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

Optimal. Leaf size=255 \[ -\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i a b d x}{2 c^2}-\frac{7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{i b^2 d \log \left (c^2 x^2+1\right )}{3 c^3}+\frac{b^2 d x}{3 c^2}+\frac{i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{i b^2 d x^2}{12 c} \]

[Out]

((I/2)*a*b*d*x)/c^2 + (b^2*d*x)/(3*c^2) + ((I/12)*b^2*d*x^2)/c - (b^2*d*ArcTan[c*x])/(3*c^3) + ((I/2)*b^2*d*x*
ArcTan[c*x])/c^2 - (b*d*x^2*(a + b*ArcTan[c*x]))/(3*c) - (I/6)*b*d*x^3*(a + b*ArcTan[c*x]) - (((7*I)/12)*d*(a
+ b*ArcTan[c*x])^2)/c^3 + (d*x^3*(a + b*ArcTan[c*x])^2)/3 + (I/4)*c*d*x^4*(a + b*ArcTan[c*x])^2 - (2*b*d*(a +
b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) - ((I/3)*b^2*d*Log[1 + c^2*x^2])/c^3 - ((I/3)*b^2*d*PolyLog[2, 1 -
2/(1 + I*c*x)])/c^3

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Rubi [A]  time = 0.491, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 14, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.609, Rules used = {4876, 4852, 4916, 321, 203, 4920, 4854, 2402, 2315, 266, 43, 4846, 260, 4884} \[ -\frac{i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{3 c^3}+\frac{i a b d x}{2 c^2}-\frac{7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}-\frac{2 b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c^3}+\frac{1}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{i b^2 d \log \left (c^2 x^2+1\right )}{3 c^3}+\frac{b^2 d x}{3 c^2}+\frac{i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{i b^2 d x^2}{12 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*a*b*d*x)/c^2 + (b^2*d*x)/(3*c^2) + ((I/12)*b^2*d*x^2)/c - (b^2*d*ArcTan[c*x])/(3*c^3) + ((I/2)*b^2*d*x*
ArcTan[c*x])/c^2 - (b*d*x^2*(a + b*ArcTan[c*x]))/(3*c) - (I/6)*b*d*x^3*(a + b*ArcTan[c*x]) - (((7*I)/12)*d*(a
+ b*ArcTan[c*x])^2)/c^3 + (d*x^3*(a + b*ArcTan[c*x])^2)/3 + (I/4)*c*d*x^4*(a + b*ArcTan[c*x])^2 - (2*b*d*(a +
b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(3*c^3) - ((I/3)*b^2*d*Log[1 + c^2*x^2])/c^3 - ((I/3)*b^2*d*PolyLog[2, 1 -
2/(1 + I*c*x)])/c^3

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 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 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 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 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]

Rubi steps

\begin{align*} \int x^2 (d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d x^2 \left (a+b \tan ^{-1}(c x)\right )^2+i c d x^3 \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+(i c d) \int x^3 \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}{4} i c d x^4 \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}{2} \left (i b c^2 d\right ) \int \frac{x^4 \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}{4} i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac{1}{2} (i b d) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac{1}{2} (i b d) \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\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{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\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}{4} i c d x^4 \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{(i b d) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^2}-\frac{(i b d) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^2}-\frac{(2 b d) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx}{3 c^2}+\frac{1}{6} \left (i b^2 c d\right ) \int \frac{x^3}{1+c^2 x^2} \, dx\\ &=\frac{i a b d x}{2 c^2}+\frac{b^2 d x}{3 c^2}-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} i c d x^4 \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 (i b^2 d\right ) \int \tan ^{-1}(c x) \, dx}{2 c^2}-\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}{12} \left (i b^2 c d\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )\\ &=\frac{i a b d x}{2 c^2}+\frac{b^2 d x}{3 c^2}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} i c d x^4 \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 (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 (i b^2 d\right ) \int \frac{x}{1+c^2 x^2} \, dx}{2 c}+\frac{1}{12} \left (i b^2 c d\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{i a b d x}{2 c^2}+\frac{b^2 d x}{3 c^2}+\frac{i b^2 d x^2}{12 c}-\frac{b^2 d \tan ^{-1}(c x)}{3 c^3}+\frac{i b^2 d x \tan ^{-1}(c x)}{2 c^2}-\frac{b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{3 c}-\frac{1}{6} i b d x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{7 i d \left (a+b \tan ^{-1}(c x)\right )^2}{12 c^3}+\frac{1}{3} d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+\frac{1}{4} i c d x^4 \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{i b^2 d \log \left (1+c^2 x^2\right )}{3 c^3}-\frac{i b^2 d \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{3 c^3}\\ \end{align*}

Mathematica [A]  time = 0.591374, size = 241, normalized size = 0.95 \[ \frac{i d \left (4 b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+3 a^2 c^4 x^4-4 i a^2 c^3 x^3-2 a b c^3 x^3+4 i a b c^2 x^2-4 i a b \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (a \left (3 c^4 x^4-4 i c^3 x^3-3\right )+b \left (-c^3 x^3+2 i c^2 x^2+3 c x+2 i\right )+4 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+6 a b c x+b^2 c^2 x^2-4 b^2 \log \left (c^2 x^2+1\right )+b^2 \left (3 c^4 x^4-4 i c^3 x^3+1\right ) \tan ^{-1}(c x)^2-4 i b^2 c x+b^2\right )}{12 c^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.094, size = 467, normalized size = 1.8 \begin{align*}{\frac{{\frac{i}{12}}d{b}^{2}{x}^{2}}{c}}+{\frac{{\frac{i}{2}}{b}^{2}dx\arctan \left ( cx \right ) }{{c}^{2}}}+{\frac{dab\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,{c}^{3}}}+{\frac{i}{2}}cdab\arctan \left ( cx \right ){x}^{4}+{\frac{d{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,{c}^{3}}}-{\frac{d{b}^{2}\arctan \left ( cx \right ){x}^{2}}{3\,c}}-{\frac{i}{6}}dab{x}^{3}-{\frac{i}{6}}d{b}^{2}\arctan \left ( cx \right ){x}^{3}+{\frac{i}{4}}cd{a}^{2}{x}^{4}-{\frac{{\frac{i}{6}}d{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{3}}}-{\frac{{\frac{i}{4}}d{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{{c}^{3}}}-{\frac{{\frac{i}{12}}d{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{3}}}+{\frac{{\frac{i}{12}}d{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{{c}^{3}}}+{\frac{{\frac{i}{6}}d{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{3}}}-{\frac{dab{x}^{2}}{3\,c}}+{\frac{2\,dab\arctan \left ( cx \right ){x}^{3}}{3}}+{\frac{d{a}^{2}{x}^{3}}{3}}-{\frac{d{b}^{2}\arctan \left ( cx \right ) }{3\,{c}^{3}}}+{\frac{d{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{3}}{3}}+{\frac{{\frac{i}{2}}abdx}{{c}^{2}}}-{\frac{{\frac{i}{3}}{b}^{2}d\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{3}}}+{\frac{{\frac{i}{6}}d{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{{c}^{3}}}+{\frac{d{b}^{2}x}{3\,{c}^{2}}}-{\frac{{\frac{i}{6}}d{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{{c}^{3}}}+{\frac{i}{4}}cd{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{4}-{\frac{{\frac{i}{2}}abd\arctan \left ( cx \right ) }{{c}^{3}}}+{\frac{{\frac{i}{6}}d{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{{c}^{3}}}-{\frac{{\frac{i}{6}}d{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*I*b^2*d*x^2/c+1/2*I*b^2*d*x*arctan(c*x)/c^2+1/3/c^3*d*a*b*ln(c^2*x^2+1)+1/2*I*c*d*a*b*arctan(c*x)*x^4+1/3
/c^3*d*b^2*arctan(c*x)*ln(c^2*x^2+1)-1/3/c*d*b^2*arctan(c*x)*x^2-1/6*I*d*a*b*x^3-1/6*I*d*b^2*arctan(c*x)*x^3+1
/4*I*c*d*a^2*x^4-1/6*I/c^3*d*b^2*dilog(-1/2*I*(c*x+I))-1/4*I/c^3*d*b^2*arctan(c*x)^2-1/12*I/c^3*d*b^2*ln(c*x-I
)^2+1/12*I/c^3*d*b^2*ln(c*x+I)^2+1/6*I/c^3*d*b^2*dilog(1/2*I*(c*x-I))-1/3/c*d*a*b*x^2+2/3*d*a*b*arctan(c*x)*x^
3+1/3*d*a^2*x^3-1/3*b^2*d*arctan(c*x)/c^3+1/3*d*b^2*arctan(c*x)^2*x^3+1/2*I*a*b*d*x/c^2-1/3*I*b^2*d*ln(c^2*x^2
+1)/c^3+1/6*I/c^3*d*b^2*ln(c^2*x^2+1)*ln(c*x-I)+1/3*b^2*d*x/c^2-1/6*I/c^3*d*b^2*ln(c^2*x^2+1)*ln(c*x+I)+1/4*I*
c*d*b^2*arctan(c*x)^2*x^4-1/2*I/c^3*d*a*b*arctan(c*x)+1/6*I/c^3*d*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/6*I/c^3*d*
b^2*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*} \frac{1}{4} i \, a^{2} c d x^{4} + \frac{1}{3} \, a^{2} d x^{3} + \frac{1}{6} i \,{\left (3 \, x^{4} \arctan \left (c x\right ) - c{\left (\frac{c^{2} x^{3} - 3 \, x}{c^{4}} + \frac{3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a b c d + \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}{192} \,{\left (12 i \, b^{2} c d x^{4} + 16 \, b^{2} d x^{3}\right )} \arctan \left (c x\right )^{2} - \frac{1}{48} \,{\left (3 \, b^{2} c d x^{4} - 4 i \, b^{2} d x^{3}\right )} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) + \frac{1}{192} \,{\left (-3 i \, b^{2} c d x^{4} - 4 \, b^{2} d x^{3}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + i \, \int -\frac{14 \, b^{2} c^{2} d x^{4} \arctan \left (c x\right ) - 36 \,{\left (b^{2} c^{3} d x^{5} + b^{2} c d x^{3}\right )} \arctan \left (c x\right )^{2} - 3 \,{\left (b^{2} c^{3} d x^{5} + b^{2} c d x^{3}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} -{\left (3 \, b^{2} c^{3} d x^{5} - 4 \, b^{2} c d x^{3} - 12 \,{\left (b^{2} c^{2} d x^{4} + b^{2} d x^{2}\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{48 \,{\left (c^{2} x^{2} + 1\right )}}\,{d x} + \int \frac{36 \,{\left (b^{2} c^{2} d x^{4} + b^{2} d x^{2}\right )} \arctan \left (c x\right )^{2} + 3 \,{\left (b^{2} c^{2} d x^{4} + b^{2} d x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + 2 \,{\left (3 \, b^{2} c^{3} d x^{5} - 4 \, b^{2} c d x^{3}\right )} \arctan \left (c x\right ) +{\left (7 \, b^{2} c^{2} d x^{4} + 12 \,{\left (b^{2} c^{3} d x^{5} + b^{2} c d x^{3}\right )} \arctan \left (c x\right )\right )} \log \left (c^{2} x^{2} + 1\right )}{48 \,{\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*(d+I*c*d*x)*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

1/4*I*a^2*c*d*x^4 + 1/3*a^2*d*x^3 + 1/6*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*
b*c*d + 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*a*b*d + 1/192*(12*I*b^2*c*d*x^4 + 16*b^2*
d*x^3)*arctan(c*x)^2 - 1/48*(3*b^2*c*d*x^4 - 4*I*b^2*d*x^3)*arctan(c*x)*log(c^2*x^2 + 1) + 1/192*(-3*I*b^2*c*d
*x^4 - 4*b^2*d*x^3)*log(c^2*x^2 + 1)^2 + I*integrate(-1/48*(14*b^2*c^2*d*x^4*arctan(c*x) - 36*(b^2*c^3*d*x^5 +
b^2*c*d*x^3)*arctan(c*x)^2 - 3*(b^2*c^3*d*x^5 + b^2*c*d*x^3)*log(c^2*x^2 + 1)^2 - (3*b^2*c^3*d*x^5 - 4*b^2*c*
d*x^3 - 12*(b^2*c^2*d*x^4 + b^2*d*x^2)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x) + integrate(1/48*(36*(
b^2*c^2*d*x^4 + b^2*d*x^2)*arctan(c*x)^2 + 3*(b^2*c^2*d*x^4 + b^2*d*x^2)*log(c^2*x^2 + 1)^2 + 2*(3*b^2*c^3*d*x
^5 - 4*b^2*c*d*x^3)*arctan(c*x) + (7*b^2*c^2*d*x^4 + 12*(b^2*c^3*d*x^5 + b^2*c*d*x^3)*arctan(c*x))*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*} \frac{1}{48} \,{\left (-3 i \, b^{2} c d x^{4} - 4 \, b^{2} d x^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\rm integral}\left (\frac{12 i \, a^{2} c^{3} d x^{5} + 12 \, a^{2} c^{2} d x^{4} + 12 i \, a^{2} c d x^{3} + 12 \, a^{2} d x^{2} -{\left (12 \, a b c^{3} d x^{5} -{\left (12 i \, a b + 3 \, b^{2}\right )} c^{2} d x^{4} + 4 \,{\left (3 \, a b + i \, b^{2}\right )} c d x^{3} - 12 i \, a b d x^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{12 \,{\left (c^{2} x^{2} + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-3*I*b^2*c*d*x^4 - 4*b^2*d*x^3)*log(-(c*x + I)/(c*x - I))^2 + integral(1/12*(12*I*a^2*c^3*d*x^5 + 12*a^2
*c^2*d*x^4 + 12*I*a^2*c*d*x^3 + 12*a^2*d*x^2 - (12*a*b*c^3*d*x^5 - (12*I*a*b + 3*b^2)*c^2*d*x^4 + 4*(3*a*b + I
*b^2)*c*d*x^3 - 12*I*a*b*d*x^2)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

[Out]

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