3.1.76 \(\int x^3 (d+i c d x)^2 (a+b \text {ArcTan}(c x))^2 \, dx\) [76]

Optimal. Leaf size=373 \[ \frac {5 a b d^2 x}{6 c^3}-\frac {3 i b^2 d^2 x}{5 c^3}+\frac {31 b^2 d^2 x^2}{180 c^2}+\frac {i b^2 d^2 x^3}{15 c}-\frac {1}{60} b^2 d^2 x^4+\frac {3 i b^2 d^2 \text {ArcTan}(c x)}{5 c^4}+\frac {5 b^2 d^2 x \text {ArcTan}(c x)}{6 c^3}+\frac {2 i b d^2 x^2 (a+b \text {ArcTan}(c x))}{5 c^2}-\frac {5 b d^2 x^3 (a+b \text {ArcTan}(c x))}{18 c}-\frac {1}{5} i b d^2 x^4 (a+b \text {ArcTan}(c x))+\frac {1}{15} b c d^2 x^5 (a+b \text {ArcTan}(c x))-\frac {49 d^2 (a+b \text {ArcTan}(c x))^2}{60 c^4}+\frac {1}{4} d^2 x^4 (a+b \text {ArcTan}(c x))^2+\frac {2}{5} i c d^2 x^5 (a+b \text {ArcTan}(c x))^2-\frac {1}{6} c^2 d^2 x^6 (a+b \text {ArcTan}(c x))^2+\frac {4 i b d^2 (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {53 b^2 d^2 \log \left (1+c^2 x^2\right )}{90 c^4}-\frac {2 b^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{5 c^4} \]

[Out]

5/6*a*b*d^2*x/c^3+4/5*I*b*d^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^4+31/180*b^2*d^2*x^2/c^2+2/5*I*b*d^2*x^2*(a+
b*arctan(c*x))/c^2-1/60*b^2*d^2*x^4+2/5*I*c*d^2*x^5*(a+b*arctan(c*x))^2+5/6*b^2*d^2*x*arctan(c*x)/c^3-3/5*I*b^
2*d^2*x/c^3-5/18*b*d^2*x^3*(a+b*arctan(c*x))/c+1/15*I*b^2*d^2*x^3/c+1/15*b*c*d^2*x^5*(a+b*arctan(c*x))-49/60*d
^2*(a+b*arctan(c*x))^2/c^4+1/4*d^2*x^4*(a+b*arctan(c*x))^2+3/5*I*b^2*d^2*arctan(c*x)/c^4-1/6*c^2*d^2*x^6*(a+b*
arctan(c*x))^2-1/5*I*b*d^2*x^4*(a+b*arctan(c*x))-53/90*b^2*d^2*ln(c^2*x^2+1)/c^4-2/5*b^2*d^2*polylog(2,1-2/(1+
I*c*x))/c^4

________________________________________________________________________________________

Rubi [A]
time = 0.70, antiderivative size = 373, normalized size of antiderivative = 1.00, number of steps used = 43, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4996, 4946, 5036, 272, 45, 4930, 266, 5004, 308, 209, 327, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {49 d^2 (a+b \text {ArcTan}(c x))^2}{60 c^4}+\frac {4 i b d^2 \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{5 c^4}-\frac {1}{6} c^2 d^2 x^6 (a+b \text {ArcTan}(c x))^2+\frac {2 i b d^2 x^2 (a+b \text {ArcTan}(c x))}{5 c^2}+\frac {2}{5} i c d^2 x^5 (a+b \text {ArcTan}(c x))^2+\frac {1}{15} b c d^2 x^5 (a+b \text {ArcTan}(c x))+\frac {1}{4} d^2 x^4 (a+b \text {ArcTan}(c x))^2-\frac {1}{5} i b d^2 x^4 (a+b \text {ArcTan}(c x))-\frac {5 b d^2 x^3 (a+b \text {ArcTan}(c x))}{18 c}+\frac {5 a b d^2 x}{6 c^3}+\frac {3 i b^2 d^2 \text {ArcTan}(c x)}{5 c^4}+\frac {5 b^2 d^2 x \text {ArcTan}(c x)}{6 c^3}-\frac {2 b^2 d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{5 c^4}-\frac {3 i b^2 d^2 x}{5 c^3}+\frac {31 b^2 d^2 x^2}{180 c^2}-\frac {53 b^2 d^2 \log \left (c^2 x^2+1\right )}{90 c^4}+\frac {i b^2 d^2 x^3}{15 c}-\frac {1}{60} b^2 d^2 x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*b*d^2*x)/(6*c^3) - (((3*I)/5)*b^2*d^2*x)/c^3 + (31*b^2*d^2*x^2)/(180*c^2) + ((I/15)*b^2*d^2*x^3)/c - (b^2
*d^2*x^4)/60 + (((3*I)/5)*b^2*d^2*ArcTan[c*x])/c^4 + (5*b^2*d^2*x*ArcTan[c*x])/(6*c^3) + (((2*I)/5)*b*d^2*x^2*
(a + b*ArcTan[c*x]))/c^2 - (5*b*d^2*x^3*(a + b*ArcTan[c*x]))/(18*c) - (I/5)*b*d^2*x^4*(a + b*ArcTan[c*x]) + (b
*c*d^2*x^5*(a + b*ArcTan[c*x]))/15 - (49*d^2*(a + b*ArcTan[c*x])^2)/(60*c^4) + (d^2*x^4*(a + b*ArcTan[c*x])^2)
/4 + ((2*I)/5)*c*d^2*x^5*(a + b*ArcTan[c*x])^2 - (c^2*d^2*x^6*(a + b*ArcTan[c*x])^2)/6 + (((4*I)/5)*b*d^2*(a +
b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/c^4 - (53*b^2*d^2*Log[1 + c^2*x^2])/(90*c^4) - (2*b^2*d^2*PolyLog[2, 1 - 2
/(1 + I*c*x)])/(5*c^4)

Rule 45

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 266

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 272

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 308

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 327

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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 4930

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

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 4996

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 5004

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 5036

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 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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)^2 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )^2+2 i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2-c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^2 \int x^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+\left (2 i c d^2\right ) \int x^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx-\left (c^2 d^2\right ) \int x^5 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{2} \left (b c d^2\right ) \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {1}{5} \left (4 i b c^2 d^2\right ) \int \frac {x^5 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx+\frac {1}{3} \left (b c^3 d^2\right ) \int \frac {x^6 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{5} \left (4 i b d^2\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {1}{5} \left (4 i b d^2\right ) \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\frac {\left (b d^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}+\frac {\left (b d^2\right ) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{2 c}+\frac {1}{3} \left (b c d^2\right ) \int x^4 \left (a+b \tan ^{-1}(c x)\right ) \, dx-\frac {1}{3} \left (b c d^2\right ) \int \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-\frac {b d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{6 c}-\frac {1}{5} i b d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{15} b c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{6} \left (b^2 d^2\right ) \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {\left (b d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c^3}-\frac {\left (b d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{2 c^3}+\frac {\left (4 i b d^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{5 c^2}-\frac {\left (4 i b d^2\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c^2}-\frac {\left (b d^2\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c}+\frac {\left (b d^2\right ) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{3 c}+\frac {1}{5} \left (i b^2 c d^2\right ) \int \frac {x^4}{1+c^2 x^2} \, dx-\frac {1}{15} \left (b^2 c^2 d^2\right ) \int \frac {x^5}{1+c^2 x^2} \, dx\\ &=\frac {a b d^2 x}{2 c^3}+\frac {2 i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {5 b d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{18 c}-\frac {1}{5} i b d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{15} b c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {13 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{20 c^4}+\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{12} \left (b^2 d^2\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {1}{9} \left (b^2 d^2\right ) \int \frac {x^3}{1+c^2 x^2} \, dx+\frac {\left (4 i b d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx}{5 c^3}+\frac {\left (b d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{3 c^3}-\frac {\left (b d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{3 c^3}+\frac {\left (b^2 d^2\right ) \int \tan ^{-1}(c x) \, dx}{2 c^3}-\frac {\left (2 i b^2 d^2\right ) \int \frac {x^2}{1+c^2 x^2} \, dx}{5 c}+\frac {1}{5} \left (i b^2 c d^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 {1}{30} \left (b^2 c^2 d^2\right ) \text {Subst}\left (\int \frac {x^2}{1+c^2 x} \, dx,x,x^2\right )\\ &=\frac {5 a b d^2 x}{6 c^3}-\frac {3 i b^2 d^2 x}{5 c^3}+\frac {i b^2 d^2 x^3}{15 c}+\frac {b^2 d^2 x \tan ^{-1}(c x)}{2 c^3}+\frac {2 i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {5 b d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{18 c}-\frac {1}{5} i b d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{15} b c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {49 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{60 c^4}+\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}+\frac {1}{18} \left (b^2 d^2\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )+\frac {1}{12} \left (b^2 d^2\right ) \text {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^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3}+\frac {\left (2 i b^2 d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c^3}-\frac {\left (4 i b^2 d^2\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^2\right ) \int \tan ^{-1}(c x) \, dx}{3 c^3}-\frac {\left (b^2 d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{2 c^2}-\frac {1}{30} \left (b^2 c^2 d^2\right ) \text {Subst}\left (\int \left (-\frac {1}{c^4}+\frac {x}{c^2}+\frac {1}{c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=\frac {5 a b d^2 x}{6 c^3}-\frac {3 i b^2 d^2 x}{5 c^3}+\frac {7 b^2 d^2 x^2}{60 c^2}+\frac {i b^2 d^2 x^3}{15 c}-\frac {1}{60} b^2 d^2 x^4+\frac {3 i b^2 d^2 \tan ^{-1}(c x)}{5 c^4}+\frac {5 b^2 d^2 x \tan ^{-1}(c x)}{6 c^3}+\frac {2 i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {5 b d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{18 c}-\frac {1}{5} i b d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{15} b c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {49 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{60 c^4}+\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {11 b^2 d^2 \log \left (1+c^2 x^2\right )}{30 c^4}+\frac {1}{18} \left (b^2 d^2\right ) \text {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 (4 b^2 d^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{5 c^4}-\frac {\left (b^2 d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 c^2}\\ &=\frac {5 a b d^2 x}{6 c^3}-\frac {3 i b^2 d^2 x}{5 c^3}+\frac {31 b^2 d^2 x^2}{180 c^2}+\frac {i b^2 d^2 x^3}{15 c}-\frac {1}{60} b^2 d^2 x^4+\frac {3 i b^2 d^2 \tan ^{-1}(c x)}{5 c^4}+\frac {5 b^2 d^2 x \tan ^{-1}(c x)}{6 c^3}+\frac {2 i b d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )}{5 c^2}-\frac {5 b d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )}{18 c}-\frac {1}{5} i b d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{15} b c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {49 d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{60 c^4}+\frac {1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {4 i b d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{5 c^4}-\frac {53 b^2 d^2 \log \left (1+c^2 x^2\right )}{90 c^4}-\frac {2 b^2 d^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{5 c^4}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 342, normalized size = 0.92 \begin {gather*} \frac {d^2 \left (108 i a b+34 b^2+150 a b c x-108 i b^2 c x+72 i a b c^2 x^2+31 b^2 c^2 x^2-50 a b c^3 x^3+12 i b^2 c^3 x^3+45 a^2 c^4 x^4-36 i a b c^4 x^4-3 b^2 c^4 x^4+72 i a^2 c^5 x^5+12 a b c^5 x^5-30 a^2 c^6 x^6-3 b^2 \left (1-15 c^4 x^4-24 i c^5 x^5+10 c^6 x^6\right ) \text {ArcTan}(c x)^2+2 b \text {ArcTan}(c x) \left (b \left (54 i+75 c x+36 i c^2 x^2-25 c^3 x^3-18 i c^4 x^4+6 c^5 x^5\right )+a \left (-75+45 c^4 x^4+72 i c^5 x^5-30 c^6 x^6\right )+72 i b \log \left (1+e^{2 i \text {ArcTan}(c x)}\right )\right )-72 i a b \log \left (1+c^2 x^2\right )-106 b^2 \log \left (1+c^2 x^2\right )+72 b^2 \text {PolyLog}\left (2,-e^{2 i \text {ArcTan}(c x)}\right )\right )}{180 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*((108*I)*a*b + 34*b^2 + 150*a*b*c*x - (108*I)*b^2*c*x + (72*I)*a*b*c^2*x^2 + 31*b^2*c^2*x^2 - 50*a*b*c^3*
x^3 + (12*I)*b^2*c^3*x^3 + 45*a^2*c^4*x^4 - (36*I)*a*b*c^4*x^4 - 3*b^2*c^4*x^4 + (72*I)*a^2*c^5*x^5 + 12*a*b*c
^5*x^5 - 30*a^2*c^6*x^6 - 3*b^2*(1 - 15*c^4*x^4 - (24*I)*c^5*x^5 + 10*c^6*x^6)*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]
*(b*(54*I + 75*c*x + (36*I)*c^2*x^2 - 25*c^3*x^3 - (18*I)*c^4*x^4 + 6*c^5*x^5) + a*(-75 + 45*c^4*x^4 + (72*I)*
c^5*x^5 - 30*c^6*x^6) + (72*I)*b*Log[1 + E^((2*I)*ArcTan[c*x])]) - (72*I)*a*b*Log[1 + c^2*x^2] - 106*b^2*Log[1
+ c^2*x^2] + 72*b^2*PolyLog[2, -E^((2*I)*ArcTan[c*x])]))/(180*c^4)

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Maple [A]
time = 0.29, size = 624, normalized size = 1.67 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(4/5*I*d^2*a*b*arctan(c*x)*c^5*x^5+5/6*a*b*c*d^2*x+5/6*b^2*c*d^2*x*arctan(c*x)-1/3*d^2*a*b*arctan(c*x)*c
^6*x^6+1/2*d^2*a*b*arctan(c*x)*c^4*x^4-1/5*I*d^2*a*b*c^4*x^4+2/5*I*d^2*a*b*c^2*x^2-1/5*I*d^2*b^2*arctan(c*x)*c
^4*x^4+2/5*I*d^2*b^2*arctan(c*x)^2*c^5*x^5+2/5*I*d^2*b^2*arctan(c*x)*c^2*x^2+1/15*d^2*b^2*arctan(c*x)*c^5*x^5-
5/18*d^2*b^2*arctan(c*x)*c^3*x^3+1/15*I*d^2*b^2*c^3*x^3-3/5*I*d^2*b^2*c*x-1/6*d^2*b^2*arctan(c*x)^2*c^6*x^6+1/
4*d^2*b^2*arctan(c*x)^2*c^4*x^4+1/15*d^2*a*b*c^5*x^5-5/18*d^2*a*b*c^3*x^3-2/5*I*d^2*a*b*ln(c^2*x^2+1)-2/5*I*d^
2*b^2*ln(c^2*x^2+1)*arctan(c*x)+3/5*I*d^2*b^2*arctan(c*x)-1/60*d^2*b^2*c^4*x^4+31/180*d^2*b^2*c^2*x^2-5/6*d^2*
a*b*arctan(c*x)+1/5*d^2*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-1/5*d^2*b^2*ln(c*x+I)*ln(c^2*x^2+1)-1/5*d^2*b^2*ln(c*x
-I)*ln(-1/2*I*(c*x+I))+1/5*d^2*b^2*ln(c*x-I)*ln(c^2*x^2+1)+d^2*a^2*(-1/6*c^6*x^6+2/5*I*c^5*x^5+1/4*c^4*x^4)-5/
12*d^2*b^2*arctan(c*x)^2+1/10*d^2*b^2*ln(c*x+I)^2-1/10*d^2*b^2*ln(c*x-I)^2+1/5*d^2*b^2*dilog(1/2*I*(c*x-I))-1/
5*d^2*b^2*dilog(-1/2*I*(c*x+I))-53/90*b^2*d^2*ln(c^2*x^2+1))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*a^2*c^2*d^2*x^6 + 2/5*I*a^2*c*d^2*x^5 + 1/4*b^2*d^2*x^4*arctan(c*x)^2 + 1/4*a^2*d^2*x^4 - 1/45*(15*x^6*ar
ctan(c*x) - c*((3*c^4*x^5 - 5*c^2*x^3 + 15*x)/c^6 - 15*arctan(c*x)/c^7))*a*b*c^2*d^2 + 1/5*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^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*d^2 - 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^2 - 1/120*(5*b^2*c^2*d^2*x^6 - 12*I*b^2*c*d^2*x^5)*arcta
n(c*x)^2 + 1/120*(-5*I*b^2*c^2*d^2*x^6 - 12*b^2*c*d^2*x^5)*arctan(c*x)*log(c^2*x^2 + 1) + 1/480*(5*b^2*c^2*d^2
*x^6 - 12*I*b^2*c*d^2*x^5)*log(c^2*x^2 + 1)^2 - integrate(-1/240*(68*b^2*c^3*d^2*x^6*arctan(c*x) - 180*(b^2*c^
4*d^2*x^7 + b^2*c^2*d^2*x^5)*arctan(c*x)^2 - 15*(b^2*c^4*d^2*x^7 + b^2*c^2*d^2*x^5)*log(c^2*x^2 + 1)^2 - 2*(5*
b^2*c^4*d^2*x^7 - 12*b^2*c^2*d^2*x^5 - 60*(b^2*c^3*d^2*x^6 + b^2*c*d^2*x^4)*arctan(c*x))*log(c^2*x^2 + 1))/(c^
2*x^2 + 1), x) + I*integrate(1/120*(180*(b^2*c^3*d^2*x^6 + b^2*c*d^2*x^4)*arctan(c*x)^2 + 15*(b^2*c^3*d^2*x^6
+ b^2*c*d^2*x^4)*log(c^2*x^2 + 1)^2 + 2*(5*b^2*c^4*d^2*x^7 - 12*b^2*c^2*d^2*x^5)*arctan(c*x) + (17*b^2*c^3*d^2
*x^6 + 30*(b^2*c^4*d^2*x^7 + b^2*c^2*d^2*x^5)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(10*b^2*c^2*d^2*x^6 - 24*I*b^2*c*d^2*x^5 - 15*b^2*d^2*x^4)*log(-(c*x + I)/(c*x - I))^2 + integral(-1/60*
(60*a^2*c^4*d^2*x^7 - 120*I*a^2*c^3*d^2*x^6 - 120*I*a^2*c*d^2*x^4 - 60*a^2*d^2*x^3 - (-60*I*a*b*c^4*d^2*x^7 -
10*(12*a*b - I*b^2)*c^3*d^2*x^6 + 24*b^2*c^2*d^2*x^5 - 15*(8*a*b + I*b^2)*c*d^2*x^4 + 60*I*a*b*d^2*x^3)*log(-(
c*x + I)/(c*x - I)))/(c^2*x^2 + 1), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

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)^2,x)

[Out]

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

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