3.239 \(\int \frac{(a+b \sinh ^{-1}(c x))^2}{x (d+c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=193 \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 x^2+1}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (c^2 x^2+1\right )}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^2}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 d^2} \]

[Out]

-((b*c*x*(a + b*ArcSinh[c*x]))/(d^2*Sqrt[1 + c^2*x^2])) + (a + b*ArcSinh[c*x])^2/(2*d^2*(1 + c^2*x^2)) - (2*(a
+ b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/d^2 + (b^2*Log[1 + c^2*x^2])/(2*d^2) - (b*(a + b*ArcSinh[c*x
])*PolyLog[2, -E^(2*ArcSinh[c*x])])/d^2 + (b*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d^2 + (b^2*P
olyLog[3, -E^(2*ArcSinh[c*x])])/(2*d^2) - (b^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d^2)

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Rubi [A]  time = 0.34, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {5755, 5720, 5461, 4182, 2531, 2282, 6589, 5687, 260} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac{b^2 \text{PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b^2 \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{c^2 x^2+1}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (c^2 x^2+1\right )}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^2}+\frac{b^2 \log \left (c^2 x^2+1\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*x*(a + b*ArcSinh[c*x]))/(d^2*Sqrt[1 + c^2*x^2])) + (a + b*ArcSinh[c*x])^2/(2*d^2*(1 + c^2*x^2)) - (2*(a
+ b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/d^2 + (b^2*Log[1 + c^2*x^2])/(2*d^2) - (b*(a + b*ArcSinh[c*x
])*PolyLog[2, -E^(2*ArcSinh[c*x])])/d^2 + (b*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d^2 + (b^2*P
olyLog[3, -E^(2*ArcSinh[c*x])])/(2*d^2) - (b^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d^2)

Rule 5755

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp
[((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p
+ 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(p + 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, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 5720

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

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 5687

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^2} \, dx &=\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac{(b c) \int \frac{a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^2}+\frac{\int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )} \, dx}{d}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac{\operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(x) \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}+\frac{\left (b^2 c^2\right ) \int \frac{x}{1+c^2 x^2} \, dx}{d^2}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 d^2}+\frac{2 \operatorname{Subst}\left (\int (a+b x)^2 \text{csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}-\frac{b^2 \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac{b c x \left (a+b \sinh ^{-1}(c x)\right )}{d^2 \sqrt{1+c^2 x^2}}+\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \log \left (1+c^2 x^2\right )}{2 d^2}-\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^2}+\frac{b^2 \text{Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}-\frac{b^2 \text{Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2}\\ \end{align*}

Mathematica [C]  time = 1.80625, size = 428, normalized size = 2.22 \[ \frac{2 a b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )\right )+a b \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )+2 b^2 \left (\sinh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \sinh ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )+\frac{1}{2} \log \left (c^2 x^2+1\right )+\frac{\sinh ^{-1}(c x)^2}{2 c^2 x^2+2}-\frac{c x \sinh ^{-1}(c x)}{\sqrt{c^2 x^2+1}}-\frac{2}{3} \sinh ^{-1}(c x)^3-\sinh ^{-1}(c x)^2 \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )+\sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\frac{i \pi ^3}{24}\right )+\frac{a^2}{c^2 x^2+1}-a^2 \log \left (c^2 x^2+1\right )+2 a^2 \log (c x)-\frac{a b \left (\sqrt{c^2 x^2+1}-i \sinh ^{-1}(c x)\right )}{c x+i}-\frac{a b \left (\sqrt{c^2 x^2+1}+i \sinh ^{-1}(c x)\right )}{c x-i}-2 a b \sinh ^{-1}(c x)^2+4 a b \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2/(1 + c^2*x^2) - (a*b*(Sqrt[1 + c^2*x^2] - I*ArcSinh[c*x]))/(I + c*x) - (a*b*(Sqrt[1 + c^2*x^2] + I*ArcSin
h[c*x]))/(-I + c*x) - 2*a*b*ArcSinh[c*x]^2 + 4*a*b*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + 2*a^2*Log[c*x] -
a^2*Log[1 + c^2*x^2] + a*b*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 + I*E^ArcSinh[c*x]]) - 4*PolyLog[2, (-I)*E^A
rcSinh[c*x]]) + a*b*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 - I*E^ArcSinh[c*x]]) - 4*PolyLog[2, I*E^ArcSinh[c*x]
]) + 2*a*b*PolyLog[2, E^(2*ArcSinh[c*x])] + 2*b^2*((I/24)*Pi^3 - (c*x*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] + ArcSin
h[c*x]^2/(2 + 2*c^2*x^2) - (2*ArcSinh[c*x]^3)/3 - ArcSinh[c*x]^2*Log[1 + E^(-2*ArcSinh[c*x])] + ArcSinh[c*x]^2
*Log[1 - E^(2*ArcSinh[c*x])] + Log[1 + c^2*x^2]/2 + ArcSinh[c*x]*PolyLog[2, -E^(-2*ArcSinh[c*x])] + ArcSinh[c*
x]*PolyLog[2, E^(2*ArcSinh[c*x])] + PolyLog[3, -E^(-2*ArcSinh[c*x])]/2 - PolyLog[3, E^(2*ArcSinh[c*x])]/2))/(2
*d^2)

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Maple [B]  time = 0.128, size = 724, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*b/d^2/(c^2*x^2+1)-a*b/d^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*b^2/d^2*arcsinh(c*x)^2/(c^2*x^2+1)+b^2/d
^2*arcsinh(c*x)/(c^2*x^2+1)-b^2/d^2*arcsinh(c*x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-b^2/d^2*arcsinh(c*x)*polylo
g(2,-(c*x+(c^2*x^2+1)^(1/2))^2)+b^2/d^2*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))+2*b^2/d^2*arcsinh(c*x)*poly
log(2,-c*x-(c^2*x^2+1)^(1/2))+b^2/d^2*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))+2*b^2/d^2*arcsinh(c*x)*polylo
g(2,c*x+(c^2*x^2+1)^(1/2))+2*a*b/d^2*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+2*a*b/d^2*polylog(2,c*x+(c^2*x^2+1)^(1/
2))-b^2/d^2*arcsinh(c*x)/(c^2*x^2+1)^(1/2)*c*x+b^2/d^2*arcsinh(c*x)/(c^2*x^2+1)*c^2*x^2-a*b/d^2/(c^2*x^2+1)^(1
/2)*c*x+a*b/d^2*c^2*x^2/(c^2*x^2+1)+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^2+2*a*b/d^2*arcsinh(c*x)*l
n(1+c*x+(c^2*x^2+1)^(1/2))+2*a*b/d^2*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+a^2/d^2*ln(c*x)-2*b^2/d^2*ln(c*x
+(c^2*x^2+1)^(1/2))+b^2/d^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/2*a^2/d^2/(c^2*x^2+1)-1/2*a^2/d^2*ln(c^2*x^2+1)-
2*b^2/d^2*polylog(3,-c*x-(c^2*x^2+1)^(1/2))-2*b^2/d^2*polylog(3,c*x+(c^2*x^2+1)^(1/2))+a*b/d^2*arcsinh(c*x)/(c
^2*x^2+1)-2*a*b/d^2*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}{c^{4} d^{2} x^{5} + 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^4*d^2*x^5 + 2*c^2*d^2*x^3 + d^2*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{4} x^{5} + 2 c^{2} x^{3} + x}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{c^{4} x^{5} + 2 c^{2} x^{3} + x}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c x \right )}}{c^{4} x^{5} + 2 c^{2} x^{3} + x}\, dx}{d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**4*x**5 + 2*c**2*x**3 + x), x) + Integral(b**2*asinh(c*x)**2/(c**4*x**5 + 2*c**2*x**3 + x),
x) + Integral(2*a*b*asinh(c*x)/(c**4*x**5 + 2*c**2*x**3 + x), x))/d**2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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