3.1257 \(\int (d+e x^2)^2 (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=442 \[ -\frac{2 i b^2 d e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{5 c^5}+\frac{i b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{4 b d e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{2 b e^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{2 b^2 d e x}{3 c^2}-\frac{2 b^2 d e \tan ^{-1}(c x)}{3 c^3}+\frac{b^2 e^2 x^3}{30 c^2}-\frac{3 b^2 e^2 x}{10 c^4}+\frac{3 b^2 e^2 \tan ^{-1}(c x)}{10 c^5} \]

[Out]

(2*b^2*d*e*x)/(3*c^2) - (3*b^2*e^2*x)/(10*c^4) + (b^2*e^2*x^3)/(30*c^2) - (2*b^2*d*e*ArcTan[c*x])/(3*c^3) + (3
*b^2*e^2*ArcTan[c*x])/(10*c^5) - (2*b*d*e*x^2*(a + b*ArcTan[c*x]))/(3*c) + (b*e^2*x^2*(a + b*ArcTan[c*x]))/(5*
c^3) - (b*e^2*x^4*(a + b*ArcTan[c*x]))/(10*c) + (I*d^2*(a + b*ArcTan[c*x])^2)/c - (((2*I)/3)*d*e*(a + b*ArcTan
[c*x])^2)/c^3 + ((I/5)*e^2*(a + b*ArcTan[c*x])^2)/c^5 + d^2*x*(a + b*ArcTan[c*x])^2 + (2*d*e*x^3*(a + b*ArcTan
[c*x])^2)/3 + (e^2*x^5*(a + b*ArcTan[c*x])^2)/5 + (2*b*d^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c - (4*b*d*
e*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) + (2*b*e^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(5*c^5) +
(I*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (((2*I)/3)*b^2*d*e*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3 + ((I/5)*
b^2*e^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^5

________________________________________________________________________________________

Rubi [A]  time = 0.689419, antiderivative size = 442, normalized size of antiderivative = 1., number of steps used = 30, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {4914, 4846, 4920, 4854, 2402, 2315, 4852, 4916, 321, 203, 302} \[ -\frac{2 i b^2 d e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 e^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{5 c^5}+\frac{i b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{4 b d e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{2 b e^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac{2 b d^2 \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{2 b^2 d e x}{3 c^2}-\frac{2 b^2 d e \tan ^{-1}(c x)}{3 c^3}+\frac{b^2 e^2 x^3}{30 c^2}-\frac{3 b^2 e^2 x}{10 c^4}+\frac{3 b^2 e^2 \tan ^{-1}(c x)}{10 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b^2*d*e*x)/(3*c^2) - (3*b^2*e^2*x)/(10*c^4) + (b^2*e^2*x^3)/(30*c^2) - (2*b^2*d*e*ArcTan[c*x])/(3*c^3) + (3
*b^2*e^2*ArcTan[c*x])/(10*c^5) - (2*b*d*e*x^2*(a + b*ArcTan[c*x]))/(3*c) + (b*e^2*x^2*(a + b*ArcTan[c*x]))/(5*
c^3) - (b*e^2*x^4*(a + b*ArcTan[c*x]))/(10*c) + (I*d^2*(a + b*ArcTan[c*x])^2)/c - (((2*I)/3)*d*e*(a + b*ArcTan
[c*x])^2)/c^3 + ((I/5)*e^2*(a + b*ArcTan[c*x])^2)/c^5 + d^2*x*(a + b*ArcTan[c*x])^2 + (2*d*e*x^3*(a + b*ArcTan
[c*x])^2)/3 + (e^2*x^5*(a + b*ArcTan[c*x])^2)/5 + (2*b*d^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c - (4*b*d*
e*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) + (2*b*e^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(5*c^5) +
(I*b^2*d^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c - (((2*I)/3)*b^2*d*e*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^3 + ((I/5)*
b^2*e^2*PolyLog[2, 1 - 2/(1 + I*c*x)])/c^5

Rule 4914

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a
+ b*ArcTan[c*x])^p, (d + e*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[q] && IGtQ[p, 0]

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 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]

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 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTan[c*x])^p)
/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps

\begin{align*} \int \left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d^2 \left (a+b \tan ^{-1}(c x)\right )^2+2 d e x^2 \left (a+b \tan ^{-1}(c x)\right )^2+e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+(2 d e) \int x^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+e^2 \int x^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\left (2 b c d^2\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{1}{3} (4 b c d e) \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac{1}{5} \left (2 b c e^2\right ) \int \frac{x^5 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\left (2 b d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx-\frac{(4 b d e) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c}+\frac{(4 b d e) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c}-\frac{\left (2 b e^2\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c}+\frac{\left (2 b e^2\right ) \int \frac{x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c}\\ &=-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c}-\left (2 b^2 d^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\frac{1}{3} \left (2 b^2 d e\right ) \int \frac{x^2}{1+c^2 x^2} \, dx-\frac{(4 b d e) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c^2}+\frac{1}{10} \left (b^2 e^2\right ) \int \frac{x^4}{1+c^2 x^2} \, dx+\frac{\left (2 b e^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c^3}-\frac{\left (2 b e^2\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c^3}\\ &=\frac{2 b^2 d e x}{3 c^2}-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c}-\frac{4 b d e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^3}+\frac{\left (2 i b^2 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c}-\frac{\left (2 b^2 d e\right ) \int \frac{1}{1+c^2 x^2} \, dx}{3 c^2}+\frac{\left (4 b^2 d e\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{3 c^2}+\frac{1}{10} \left (b^2 e^2\right ) \int \left (-\frac{1}{c^4}+\frac{x^2}{c^2}+\frac{1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx+\frac{\left (2 b e^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{5 c^4}-\frac{\left (b^2 e^2\right ) \int \frac{x^2}{1+c^2 x^2} \, dx}{5 c^2}\\ &=\frac{2 b^2 d e x}{3 c^2}-\frac{3 b^2 e^2 x}{10 c^4}+\frac{b^2 e^2 x^3}{30 c^2}-\frac{2 b^2 d e \tan ^{-1}(c x)}{3 c^3}-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c}-\frac{4 b d e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^3}+\frac{2 b e^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{i b^2 d^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c}-\frac{\left (4 i b^2 d e\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{3 c^3}+\frac{\left (b^2 e^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{10 c^4}+\frac{\left (b^2 e^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{5 c^4}-\frac{\left (2 b^2 e^2\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{5 c^4}\\ &=\frac{2 b^2 d e x}{3 c^2}-\frac{3 b^2 e^2 x}{10 c^4}+\frac{b^2 e^2 x^3}{30 c^2}-\frac{2 b^2 d e \tan ^{-1}(c x)}{3 c^3}+\frac{3 b^2 e^2 \tan ^{-1}(c x)}{10 c^5}-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c}-\frac{4 b d e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^3}+\frac{2 b e^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{i b^2 d^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c}-\frac{2 i b^2 d e \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{\left (2 i b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{5 c^5}\\ &=\frac{2 b^2 d e x}{3 c^2}-\frac{3 b^2 e^2 x}{10 c^4}+\frac{b^2 e^2 x^3}{30 c^2}-\frac{2 b^2 d e \tan ^{-1}(c x)}{3 c^3}+\frac{3 b^2 e^2 \tan ^{-1}(c x)}{10 c^5}-\frac{2 b d e x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{b e^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}-\frac{b e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{i d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac{2 i d e \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}+\frac{i e^2 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2}{3} d e x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{5} e^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{2 b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c}-\frac{4 b d e \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{3 c^3}+\frac{2 b e^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{5 c^5}+\frac{i b^2 d^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c}-\frac{2 i b^2 d e \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 e^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{5 c^5}\\ \end{align*}

Mathematica [A]  time = 1.08642, size = 391, normalized size = 0.88 \[ \frac{-2 i b^2 \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+30 a^2 c^5 d^2 x+20 a^2 c^5 d e x^3+6 a^2 c^5 e^2 x^5+b \tan ^{-1}(c x) \left (4 a c^5 x \left (15 d^2+10 d e x^2+3 e^2 x^4\right )+4 b \left (15 c^4 d^2-10 c^2 d e+3 e^2\right ) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-b e \left (c^2 x^2+1\right ) \left (c^2 \left (20 d+3 e x^2\right )-9 e\right )\right )-30 a b c^4 d^2 \log \left (c^2 x^2+1\right )-20 a b c^4 d e x^2+20 a b c^2 d e \log \left (c^2 x^2+1\right )-3 a b c^4 e^2 x^4+6 a b c^2 e^2 x^2-6 a b e^2 \log \left (c^2 x^2+1\right )+9 a b e^2+2 b^2 \tan ^{-1}(c x)^2 \left (c^5 \left (15 d^2 x+10 d e x^3+3 e^2 x^5\right )-15 i c^4 d^2+10 i c^2 d e-3 i e^2\right )+20 b^2 c^3 d e x+b^2 c^3 e^2 x^3-9 b^2 c e^2 x}{30 c^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(9*a*b*e^2 + 30*a^2*c^5*d^2*x + 20*b^2*c^3*d*e*x - 9*b^2*c*e^2*x - 20*a*b*c^4*d*e*x^2 + 6*a*b*c^2*e^2*x^2 + 20
*a^2*c^5*d*e*x^3 + b^2*c^3*e^2*x^3 - 3*a*b*c^4*e^2*x^4 + 6*a^2*c^5*e^2*x^5 + 2*b^2*((-15*I)*c^4*d^2 + (10*I)*c
^2*d*e - (3*I)*e^2 + c^5*(15*d^2*x + 10*d*e*x^3 + 3*e^2*x^5))*ArcTan[c*x]^2 + b*ArcTan[c*x]*(4*a*c^5*x*(15*d^2
+ 10*d*e*x^2 + 3*e^2*x^4) - b*e*(1 + c^2*x^2)*(-9*e + c^2*(20*d + 3*e*x^2)) + 4*b*(15*c^4*d^2 - 10*c^2*d*e +
3*e^2)*Log[1 + E^((2*I)*ArcTan[c*x])]) - 30*a*b*c^4*d^2*Log[1 + c^2*x^2] + 20*a*b*c^2*d*e*Log[1 + c^2*x^2] - 6
*a*b*e^2*Log[1 + c^2*x^2] - (2*I)*b^2*(15*c^4*d^2 - 10*c^2*d*e + 3*e^2)*PolyLog[2, -E^((2*I)*ArcTan[c*x])])/(3
0*c^5)

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Maple [B]  time = 0.129, size = 1005, normalized size = 2.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*I/c*b^2*ln(c*x+I)^2*d^2+1/2*I/c*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))*d^2-1/2*I/c*b^2*ln(c*x-I)*ln(c^2*x^2+1)*
d^2-2/3/c*b^2*arctan(c*x)*x^2*d*e+1/10*I/c^5*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))*e^2+1/3*I/c^3*b^2*ln(c*x-I)*ln(c
^2*x^2+1)*d*e+1/3*I/c^3*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))*d*e-1/3*I/c^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))*d*e-1/3
*I/c^3*b^2*ln(c*x+I)*ln(c^2*x^2+1)*d*e+1/5*a^2*x^5*e^2+a^2*d^2*x-3/10*b^2*e^2*x/c^4+1/30*b^2*e^2*x^3/c^2+3/10*
b^2*e^2*arctan(c*x)/c^5+1/5/c^3*b^2*arctan(c*x)*x^2*e^2-1/5/c^5*a*b*ln(c^2*x^2+1)*e^2-1/10/c*a*b*x^4*e^2+1/5/c
^3*a*b*x^2*e^2+2/3*b^2*arctan(c*x)^2*x^3*d*e+2/5*a*b*arctan(c*x)*x^5*e^2+2*a*b*arctan(c*x)*d^2*x+1/10*I/c^5*b^
2*dilog(-1/2*I*(c*x+I))*e^2+1/20*I/c^5*b^2*ln(c*x-I)^2*e^2-1/c*a*b*ln(c^2*x^2+1)*d^2-1/10/c*b^2*arctan(c*x)*x^
4*e^2-1/10*I/c^5*b^2*dilog(1/2*I*(c*x-I))*e^2-1/10*I/c^5*b^2*ln(c*x-I)*ln(c^2*x^2+1)*e^2-1/3*I/c^3*b^2*dilog(-
1/2*I*(c*x+I))*d*e-1/6*I/c^3*b^2*ln(c*x-I)^2*d*e+1/6*I/c^3*b^2*ln(c*x+I)^2*d*e+1/3*I/c^3*b^2*dilog(1/2*I*(c*x-
I))*d*e+1/2*I/c*b^2*ln(c*x+I)*ln(c^2*x^2+1)*d^2-1/2*I/c*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))*d^2-1/10*I/c^5*b^2*ln(
c*x+I)*ln(1/2*I*(c*x-I))*e^2+1/10*I/c^5*b^2*ln(c*x+I)*ln(c^2*x^2+1)*e^2+4/3*a*b*arctan(c*x)*x^3*d*e+2/3/c^3*a*
b*ln(c^2*x^2+1)*d*e+2/3/c^3*b^2*arctan(c*x)*ln(c^2*x^2+1)*d*e-2/3/c*a*b*x^2*d*e-1/2*I/c*b^2*dilog(1/2*I*(c*x-I
))*d^2-1/c*b^2*arctan(c*x)*ln(c^2*x^2+1)*d^2+2/3*a^2*x^3*d*e+b^2*arctan(c*x)^2*d^2*x+1/4*I/c*b^2*ln(c*x-I)^2*d
^2+1/2*I/c*b^2*dilog(-1/2*I*(c*x+I))*d^2-1/5/c^5*b^2*arctan(c*x)*ln(c^2*x^2+1)*e^2-1/20*I/c^5*b^2*ln(c*x+I)^2*
e^2+1/5*b^2*arctan(c*x)^2*x^5*e^2+2/3*b^2*d*e*x/c^2-2/3*b^2*d*e*arctan(c*x)/c^3

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

[Out]

1/5*a^2*e^2*x^5 + 2/3*a^2*d*e*x^3 + 180*b^2*c^2*e^2*integrate(1/240*x^6*arctan(c*x)^2/(c^2*x^2 + 1), x) + 15*b
^2*c^2*e^2*integrate(1/240*x^6*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 12*b^2*c^2*e^2*integrate(1/240*x^6*log(c
^2*x^2 + 1)/(c^2*x^2 + 1), x) + 360*b^2*c^2*d*e*integrate(1/240*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) + 30*b^2*c
^2*d*e*integrate(1/240*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 40*b^2*c^2*d*e*integrate(1/240*x^4*log(c^2*x
^2 + 1)/(c^2*x^2 + 1), x) + 180*b^2*c^2*d^2*integrate(1/240*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 15*b^2*c^2*d
^2*integrate(1/240*x^2*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 60*b^2*c^2*d^2*integrate(1/240*x^2*log(c^2*x^2 +
1)/(c^2*x^2 + 1), x) + 1/4*b^2*d^2*arctan(c*x)^3/c - 24*b^2*c*e^2*integrate(1/240*x^5*arctan(c*x)/(c^2*x^2 +
1), x) - 80*b^2*c*d*e*integrate(1/240*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - 120*b^2*c*d^2*integrate(1/240*x*arct
an(c*x)/(c^2*x^2 + 1), x) + 2/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*d*e + 1/10*(4*x^5
*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^2 + 1)/c^6))*a*b*e^2 + a^2*d^2*x + 180*b^2*e^2*integrate
(1/240*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) + 15*b^2*e^2*integrate(1/240*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1),
x) + 360*b^2*d*e*integrate(1/240*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x) + 30*b^2*d*e*integrate(1/240*x^2*log(c^2*
x^2 + 1)^2/(c^2*x^2 + 1), x) + 15*b^2*d^2*integrate(1/240*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^2/c + 1/60*(3*b^2*e^2*x^5 + 10*b^2*d*e*x^3 + 15*b^2*d^2*x)*arctan(c*x)^2 - 1/2
40*(3*b^2*e^2*x^5 + 10*b^2*d*e*x^3 + 15*b^2*d^2*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^{2} x^{4} + 2 \, a^{2} d e x^{2} + a^{2} d^{2} +{\left (b^{2} e^{2} x^{4} + 2 \, b^{2} d e x^{2} + b^{2} d^{2}\right )} \arctan \left (c x\right )^{2} + 2 \,{\left (a b e^{2} x^{4} + 2 \, a b d e x^{2} + a b d^{2}\right )} \arctan \left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*e^2*x^4 + 2*a^2*d*e*x^2 + a^2*d^2 + (b^2*e^2*x^4 + 2*b^2*d*e*x^2 + b^2*d^2)*arctan(c*x)^2 + 2*(a*
b*e^2*x^4 + 2*a*b*d*e*x^2 + a*b*d^2)*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^{2}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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