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

Optimal. Leaf size=381 \[ \frac {3 b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (c^2 x^2+1\right )}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (c^2 x^2+1\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (c^2 x^2+1\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^3}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b^2 c^2}{12 d^3 \left (c^2 x^2+1\right )}-\frac {7 b^2 c^2 \log \left (c^2 x^2+1\right )}{6 d^3}+\frac {b^2 c^2 \log (x)}{d^3} \]

[Out]

1/12*b^2*c^2/d^3/(c^2*x^2+1)-b*c*(a+b*arcsinh(c*x))/d^3/x/(c^2*x^2+1)^(3/2)-5/6*b*c^3*x*(a+b*arcsinh(c*x))/d^3
/(c^2*x^2+1)^(3/2)-3/4*c^2*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)^2-1/2*(a+b*arcsinh(c*x))^2/d^3/x^2/(c^2*x^2+1)
^2-3/2*c^2*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)+6*c^2*(a+b*arcsinh(c*x))^2*arctanh((c*x+(c^2*x^2+1)^(1/2))^2)/
d^3+b^2*c^2*ln(x)/d^3-7/6*b^2*c^2*ln(c^2*x^2+1)/d^3+3*b*c^2*(a+b*arcsinh(c*x))*polylog(2,-(c*x+(c^2*x^2+1)^(1/
2))^2)/d^3-3*b*c^2*(a+b*arcsinh(c*x))*polylog(2,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-3/2*b^2*c^2*polylog(3,-(c*x+(c^
2*x^2+1)^(1/2))^2)/d^3+3/2*b^2*c^2*polylog(3,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+4/3*b*c^3*x*(a+b*arcsinh(c*x))/d^3
/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.80, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 19, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.731, Rules used = {5747, 5755, 5720, 5461, 4182, 2531, 2282, 6589, 5687, 260, 5690, 261, 271, 192, 191, 5732, 12, 1251, 893} \[ \frac {3 b c^2 \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 b c^2 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (c^2 x^2+1\right )}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (c^2 x^2+1\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (c^2 x^2+1\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^3}+\frac {b^2 c^2}{12 d^3 \left (c^2 x^2+1\right )}-\frac {7 b^2 c^2 \log \left (c^2 x^2+1\right )}{6 d^3}+\frac {b^2 c^2 \log (x)}{d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2)/(12*d^3*(1 + c^2*x^2)) - (b*c*(a + b*ArcSinh[c*x]))/(d^3*x*(1 + c^2*x^2)^(3/2)) - (5*b*c^3*x*(a + b*
ArcSinh[c*x]))/(6*d^3*(1 + c^2*x^2)^(3/2)) + (4*b*c^3*x*(a + b*ArcSinh[c*x]))/(3*d^3*Sqrt[1 + c^2*x^2]) - (3*c
^2*(a + b*ArcSinh[c*x])^2)/(4*d^3*(1 + c^2*x^2)^2) - (a + b*ArcSinh[c*x])^2/(2*d^3*x^2*(1 + c^2*x^2)^2) - (3*c
^2*(a + b*ArcSinh[c*x])^2)/(2*d^3*(1 + c^2*x^2)) + (6*c^2*(a + b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/
d^3 + (b^2*c^2*Log[x])/d^3 - (7*b^2*c^2*Log[1 + c^2*x^2])/(6*d^3) + (3*b*c^2*(a + b*ArcSinh[c*x])*PolyLog[2, -
E^(2*ArcSinh[c*x])])/d^3 - (3*b*c^2*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d^3 - (3*b^2*c^2*Poly
Log[3, -E^(2*ArcSinh[c*x])])/(2*d^3) + (3*b^2*c^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d^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 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

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]

Rule 261

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

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

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 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 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 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 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 5690

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

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 5732

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(1 + c^2*x^2)^p, x]}, Dist[d^p*(a + b*ArcSinh[c*x]), u, x] - Dist[b*c*d^p, Int[SimplifyIntegrand[u/Sqrt[1 +
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p - 1/2] && (IGtQ[(m + 1)/2,
0] || ILtQ[(m + 2*p + 3)/2, 0]) && NeQ[p, -2^(-1)] && GtQ[d, 0]

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 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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^3 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {8 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {\left (b^2 c^2\right ) \int \frac {-3-12 c^2 x^2-8 c^4 x^4}{3 x \left (1+c^2 x^2\right )^2} \, dx}{d^3}+\frac {\left (3 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}-\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {8 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (b^2 c^2\right ) \int \frac {-3-12 c^2 x^2-8 c^4 x^4}{x \left (1+c^2 x^2\right )^2} \, dx}{3 d^3}+\frac {\left (b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}+\frac {\left (3 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac {\left (b^2 c^4\right ) \int \frac {x}{\left (1+c^2 x^2\right )^2} \, dx}{2 d^3}-\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )} \, dx}{d^2}\\ &=\frac {b^2 c^2}{4 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (3 c^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {-3-12 c^2 x-8 c^4 x^2}{x \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (b^2 c^4\right ) \int \frac {x}{1+c^2 x^2} \, dx}{d^3}-\frac {\left (3 b^2 c^4\right ) \int \frac {x}{1+c^2 x^2} \, dx}{d^3}\\ &=\frac {b^2 c^2}{4 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 b^2 c^2 \log \left (1+c^2 x^2\right )}{d^3}-\frac {\left (6 c^2\right ) \operatorname {Subst}\left (\int (a+b x)^2 \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \left (-\frac {3}{x}-\frac {c^2}{\left (1+c^2 x\right )^2}-\frac {5 c^2}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {\left (6 b c^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (6 b c^2\right ) \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {\left (3 b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

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Mathematica [C]  time = 9.46, size = 759, normalized size = 1.99 \[ -\frac {a^2 c^2}{d^3 \left (c^2 x^2+1\right )}-\frac {a^2 c^2}{4 d^3 \left (c^2 x^2+1\right )^2}+\frac {3 a^2 c^2 \log \left (c^2 x^2+1\right )}{2 d^3}-\frac {3 a^2 c^2 \log (x)}{d^3}-\frac {a^2}{2 d^3 x^2}+\frac {2 a b \left (\frac {3}{2} c^3 \left (\frac {2 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{c}-\frac {\sinh ^{-1}(c x)^2}{2 c}+\frac {2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{c}\right )+\frac {3}{2} c^3 \left (\frac {2 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{c}-\frac {\sinh ^{-1}(c x)^2}{2 c}+\frac {2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )}{c}\right )-3 c^2 \left (\frac {1}{2} \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )-\frac {1}{2} \sinh ^{-1}(c x)^2+\sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )\right )-\frac {c^2 \left (-3 \sinh ^{-1}(c x)+(-c x+2 i) \sqrt {c^2 x^2+1}\right )}{48 (c x-i)^2}-\frac {9 i c^2 \left (\sqrt {c^2 x^2+1}+i \sinh ^{-1}(c x)\right )}{16 (-1-i c x)}+\frac {c^2 \left (3 \sinh ^{-1}(c x)+\sqrt {c^2 x^2+1} (c x+2 i)\right )}{48 (c x+i)^2}-\frac {c x \sqrt {c^2 x^2+1}+\sinh ^{-1}(c x)}{2 x^2}-\frac {9 i c^3 \left (\sinh ^{-1}(c x)+i \sqrt {c^2 x^2+1}\right )}{16 \left (c^2 x+i c\right )}\right )}{d^3}+\frac {b^2 c^2 \left (\frac {1}{24} \left (\frac {2}{c^2 x^2+1}-56 \log \left (\sqrt {c^2 x^2+1}\right )-\frac {24 \sinh ^{-1}(c x)^2}{c^2 x^2+1}-\frac {12 \sinh ^{-1}(c x)^2}{c^2 x^2}-\frac {6 \sinh ^{-1}(c x)^2}{\left (c^2 x^2+1\right )^2}-\frac {24 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{c x}+\frac {56 c x \sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {4 c x \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}-36 \text {Li}_3\left (-e^{-2 \sinh ^{-1}(c x)}\right )+36 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )+24 \log (c x)+48 \sinh ^{-1}(c x)^3+72 \sinh ^{-1}(c x)^2 \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )-72 \sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )-3 i \pi ^3\right )-3 \sinh ^{-1}(c x) \text {Li}_2\left (-e^{-2 \sinh ^{-1}(c x)}\right )-3 \sinh ^{-1}(c x) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )\right )}{d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*a^2/(d^3*x^2) - (a^2*c^2)/(4*d^3*(1 + c^2*x^2)^2) - (a^2*c^2)/(d^3*(1 + c^2*x^2)) - (3*a^2*c^2*Log[x])/d^
3 + (3*a^2*c^2*Log[1 + c^2*x^2])/(2*d^3) + (2*a*b*(-1/48*(c^2*((2*I - c*x)*Sqrt[1 + c^2*x^2] - 3*ArcSinh[c*x])
)/(-I + c*x)^2 - (((9*I)/16)*c^2*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(-1 - I*c*x) - (((9*I)/16)*c^3*(I*Sqrt[
1 + c^2*x^2] + ArcSinh[c*x]))/(I*c + c^2*x) - (c*x*Sqrt[1 + c^2*x^2] + ArcSinh[c*x])/(2*x^2) + (c^2*((2*I + c*
x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x]))/(48*(I + c*x)^2) + (3*c^3*(-1/2*ArcSinh[c*x]^2/c + (2*ArcSinh[c*x]*Log
[1 + I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, (-I)*E^ArcSinh[c*x]])/c))/2 + (3*c^3*(-1/2*ArcSinh[c*x]^2/c + (2*Arc
Sinh[c*x]*Log[1 - I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, I*E^ArcSinh[c*x]])/c))/2 - 3*c^2*(-1/2*ArcSinh[c*x]^2 +
ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + PolyLog[2, E^(2*ArcSinh[c*x])]/2)))/d^3 + (b^2*c^2*(-3*ArcSinh[c*x
]*PolyLog[2, -E^(-2*ArcSinh[c*x])] - 3*ArcSinh[c*x]*PolyLog[2, E^(2*ArcSinh[c*x])] + ((-3*I)*Pi^3 + 2/(1 + c^2
*x^2) + (4*c*x*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) + (56*c*x*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] - (24*Sqrt[1 + c^2*
x^2]*ArcSinh[c*x])/(c*x) - (12*ArcSinh[c*x]^2)/(c^2*x^2) - (6*ArcSinh[c*x]^2)/(1 + c^2*x^2)^2 - (24*ArcSinh[c*
x]^2)/(1 + c^2*x^2) + 48*ArcSinh[c*x]^3 + 72*ArcSinh[c*x]^2*Log[1 + E^(-2*ArcSinh[c*x])] - 72*ArcSinh[c*x]^2*L
og[1 - E^(2*ArcSinh[c*x])] + 24*Log[c*x] - 56*Log[Sqrt[1 + c^2*x^2]] - 36*PolyLog[3, -E^(-2*ArcSinh[c*x])] + 3
6*PolyLog[3, E^(2*ArcSinh[c*x])])/24))/d^3

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*x^
3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} + d\right )}^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.43, size = 1436, normalized size = 3.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b^2*c^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-4/3*c^6*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*x^4-8/3*c^4*a*b/d
^3/(c^4*x^4+2*c^2*x^2+1)*x^2-4/3*c^6*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*x^4-3/2*c^4*b^2/d^3/(c^4*x^4+2
*c^2*x^2+1)*arcsinh(c*x)^2*x^2-8/3*c^4*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*x^2-6*c^2*a*b/d^3*arcsinh(c*
x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-6*c^2*a*b/d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-9/2*c^2*a*b/d^3/(c^4*x^4
+2*c^2*x^2+1)*arcsinh(c*x)+6*c^2*a*b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-a*b/d^3/(c^4*x^4+2*c^2*x
^2+1)/x^2*arcsinh(c*x)+3*c^2*b^2/d^3*arcsinh(c*x)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)-9/4*c^2*b^2/d^3/(c^4*x
^4+2*c^2*x^2+1)*arcsinh(c*x)^2-3*c^2*b^2/d^3*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))-6*c^2*b^2/d^3*arcsinh(
c*x)*polylog(2,c*x+(c^2*x^2+1)^(1/2))+3*c^2*b^2/d^3*arcsinh(c*x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-4/3*c^2*b^2
/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)-4/3*c^2*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)+1/12*c^4*b^2/d^3/(c^4*x^4+2*c^2*
x^2+1)*x^2-1/2*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)/x^2*arcsinh(c*x)^2+3*c^2*a*b/d^3*polylog(2,-(c*x+(c^2*x^2+1)^(1/2
))^2)-6*c^2*a*b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-6*c^2*a*b/d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))-3*c^2*b^2
/d^3*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))-6*c^2*b^2/d^3*arcsinh(c*x)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-1
/2*a^2/d^3/x^2-1/4*c^2*a^2/d^3/(c^2*x^2+1)^2-c^2*a^2/d^3/(c^2*x^2+1)+1/12*c^2*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)-3*
c^2*a^2/d^3*ln(c*x)+3/2*c^2*a^2/d^3*ln(c^2*x^2+1)+8/3*c^2*b^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2))+6*c^2*b^2/d^3*poly
log(3,-c*x-(c^2*x^2+1)^(1/2))+6*c^2*b^2/d^3*polylog(3,c*x+(c^2*x^2+1)^(1/2))+c^2*b^2/d^3*ln(c*x+(c^2*x^2+1)^(1
/2)-1)+c^2*b^2/d^3*ln(1+c*x+(c^2*x^2+1)^(1/2))-7/3*c^2*b^2/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-3*c^4*a*b/d^3/(
c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*x^2+1/2*c^3*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*x*(c^2*x^2+1)^(1/2)+4/3*c^5*a*b/d^
3/(c^4*x^4+2*c^2*x^2+1)*x^3*(c^2*x^2+1)^(1/2)-c*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)/x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)
-c*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)/x*(c^2*x^2+1)^(1/2)+4/3*c^5*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*(c^2*x
^2+1)^(1/2)*x^3+1/2*c^3*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{4} \, a^{2} {\left (\frac {6 \, c^{4} x^{4} + 9 \, c^{2} x^{2} + 2}{c^{4} d^{3} x^{6} + 2 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} - \frac {6 \, c^{2} \log \left (c^{2} x^{2} + 1\right )}{d^{3}} + \frac {12 \, c^{2} \log \relax (x)}{d^{3}}\right )} + \int \frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}} + \frac {2 \, a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{9} + 3 \, c^{4} d^{3} x^{7} + 3 \, c^{2} d^{3} x^{5} + d^{3} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*((6*c^4*x^4 + 9*c^2*x^2 + 2)/(c^4*d^3*x^6 + 2*c^2*d^3*x^4 + d^3*x^2) - 6*c^2*log(c^2*x^2 + 1)/d^3 + 1
2*c^2*log(x)/d^3) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5
+ d^3*x^3) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{x^3\,{\left (d\,c^2\,x^2+d\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2/(x^3*(d + c^2*d*x^2)^3),x)

[Out]

int((a + b*asinh(c*x))^2/(x^3*(d + c^2*d*x^2)^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**6*x**9 + 3*c**4*x**7 + 3*c**2*x**5 + x**3), x) + Integral(b**2*asinh(c*x)**2/(c**6*x**9 + 3
*c**4*x**7 + 3*c**2*x**5 + x**3), x) + Integral(2*a*b*asinh(c*x)/(c**6*x**9 + 3*c**4*x**7 + 3*c**2*x**5 + x**3
), x))/d**3

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