3.1.86 \(\int x (d+i c d x)^3 (a+b \text {ArcTan}(c x))^2 \, dx\) [86]

Optimal. Leaf size=307 \[ -\frac {5 a b d^3 x}{2 c}+\frac {13 i b^2 d^3 x}{10 c}-\frac {1}{4} b^2 d^3 x^2-\frac {1}{30} i b^2 c d^3 x^3-\frac {13 i b^2 d^3 \text {ArcTan}(c x)}{10 c^2}-\frac {5 b^2 d^3 x \text {ArcTan}(c x)}{2 c}-\frac {6}{5} i b d^3 x^2 (a+b \text {ArcTan}(c x))+\frac {1}{2} b c d^3 x^3 (a+b \text {ArcTan}(c x))+\frac {1}{10} i b c^2 d^3 x^4 (a+b \text {ArcTan}(c x))+\frac {d^3 (1+i c x)^4 (a+b \text {ArcTan}(c x))^2}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \text {ArcTan}(c x))^2}{5 c^2}-\frac {12 i b d^3 (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1-i c x}\right )}{5 c^2}+\frac {3 b^2 d^3 \log \left (1+c^2 x^2\right )}{2 c^2}-\frac {6 b^2 d^3 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{5 c^2} \]

[Out]

-5/2*a*b*d^3*x/c+13/10*I*b^2*d^3*x/c-1/4*b^2*d^3*x^2-1/30*I*b^2*c*d^3*x^3-13/10*I*b^2*d^3*arctan(c*x)/c^2-5/2*
b^2*d^3*x*arctan(c*x)/c-6/5*I*b*d^3*x^2*(a+b*arctan(c*x))+1/2*b*c*d^3*x^3*(a+b*arctan(c*x))+1/10*I*b*c^2*d^3*x
^4*(a+b*arctan(c*x))+1/4*d^3*(1+I*c*x)^4*(a+b*arctan(c*x))^2/c^2-1/5*d^3*(1+I*c*x)^5*(a+b*arctan(c*x))^2/c^2-1
2/5*I*b*d^3*(a+b*arctan(c*x))*ln(2/(1-I*c*x))/c^2+3/2*b^2*d^3*ln(c^2*x^2+1)/c^2-6/5*b^2*d^3*polylog(2,1-2/(1-I
*c*x))/c^2

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Rubi [A]
time = 0.45, antiderivative size = 307, normalized size of antiderivative = 1.00, number of steps used = 38, number of rules used = 14, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {4996, 4974, 4930, 266, 4946, 327, 209, 272, 45, 1600, 4964, 2449, 2352, 308} \begin {gather*} \frac {1}{10} i b c^2 d^3 x^4 (a+b \text {ArcTan}(c x))-\frac {d^3 (1+i c x)^5 (a+b \text {ArcTan}(c x))^2}{5 c^2}+\frac {d^3 (1+i c x)^4 (a+b \text {ArcTan}(c x))^2}{4 c^2}-\frac {12 i b d^3 \log \left (\frac {2}{1-i c x}\right ) (a+b \text {ArcTan}(c x))}{5 c^2}+\frac {1}{2} b c d^3 x^3 (a+b \text {ArcTan}(c x))-\frac {6}{5} i b d^3 x^2 (a+b \text {ArcTan}(c x))-\frac {5 a b d^3 x}{2 c}-\frac {13 i b^2 d^3 \text {ArcTan}(c x)}{10 c^2}-\frac {5 b^2 d^3 x \text {ArcTan}(c x)}{2 c}-\frac {6 b^2 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{5 c^2}+\frac {3 b^2 d^3 \log \left (c^2 x^2+1\right )}{2 c^2}-\frac {1}{30} i b^2 c d^3 x^3+\frac {13 i b^2 d^3 x}{10 c}-\frac {1}{4} b^2 d^3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a*b*d^3*x)/(2*c) + (((13*I)/10)*b^2*d^3*x)/c - (b^2*d^3*x^2)/4 - (I/30)*b^2*c*d^3*x^3 - (((13*I)/10)*b^2*d
^3*ArcTan[c*x])/c^2 - (5*b^2*d^3*x*ArcTan[c*x])/(2*c) - ((6*I)/5)*b*d^3*x^2*(a + b*ArcTan[c*x]) + (b*c*d^3*x^3
*(a + b*ArcTan[c*x]))/2 + (I/10)*b*c^2*d^3*x^4*(a + b*ArcTan[c*x]) + (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x])^2)
/(4*c^2) - (d^3*(1 + I*c*x)^5*(a + b*ArcTan[c*x])^2)/(5*c^2) - (((12*I)/5)*b*d^3*(a + b*ArcTan[c*x])*Log[2/(1
- I*c*x)])/c^2 + (3*b^2*d^3*Log[1 + c^2*x^2])/(2*c^2) - (6*b^2*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)])/(5*c^2)

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 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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 4974

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a
+ b*ArcTan[c*x])^p/(e*(q + 1))), x] - Dist[b*c*(p/(e*(q + 1))), Int[ExpandIntegrand[(a + b*ArcTan[c*x])^(p -
1), (d + e*x)^(q + 1)/(1 + c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] && N
eQ[q, -1]

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])

Rubi steps

\begin {align*} \int x (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx &=\int \left (\frac {i (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-\frac {i (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{c d}\right ) \, dx\\ &=\frac {i \int (d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c}-\frac {i \int (d+i c d x)^4 \left (a+b \tan ^{-1}(c x)\right )^2 \, dx}{c d}\\ &=\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}+\frac {(2 b) \int \left (-15 d^5 \left (a+b \tan ^{-1}(c x)\right )-11 i c d^5 x \left (a+b \tan ^{-1}(c x)\right )+5 c^2 d^5 x^2 \left (a+b \tan ^{-1}(c x)\right )+i c^3 d^5 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {16 i \left (i d^5-c d^5 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{5 c d^2}-\frac {b \int \left (-7 d^4 \left (a+b \tan ^{-1}(c x)\right )-4 i c d^4 x \left (a+b \tan ^{-1}(c x)\right )+c^2 d^4 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {8 i \left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2}\right ) \, dx}{2 c d}\\ &=\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac {(32 i b) \int \frac {\left (i d^5-c d^5 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{5 c d^2}+\frac {(4 i b) \int \frac {\left (i d^4-c d^4 x\right ) \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d}+\left (2 i b d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx-\frac {1}{5} \left (22 i b d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {\left (7 b d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{2 c}-\frac {\left (6 b d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c}-\frac {1}{2} \left (b c d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\left (2 b c d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx+\frac {1}{5} \left (2 i b c^2 d^3\right ) \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=-\frac {5 a b d^3 x}{2 c}-\frac {6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac {(32 i b) \int \frac {a+b \tan ^{-1}(c x)}{-\frac {i}{d^5}-\frac {c x}{d^5}} \, dx}{5 c d^2}+\frac {(4 i b) \int \frac {a+b \tan ^{-1}(c x)}{-\frac {i}{d^4}-\frac {c x}{d^4}} \, dx}{c d}+\frac {\left (7 b^2 d^3\right ) \int \tan ^{-1}(c x) \, dx}{2 c}-\frac {\left (6 b^2 d^3\right ) \int \tan ^{-1}(c x) \, dx}{c}-\left (i b^2 c d^3\right ) \int \frac {x^2}{1+c^2 x^2} \, dx+\frac {1}{5} \left (11 i b^2 c d^3\right ) \int \frac {x^2}{1+c^2 x^2} \, dx+\frac {1}{6} \left (b^2 c^2 d^3\right ) \int \frac {x^3}{1+c^2 x^2} \, dx-\frac {1}{3} \left (2 b^2 c^2 d^3\right ) \int \frac {x^3}{1+c^2 x^2} \, dx-\frac {1}{10} \left (i b^2 c^3 d^3\right ) \int \frac {x^4}{1+c^2 x^2} \, dx\\ &=-\frac {5 a b d^3 x}{2 c}+\frac {6 i b^2 d^3 x}{5 c}-\frac {5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac {6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac {12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{5 c^2}-\frac {1}{2} \left (7 b^2 d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx+\left (6 b^2 d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx+\frac {\left (i b^2 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{c}-\frac {\left (11 i b^2 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{5 c}-\frac {\left (4 i b^2 d^3\right ) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{c}+\frac {\left (32 i b^2 d^3\right ) \int \frac {\log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{5 c}+\frac {1}{12} \left (b^2 c^2 d^3\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )-\frac {1}{3} \left (b^2 c^2 d^3\right ) \text {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )-\frac {1}{10} \left (i b^2 c^3 d^3\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 {5 a b d^3 x}{2 c}+\frac {13 i b^2 d^3 x}{10 c}-\frac {1}{30} i b^2 c d^3 x^3-\frac {6 i b^2 d^3 \tan ^{-1}(c x)}{5 c^2}-\frac {5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac {6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac {12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{5 c^2}+\frac {5 b^2 d^3 \log \left (1+c^2 x^2\right )}{4 c^2}+\frac {\left (4 b^2 d^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )}{c^2}-\frac {\left (32 b^2 d^3\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-i c x}\right )}{5 c^2}-\frac {\left (i b^2 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{10 c}+\frac {1}{12} \left (b^2 c^2 d^3\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 {1}{3} \left (b^2 c^2 d^3\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 {5 a b d^3 x}{2 c}+\frac {13 i b^2 d^3 x}{10 c}-\frac {1}{4} b^2 d^3 x^2-\frac {1}{30} i b^2 c d^3 x^3-\frac {13 i b^2 d^3 \tan ^{-1}(c x)}{10 c^2}-\frac {5 b^2 d^3 x \tan ^{-1}(c x)}{2 c}-\frac {6}{5} i b d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} b c d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{10} i b c^2 d^3 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 c^2}-\frac {d^3 (1+i c x)^5 \left (a+b \tan ^{-1}(c x)\right )^2}{5 c^2}-\frac {12 i b d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{5 c^2}+\frac {3 b^2 d^3 \log \left (1+c^2 x^2\right )}{2 c^2}-\frac {6 b^2 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{5 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 325, normalized size = 1.06 \begin {gather*} \frac {d^3 \left (-18 i a b-15 b^2-150 a b c x+78 i b^2 c x+30 a^2 c^2 x^2-72 i a b c^2 x^2-15 b^2 c^2 x^2+60 i a^2 c^3 x^3+30 a b c^3 x^3-2 i b^2 c^3 x^3-45 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 (1-4 i c x) (-i+c x)^4 \text {ArcTan}(c x)^2+6 b \text {ArcTan}(c x) \left (b \left (-13 i-25 c x-12 i c^2 x^2+5 c^3 x^3+i c^4 x^4\right )+a \left (25+10 c^2 x^2+20 i c^3 x^3-15 c^4 x^4-4 i c^5 x^5\right )-24 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 )+90 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 )}{60 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d^3*((-18*I)*a*b - 15*b^2 - 150*a*b*c*x + (78*I)*b^2*c*x + 30*a^2*c^2*x^2 - (72*I)*a*b*c^2*x^2 - 15*b^2*c^2*x
^2 + (60*I)*a^2*c^3*x^3 + 30*a*b*c^3*x^3 - (2*I)*b^2*c^3*x^3 - 45*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 - (4*I)*c*x)*(-I + c*x)^4*ArcTan[c*x]^2 + 6*b*ArcTan[c*x]*(b*(-13*I - 25*c*x - (12*I)*c^2*
x^2 + 5*c^3*x^3 + I*c^4*x^4) + a*(25 + 10*c^2*x^2 + (20*I)*c^3*x^3 - 15*c^4*x^4 - (4*I)*c^5*x^5) - (24*I)*b*Lo
g[1 + E^((2*I)*ArcTan[c*x])]) + (72*I)*a*b*Log[1 + c^2*x^2] + 90*b^2*Log[1 + c^2*x^2] - 72*b^2*PolyLog[2, -E^(
(2*I)*ArcTan[c*x])]))/(60*c^2)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 625 vs. \(2 (269 ) = 538\).
time = 0.23, size = 626, normalized size = 2.04 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(-2/5*I*d^3*a*b*arctan(c*x)*c^5*x^5+2*I*d^3*a*b*arctan(c*x)*c^3*x^3-5/2*b^2*c*d^3*x*arctan(c*x)+1/2*d^3*
a*b*c^3*x^3-3/4*d^3*b^2*arctan(c*x)^2*c^4*x^4+1/2*d^3*b^2*arctan(c*x)*c^3*x^3-13/10*I*d^3*b^2*arctan(c*x)+6/5*
I*d^3*b^2*ln(c^2*x^2+1)*arctan(c*x)+1/2*d^3*b^2*arctan(c*x)^2*c^2*x^2-1/30*I*d^3*b^2*c^3*x^3+13/10*I*d^3*b^2*c
*x+6/5*I*d^3*a*b*ln(c^2*x^2+1)+3/5*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/4*d^3*b^2*c^2*x^2+5/2*d^3*a*b*arctan
(c*x)+3/5*d^3*b^2*ln(c*x+I)*ln(c^2*x^2+1)-3/5*d^3*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-3/5*d^3*b^2*ln(c*x-I)*ln(c^2
*x^2+1)-3/2*d^3*a*b*arctan(c*x)*c^4*x^4+1/10*I*d^3*a*b*c^4*x^4-6/5*I*d^3*a*b*c^2*x^2+d^3*a*b*arctan(c*x)*c^2*x
^2+I*d^3*b^2*arctan(c*x)^2*c^3*x^3+1/10*I*d^3*b^2*arctan(c*x)*c^4*x^4-1/5*I*d^3*b^2*arctan(c*x)^2*c^5*x^5-6/5*
I*d^3*b^2*arctan(c*x)*c^2*x^2-3/5*d^3*b^2*dilog(1/2*I*(c*x-I))+3/5*d^3*b^2*dilog(-1/2*I*(c*x+I))+5/4*d^3*b^2*a
rctan(c*x)^2-3/10*d^3*b^2*ln(c*x+I)^2+3/10*d^3*b^2*ln(c*x-I)^2+d^3*a^2*(-1/5*I*c^5*x^5-3/4*c^4*x^4+I*c^3*x^3+1
/2*c^2*x^2)-5/2*a*b*c*d^3*x+3/2*b^2*d^3*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*(d+I*c*d*x)^3*(a+b*arctan(c*x))^2,x, algorithm="maxima")

[Out]

-1/5*I*a^2*c^3*d^3*x^5 - 3/4*a^2*c^2*d^3*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^3*d^3 + I*a^2*c*d^3*x^3 + 1/2*b^2*d^3*x^2*arctan(c*x)^2 - 1/2*(3*x^4*arctan(c*x) - c*((
c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*a*b*c^2*d^3 + I*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^
4))*a*b*c*d^3 + 1/2*a^2*d^3*x^2 + (x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*d^3 - 1/2*(2*c*(x/c^2 -
arctan(c*x)/c^3)*arctan(c*x) + (arctan(c*x)^2 - log(c^2*x^2 + 1))/c^2)*b^2*d^3 + 1/80*(-4*I*b^2*c^3*d^3*x^5 -
15*b^2*c^2*d^3*x^4 + 20*I*b^2*c*d^3*x^3)*arctan(c*x)^2 + 1/80*(4*b^2*c^3*d^3*x^5 - 15*I*b^2*c^2*d^3*x^4 - 20*b
^2*c*d^3*x^3)*arctan(c*x)*log(c^2*x^2 + 1) - 1/320*(-4*I*b^2*c^3*d^3*x^5 - 15*b^2*c^2*d^3*x^4 + 20*I*b^2*c*d^3
*x^3)*log(c^2*x^2 + 1)^2 - I*integrate(1/80*(60*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*x^4 - 3*b^2*c*d^3*x^2)*arctan
(c*x)^2 + 5*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*x^4 - 3*b^2*c*d^3*x^2)*log(c^2*x^2 + 1)^2 - 2*(19*b^2*c^4*d^3*x^5
- 20*b^2*c^2*d^3*x^3)*arctan(c*x) + (4*b^2*c^5*d^3*x^6 - 35*b^2*c^3*d^3*x^4 - 60*(b^2*c^4*d^3*x^5 + b^2*c^2*d
^3*x^3)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x) - integrate(1/80*(180*(b^2*c^4*d^3*x^5 + b^2*c^2*d^3*
x^3)*arctan(c*x)^2 + 15*(b^2*c^4*d^3*x^5 + b^2*c^2*d^3*x^3)*log(c^2*x^2 + 1)^2 + 2*(4*b^2*c^5*d^3*x^6 - 35*b^2
*c^3*d^3*x^4)*arctan(c*x) + (19*b^2*c^4*d^3*x^5 - 20*b^2*c^2*d^3*x^3 + 20*(b^2*c^5*d^3*x^6 - 2*b^2*c^3*d^3*x^4
- 3*b^2*c*d^3*x^2)*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*(d+I*c*d*x)^3*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

1/80*(4*I*b^2*c^3*d^3*x^5 + 15*b^2*c^2*d^3*x^4 - 20*I*b^2*c*d^3*x^3 - 10*b^2*d^3*x^2)*log(-(c*x + I)/(c*x - I)
)^2 + integral(1/20*(-20*I*a^2*c^5*d^3*x^6 - 60*a^2*c^4*d^3*x^5 + 40*I*a^2*c^3*d^3*x^4 - 40*a^2*c^2*d^3*x^3 +
60*I*a^2*c*d^3*x^2 + 20*a^2*d^3*x + (20*a*b*c^5*d^3*x^6 - 4*(15*I*a*b + b^2)*c^4*d^3*x^5 - 5*(8*a*b - 3*I*b^2)
*c^3*d^3*x^4 - 20*(2*I*a*b - b^2)*c^2*d^3*x^3 - 10*(6*a*b + I*b^2)*c*d^3*x^2 + 20*I*a*b*d^3*x)*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*(d+I*c*d*x)**3*(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*(d+I*c*d*x)^3*(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\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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