3.1248 \(\int x^2 (d+e x^2) (a+b \tan ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=323 \[ -\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{5 c^5}-\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{b e x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac{i e \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{2 b e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{b^2 d x}{3 c^2}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{b^2 e x^3}{30 c^2}-\frac{3 b^2 e x}{10 c^4}+\frac{3 b^2 e \tan ^{-1}(c x)}{10 c^5} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.590095, antiderivative size = 323, normalized size of antiderivative = 1., number of steps used = 25, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {4980, 4852, 4916, 321, 203, 4920, 4854, 2402, 2315, 302} \[ -\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i b^2 e \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{5 c^5}-\frac{i d \left (a+b \tan ^{-1}(c x)\right )^2}{3 c^3}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{b e x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^3}+\frac{i e \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^5}+\frac{2 b e \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^5}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}+\frac{1}{5} e x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{b e x^4 \left (a+b \tan ^{-1}(c x)\right )}{10 c}+\frac{b^2 d x}{3 c^2}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{b^2 e x^3}{30 c^2}-\frac{3 b^2 e x}{10 c^4}+\frac{3 b^2 e \tan ^{-1}(c x)}{10 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4980

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

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

Mathematica [A]  time = 0.809411, size = 287, normalized size = 0.89 \[ \frac{2 i b^2 \left (5 c^2 d-3 e\right ) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+10 a^2 c^5 d x^3+6 a^2 c^5 e x^5-b \tan ^{-1}(c x) \left (-4 a c^5 x^3 \left (5 d+3 e x^2\right )+b \left (c^2 x^2+1\right ) \left (c^2 \left (10 d+3 e x^2\right )-9 e\right )+4 b \left (5 c^2 d-3 e\right ) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )-10 a b c^4 d x^2+10 a b c^2 d \log \left (c^2 x^2+1\right )-3 a b c^4 e x^4+6 a b c^2 e x^2-6 a b e \log \left (c^2 x^2+1\right )+9 a b e+2 b^2 \tan ^{-1}(c x)^2 \left (c^5 \left (5 d x^3+3 e x^5\right )+5 i c^2 d-3 i e\right )+10 b^2 c^3 d x+b^2 c^3 e x^3-9 b^2 c e x}{30 c^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.132, size = 667, normalized size = 2.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*I/c^3*b^2*ln(c*x-I)^2*d+1/20*I/c^5*b^2*ln(c*x-I)^2*e-1/10*I/c^5*b^2*dilog(1/2*I*(c*x-I))*e-1/6*I/c^3*b^2
*dilog(-1/2*I*(c*x+I))*d-1/10/c*a*b*e*x^4+1/5/c^3*b^2*arctan(c*x)*x^2*e+1/3/c^3*b^2*arctan(c*x)*ln(c^2*x^2+1)*
d-1/3/c*b^2*arctan(c*x)*d*x^2-1/5/c^5*a*b*ln(c^2*x^2+1)*e+1/5/c^3*a*b*x^2*e-3/10*b^2*e*x/c^4+1/30*b^2*e*x^3/c^
2+3/10*b^2*e*arctan(c*x)/c^5+1/10*I/c^5*b^2*ln(c*x+I)*ln(c^2*x^2+1)*e+1/10*I/c^5*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+
I))*e+1/6*I/c^3*b^2*ln(c*x-I)*ln(c^2*x^2+1)*d+1/6*I/c^3*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))*d-1/6*I/c^3*b^2*ln(c*x
-I)*ln(-1/2*I*(c*x+I))*d-1/10*I/c^5*b^2*ln(c*x-I)*ln(c^2*x^2+1)*e-1/10*I/c^5*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))*e
-1/6*I/c^3*b^2*ln(c*x+I)*ln(c^2*x^2+1)*d+1/5*a^2*e*x^5+1/3*a^2*d*x^3-1/3*b^2*d*arctan(c*x)/c^3+1/5*b^2*arctan(
c*x)^2*e*x^5+1/3*b^2*arctan(c*x)^2*d*x^3-1/5/c^5*b^2*arctan(c*x)*ln(c^2*x^2+1)*e+1/3*b^2*d*x/c^2-1/10/c*b^2*ar
ctan(c*x)*e*x^4-1/3/c*a*b*d*x^2+2/3*a*b*arctan(c*x)*d*x^3+2/5*a*b*arctan(c*x)*e*x^5+1/3/c^3*a*b*ln(c^2*x^2+1)*
d-1/20*I/c^5*b^2*ln(c*x+I)^2*e+1/12*I/c^3*b^2*ln(c*x+I)^2*d+1/10*I/c^5*b^2*dilog(-1/2*I*(c*x+I))*e+1/6*I/c^3*b
^2*dilog(1/2*I*(c*x-I))*d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{5} \, a^{2} e x^{5} + \frac{1}{3} \, a^{2} d x^{3} + \frac{1}{3} \,{\left (2 \, x^{3} \arctan \left (c x\right ) - c{\left (\frac{x^{2}}{c^{2}} - \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} a b d + \frac{1}{10} \,{\left (4 \, x^{5} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac{2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a b e + \frac{1}{60} \,{\left (3 \, b^{2} e x^{5} + 5 \, b^{2} d x^{3}\right )} \arctan \left (c x\right )^{2} - \frac{1}{240} \,{\left (3 \, b^{2} e x^{5} + 5 \, b^{2} d x^{3}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + \int \frac{180 \,{\left (b^{2} c^{2} e x^{6} + b^{2} d x^{2} +{\left (b^{2} c^{2} d + b^{2} e\right )} x^{4}\right )} \arctan \left (c x\right )^{2} + 15 \,{\left (b^{2} c^{2} e x^{6} + b^{2} d x^{2} +{\left (b^{2} c^{2} d + b^{2} e\right )} x^{4}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - 8 \,{\left (3 \, b^{2} c e x^{5} + 5 \, b^{2} c d x^{3}\right )} \arctan \left (c x\right ) + 4 \,{\left (3 \, b^{2} c^{2} e x^{6} + 5 \, b^{2} c^{2} d x^{4}\right )} \log \left (c^{2} x^{2} + 1\right )}{240 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^2*e*x^5 + 1/3*a^2*d*x^3 + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*d + 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 + 1/60*(3*b^2*e*x^5 + 5*b^2*d*x^3)*
arctan(c*x)^2 - 1/240*(3*b^2*e*x^5 + 5*b^2*d*x^3)*log(c^2*x^2 + 1)^2 + integrate(1/240*(180*(b^2*c^2*e*x^6 + b
^2*d*x^2 + (b^2*c^2*d + b^2*e)*x^4)*arctan(c*x)^2 + 15*(b^2*c^2*e*x^6 + b^2*d*x^2 + (b^2*c^2*d + b^2*e)*x^4)*l
og(c^2*x^2 + 1)^2 - 8*(3*b^2*c*e*x^5 + 5*b^2*c*d*x^3)*arctan(c*x) + 4*(3*b^2*c^2*e*x^6 + 5*b^2*c^2*d*x^4)*log(
c^2*x^2 + 1))/(c^2*x^2 + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (a^{2} e x^{4} + a^{2} d x^{2} +{\left (b^{2} e x^{4} + b^{2} d x^{2}\right )} \arctan \left (c x\right )^{2} + 2 \,{\left (a b e x^{4} + a b d x^{2}\right )} \arctan \left (c x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(a^2*e*x^4 + a^2*d*x^2 + (b^2*e*x^4 + b^2*d*x^2)*arctan(c*x)^2 + 2*(a*b*e*x^4 + a*b*d*x^2)*arctan(c*x)
, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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