3.3.43 \(\int \frac {x^4 (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^3} \, dx\) [243]

Optimal. Leaf size=320 \[ -\frac {b^2 x}{12 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {1+c^2 x^2}}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (1+c^2 x^2\right )^2}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (1+c^2 x^2\right )}+\frac {3 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {7 b^2 \text {ArcTan}(c x)}{6 c^5 d^3}-\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}+\frac {3 i b^2 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {3 i b^2 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3} \]

[Out]

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

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Rubi [A]
time = 0.37, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5810, 5789, 4265, 2611, 2320, 6724, 5798, 209, 272, 45, 5804, 12, 393} \begin {gather*} \frac {3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^5 d^3}-\frac {3 i b \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}+\frac {3 i b \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3}-\frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 c^2 d^3 \left (c^2 x^2+1\right )^2}-\frac {5 b \left (a+b \sinh ^{-1}(c x)\right )}{4 c^5 d^3 \sqrt {c^2 x^2+1}}+\frac {b \left (a+b \sinh ^{-1}(c x)\right )}{6 c^5 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {3 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 c^4 d^3 \left (c^2 x^2+1\right )}+\frac {7 b^2 \text {ArcTan}(c x)}{6 c^5 d^3}+\frac {3 i b^2 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {3 i b^2 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 c^5 d^3}-\frac {b^2 x}{12 c^4 d^3 \left (c^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^3,x]

[Out]

-1/12*(b^2*x)/(c^4*d^3*(1 + c^2*x^2)) + (b*(a + b*ArcSinh[c*x]))/(6*c^5*d^3*(1 + c^2*x^2)^(3/2)) - (5*b*(a + b
*ArcSinh[c*x]))/(4*c^5*d^3*Sqrt[1 + c^2*x^2]) - (x^3*(a + b*ArcSinh[c*x])^2)/(4*c^2*d^3*(1 + c^2*x^2)^2) - (3*
x*(a + b*ArcSinh[c*x])^2)/(8*c^4*d^3*(1 + c^2*x^2)) + (3*(a + b*ArcSinh[c*x])^2*ArcTan[E^ArcSinh[c*x]])/(4*c^5
*d^3) + (7*b^2*ArcTan[c*x])/(6*c^5*d^3) - (((3*I)/4)*b*(a + b*ArcSinh[c*x])*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(
c^5*d^3) + (((3*I)/4)*b*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/(c^5*d^3) + (((3*I)/4)*b^2*PolyLog[
3, (-I)*E^ArcSinh[c*x]])/(c^5*d^3) - (((3*I)/4)*b^2*PolyLog[3, I*E^ArcSinh[c*x]])/(c^5*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 45

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 209

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

Rule 272

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 393

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

Rule 2320

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 2611

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 4265

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 5789

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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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 5804

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

Rule 5810

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/(2*c*(p + 1)))*Simp[
(d + e*x^2)^p/(1 + c^2*x^2)^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}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && IGtQ[m, 1]

Rule 6724

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

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Mathematica [A]
time = 2.14, size = 552, normalized size = 1.72 \begin {gather*} \frac {\frac {6 a^2 c x}{\left (1+c^2 x^2\right )^2}-\frac {15 a^2 c x}{1+c^2 x^2}+\frac {15 a b \left (\sqrt {1+c^2 x^2}-i \sinh ^{-1}(c x)\right )}{-1+i c x}+\frac {15 a b \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{-1-i c x}-\frac {i a b \left ((-2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{(-i+c x)^2}+\frac {i a b \left ((2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{(i+c x)^2}+9 a^2 \text {ArcTan}(c x)+\frac {9}{2} i 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 )-\frac {9}{2} i 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 )+b^2 \left (-\frac {2 c x}{1+c^2 x^2}+\frac {4 \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}}-\frac {30 \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}+\frac {6 c x \sinh ^{-1}(c x)^2}{\left (1+c^2 x^2\right )^2}-\frac {15 c x \sinh ^{-1}(c x)^2}{1+c^2 x^2}+56 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-9 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+9 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-18 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+18 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )-18 i \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )+18 i \text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )\right )}{24 c^5 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^3,x]

[Out]

((6*a^2*c*x)/(1 + c^2*x^2)^2 - (15*a^2*c*x)/(1 + c^2*x^2) + (15*a*b*(Sqrt[1 + c^2*x^2] - I*ArcSinh[c*x]))/(-1
+ I*c*x) + (15*a*b*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(-1 - I*c*x) - (I*a*b*((-2*I + c*x)*Sqrt[1 + c^2*x^2]
+ 3*ArcSinh[c*x]))/(-I + c*x)^2 + (I*a*b*((2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x]))/(I + c*x)^2 + 9*a^
2*ArcTan[c*x] + ((9*I)/2)*a*b*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 + I*E^ArcSinh[c*x]]) - 4*PolyLog[2, (-I)*E
^ArcSinh[c*x]]) - ((9*I)/2)*a*b*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 - I*E^ArcSinh[c*x]]) - 4*PolyLog[2, I*E^
ArcSinh[c*x]]) + b^2*((-2*c*x)/(1 + c^2*x^2) + (4*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) - (30*ArcSinh[c*x])/Sqrt[1
+ c^2*x^2] + (6*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2)^2 - (15*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2) + 56*ArcTan[Tanh[
ArcSinh[c*x]/2]] - (9*I)*ArcSinh[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] + (9*I)*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c
*x]] - (18*I)*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x]] + (18*I)*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]]
- (18*I)*PolyLog[3, (-I)/E^ArcSinh[c*x]] + (18*I)*PolyLog[3, I/E^ArcSinh[c*x]]))/(24*c^5*d^3)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {x^{4} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (c^{2} d \,x^{2}+d \right )^{3}}\, dx\]

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)^3,x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^3,x, algorithm="maxima")

[Out]

-1/8*a^2*((5*c^2*x^3 + 3*x)/(c^8*d^3*x^4 + 2*c^6*d^3*x^2 + c^4*d^3) - 3*arctan(c*x)/(c^5*d^3)) + integrate(b^2
*x^4*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3) + 2*a*b*x^4*log(c*x +
sqrt(c^2*x^2 + 1))/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^3,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^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*
x^2 + d^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} x^{4}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{6} + 3 c^{4} x^{4} + 3 c^{2} x^{2} + 1}\, dx}{d^{3}} \end {gather*}

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)**3,x)

[Out]

(Integral(a**2*x**4/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(b**2*x**4*asinh(c*x)**2/(c**6*x
**6 + 3*c**4*x**4 + 3*c**2*x**2 + 1), x) + Integral(2*a*b*x**4*asinh(c*x)/(c**6*x**6 + 3*c**4*x**4 + 3*c**2*x*
*2 + 1), x))/d**3

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d\,c^2\,x^2+d\right )}^3} \,d x \end {gather*}

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)^3,x)

[Out]

int((x^4*(a + b*asinh(c*x))^2)/(d + c^2*d*x^2)^3, x)

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