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

Optimal. Leaf size=389 \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{2 b^2 c \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{2 b^2 c \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{15 i b^2 c \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{15 i b^2 c \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{c^2 x^2+1}}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (c^2 x^2+1\right )}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (c^2 x^2+1\right )^2}-\frac{15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3}-\frac{4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac{b^2 c^2 x}{12 d^3 \left (c^2 x^2+1\right )}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3} \]

[Out]

(b^2*c^2*x)/(12*d^3*(1 + c^2*x^2)) - (b*c*(a + b*ArcSinh[c*x]))/(6*d^3*(1 + c^2*x^2)^(3/2)) - (7*b*c*(a + b*Ar
cSinh[c*x]))/(4*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])^2/(d^3*x*(1 + c^2*x^2)^2) - (5*c^2*x*(a + b*ArcS
inh[c*x])^2)/(4*d^3*(1 + c^2*x^2)^2) - (15*c^2*x*(a + b*ArcSinh[c*x])^2)/(8*d^3*(1 + c^2*x^2)) - (15*c*(a + b*
ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(4*d^3) + (11*b^2*c*ArcTan[c*x])/(6*d^3) - (4*b*c*(a + b*ArcSinh[c*x])
*ArcTanh[E^ArcSinh[c*x]])/d^3 - (2*b^2*c*PolyLog[2, -E^ArcSinh[c*x]])/d^3 + (((15*I)/4)*b*c*(a + b*ArcSinh[c*x
])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 - (((15*I)/4)*b*c*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d
^3 + (2*b^2*c*PolyLog[2, E^ArcSinh[c*x]])/d^3 - (((15*I)/4)*b^2*c*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d^3 + (((15
*I)/4)*b^2*c*PolyLog[3, I*E^ArcSinh[c*x]])/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.766648, antiderivative size = 389, normalized size of antiderivative = 1., number of steps used = 27, number of rules used = 15, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.577, Rules used = {5747, 5690, 5693, 4180, 2531, 2282, 6589, 5717, 203, 199, 5755, 5760, 4182, 2279, 2391} \[ \frac{15 i b c \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{15 i b c \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}-\frac{2 b^2 c \text{PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{2 b^2 c \text{PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{15 i b^2 c \text{PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{15 i b^2 c \text{PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{c^2 x^2+1}}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (c^2 x^2+1\right )}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (c^2 x^2+1\right )^2}-\frac{15 c \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3}-\frac{4 b c \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac{b^2 c^2 x}{12 d^3 \left (c^2 x^2+1\right )}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2*x)/(12*d^3*(1 + c^2*x^2)) - (b*c*(a + b*ArcSinh[c*x]))/(6*d^3*(1 + c^2*x^2)^(3/2)) - (7*b*c*(a + b*Ar
cSinh[c*x]))/(4*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])^2/(d^3*x*(1 + c^2*x^2)^2) - (5*c^2*x*(a + b*ArcS
inh[c*x])^2)/(4*d^3*(1 + c^2*x^2)^2) - (15*c^2*x*(a + b*ArcSinh[c*x])^2)/(8*d^3*(1 + c^2*x^2)) - (15*c*(a + b*
ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(4*d^3) + (11*b^2*c*ArcTan[c*x])/(6*d^3) - (4*b*c*(a + b*ArcSinh[c*x])
*ArcTanh[E^ArcSinh[c*x]])/d^3 - (2*b^2*c*PolyLog[2, -E^ArcSinh[c*x]])/d^3 + (((15*I)/4)*b*c*(a + b*ArcSinh[c*x
])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/d^3 - (((15*I)/4)*b*c*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d
^3 + (2*b^2*c*PolyLog[2, E^ArcSinh[c*x]])/d^3 - (((15*I)/4)*b^2*c*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d^3 + (((15
*I)/4)*b^2*c*PolyLog[3, I*E^ArcSinh[c*x]])/d^3

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 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 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 5717

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 199

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 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 )^3} \, dx &=-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\left (5 c^2\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^3} \, dx+\frac{(2 b c) \int \frac{a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=\frac{2 b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac{(2 b c) \int \frac{a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac{\left (2 b^2 c^2\right ) \int \frac{1}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^3}+\frac{\left (5 b c^3\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}-\frac{\left (15 c^2\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac{b^2 c^2 x}{3 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac{2 b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac{(2 b c) \int \frac{a+b \sinh ^{-1}(c x)}{x \sqrt{1+c^2 x^2}} \, dx}{d^3}-\frac{\left (b^2 c^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 d^3}+\frac{\left (5 b^2 c^2\right ) \int \frac{1}{\left (1+c^2 x^2\right )^2} \, dx}{6 d^3}-\frac{\left (2 b^2 c^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{d^3}+\frac{\left (15 b c^3\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{4 d^3}-\frac{\left (15 c^2\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac{7 b^2 c \tan ^{-1}(c x)}{3 d^3}-\frac{(15 c) \operatorname{Subst}\left (\int (a+b x)^2 \text{sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}+\frac{(2 b c) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac{\left (5 b^2 c^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{12 d^3}+\frac{\left (15 b^2 c^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{4 d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{(15 i b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\frac{(15 i b c) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\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^3}+\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^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}+\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\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^3}+\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^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{2 b^2 c \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{2 b^2 c \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{\left (15 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ &=\frac{b^2 c^2 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac{b c \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac{7 b c \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3 \sqrt{1+c^2 x^2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{d^3 x \left (1+c^2 x^2\right )^2}-\frac{5 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac{15 c^2 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac{15 c \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{11 b^2 c \tan ^{-1}(c x)}{6 d^3}-\frac{4 b c \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{2 b^2 c \text{Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{d^3}+\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac{15 i b c \left (a+b \sinh ^{-1}(c x)\right ) \text{Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{2 b^2 c \text{Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{d^3}-\frac{15 i b^2 c \text{Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac{15 i b^2 c \text{Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ \end{align*}

Mathematica [A]  time = 7.61938, size = 716, normalized size = 1.84 \[ \frac{2 a b c \left (\frac{15}{16} i \left (2 \text{PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )-\frac{1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-\frac{15}{16} i \left (2 \text{PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )-\frac{1}{2} \sinh ^{-1}(c x)^2+2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )+\frac{7 \left (\sqrt{c^2 x^2+1}+i \sinh ^{-1}(c x)\right )}{16 (-1-i c x)}-\frac{7 \left (\sinh ^{-1}(c x)+i \sqrt{c^2 x^2+1}\right )}{16 (c x+i)}+\frac{i \left (3 \sinh ^{-1}(c x)+\sqrt{c^2 x^2+1} (c x-2 i)\right )}{48 (c x-i)^2}-\frac{i \left (3 \sinh ^{-1}(c x)+\sqrt{c^2 x^2+1} (c x+2 i)\right )}{48 (c x+i)^2}-\tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )-\frac{\sinh ^{-1}(c x)}{c x}\right )}{d^3}+\frac{b^2 c \left (90 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )-90 i \sinh ^{-1}(c x) \text{PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+48 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-48 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )+90 i \text{PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )-90 i \text{PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )+\frac{2 c x}{c^2 x^2+1}-\frac{21 c x \sinh ^{-1}(c x)^2}{c^2 x^2+1}-\frac{6 c x \sinh ^{-1}(c x)^2}{\left (c^2 x^2+1\right )^2}-\frac{42 \sinh ^{-1}(c x)}{\sqrt{c^2 x^2+1}}-\frac{4 \sinh ^{-1}(c x)}{\left (c^2 x^2+1\right )^{3/2}}+45 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-45 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+48 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-48 \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+12 \sinh ^{-1}(c x)^2 \tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )-12 \sinh ^{-1}(c x)^2 \coth \left (\frac{1}{2} \sinh ^{-1}(c x)\right )+88 \tan ^{-1}\left (\tanh \left (\frac{1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{24 d^3}-\frac{7 a^2 c^2 x}{8 d^3 \left (c^2 x^2+1\right )}-\frac{a^2 c^2 x}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac{15 a^2 c \tan ^{-1}(c x)}{8 d^3}-\frac{a^2}{d^3 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(a^2/(d^3*x)) - (a^2*c^2*x)/(4*d^3*(1 + c^2*x^2)^2) - (7*a^2*c^2*x)/(8*d^3*(1 + c^2*x^2)) - (15*a^2*c*ArcTan[
c*x])/(8*d^3) + (2*a*b*c*((7*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(16*(-1 - I*c*x)) - ArcSinh[c*x]/(c*x) - (7
*(I*Sqrt[1 + c^2*x^2] + ArcSinh[c*x]))/(16*(I + c*x)) + ((I/48)*((-2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*
x]))/(-I + c*x)^2 - ((I/48)*((2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x]))/(I + c*x)^2 - ArcTanh[Sqrt[1 + c
^2*x^2]] + ((15*I)/16)*(-ArcSinh[c*x]^2/2 + 2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] + 2*PolyLog[2, (-I)*E^Arc
Sinh[c*x]]) - ((15*I)/16)*(-ArcSinh[c*x]^2/2 + 2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] + 2*PolyLog[2, I*E^Arc
Sinh[c*x]])))/d^3 + (b^2*c*((2*c*x)/(1 + c^2*x^2) - (4*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) - (42*ArcSinh[c*x])/S
qrt[1 + c^2*x^2] - (6*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2)^2 - (21*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2) + 88*ArcTan[
Tanh[ArcSinh[c*x]/2]] - 12*ArcSinh[c*x]^2*Coth[ArcSinh[c*x]/2] + 48*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] +
(45*I)*ArcSinh[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] - (45*I)*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*x]] - 48*ArcSinh
[c*x]*Log[1 + E^(-ArcSinh[c*x])] + 48*PolyLog[2, -E^(-ArcSinh[c*x])] + (90*I)*ArcSinh[c*x]*PolyLog[2, (-I)/E^A
rcSinh[c*x]] - (90*I)*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] - 48*PolyLog[2, E^(-ArcSinh[c*x])] + (90*I)*Po
lyLog[3, (-I)/E^ArcSinh[c*x]] - (90*I)*PolyLog[3, I/E^ArcSinh[c*x]] + 12*ArcSinh[c*x]^2*Tanh[ArcSinh[c*x]/2]))
/(24*d^3)

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Maple [F]  time = 0.36, 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 ) ^{3}}}\, 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)^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{8} \, a^{2}{\left (\frac{15 \, c^{4} x^{4} + 25 \, c^{2} x^{2} + 8}{c^{4} d^{3} x^{5} + 2 \, c^{2} d^{3} x^{3} + d^{3} x} + \frac{15 \, c \arctan \left (c x\right )}{d^{3}}\right )} + \int \frac{b^{2} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2}}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} x^{2}} + \frac{2 \, a b \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )}{c^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} 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)^3,x, algorithm="maxima")

[Out]

-1/8*a^2*((15*c^4*x^4 + 25*c^2*x^2 + 8)/(c^4*d^3*x^5 + 2*c^2*d^3*x^3 + d^3*x) + 15*c*arctan(c*x)/d^3) + integr
ate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^6*d^3*x^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*x^2) + 2*a*b*log(c*x
+ sqrt(c^2*x^2 + 1))/(c^6*d^3*x^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*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^{6} d^{3} x^{8} + 3 \, c^{4} d^{3} x^{6} + 3 \, c^{2} d^{3} x^{4} + d^{3} 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)^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^8 + 3*c^4*d^3*x^6 + 3*c^2*d^3*x^4 + d^3*x^
2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a^{2}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac{b^{2} \operatorname{asinh}^{2}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx + \int \frac{2 a b \operatorname{asinh}{\left (c x \right )}}{c^{6} x^{8} + 3 c^{4} x^{6} + 3 c^{2} x^{4} + x^{2}}\, dx}{d^{3}} \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)**3,x)

[Out]

(Integral(a**2/(c**6*x**8 + 3*c**4*x**6 + 3*c**2*x**4 + x**2), x) + Integral(b**2*asinh(c*x)**2/(c**6*x**8 + 3
*c**4*x**6 + 3*c**2*x**4 + x**2), x) + Integral(2*a*b*asinh(c*x)/(c**6*x**8 + 3*c**4*x**6 + 3*c**2*x**4 + x**2
), x))/d**3

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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 )}^{3} 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)^3,x, algorithm="giac")

[Out]

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