3.134 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))}{d+e x} \, dx\)

Optimal. Leaf size=297 \[ -\frac{i b d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^4}+\frac{i b d^3 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^4}+\frac{d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^4}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^4}-\frac{d x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e^2}+\frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 e}+\frac{a d^2 x}{e^3}-\frac{b d^2 \log \left (c^2 x^2+1\right )}{2 c e^3}-\frac{b d \tan ^{-1}(c x)}{2 c^2 e^2}+\frac{b \log \left (c^2 x^2+1\right )}{6 c^3 e}+\frac{b d^2 x \tan ^{-1}(c x)}{e^3}+\frac{b d x}{2 c e^2}-\frac{b x^2}{6 c e} \]

[Out]

(a*d^2*x)/e^3 + (b*d*x)/(2*c*e^2) - (b*x^2)/(6*c*e) - (b*d*ArcTan[c*x])/(2*c^2*e^2) + (b*d^2*x*ArcTan[c*x])/e^
3 - (d*x^2*(a + b*ArcTan[c*x]))/(2*e^2) + (x^3*(a + b*ArcTan[c*x]))/(3*e) + (d^3*(a + b*ArcTan[c*x])*Log[2/(1
- I*c*x)])/e^4 - (d^3*(a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e^4 - (b*d^2*Log[1 +
c^2*x^2])/(2*c*e^3) + (b*Log[1 + c^2*x^2])/(6*c^3*e) - ((I/2)*b*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^4 + ((I/
2)*b*d^3*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e^4

________________________________________________________________________________________

Rubi [A]  time = 0.270136, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.632, Rules used = {4876, 4846, 260, 4852, 321, 203, 266, 43, 4856, 2402, 2315, 2447} \[ -\frac{i b d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^4}+\frac{i b d^3 \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^4}+\frac{d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^4}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^4}-\frac{d x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 e^2}+\frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 e}+\frac{a d^2 x}{e^3}-\frac{b d^2 \log \left (c^2 x^2+1\right )}{2 c e^3}-\frac{b d \tan ^{-1}(c x)}{2 c^2 e^2}+\frac{b \log \left (c^2 x^2+1\right )}{6 c^3 e}+\frac{b d^2 x \tan ^{-1}(c x)}{e^3}+\frac{b d x}{2 c e^2}-\frac{b x^2}{6 c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*d^2*x)/e^3 + (b*d*x)/(2*c*e^2) - (b*x^2)/(6*c*e) - (b*d*ArcTan[c*x])/(2*c^2*e^2) + (b*d^2*x*ArcTan[c*x])/e^
3 - (d*x^2*(a + b*ArcTan[c*x]))/(2*e^2) + (x^3*(a + b*ArcTan[c*x]))/(3*e) + (d^3*(a + b*ArcTan[c*x])*Log[2/(1
- I*c*x)])/e^4 - (d^3*(a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e^4 - (b*d^2*Log[1 +
c^2*x^2])/(2*c*e^3) + (b*Log[1 + c^2*x^2])/(6*c^3*e) - ((I/2)*b*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/e^4 + ((I/
2)*b*d^3*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x))])/e^4

Rule 4876

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[Ex
pandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p,
0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 321

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

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 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 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 4856

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])*Log[2/(1 -
I*c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d
+ e*x))/((c*d + I*e)*(1 - I*c*x))]/(1 + c^2*x^2), x], x] + Simp[((a + b*ArcTan[c*x])*Log[(2*c*(d + e*x))/((c*d
+ I*e)*(1 - I*c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

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

Mathematica [A]  time = 3.28679, size = 484, normalized size = 1.63 \[ -\frac{-3 i b d^3 \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+3 i b d^3 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-6 a d^2 e x+6 a d^3 \log (d+e x)+3 a d e^2 x^2-2 a e^3 x^3-\frac{3 b d^2 e \sqrt{\frac{c^2 d^2}{e^2}+1} \tan ^{-1}(c x)^2 e^{i \tan ^{-1}\left (\frac{c d}{e}\right )}}{c}+\frac{3 b d^2 e \log \left (c^2 x^2+1\right )}{c}+\frac{3}{2} \pi b d^3 \log \left (c^2 x^2+1\right )+\frac{3 b d e^2 \tan ^{-1}(c x)}{c^2}-\frac{b e^3 \log \left (c^2 x^2+1\right )}{c^3}+\frac{b e^3}{c^3}-6 i b d^3 \tan ^{-1}(c x) \tan ^{-1}\left (\frac{c d}{e}\right )+\frac{3 b d^2 e \tan ^{-1}(c x)^2}{c}-6 b d^2 e x \tan ^{-1}(c x)+6 b d^3 \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+6 b d^3 \tan ^{-1}(c x) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-6 b d^3 \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )\right )+3 i b d^3 \tan ^{-1}(c x)^2+3 i \pi b d^3 \tan ^{-1}(c x)+3 \pi b d^3 \log \left (1+e^{-2 i \tan ^{-1}(c x)}\right )-6 b d^3 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+3 b d e^2 x^2 \tan ^{-1}(c x)-\frac{3 b d e^2 x}{c}+\frac{b e^3 x^2}{c}-2 b e^3 x^3 \tan ^{-1}(c x)}{6 e^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((b*e^3)/c^3 - 6*a*d^2*e*x - (3*b*d*e^2*x)/c + 3*a*d*e^2*x^2 + (b*e^3*x^2)/c - 2*a*e^3*x^3 + (3*b*d*e^2*ArcTa
n[c*x])/c^2 + (3*I)*b*d^3*Pi*ArcTan[c*x] - 6*b*d^2*e*x*ArcTan[c*x] + 3*b*d*e^2*x^2*ArcTan[c*x] - 2*b*e^3*x^3*A
rcTan[c*x] - (6*I)*b*d^3*ArcTan[(c*d)/e]*ArcTan[c*x] + (3*I)*b*d^3*ArcTan[c*x]^2 + (3*b*d^2*e*ArcTan[c*x]^2)/c
- (3*b*d^2*Sqrt[1 + (c^2*d^2)/e^2]*e*E^(I*ArcTan[(c*d)/e])*ArcTan[c*x]^2)/c + 3*b*d^3*Pi*Log[1 + E^((-2*I)*Ar
cTan[c*x])] - 6*b*d^3*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + 6*b*d^3*ArcTan[(c*d)/e]*Log[1 - E^((2*I)*(A
rcTan[(c*d)/e] + ArcTan[c*x]))] + 6*b*d^3*ArcTan[c*x]*Log[1 - E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))] + 6*a
*d^3*Log[d + e*x] + (3*b*d^2*e*Log[1 + c^2*x^2])/c - (b*e^3*Log[1 + c^2*x^2])/c^3 + (3*b*d^3*Pi*Log[1 + c^2*x^
2])/2 - 6*b*d^3*ArcTan[(c*d)/e]*Log[Sin[ArcTan[(c*d)/e] + ArcTan[c*x]]] + (3*I)*b*d^3*PolyLog[2, -E^((2*I)*Arc
Tan[c*x])] - (3*I)*b*d^3*PolyLog[2, E^((2*I)*(ArcTan[(c*d)/e] + ArcTan[c*x]))])/(6*e^4)

________________________________________________________________________________________

Maple [A]  time = 0.053, size = 394, normalized size = 1.3 \begin{align*}{\frac{{x}^{3}a}{3\,e}}-{\frac{ad{x}^{2}}{2\,{e}^{2}}}+{\frac{a{d}^{2}x}{{e}^{3}}}-{\frac{a{d}^{3}\ln \left ( ecx+dc \right ) }{{e}^{4}}}+{\frac{b{x}^{3}\arctan \left ( cx \right ) }{3\,e}}-{\frac{\arctan \left ( cx \right ) bd{x}^{2}}{2\,{e}^{2}}}+{\frac{b{d}^{2}x\arctan \left ( cx \right ) }{{e}^{3}}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) \ln \left ( ecx+dc \right ) }{{e}^{4}}}-{\frac{b\ln \left ({c}^{2}{d}^{2}-2\, \left ( ecx+dc \right ) cd+ \left ( ecx+dc \right ) ^{2}+{e}^{2} \right ){d}^{2}}{2\,c{e}^{3}}}-{\frac{\arctan \left ( cx \right ) bd}{2\,{c}^{2}{e}^{2}}}+{\frac{b\ln \left ({c}^{2}{d}^{2}-2\, \left ( ecx+dc \right ) cd+ \left ( ecx+dc \right ) ^{2}+{e}^{2} \right ) }{6\,{c}^{3}e}}+{\frac{bdx}{2\,c{e}^{2}}}+{\frac{2\,b{d}^{2}}{3\,c{e}^{3}}}-{\frac{b{x}^{2}}{6\,ce}}-{\frac{{\frac{i}{2}}b{d}^{3}}{{e}^{4}}{\it dilog} \left ({\frac{ie-ecx}{dc+ie}} \right ) }+{\frac{{\frac{i}{2}}b{d}^{3}\ln \left ( ecx+dc \right ) }{{e}^{4}}\ln \left ({\frac{ie+ecx}{ie-dc}} \right ) }+{\frac{{\frac{i}{2}}b{d}^{3}}{{e}^{4}}{\it dilog} \left ({\frac{ie+ecx}{ie-dc}} \right ) }-{\frac{{\frac{i}{2}}b{d}^{3}\ln \left ( ecx+dc \right ) }{{e}^{4}}\ln \left ({\frac{ie-ecx}{dc+ie}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctan(c*x))/(e*x+d),x)

[Out]

1/3*a/e*x^3-1/2*a/e^2*d*x^2+a*d^2*x/e^3-a*d^3/e^4*ln(c*e*x+c*d)+1/3*b*arctan(c*x)/e*x^3-1/2*b*arctan(c*x)/e^2*
d*x^2+b*d^2*x*arctan(c*x)/e^3-b*arctan(c*x)*d^3/e^4*ln(c*e*x+c*d)-1/2/c*b/e^3*ln(c^2*d^2-2*(c*e*x+c*d)*c*d+(c*
e*x+c*d)^2+e^2)*d^2-1/2*b*d*arctan(c*x)/c^2/e^2+1/6/c^3*b/e*ln(c^2*d^2-2*(c*e*x+c*d)*c*d+(c*e*x+c*d)^2+e^2)+1/
2*b*d*x/c/e^2+2/3/c*b*d^2/e^3-1/6*b*x^2/c/e-1/2*I*b/e^4*d^3*dilog((I*e-e*c*x)/(d*c+I*e))+1/2*I*b/e^4*d^3*ln(c*
e*x+c*d)*ln((I*e+e*c*x)/(I*e-d*c))+1/2*I*b/e^4*d^3*dilog((I*e+e*c*x)/(I*e-d*c))-1/2*I*b/e^4*d^3*ln(c*e*x+c*d)*
ln((I*e-e*c*x)/(d*c+I*e))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \, a{\left (\frac{6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac{2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} + 2 \, b \int \frac{x^{3} \arctan \left (c x\right )}{2 \,{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \arctan \left (c x\right ) + a x^{3}}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^3*arctan(c*x) + a*x^3)/(e*x + d), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x))/(e*x+d),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )} x^{3}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctan(c*x) + a)*x^3/(e*x + d), x)