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

Optimal. Leaf size=287 \[ -\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {2 i b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^4}+\frac {a b d x}{2 c^3}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{5 c^4}+\frac {3 i b^2 d \tan ^{-1}(c x)}{10 c^4}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x \tan ^{-1}(c x)}{2 c^3}+\frac {b^2 d x^2}{12 c^2}-\frac {b^2 d \log \left (c^2 x^2+1\right )}{3 c^4}+\frac {i b^2 d x^3}{30 c} \]

[Out]

1/2*a*b*d*x/c^3-3/10*I*b^2*d*x/c^3+1/12*b^2*d*x^2/c^2+1/30*I*b^2*d*x^3/c+3/10*I*b^2*d*arctan(c*x)/c^4+1/2*b^2*
d*x*arctan(c*x)/c^3+1/5*I*b*d*x^2*(a+b*arctan(c*x))/c^2-1/6*b*d*x^3*(a+b*arctan(c*x))/c-1/10*I*b*d*x^4*(a+b*ar
ctan(c*x))-9/20*d*(a+b*arctan(c*x))^2/c^4+1/4*d*x^4*(a+b*arctan(c*x))^2+1/5*I*c*d*x^5*(a+b*arctan(c*x))^2+2/5*
I*b*d*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4-1/3*b^2*d*ln(c^2*x^2+1)/c^4-1/5*b^2*d*polylog(2,1-2/(1+I*c*x))/c^4

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Rubi [A]  time = 0.60, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 15, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {4876, 4852, 4916, 266, 43, 4846, 260, 4884, 302, 203, 321, 4920, 4854, 2402, 2315} \[ -\frac {b^2 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^4}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}+\frac {a b d x}{2 c^3}-\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {2 i b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{5 c^4}+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}+\frac {b^2 d x^2}{12 c^2}-\frac {b^2 d \log \left (c^2 x^2+1\right )}{3 c^4}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x \tan ^{-1}(c x)}{2 c^3}+\frac {3 i b^2 d \tan ^{-1}(c x)}{10 c^4}+\frac {i b^2 d x^3}{30 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*d*x)/(2*c^3) - (((3*I)/10)*b^2*d*x)/c^3 + (b^2*d*x^2)/(12*c^2) + ((I/30)*b^2*d*x^3)/c + (((3*I)/10)*b^2*d
*ArcTan[c*x])/c^4 + (b^2*d*x*ArcTan[c*x])/(2*c^3) + ((I/5)*b*d*x^2*(a + b*ArcTan[c*x]))/c^2 - (b*d*x^3*(a + b*
ArcTan[c*x]))/(6*c) - (I/10)*b*d*x^4*(a + b*ArcTan[c*x]) - (9*d*(a + b*ArcTan[c*x])^2)/(20*c^4) + (d*x^4*(a +
b*ArcTan[c*x])^2)/4 + (I/5)*c*d*x^5*(a + b*ArcTan[c*x])^2 + (((2*I)/5)*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*
x)])/c^4 - (b^2*d*Log[1 + c^2*x^2])/(3*c^4) - (b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x)])/(5*c^4)

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 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 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 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 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]

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 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 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 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 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 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 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 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 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 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]

Rubi steps

\begin {align*} \int x^3 (d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d x^3 \left (a+b \tan ^{-1}(c x)\right )^2+i c d x^4 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+(i c d) \int x^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} (b c d) \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {1}{5} \left (2 i b c^2 d\right ) \int \frac {x^5 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{5} (2 i b d) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {1}{5} (2 i b d) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {(b d) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac {(b d) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{6} \left (b^2 d\right ) \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {(b d) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac {(b d) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}+\frac {(2 i b d) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c^2}-\frac {(2 i b d) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c^2}+\frac {1}{10} \left (i b^2 c d\right ) \int \frac {x^4}{1+c^2 x^2} \, dx\\ &=\frac {a b d x}{2 c^3}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{12} \left (b^2 d\right ) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {(2 i b d) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{5 c^3}+\frac {\left (b^2 d\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}-\frac {\left (i b^2 d\right ) \int \frac {x^2}{1+c^2 x^2} \, dx}{5 c}+\frac {1}{10} \left (i b^2 c d\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 {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {i b^2 d x^3}{30 c}+\frac {b^2 d x \tan ^{-1}(c x)}{2 c^3}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}+\frac {1}{12} \left (b^2 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 {\left (i b^2 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{10 c^3}+\frac {\left (i b^2 d\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (2 i b^2 d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (b^2 d\right ) \int \frac {x}{1+c^2 x^2} \, dx}{2 c^2}\\ &=\frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {i b^2 d x^3}{30 c}+\frac {3 i b^2 d \tan ^{-1}(c x)}{10 c^4}+\frac {b^2 d x \tan ^{-1}(c x)}{2 c^3}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{3 c^4}-\frac {\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{5 c^4}\\ &=\frac {a b d x}{2 c^3}-\frac {3 i b^2 d x}{10 c^3}+\frac {b^2 d x^2}{12 c^2}+\frac {i b^2 d x^3}{30 c}+\frac {3 i b^2 d \tan ^{-1}(c x)}{10 c^4}+\frac {b^2 d x \tan ^{-1}(c x)}{2 c^3}+\frac {i b d x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {b d x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{10} i b d x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {9 d \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{5} i c d x^5 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {b^2 d \log \left (1+c^2 x^2\right )}{3 c^4}-\frac {b^2 d \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{5 c^4}\\ \end {align*}

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Mathematica [A]  time = 0.86, size = 285, normalized size = 0.99 \[ \frac {d \left (12 i a^2 c^5 x^5+15 a^2 c^4 x^4-6 i a b c^4 x^4-10 a b c^3 x^3+12 i a b c^2 x^2-12 i a b \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (3 a \left (4 i c^5 x^5+5 c^4 x^4-5\right )+b \left (-3 i c^4 x^4-5 c^3 x^3+6 i c^2 x^2+15 c x+9 i\right )+12 i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+30 a b c x+18 i a b+2 i b^2 c^3 x^3+5 b^2 c^2 x^2-20 b^2 \log \left (c^2 x^2+1\right )+3 b^2 \left (4 i c^5 x^5+5 c^4 x^4-1\right ) \tan ^{-1}(c x)^2+12 b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-18 i b^2 c x+5 b^2\right )}{60 c^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d*((18*I)*a*b + 5*b^2 + 30*a*b*c*x - (18*I)*b^2*c*x + (12*I)*a*b*c^2*x^2 + 5*b^2*c^2*x^2 - 10*a*b*c^3*x^3 + (
2*I)*b^2*c^3*x^3 + 15*a^2*c^4*x^4 - (6*I)*a*b*c^4*x^4 + (12*I)*a^2*c^5*x^5 + 3*b^2*(-1 + 5*c^4*x^4 + (4*I)*c^5
*x^5)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*(b*(9*I + 15*c*x + (6*I)*c^2*x^2 - 5*c^3*x^3 - (3*I)*c^4*x^4) + 3*a*(-5
+ 5*c^4*x^4 + (4*I)*c^5*x^5) + (12*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (12*I)*a*b*Log[1 + c^2*x^2] - 20*b^2
*Log[1 + c^2*x^2] + 12*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(60*c^4)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \frac {1}{80} \, {\left (-4 i \, b^{2} c d x^{5} - 5 \, b^{2} d x^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + {\rm integral}\left (\frac {20 i \, a^{2} c^{3} d x^{6} + 20 \, a^{2} c^{2} d x^{5} + 20 i \, a^{2} c d x^{4} + 20 \, a^{2} d x^{3} - {\left (20 \, a b c^{3} d x^{6} + 4 \, {\left (-5 i \, a b - b^{2}\right )} c^{2} d x^{5} + {\left (20 \, a b + 5 i \, b^{2}\right )} c d x^{4} - 20 i \, a b d x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{20 \, {\left (c^{2} x^{2} + 1\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/80*(-4*I*b^2*c*d*x^5 - 5*b^2*d*x^4)*log(-(c*x + I)/(c*x - I))^2 + integral(1/20*(20*I*a^2*c^3*d*x^6 + 20*a^2
*c^2*d*x^5 + 20*I*a^2*c*d*x^4 + 20*a^2*d*x^3 - (20*a*b*c^3*d*x^6 + 4*(-5*I*a*b - b^2)*c^2*d*x^5 + (20*a*b + 5*
I*b^2)*c*d*x^4 - 20*I*a*b*d*x^3)*log(-(c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.10, size = 499, normalized size = 1.74 \[ \frac {2 i c d a b \arctan \left (c x \right ) x^{5}}{5}+\frac {d \,a^{2} x^{4}}{4}+\frac {d a b \arctan \left (c x \right ) x^{4}}{2}-\frac {d \,b^{2} \arctan \left (c x \right ) x^{3}}{6 c}-\frac {d a b \arctan \left (c x \right )}{2 c^{4}}-\frac {3 i b^{2} d x}{10 c^{3}}-\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )}{10 c^{4}}-\frac {i d a b \,x^{4}}{10}-\frac {i d \,b^{2} \arctan \left (c x \right ) x^{4}}{10}+\frac {d \,b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )}{10 c^{4}}+\frac {i b^{2} d \,x^{3}}{30 c}+\frac {3 i b^{2} d \arctan \left (c x \right )}{10 c^{4}}+\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )}{10 c^{4}}-\frac {d \,b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{10 c^{4}}+\frac {i c d \,a^{2} x^{5}}{5}+\frac {a b d x}{2 c^{3}}+\frac {b^{2} d x \arctan \left (c x \right )}{2 c^{3}}+\frac {b^{2} d \,x^{2}}{12 c^{2}}+\frac {i c d \,b^{2} \arctan \left (c x \right )^{2} x^{5}}{5}-\frac {i d \,b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )}{5 c^{4}}-\frac {b^{2} d \ln \left (c^{2} x^{2}+1\right )}{3 c^{4}}+\frac {i d a b \,x^{2}}{5 c^{2}}+\frac {i d \,b^{2} \arctan \left (c x \right ) x^{2}}{5 c^{2}}-\frac {i d a b \ln \left (c^{2} x^{2}+1\right )}{5 c^{4}}-\frac {d a b \,x^{3}}{6 c}-\frac {d \,b^{2} \ln \left (c x -i\right )^{2}}{20 c^{4}}+\frac {d \,b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )}{10 c^{4}}-\frac {d \,b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{10 c^{4}}+\frac {d \,b^{2} \ln \left (c x +i\right )^{2}}{20 c^{4}}+\frac {d \,b^{2} \arctan \left (c x \right )^{2} x^{4}}{4}-\frac {d \,b^{2} \arctan \left (c x \right )^{2}}{4 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{5} i \, a^{2} c d x^{5} + \frac {1}{4} \, b^{2} d x^{4} \arctan \left (c x\right )^{2} + \frac {1}{4} \, a^{2} d x^{4} + \frac {1}{10} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} a b c d + \frac {1}{80} i \, {\left (4 \, x^{5} \arctan \left (c x\right )^{2} - x^{5} \log \left (c^{2} x^{2} + 1\right )^{2} + 80 \, \int \frac {4 \, c^{2} x^{6} \log \left (c^{2} x^{2} + 1\right ) - 8 \, c x^{5} \arctan \left (c x\right ) + 60 \, {\left (c^{2} x^{6} + x^{4}\right )} \arctan \left (c x\right )^{2} + 5 \, {\left (c^{2} x^{6} + x^{4}\right )} \log \left (c^{2} x^{2} + 1\right )^{2}}{80 \, {\left (c^{2} x^{2} + 1\right )}}\,{d x}\right )} b^{2} c d + \frac {1}{6} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} a b d - \frac {1}{12} \, {\left (2 \, c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )} \arctan \left (c x\right ) - \frac {c^{2} x^{2} + 3 \, \arctan \left (c x\right )^{2} - 4 \, \log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )} b^{2} d \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*I*a^2*c*d*x^5 + 1/4*b^2*d*x^4*arctan(c*x)^2 + 1/4*a^2*d*x^4 + 1/10*I*(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*c*d + 1/80*I*(4*x^5*arctan(c*x)^2 - x^5*log(c^2*x^2 + 1)^2 + 80*integr
ate(1/80*(4*c^2*x^6*log(c^2*x^2 + 1) - 8*c*x^5*arctan(c*x) + 60*(c^2*x^6 + x^4)*arctan(c*x)^2 + 5*(c^2*x^6 + x
^4)*log(c^2*x^2 + 1)^2)/(c^2*x^2 + 1), x))*b^2*c*d + 1/6*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arcta
n(c*x)/c^5))*a*b*d - 1/12*(2*c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5)*arctan(c*x) - (c^2*x^2 + 3*arctan(c*x
)^2 - 4*log(c^2*x^2 + 1))/c^4)*b^2*d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,\left (d+c\,d\,x\,1{}\mathrm {i}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atan(c*x))^2*(d + c*d*x*1i),x)

[Out]

int(x^3*(a + b*atan(c*x))^2*(d + c*d*x*1i), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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