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

Optimal. Leaf size=204 \[ \frac{2 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{2 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{2 b^2 c \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b^2 c \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}-\frac{4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d} \]

[Out]

-((a + b*ArcSinh[c*x])^2/(d*x)) - (2*c*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/d - (4*b*c*(a + b*ArcSin
h[c*x])*ArcTanh[E^ArcSinh[c*x]])/d - (2*b^2*c*PolyLog[2, -E^ArcSinh[c*x]])/d + ((2*I)*b*c*(a + b*ArcSinh[c*x])
*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d - ((2*I)*b*c*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d + (2*b^2
*c*PolyLog[2, E^ArcSinh[c*x]])/d - ((2*I)*b^2*c*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d + ((2*I)*b^2*c*PolyLog[3, I
*E^ArcSinh[c*x]])/d

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Rubi [A]  time = 0.335271, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {5747, 5693, 4180, 2531, 2282, 6589, 5760, 4182, 2279, 2391} \[ \frac{2 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{2 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d}-\frac{2 b^2 c \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b^2 c \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d}-\frac{4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b*ArcSinh[c*x])^2/(d*x)) - (2*c*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/d - (4*b*c*(a + b*ArcSin
h[c*x])*ArcTanh[E^ArcSinh[c*x]])/d - (2*b^2*c*PolyLog[2, -E^ArcSinh[c*x]])/d + ((2*I)*b*c*(a + b*ArcSinh[c*x])
*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d - ((2*I)*b*c*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d + (2*b^2
*c*PolyLog[2, E^ArcSinh[c*x]])/d - ((2*I)*b^2*c*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d + ((2*I)*b^2*c*PolyLog[3, I
*E^ArcSinh[c*x]])/d

Rule 5747

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)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5693

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

Rule 4180

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c
+ d*x)^m*ArcTanh[E^(-(I*e) + f*fz*x)/E^(I*k*Pi)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*
Log[1 - E^(-(I*e) + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e)
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && 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 5760

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

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-c^2 \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx+\frac{(2 b c) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{d}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{c \operatorname{Subst}\left (\int (a+b x)^2 \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{(2 b c) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{(2 i b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{(2 i b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}-\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 b^2 c \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d}\\ &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d x}-\frac{2 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 b^2 c \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 b^2 c \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d}-\frac{2 i b^2 c \text{Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d}+\frac{2 i b^2 c \text{Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 1.06068, size = 363, normalized size = 1.78 \[ -\frac{\frac{1}{2} i a b c \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )\right )-\frac{1}{2} i a b c \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )-b^2 c \left (2 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-2 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+2 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-2 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+2 i \text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-2 i \text{PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )-\frac{\sinh ^{-1}(c x)^2}{c x}+i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+2 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )\right )+a^2 c \tan ^{-1}(c x)+\frac{a^2}{x}+2 a b c \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )+\frac{2 a b \sinh ^{-1}(c x)}{x}}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a^2/x + (2*a*b*ArcSinh[c*x])/x + a^2*c*ArcTan[c*x] + 2*a*b*c*ArcTanh[Sqrt[1 + c^2*x^2]] + (I/2)*a*b*c*(ArcS
inh[c*x]*(ArcSinh[c*x] - 4*Log[1 + I*E^ArcSinh[c*x]]) - 4*PolyLog[2, (-I)*E^ArcSinh[c*x]]) - (I/2)*a*b*c*(ArcS
inh[c*x]*(ArcSinh[c*x] - 4*Log[1 - I*E^ArcSinh[c*x]]) - 4*PolyLog[2, I*E^ArcSinh[c*x]]) - b^2*c*(-(ArcSinh[c*x
]^2/(c*x)) + 2*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] + I*ArcSinh[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] - I*ArcSin
h[c*x]^2*Log[1 + I/E^ArcSinh[c*x]] - 2*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 2*PolyLog[2, -E^(-ArcSinh[c*x
])] + (2*I)*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (2*I)*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] - 2
*PolyLog[2, E^(-ArcSinh[c*x])] + (2*I)*PolyLog[3, (-I)/E^ArcSinh[c*x]] - (2*I)*PolyLog[3, I/E^ArcSinh[c*x]]))/
d)

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Maple [F]  time = 0.201, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2}}{{x}^{2} \left ({c}^{2}d{x}^{2}+d \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*(c*arctan(c*x)/d + 1/(d*x)) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^2*d*x^4 + d*x^2) + 2*a*b*lo
g(c*x + sqrt(c^2*x^2 + 1))/(c^2*d*x^4 + d*x^2), 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^{2} d x^{4} + d x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**2*x**4 + x**2), x) + Integral(b**2*asinh(c*x)**2/(c**2*x**4 + x**2), x) + Integral(2*a*b*as
inh(c*x)/(c**2*x**4 + x**2), x))/d

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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 )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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