3.234 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^2} \, dx\)

Optimal. Leaf size=279 \[ \frac {3 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}-\frac {3 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}-\frac {3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac {2 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2 \sqrt {c^2 x^2+1}}-\frac {3 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {3 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}-\frac {b^2 \tan ^{-1}(c x)}{c^5 d^2}+\frac {2 b^2 x}{c^4 d^2} \]

[Out]

2*b^2*x/c^4/d^2+3/2*x*(a+b*arcsinh(c*x))^2/c^4/d^2-1/2*x^3*(a+b*arcsinh(c*x))^2/c^2/d^2/(c^2*x^2+1)-3*(a+b*arc
sinh(c*x))^2*arctan(c*x+(c^2*x^2+1)^(1/2))/c^5/d^2-b^2*arctan(c*x)/c^5/d^2+3*I*b*(a+b*arcsinh(c*x))*polylog(2,
-I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d^2-3*I*b*(a+b*arcsinh(c*x))*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d^2-3*I*
b^2*polylog(3,-I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d^2+3*I*b^2*polylog(3,I*(c*x+(c^2*x^2+1)^(1/2)))/c^5/d^2+b*(a+b*
arcsinh(c*x))/c^5/d^2/(c^2*x^2+1)^(1/2)-2*b*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c^5/d^2

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Rubi [A]  time = 0.53, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {5751, 5767, 5693, 4180, 2531, 2282, 6589, 5717, 8, 266, 43, 5732, 388, 205} \[ \frac {3 i b \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}-\frac {3 i b \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}-\frac {3 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}+\frac {3 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{c^5 d^2}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^2 d^2 \left (c^2 x^2+1\right )}-\frac {2 b \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d^2 \sqrt {c^2 x^2+1}}+\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac {3 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d^2}+\frac {2 b^2 x}{c^4 d^2}-\frac {b^2 \tan ^{-1}(c x)}{c^5 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^2,x]

[Out]

(2*b^2*x)/(c^4*d^2) + (b*(a + b*ArcSinh[c*x]))/(c^5*d^2*Sqrt[1 + c^2*x^2]) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*Arc
Sinh[c*x]))/(c^5*d^2) + (3*x*(a + b*ArcSinh[c*x])^2)/(2*c^4*d^2) - (x^3*(a + b*ArcSinh[c*x])^2)/(2*c^2*d^2*(1
+ c^2*x^2)) - (3*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(c^5*d^2) - (b^2*ArcTan[c*x])/(c^5*d^2) + ((3*
I)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(c^5*d^2) - ((3*I)*b*(a + b*ArcSinh[c*x])*PolyLog[2
, I*E^ArcSinh[c*x]])/(c^5*d^2) - ((3*I)*b^2*PolyLog[3, (-I)*E^ArcSinh[c*x]])/(c^5*d^2) + ((3*I)*b^2*PolyLog[3,
I*E^ArcSinh[c*x]])/(c^5*d^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 205

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

Rule 266

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 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 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 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 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 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 5751

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

Rule 5767

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

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

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Mathematica [A]  time = 2.23, size = 482, normalized size = 1.73 \[ \frac {-\frac {3 a^2 \tan ^{-1}(c x)}{c^5}+\frac {2 a^2 x}{c^4}+\frac {a^2 x}{c^6 x^2+c^4}-\frac {2 a b \left (-2 c^3 x^3 \sinh ^{-1}(c x)-3 i \left (c^2 x^2+1\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )+3 i \left (c^2 x^2+1\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )+2 c^2 x^2 \sqrt {c^2 x^2+1}+\sqrt {c^2 x^2+1}+3 i c^2 x^2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-3 i c^2 x^2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )-3 c x \sinh ^{-1}(c x)+3 i \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )-3 i \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )}{c^7 x^2+c^5}+\frac {2 b^2 \left (\frac {c x \sinh ^{-1}(c x)^2}{2 c^2 x^2+2}-2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)+\frac {\sinh ^{-1}(c x)}{\sqrt {c^2 x^2+1}}+\frac {1}{2} i \left (6 \sinh ^{-1}(c x) \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-6 \sinh ^{-1}(c x) \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+6 \text {Li}_3\left (-i e^{-\sinh ^{-1}(c x)}\right )-6 \text {Li}_3\left (i e^{-\sinh ^{-1}(c x)}\right )+3 \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-3 \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )+c x \left (\sinh ^{-1}(c x)^2+2\right )\right )}{c^5}}{2 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^2,x]

[Out]

((2*a^2*x)/c^4 + (a^2*x)/(c^4 + c^6*x^2) - (3*a^2*ArcTan[c*x])/c^5 - (2*a*b*(Sqrt[1 + c^2*x^2] + 2*c^2*x^2*Sqr
t[1 + c^2*x^2] - 3*c*x*ArcSinh[c*x] - 2*c^3*x^3*ArcSinh[c*x] + (3*I)*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] +
(3*I)*c^2*x^2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]] - (3*I)*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - (3*I)*c^
2*x^2*ArcSinh[c*x]*Log[1 + I*E^ArcSinh[c*x]] - (3*I)*(1 + c^2*x^2)*PolyLog[2, (-I)*E^ArcSinh[c*x]] + (3*I)*(1
+ c^2*x^2)*PolyLog[2, I*E^ArcSinh[c*x]]))/(c^5 + c^7*x^2) + (2*b^2*(ArcSinh[c*x]/Sqrt[1 + c^2*x^2] - 2*Sqrt[1
+ c^2*x^2]*ArcSinh[c*x] + (c*x*ArcSinh[c*x]^2)/(2 + 2*c^2*x^2) + c*x*(2 + ArcSinh[c*x]^2) + (I/2)*((4*I)*ArcTa
n[Tanh[ArcSinh[c*x]/2]] + 3*ArcSinh[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] - 3*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*
x]] + 6*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x]] - 6*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] + 6*PolyLog
[3, (-I)/E^ArcSinh[c*x]] - 6*PolyLog[3, I/E^ArcSinh[c*x]])))/c^5)/(2*d^2)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {arsinh}\left (c x\right ) + a^{2} x^{4}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^2,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^2,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (c^{2} d \,x^{2}+d \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^2,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^2,x)

[Out]

int((x^4*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{4}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{4} + 2 c^{2} x^{2} + 1}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**2,x)

[Out]

(Integral(a**2*x**4/(c**4*x**4 + 2*c**2*x**2 + 1), x) + Integral(b**2*x**4*asinh(c*x)**2/(c**4*x**4 + 2*c**2*x
**2 + 1), x) + Integral(2*a*b*x**4*asinh(c*x)/(c**4*x**4 + 2*c**2*x**2 + 1), x))/d**2

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