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

Optimal. Leaf size=213 \[ \frac{\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^2}+\frac{9 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^2}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^2}-\frac{\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^2}-\frac{9 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^2}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^2}-\frac{x \left (c^2 x^2+1\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

[Out]

-((x*(1 + c^2*x^2)^2)/(b*c*(a + b*ArcSinh[c*x]))) + (Cosh[a/b]*CoshIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^
2) + (9*Cosh[(3*a)/b]*CoshIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^2) + (5*Cosh[(5*a)/b]*CoshIntegral[(
5*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^2) - (Sinh[a/b]*SinhIntegral[(a + b*ArcSinh[c*x])/b])/(8*b^2*c^2) - (9*S
inh[(3*a)/b]*SinhIntegral[(3*(a + b*ArcSinh[c*x]))/b])/(16*b^2*c^2) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*(a + b*
ArcSinh[c*x]))/b])/(16*b^2*c^2)

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Rubi [A]  time = 0.732221, antiderivative size = 209, normalized size of antiderivative = 0.98, number of steps used = 22, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {5777, 5699, 3312, 3303, 3298, 3301, 5779, 5448} \[ \frac{\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^2}+\frac{9 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}-\frac{\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^2}-\frac{9 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}-\frac{x \left (c^2 x^2+1\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*(1 + c^2*x^2)^2)/(b*c*(a + b*ArcSinh[c*x]))) + (Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]])/(8*b^2*c^2) +
(9*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2*c^2) + (5*Cosh[(5*a)/b]*CoshIntegral[(5*a)/b
+ 5*ArcSinh[c*x]])/(16*b^2*c^2) - (Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]])/(8*b^2*c^2) - (9*Sinh[(3*a)/b]
*SinhIntegral[(3*a)/b + 3*ArcSinh[c*x]])/(16*b^2*c^2) - (5*Sinh[(5*a)/b]*SinhIntegral[(5*a)/b + 5*ArcSinh[c*x]
])/(16*b^2*c^2)

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 5699

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

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

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]

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]

Rubi steps

\begin{align*} \int \frac{x \left (1+c^2 x^2\right )^{3/2}}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\int \frac{1+c^2 x^2}{a+b \sinh ^{-1}(c x)} \, dx}{b c}+\frac{(5 c) \int \frac{x^2 \left (1+c^2 x^2\right )}{a+b \sinh ^{-1}(c x)} \, dx}{b}\\ &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{\cosh ^3(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh ^3(x) \sinh ^2(x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\operatorname{Subst}\left (\int \left (\frac{3 \cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}+\frac{5 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{8 (a+b x)}+\frac{\cosh (3 x)}{16 (a+b x)}+\frac{\cosh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}-\frac{5 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}\\ &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac{\left (5 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{\left (3 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}+\frac{\cosh \left (\frac{3 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}+\frac{\left (5 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\left (5 \cosh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\left (5 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}-\frac{\left (3 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}-\frac{\sinh \left (\frac{3 a}{b}\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}-\frac{\left (5 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}-\frac{\left (5 \sinh \left (\frac{5 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}\\ &=-\frac{x \left (1+c^2 x^2\right )^2}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac{\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^2}+\frac{9 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}+\frac{5 \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}-\frac{\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^2}-\frac{9 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}-\frac{5 \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (\frac{5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.607108, size = 295, normalized size = 1.38 \[ -\frac{-2 \cosh \left (\frac{a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text{Chi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )-9 \cosh \left (\frac{3 a}{b}\right ) \left (a+b \sinh ^{-1}(c x)\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-5 a \cosh \left (\frac{5 a}{b}\right ) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )-5 b \cosh \left (\frac{5 a}{b}\right ) \sinh ^{-1}(c x) \text{Chi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+2 a \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+2 b \sinh \left (\frac{a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (\frac{a}{b}+\sinh ^{-1}(c x)\right )+9 a \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+9 b \sinh \left (\frac{3 a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (3 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+5 a \sinh \left (\frac{5 a}{b}\right ) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+5 b \sinh \left (\frac{5 a}{b}\right ) \sinh ^{-1}(c x) \text{Shi}\left (5 \left (\frac{a}{b}+\sinh ^{-1}(c x)\right )\right )+16 b c^5 x^5+32 b c^3 x^3+16 b c x}{16 b^2 c^2 \left (a+b \sinh ^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(16*b*c*x + 32*b*c^3*x^3 + 16*b*c^5*x^5 - 2*(a + b*ArcSinh[c*x])*Cosh[a/b]*CoshIntegral[a/b + ArcSinh[c*x]] -
9*(a + b*ArcSinh[c*x])*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcSinh[c*x])] - 5*a*Cosh[(5*a)/b]*CoshIntegral[5*
(a/b + ArcSinh[c*x])] - 5*b*ArcSinh[c*x]*Cosh[(5*a)/b]*CoshIntegral[5*(a/b + ArcSinh[c*x])] + 2*a*Sinh[a/b]*Si
nhIntegral[a/b + ArcSinh[c*x]] + 2*b*ArcSinh[c*x]*Sinh[a/b]*SinhIntegral[a/b + ArcSinh[c*x]] + 9*a*Sinh[(3*a)/
b]*SinhIntegral[3*(a/b + ArcSinh[c*x])] + 9*b*ArcSinh[c*x]*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSinh[c*x])]
+ 5*a*Sinh[(5*a)/b]*SinhIntegral[5*(a/b + ArcSinh[c*x])] + 5*b*ArcSinh[c*x]*Sinh[(5*a)/b]*SinhIntegral[5*(a/b
+ ArcSinh[c*x])])/(16*b^2*c^2*(a + b*ArcSinh[c*x]))

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Maple [B]  time = 0.233, size = 633, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(16*c^5*x^5-16*c^4*x^4*(c^2*x^2+1)^(1/2)+20*c^3*x^3-12*c^2*x^2*(c^2*x^2+1)^(1/2)+5*c*x-(c^2*x^2+1)^(1/2)
)/c^2/b/(a+b*arcsinh(c*x))-5/32/c^2/b^2*exp(5*a/b)*Ei(1,5*arcsinh(c*x)+5*a/b)-3/32*(4*c^3*x^3-4*c^2*x^2*(c^2*x
^2+1)^(1/2)+3*c*x-(c^2*x^2+1)^(1/2))/c^2/b/(a+b*arcsinh(c*x))-9/32/c^2/b^2*exp(3*a/b)*Ei(1,3*arcsinh(c*x)+3*a/
b)-1/16*(c*x-(c^2*x^2+1)^(1/2))/c^2/b/(a+b*arcsinh(c*x))-1/16/c^2/b^2*exp(a/b)*Ei(1,arcsinh(c*x)+a/b)-1/16/c^2
/b^2*(arcsinh(c*x)*Ei(1,-arcsinh(c*x)-a/b)*exp(-a/b)*b+Ei(1,-arcsinh(c*x)-a/b)*exp(-a/b)*a+x*b*c+(c^2*x^2+1)^(
1/2)*b)/(a+b*arcsinh(c*x))-3/32/c^2/b^2*(4*x^3*b*c^3+4*(c^2*x^2+1)^(1/2)*x^2*b*c^2+3*arcsinh(c*x)*exp(-3*a/b)*
Ei(1,-3*arcsinh(c*x)-3*a/b)*b+3*exp(-3*a/b)*Ei(1,-3*arcsinh(c*x)-3*a/b)*a+3*x*b*c+(c^2*x^2+1)^(1/2)*b)/(a+b*ar
csinh(c*x))-1/32/c^2/b^2*(16*x^5*b*c^5+16*(c^2*x^2+1)^(1/2)*x^4*b*c^4+20*x^3*b*c^3+12*(c^2*x^2+1)^(1/2)*x^2*b*
c^2+5*arcsinh(c*x)*exp(-5*a/b)*Ei(1,-5*arcsinh(c*x)-5*a/b)*b+5*exp(-5*a/b)*Ei(1,-5*arcsinh(c*x)-5*a/b)*a+5*x*b
*c+(c^2*x^2+1)^(1/2)*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^{5} + 2 \, c^{2} x^{3} + x\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (c^{5} x^{6} + 2 \, c^{3} x^{4} + c x^{2}\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{5 \,{\left (c^{5} x^{5} + c^{3} x^{3}\right )}{\left (c^{2} x^{2} + 1\right )}^{\frac{3}{2}} +{\left (10 \, c^{6} x^{6} + 17 \, c^{4} x^{4} + 8 \, c^{2} x^{2} + 1\right )}{\left (c^{2} x^{2} + 1\right )} +{\left (5 \, c^{7} x^{7} + 12 \, c^{5} x^{5} + 9 \, c^{3} x^{3} + 2 \, c x\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*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x, algorithm="maxima")

[Out]

-((c^4*x^5 + 2*c^2*x^3 + x)*(c^2*x^2 + 1) + (c^5*x^6 + 2*c^3*x^4 + c*x^2)*sqrt(c^2*x^2 + 1))/(a*b*c^3*x^2 + sq
rt(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((5*(c^5*x^5 + c^3*x^3)*(c^2*x^2 + 1)^(3/2) + (10*c^6*x^6 + 17*c^4*x^4 + 8*c^2*x^2 + 1)*(c^
2*x^2 + 1) + (5*c^7*x^7 + 12*c^5*x^5 + 9*c^3*x^3 + 2*c*x)*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^{3} + x\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*(c^2*x^2+1)^(3/2)/(a+b*arcsinh(c*x))^2,x, algorithm="fricas")

[Out]

integral((c^2*x^3 + x)*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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (c^{2} x^{2} + 1\right )^{\frac{3}{2}}}{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}{{\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*(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/(b*arcsinh(c*x) + a)^2, x)