3} x^{5} + c x^{3}\right )} \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 +{\left (b^{2} c^{3} x^{2} + \sqrt{c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )} + \int \frac{{\left (6 \, c^{5} x^{6} + 7 \, c^{3} x^{4} + c x^{2}\right )}{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} + 2 \,{\left (6 \, c^{6} x^{7} + 11 \, c^{4} x^{5} + 6 \, c^{2} x^{3} + x\right )}{\left (c^{2} x^{2} + 1\right )} + 3 \,{\left (2 \, c^{7} x^{8} + 5 \, c^{5} x^{6} + 4 \, c^{3} x^{4} + c x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{a b c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c +{\left (b^{2} c^{5} x^{4} +{\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \,{\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right ) + 2 \,{\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt{c^{2} x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

-((c^4*x^6 + 2*c^2*x^4 + x^2)*(c^2*x^2 + 1) + (c^5*x^7 + 2*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(((6*c^5*x^6 + 7*c^3*x^4 + c*x^2)*(c^2*x^2 + 1)^(3/2) + 2*(6*c^6*x^7 + 11*c^4*x^5 + 6*c^2
*x^3 + x)*(c^2*x^2 + 1) + 3*(2*c^7*x^8 + 5*c^5*x^6 + 4*c^3*x^4 + 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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (c^{2} x^{4} + x^{2}\right )} \sqrt{c^{2} x^{2} + 1}}{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

integral((c^2*x^4 + x^2)*sqrt(c^2*x^2 + 1)/(b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c**2*x**2+1)**(3/2)/(a+b*asinh(c*x))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} x^{2}}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x, algorithm="giac")

[Out]

integrate((c^2*x^2 + 1)^(3/2)*x^2/(b*arcsinh(c*x) + a)^2, x)