3.9 \(\int (d+e x)^3 (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=376 \[ \frac{i b^2 d (c d-e) (c d+e) \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3}-\frac{a b e x \left (6 c^2 d^2-e^2\right )}{2 c^3}-\frac{\left (-6 c^2 d^2 e^2+c^4 d^4+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}+\frac{2 b d (c d-e) (c d+e) \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (c^2 x^2+1\right )}{4 c^4}-\frac{b^2 e x \left (6 c^2 d^2-e^2\right ) \tan ^{-1}(c x)}{2 c^3}+\frac{b^2 d e^2 x}{c^2}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 e^3 \log \left (c^2 x^2+1\right )}{12 c^4} \]

[Out]

(b^2*d*e^2*x)/c^2 - (a*b*e*(6*c^2*d^2 - e^2)*x)/(2*c^3) + (b^2*e^3*x^2)/(12*c^2) - (b^2*d*e^2*ArcTan[c*x])/c^3
- (b^2*e*(6*c^2*d^2 - e^2)*x*ArcTan[c*x])/(2*c^3) - (b*d*e^2*x^2*(a + b*ArcTan[c*x]))/c - (b*e^3*x^3*(a + b*A
rcTan[c*x]))/(6*c) + (I*d*(c*d - e)*(c*d + e)*(a + b*ArcTan[c*x])^2)/c^3 - ((c^4*d^4 - 6*c^2*d^2*e^2 + e^4)*(a
+ b*ArcTan[c*x])^2)/(4*c^4*e) + ((d + e*x)^4*(a + b*ArcTan[c*x])^2)/(4*e) + (2*b*d*(c*d - e)*(c*d + e)*(a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c^3 - (b^2*e^3*Log[1 + c^2*x^2])/(12*c^4) + (b^2*e*(6*c^2*d^2 - e^2)*Log[1 +
c^2*x^2])/(4*c^4) + (I*b^2*d*(c*d - e)*(c*d + e)*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3

________________________________________________________________________________________

Rubi [A]  time = 0.573163, antiderivative size = 376, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 14, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.778, Rules used = {4864, 4846, 260, 4852, 321, 203, 266, 43, 4984, 4884, 4920, 4854, 2402, 2315} \[ \frac{i b^2 d (c d-e) (c d+e) \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3}-\frac{a b e x \left (6 c^2 d^2-e^2\right )}{2 c^3}-\frac{\left (-6 c^2 d^2 e^2+c^4 d^4+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}+\frac{2 b d (c d-e) (c d+e) \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (c^2 x^2+1\right )}{4 c^4}-\frac{b^2 e x \left (6 c^2 d^2-e^2\right ) \tan ^{-1}(c x)}{2 c^3}+\frac{b^2 d e^2 x}{c^2}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 e^3 \log \left (c^2 x^2+1\right )}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*d*e^2*x)/c^2 - (a*b*e*(6*c^2*d^2 - e^2)*x)/(2*c^3) + (b^2*e^3*x^2)/(12*c^2) - (b^2*d*e^2*ArcTan[c*x])/c^3
- (b^2*e*(6*c^2*d^2 - e^2)*x*ArcTan[c*x])/(2*c^3) - (b*d*e^2*x^2*(a + b*ArcTan[c*x]))/c - (b*e^3*x^3*(a + b*A
rcTan[c*x]))/(6*c) + (I*d*(c*d - e)*(c*d + e)*(a + b*ArcTan[c*x])^2)/c^3 - ((c^4*d^4 - 6*c^2*d^2*e^2 + e^4)*(a
+ b*ArcTan[c*x])^2)/(4*c^4*e) + ((d + e*x)^4*(a + b*ArcTan[c*x])^2)/(4*e) + (2*b*d*(c*d - e)*(c*d + e)*(a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c^3 - (b^2*e^3*Log[1 + c^2*x^2])/(12*c^4) + (b^2*e*(6*c^2*d^2 - e^2)*Log[1 +
c^2*x^2])/(4*c^4) + (I*b^2*d*(c*d - e)*(c*d + e)*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3

Rule 4864

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

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 4984

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> I
nt[ExpandIntegrand[(a + b*ArcTan[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& IGtQ[p, 0] && EqQ[e, c^2*d] && IGtQ[m, 0]

Rule 4884

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

Rule 4920

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

Rule 4854

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

Rubi steps

\begin{align*} \int (d+e x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{(b c) \int \left (\frac{e^2 \left (6 c^2 d^2-e^2\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4}+\frac{4 d e^3 x \left (a+b \tan ^{-1}(c x)\right )}{c^2}+\frac{e^4 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c^2}+\frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4+4 c^2 d (c d-e) e (c d+e) x\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 e}\\ &=\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{b \int \frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4+4 c^2 d (c d-e) e (c d+e) x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c^3 e}-\frac{\left (2 b d e^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}-\frac{\left (b e^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}-\frac{\left (b e \left (6 c^2 d^2-e^2\right )\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}\\ &=-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{b \int \left (\frac{c^4 d^4 \left (1+\frac{-6 c^2 d^2 e^2+e^4}{c^4 d^4}\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}+\frac{4 c^2 d (c d-e) e (c d+e) x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{2 c^3 e}+\left (b^2 d e^2\right ) \int \frac{x^2}{1+c^2 x^2} \, dx+\frac{1}{6} \left (b^2 e^3\right ) \int \frac{x^3}{1+c^2 x^2} \, dx-\frac{\left (b^2 e \left (6 c^2 d^2-e^2\right )\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}\\ &=\frac{b^2 d e^2 x}{c^2}-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}-\frac{b^2 e \left (6 c^2 d^2-e^2\right ) x \tan ^{-1}(c x)}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}-\frac{\left (b^2 d e^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{c^2}+\frac{1}{12} \left (b^2 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )-\frac{(2 b d (c d-e) (c d+e)) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c}+\frac{\left (b^2 e \left (6 c^2 d^2-e^2\right )\right ) \int \frac{x}{1+c^2 x^2} \, dx}{2 c^2}-\frac{\left (b \left (c^4 d^4-6 c^2 d^2 e^2+e^4\right )\right ) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3 e}\\ &=\frac{b^2 d e^2 x}{c^2}-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}-\frac{b^2 e \left (6 c^2 d^2-e^2\right ) x \tan ^{-1}(c x)}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac{1}{12} \left (b^2 e^3\right ) \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 )+\frac{(2 b d (c d-e) (c d+e)) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{c^2}\\ &=\frac{b^2 d e^2 x}{c^2}-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}-\frac{b^2 e \left (6 c^2 d^2-e^2\right ) x \tan ^{-1}(c x)}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}+\frac{2 b d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3}-\frac{b^2 e^3 \log \left (1+c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (1+c^2 x^2\right )}{4 c^4}-\frac{\left (2 b^2 d (c d-e) (c d+e)\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2}\\ &=\frac{b^2 d e^2 x}{c^2}-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}-\frac{b^2 e \left (6 c^2 d^2-e^2\right ) x \tan ^{-1}(c x)}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}+\frac{2 b d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3}-\frac{b^2 e^3 \log \left (1+c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac{\left (2 i b^2 d (c d-e) (c d+e)\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^3}\\ &=\frac{b^2 d e^2 x}{c^2}-\frac{a b e \left (6 c^2 d^2-e^2\right ) x}{2 c^3}+\frac{b^2 e^3 x^2}{12 c^2}-\frac{b^2 d e^2 \tan ^{-1}(c x)}{c^3}-\frac{b^2 e \left (6 c^2 d^2-e^2\right ) x \tan ^{-1}(c x)}{2 c^3}-\frac{b d e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac{b e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac{i d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3}-\frac{\left (c^4 d^4-6 c^2 d^2 e^2+e^4\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^4 e}+\frac{(d+e x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 e}+\frac{2 b d (c d-e) (c d+e) \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3}-\frac{b^2 e^3 \log \left (1+c^2 x^2\right )}{12 c^4}+\frac{b^2 e \left (6 c^2 d^2-e^2\right ) \log \left (1+c^2 x^2\right )}{4 c^4}+\frac{i b^2 d (c d-e) (c d+e) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3}\\ \end{align*}

Mathematica [A]  time = 0.946089, size = 472, normalized size = 1.26 \[ \frac{-12 i b^2 c d \left (c^2 d^2-e^2\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+18 a^2 c^4 d^2 e x^2+12 a^2 c^4 d^3 x+12 a^2 c^4 d e^2 x^3+3 a^2 c^4 e^3 x^4+2 b \tan ^{-1}(c x) \left (3 a \left (c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+6 c^2 d^2 e-e^3\right )-b c e \left (18 c^2 d^2 x+6 d \left (c^2 e x^2+e\right )+e^2 x \left (c^2 x^2-3\right )\right )+12 b c d \left (c^2 d^2-e^2\right ) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-36 a b c^3 d^2 e x-12 a b c^3 d^3 \log \left (c^2 x^2+1\right )-12 a b c^3 d e^2 x^2+12 a b c d e^2 \log \left (c^2 x^2+1\right )-2 a b c^3 e^3 x^3+6 a b c e^3 x+3 b^2 \tan ^{-1}(c x)^2 \left (c^4 x \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+6 c^2 d^2 e-4 i c^3 d^3+4 i c d e^2-e^3\right )+18 b^2 c^2 d^2 e \log \left (c^2 x^2+1\right )+12 b^2 c^2 d e^2 x+b^2 c^2 e^3 x^2-4 b^2 e^3 \log \left (c^2 x^2+1\right )+b^2 e^3}{12 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b^2*e^3 + 12*a^2*c^4*d^3*x - 36*a*b*c^3*d^2*e*x + 12*b^2*c^2*d*e^2*x + 6*a*b*c*e^3*x + 18*a^2*c^4*d^2*e*x^2 -
12*a*b*c^3*d*e^2*x^2 + b^2*c^2*e^3*x^2 + 12*a^2*c^4*d*e^2*x^3 - 2*a*b*c^3*e^3*x^3 + 3*a^2*c^4*e^3*x^4 + 3*b^2
*((-4*I)*c^3*d^3 + 6*c^2*d^2*e + (4*I)*c*d*e^2 - e^3 + c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3))*ArcT
an[c*x]^2 + 2*b*ArcTan[c*x]*(-(b*c*e*(18*c^2*d^2*x + e^2*x*(-3 + c^2*x^2) + 6*d*(e + c^2*e*x^2))) + 3*a*(6*c^2
*d^2*e - e^3 + c^4*x*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3)) + 12*b*c*d*(c^2*d^2 - e^2)*Log[1 + E^((2*I)*
ArcTan[c*x])]) - 12*a*b*c^3*d^3*Log[1 + c^2*x^2] + 18*b^2*c^2*d^2*e*Log[1 + c^2*x^2] + 12*a*b*c*d*e^2*Log[1 +
c^2*x^2] - 4*b^2*e^3*Log[1 + c^2*x^2] - (12*I)*b^2*c*d*(c^2*d^2 - e^2)*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(12
*c^4)

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Maple [B]  time = 0.075, size = 948, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*arctan(c*x))^2,x)

[Out]

-1/3*b^2*e^3*ln(c^2*x^2+1)/c^4+1/12*b^2*e^3*x^2/c^2+2*a*b*arctan(c*x)*x*d^3+3/2*b^2*e*arctan(c*x)^2*x^2*d^2+b^
2*e^2*arctan(c*x)^2*x^3*d+1/2*a*b*e^3*arctan(c*x)*x^4-1/4*I/c*b^2*d^3*ln(c*x+I)^2-1/2*I/c*b^2*d^3*dilog(1/2*I*
(c*x-I))+1/4*I/c*b^2*d^3*ln(c*x-I)^2+1/2*I/c*b^2*d^3*dilog(-1/2*I*(c*x+I))-1/2/c^4*a*b*e^3*arctan(c*x)+3/2/c^2
*b^2*e*arctan(c*x)^2*d^2+3/2/c^2*b^2*e*ln(c^2*x^2+1)*d^2+1/2/c^3*b^2*e^3*arctan(c*x)*x-1/6/c*b^2*e^3*arctan(c*
x)*x^3-1/c*b^2*arctan(c*x)*ln(c^2*x^2+1)*d^3-1/c*a*b*ln(c^2*x^2+1)*d^3-1/6/c*a*b*e^3*x^3+1/2*a*b/c^3*e^3*x+1/4
*a^2*e^3*x^4+a^2*x*d^3+3*a*b*e*arctan(c*x)*x^2*d^2+2*a*b*e^2*arctan(c*x)*x^3*d+1/4*I/c^3*b^2*e^2*d*ln(c*x+I)^2
-1/4*I/c^3*b^2*e^2*d*ln(c*x-I)^2+1/2*I/c*b^2*d^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))+b^2*d*e^2*x/c^2-b^2*d*e^2*arctan
(c*x)/c^3+1/2*I/c*b^2*d^3*ln(c^2*x^2+1)*ln(c*x+I)-1/2*I/c*b^2*d^3*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/2*I/c*b^2*d^3*
ln(c^2*x^2+1)*ln(c*x-I)+1/2*I/c^3*b^2*e^2*d*dilog(1/2*I*(c*x-I))-1/2*I/c^3*b^2*e^2*d*dilog(-1/2*I*(c*x+I))-1/c
*a*b*e^2*d*x^2-3*a*b/c*e*d^2*x-1/c*b^2*e^2*arctan(c*x)*d*x^2-3/c*b^2*e*arctan(c*x)*d^2*x+3/c^2*a*b*e*arctan(c*
x)*d^2+1/c^3*b^2*e^2*arctan(c*x)*ln(c^2*x^2+1)*d+1/c^3*a*b*e^2*ln(c^2*x^2+1)*d+1/4*a^2/e*d^4+3/2*a^2*e*x^2*d^2
+a^2*e^2*x^3*d+b^2*arctan(c*x)^2*x*d^3+1/4*b^2*e^3*arctan(c*x)^2*x^4-1/4/c^4*b^2*e^3*arctan(c*x)^2+1/2*I/c^3*b
^2*e^2*d*ln(c^2*x^2+1)*ln(c*x-I)-1/2*I/c^3*b^2*e^2*d*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/2*I/c^3*b^2*e^2*d*ln(c^2*x
^2+1)*ln(c*x+I)+1/2*I/c^3*b^2*e^2*d*ln(c*x+I)*ln(1/2*I*(c*x-I))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*e^3*x^4 + a^2*d*e^2*x^3 + 12*b^2*c^2*e^3*integrate(1/16*x^5*arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*c^2*
e^3*integrate(1/16*x^5*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 36*b^2*c^2*d*e^2*integrate(1/16*x^4*arctan(c*x)^
2/(c^2*x^2 + 1), x) + b^2*c^2*e^3*integrate(1/16*x^5*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3*b^2*c^2*d*e^2*inte
grate(1/16*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 36*b^2*c^2*d^2*e*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x
^2 + 1), x) + 4*b^2*c^2*d*e^2*integrate(1/16*x^4*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3*b^2*c^2*d^2*e*integrat
e(1/16*x^3*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*d^3*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^2 + 1
), x) + 6*b^2*c^2*d^2*e*integrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + b^2*c^2*d^3*integrate(1/16*x^2
*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 4*b^2*c^2*d^3*integrate(1/16*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) +
3/2*a^2*d^2*e*x^2 + 1/4*b^2*d^3*arctan(c*x)^3/c - 2*b^2*c*e^3*integrate(1/16*x^4*arctan(c*x)/(c^2*x^2 + 1), x)
- 8*b^2*c*d*e^2*integrate(1/16*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - 12*b^2*c*d^2*e*integrate(1/16*x^2*arctan(c
*x)/(c^2*x^2 + 1), x) - 8*b^2*c*d^3*integrate(1/16*x*arctan(c*x)/(c^2*x^2 + 1), x) + 3*(x^2*arctan(c*x) - c*(x
/c^2 - arctan(c*x)/c^3))*a*b*d^2*e + (2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*d*e^2 + 1/6*
(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*b*e^3 + a^2*d^3*x + 12*b^2*e^3*integrate(1
/16*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) + b^2*e^3*integrate(1/16*x^3*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 36
*b^2*d*e^2*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*d*e^2*integrate(1/16*x^2*log(c^2*x^2 + 1
)^2/(c^2*x^2 + 1), x) + 36*b^2*d^2*e*integrate(1/16*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^2*d^2*e*integrate(
1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + b^2*d^3*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2
*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d^3/c + 1/16*(b^2*e^3*x^4 + 4*b^2*d*e^2*x^3 + 6*b^2*d^2*e*x^2 + 4*b^2
*d^3*x)*arctan(c*x)^2 - 1/64*(b^2*e^3*x^4 + 4*b^2*d*e^2*x^3 + 6*b^2*d^2*e*x^2 + 4*b^2*d^3*x)*log(c^2*x^2 + 1)^
2

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{2} e^{3} x^{3} + 3 \, a^{2} d e^{2} x^{2} + 3 \, a^{2} d^{2} e x + a^{2} d^{3} +{\left (b^{2} e^{3} x^{3} + 3 \, b^{2} d e^{2} x^{2} + 3 \, b^{2} d^{2} e x + b^{2} d^{3}\right )} \arctan \left (c x\right )^{2} + 2 \,{\left (a b e^{3} x^{3} + 3 \, a b d e^{2} x^{2} + 3 \, a b d^{2} e x + a b d^{3}\right )} \arctan \left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*e^3*x^3 + 3*a^2*d*e^2*x^2 + 3*a^2*d^2*e*x + a^2*d^3 + (b^2*e^3*x^3 + 3*b^2*d*e^2*x^2 + 3*b^2*d^2*
e*x + b^2*d^3)*arctan(c*x)^2 + 2*(a*b*e^3*x^3 + 3*a*b*d*e^2*x^2 + 3*a*b*d^2*e*x + a*b*d^3)*arctan(c*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*atan(c*x))**2,x)

[Out]

Integral((a + b*atan(c*x))**2*(d + e*x)**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x + d)^3*(b*arctan(c*x) + a)^2, x)