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

Optimal. Leaf size=298 \[ -\frac {4 i b^2 d^2 \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {6 i b^2 d^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-a b^2 d^2 x+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {7 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac {b^3 d^2 \log \left (c^2 x^2+1\right )}{2 c}+\frac {3 b^3 d^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c}+\frac {2 b^3 d^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{c}+b^3 \left (-d^2\right ) x \tan ^{-1}(c x) \]

[Out]

-a*b^2*d^2*x-b^3*d^2*x*arctan(c*x)+7/2*b*d^2*(a+b*arctan(c*x))^2/c-3*I*b*d^2*x*(a+b*arctan(c*x))^2+1/2*b*c*d^2
*x^2*(a+b*arctan(c*x))^2-1/3*I*d^2*(1+I*c*x)^3*(a+b*arctan(c*x))^3/c+4*b*d^2*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x
))/c-6*I*b^2*d^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c+1/2*b^3*d^2*ln(c^2*x^2+1)/c-4*I*b^2*d^2*(a+b*arctan(c*x))
*polylog(2,1-2/(1-I*c*x))/c+3*b^3*d^2*polylog(2,1-2/(1+I*c*x))/c+2*b^3*d^2*polylog(3,1-2/(1-I*c*x))/c

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Rubi [A]  time = 0.48, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {4864, 4846, 4920, 4854, 2402, 2315, 4852, 4916, 260, 4884, 1586, 4992, 6610} \[ -\frac {4 i b^2 d^2 \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac {3 b^3 d^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c}+\frac {2 b^3 d^2 \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{c}-\frac {6 i b^2 d^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-a b^2 d^2 x+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2+\frac {7 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac {b^3 d^2 \log \left (c^2 x^2+1\right )}{2 c}+b^3 \left (-d^2\right ) x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1586

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 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 4864

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 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]

Rule 4992

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(I*(a + b*ArcT
an[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p*I)/2, Int[((a + b*ArcTan[c*x])^(p - 1)*PolyLog[2, 1 - u]
)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*I
)/(I + c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin {align*} \int (d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {(i b) \int \left (-3 d^3 \left (a+b \tan ^{-1}(c x)\right )^2-i c d^3 x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {4 i \left (i d^3-c d^3 x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}\right ) \, dx}{d}\\ &=-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {(4 b) \int \frac {\left (i d^3-c d^3 x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{d}-\left (3 i b d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx+\left (b c d^2\right ) \int x \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {(4 b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-\frac {i}{d^3}-\frac {c x}{d^3}} \, dx}{d}+\left (6 i b^2 c d^2\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx-\left (b^2 c^2 d^2\right ) \int \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {3 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\left (6 i b^2 d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx-\left (b^2 d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx+\left (b^2 d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx-\left (8 b^2 d^2\right ) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-a b^2 d^2 x+\frac {7 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {6 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {4 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\left (4 i b^3 d^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx+\left (6 i b^3 d^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (b^3 d^2\right ) \int \tan ^{-1}(c x) \, dx\\ &=-a b^2 d^2 x-b^3 d^2 x \tan ^{-1}(c x)+\frac {7 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {6 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {4 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\frac {2 b^3 d^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{c}+\frac {\left (6 b^3 d^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c}+\left (b^3 c d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx\\ &=-a b^2 d^2 x-b^3 d^2 x \tan ^{-1}(c x)+\frac {7 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-3 i b d^2 x \left (a+b \tan ^{-1}(c x)\right )^2+\frac {1}{2} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^3}{3 c}+\frac {4 b d^2 \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {6 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}+\frac {b^3 d^2 \log \left (1+c^2 x^2\right )}{2 c}-\frac {4 i b^2 d^2 \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\frac {3 b^3 d^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c}+\frac {2 b^3 d^2 \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{c}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 528, normalized size = 1.77 \[ -\frac {d^2 \left (2 a^3 c^3 x^3-6 i a^3 c^2 x^2-6 a^3 c x+6 a^2 b c^3 x^3 \tan ^{-1}(c x)-3 a^2 b c^2 x^2+12 a^2 b \log \left (c^2 x^2+1\right )-18 i a^2 b c^2 x^2 \tan ^{-1}(c x)+18 i a^2 b c x-18 i a^2 b \tan ^{-1}(c x)-18 a^2 b c x \tan ^{-1}(c x)+6 a b^2 c^3 x^3 \tan ^{-1}(c x)^2-18 i a b^2 \log \left (c^2 x^2+1\right )-18 i a b^2 c^2 x^2 \tan ^{-1}(c x)^2-6 a b^2 c^2 x^2 \tan ^{-1}(c x)+6 b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (4 i a+4 i b \tan ^{-1}(c x)+3 b\right )+6 a b^2 c x+6 i a b^2 \tan ^{-1}(c x)^2-18 a b^2 c x \tan ^{-1}(c x)^2-6 a b^2 \tan ^{-1}(c x)+36 i a b^2 c x \tan ^{-1}(c x)-48 a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+2 b^3 c^3 x^3 \tan ^{-1}(c x)^3-3 b^3 \log \left (c^2 x^2+1\right )-6 i b^3 c^2 x^2 \tan ^{-1}(c x)^3-3 b^3 c^2 x^2 \tan ^{-1}(c x)^2-12 b^3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+2 i b^3 \tan ^{-1}(c x)^3-6 b^3 c x \tan ^{-1}(c x)^3+15 b^3 \tan ^{-1}(c x)^2+18 i b^3 c x \tan ^{-1}(c x)^2+6 b^3 c x \tan ^{-1}(c x)-24 b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+36 i b^3 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{6 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(d^2*(-6*a^3*c*x + (18*I)*a^2*b*c*x + 6*a*b^2*c*x - (6*I)*a^3*c^2*x^2 - 3*a^2*b*c^2*x^2 + 2*a^3*c^3*x^3 -
(18*I)*a^2*b*ArcTan[c*x] - 6*a*b^2*ArcTan[c*x] - 18*a^2*b*c*x*ArcTan[c*x] + (36*I)*a*b^2*c*x*ArcTan[c*x] + 6*
b^3*c*x*ArcTan[c*x] - (18*I)*a^2*b*c^2*x^2*ArcTan[c*x] - 6*a*b^2*c^2*x^2*ArcTan[c*x] + 6*a^2*b*c^3*x^3*ArcTan[
c*x] + (6*I)*a*b^2*ArcTan[c*x]^2 + 15*b^3*ArcTan[c*x]^2 - 18*a*b^2*c*x*ArcTan[c*x]^2 + (18*I)*b^3*c*x*ArcTan[c
*x]^2 - (18*I)*a*b^2*c^2*x^2*ArcTan[c*x]^2 - 3*b^3*c^2*x^2*ArcTan[c*x]^2 + 6*a*b^2*c^3*x^3*ArcTan[c*x]^2 + (2*
I)*b^3*ArcTan[c*x]^3 - 6*b^3*c*x*ArcTan[c*x]^3 - (6*I)*b^3*c^2*x^2*ArcTan[c*x]^3 + 2*b^3*c^3*x^3*ArcTan[c*x]^3
- 48*a*b^2*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])] + (36*I)*b^3*ArcTan[c*x]*Log[1 + E^((2*I)*ArcTan[c*x])]
- 24*b^3*ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + 12*a^2*b*Log[1 + c^2*x^2] - (18*I)*a*b^2*Log[1 + c^2*
x^2] - 3*b^3*Log[1 + c^2*x^2] + 6*b^2*((4*I)*a + 3*b + (4*I)*b*ArcTan[c*x])*PolyLog[2, -E^((2*I)*ArcTan[c*x])]
- 12*b^3*PolyLog[3, -E^((2*I)*ArcTan[c*x])]))/c

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ \frac {1}{24} \, {\left (i \, b^{3} c^{2} d^{2} x^{3} + 3 \, b^{3} c d^{2} x^{2} - 3 i \, b^{3} d^{2} x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{3} + {\rm integral}\left (-\frac {4 \, a^{3} c^{4} d^{2} x^{4} - 8 i \, a^{3} c^{3} d^{2} x^{3} - 8 i \, a^{3} c d^{2} x - 4 \, a^{3} d^{2} - {\left (3 \, a b^{2} c^{4} d^{2} x^{4} + 3 i \, b^{3} c^{2} d^{2} x^{2} + {\left (-6 i \, a b^{2} - b^{3}\right )} c^{3} d^{2} x^{3} - 3 \, a b^{2} d^{2} - 3 \, {\left (2 i \, a b^{2} - b^{3}\right )} c d^{2} x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (-6 i \, a^{2} b c^{4} d^{2} x^{4} - 12 \, a^{2} b c^{3} d^{2} x^{3} - 12 \, a^{2} b c d^{2} x + 6 i \, a^{2} b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, {\left (c^{2} x^{2} + 1\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(I*b^3*c^2*d^2*x^3 + 3*b^3*c*d^2*x^2 - 3*I*b^3*d^2*x)*log(-(c*x + I)/(c*x - I))^3 + integral(-1/4*(4*a^3*
c^4*d^2*x^4 - 8*I*a^3*c^3*d^2*x^3 - 8*I*a^3*c*d^2*x - 4*a^3*d^2 - (3*a*b^2*c^4*d^2*x^4 + 3*I*b^3*c^2*d^2*x^2 +
(-6*I*a*b^2 - b^3)*c^3*d^2*x^3 - 3*a*b^2*d^2 - 3*(2*I*a*b^2 - b^3)*c*d^2*x)*log(-(c*x + I)/(c*x - I))^2 - (-6
*I*a^2*b*c^4*d^2*x^4 - 12*a^2*b*c^3*d^2*x^3 - 12*a^2*b*c*d^2*x + 6*I*a^2*b*d^2)*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((d+I*c*d*x)^2*(a+b*arctan(c*x))^3,x, algorithm="giac")

[Out]

sage0*x

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maple [C]  time = 6.23, size = 1815, normalized size = 6.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/2/c*d^2*b^3*arctan(c*x)^2-6/c*d^2*b^3*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-1/3*I/c*d^2*a^3-1/3*c^2*x^3*a^
3*d^2-1/c*d^2*b^3*ln((1+I*c*x)^2/(c^2*x^2+1)+1)-6/c*d^2*b^3*dilog(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-a*b^2*d^2*x
-b^3*d^2*x*arctan(c*x)-4/c*a*b^2*arctan(c*x)*ln(c^2*x^2+1)*d^2-2/c*b^3*arctan(c*x)^2*ln(c^2*x^2+1)*d^2+4/c*b^3
*ln((1+I*c*x)/(c^2*x^2+1)^(1/2))*arctan(c*x)^2*d^2+4/c*b^3*d^2*ln(2)*arctan(c*x)^2-2/c*a^2*b*ln(c^2*x^2+1)*d^2
+3*a*b^2*arctan(c*x)^2*x*d^2+3*a^2*b*arctan(c*x)*x*d^2-c^2*d^2*a^2*b*arctan(c*x)*x^3+I*c*d^2*b^3*arctan(c*x)^3
*x^2+I/c*d^2*a*b^2*ln(c*x-I)^2-6*I*d^2*a*b^2*arctan(c*x)*x+3*I/c*d^2*a^2*b*arctan(c*x)+3*I/c*d^2*a*b^2*arctan(
c*x)^2-I/c*d^2*a*b^2*ln(I+c*x)^2+3*I/c*d^2*a*b^2*ln(c^2*x^2+1)-2*I/c*d^2*a*b^2*dilog(1/2*I*(c*x-I))+2*I/c*d^2*
a*b^2*dilog(-1/2*I*(I+c*x))-6*I/c*d^2*b^3*arctan(c*x)*ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-6*I/c*d^2*b^3*arctan
(c*x)*ln(1-I*(1+I*c*x)/(c^2*x^2+1)^(1/2))-4*I/c*d^2*b^3*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+c*d^2*
a*b^2*arctan(c*x)*x^2-c^2*d^2*a*b^2*arctan(c*x)^2*x^3-I/c*d^2*b^3*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csg
n(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*arctan(c*x)^2-1/3*I
/c*d^2*b^3*arctan(c*x)^3-3*I*d^2*b^3*arctan(c*x)^2*x-3*I*d^2*a^2*b*x+1/2*c*d^2*a^2*b*x^2+I/c*d^2*b^3*arctan(c*
x)+1/c*d^2*a*b^2*arctan(c*x)+1/2*c*d^2*b^3*arctan(c*x)^2*x^2-1/3*c^2*d^2*b^3*arctan(c*x)^3*x^3+I*c*x^2*a^3*d^2
+b^3*arctan(c*x)^3*x*d^2+2/c*b^3*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))*d^2+I/c*d^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c
^2*x^2+1)+1)^2)^3*arctan(c*x)^2+2*I/c*d^2*a*b^2*ln(I+c*x)*ln(c^2*x^2+1)+3*I*c*d^2*a^2*b*arctan(c*x)*x^2+3*I*c*
d^2*a*b^2*arctan(c*x)^2*x^2-2*I/c*d^2*a*b^2*ln(c*x-I)*ln(c^2*x^2+1)-2*I/c*d^2*a*b^2*ln(I+c*x)*ln(1/2*I*(c*x-I)
)+2*I/c*d^2*a*b^2*ln(c*x-I)*ln(-1/2*I*(I+c*x))-I/c*d^2*b^3*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^3*arctan(c*x)^2-
I/c*d^2*b^3*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^3*arctan(c*x)^2+I/c*d^2*b^3*Pi*cs
gn(I/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*arctan(c*x
)^2+I/c*d^2*b^3*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)^
2)^2*arctan(c*x)^2+I/c*d^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))^2*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)*
arctan(c*x)^2-2*I/c*d^2*b^3*Pi*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*((1+I*c*x)^2/(c^2*x^2+1)+1)^2)^2*arc
tan(c*x)^2+2*I/c*d^2*b^3*Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))^2*arctan(c*x)^
2-I/c*d^2*b^3*Pi*csgn(I*(1+I*c*x)/(c^2*x^2+1)^(1/2))^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1))*arctan(c*x)^2+a^3*x*d^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^3*c^2*d^2*x^3 - 28*b^3*c^4*d^2*integrate(1/32*x^4*arctan(c*x)^3/(c^2*x^2 + 1), x) - 3*b^3*c^4*d^2*integ
rate(1/32*x^4*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) - 96*a*b^2*c^4*d^2*integrate(1/32*x^4*arctan(c*
x)^2/(c^2*x^2 + 1), x) - 4*b^3*c^4*d^2*integrate(1/32*x^4*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 12*
b^3*c^3*d^2*integrate(1/32*x^3*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c^3*d^2*integrate(1/32*x
^3*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) + 16*b^3*c^3*d^2*integrate(1/32*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) -
4*b^3*c^3*d^2*integrate(1/32*x^3*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) - 1/2*(2*x^3*arctan(c*x) - c*(x^2/c^2 -
log(c^2*x^2 + 1)/c^4))*a^2*b*c^2*d^2 + I*a^3*c*d^2*x^2 + 7/32*b^3*d^2*arctan(c*x)^4/c + 24*b^3*c^2*d^2*integra
te(1/32*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3*I*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3
))*a^2*b*c*d^2 + a*b^2*d^2*arctan(c*x)^3/c + 12*b^3*c*d^2*integrate(1/32*x*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2
*x^2 + 1), x) - b^3*c*d^2*integrate(1/32*x*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) - 12*b^3*c*d^2*integrate(1/32*
x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 3*b^3*c*d^2*integrate(1/32*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + a^3*d^
2*x + 3*b^3*d^2*integrate(1/32*arctan(c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 3/2*(2*c*x*arctan(c*x) - log
(c^2*x^2 + 1))*a^2*b*d^2/c - 1/192*(8*b^3*c^2*d^2*x^3 - 24*I*b^3*c*d^2*x^2 - 24*b^3*d^2*x)*arctan(c*x)^3 - 1/1
92*(12*I*b^3*c^2*d^2*x^3 + 36*b^3*c*d^2*x^2 - 36*I*b^3*d^2*x)*arctan(c*x)^2*log(c^2*x^2 + 1) + 1/192*(6*b^3*c^
2*d^2*x^3 - 18*I*b^3*c*d^2*x^2 - 18*b^3*d^2*x)*arctan(c*x)*log(c^2*x^2 + 1)^2 - 1/192*(-I*b^3*c^2*d^2*x^3 - 3*
b^3*c*d^2*x^2 + 3*I*b^3*d^2*x)*log(c^2*x^2 + 1)^3 - I*integrate(-1/64*(112*(b^3*c^3*d^2*x^3 + b^3*c*d^2*x)*arc
tan(c*x)^3 - (b^3*c^4*d^2*x^4 - b^3*d^2)*log(c^2*x^2 + 1)^3 + 8*(b^3*c^4*d^2*x^4 + 48*a*b^2*c^3*d^2*x^3 - 6*b^
3*c^2*d^2*x^2 + 48*a*b^2*c*d^2*x)*arctan(c*x)^2 - 2*(b^3*c^4*d^2*x^4 - 6*b^3*c^2*d^2*x^2 - 6*(b^3*c^3*d^2*x^3
+ b^3*c*d^2*x)*arctan(c*x))*log(c^2*x^2 + 1)^2 + 4*(3*(b^3*c^4*d^2*x^4 - b^3*d^2)*arctan(c*x)^2 + 2*(4*b^3*c^3
*d^2*x^3 - 3*b^3*c*d^2*x)*arctan(c*x))*log(c^2*x^2 + 1))/(c^2*x^2 + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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