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

Optimal. Leaf size=220 \[ -\frac{3 i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{3 b^3 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 c}-\frac{3 i b^2 d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac{3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac{3 b d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.339436, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4864, 4846, 4920, 4854, 2402, 2315, 1586, 4884, 4992, 6610} \[ -\frac{3 i b^2 d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{3 b^3 d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c}+\frac{3 b^3 d \text{PolyLog}\left (3,1-\frac{2}{1-i c x}\right )}{2 c}-\frac{3 i b^2 d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac{3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac{3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac{i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac{3 b d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.467632, size = 367, normalized size = 1.67 \[ \frac{i d \left (-3 b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right ) \left (2 a+2 b \tan ^{-1}(c x)-i b\right )-3 i b^3 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+3 i a^2 b \log \left (c^2 x^2+1\right )+3 a^2 b c^2 x^2 \tan ^{-1}(c x)-3 a^2 b c x+3 a^2 b \tan ^{-1}(c x)-6 i a^2 b c x \tan ^{-1}(c x)+a^3 c^2 x^2-2 i a^3 c x+3 a b^2 \log \left (c^2 x^2+1\right )+3 a b^2 c^2 x^2 \tan ^{-1}(c x)^2-3 a b^2 \tan ^{-1}(c x)^2-6 i a b^2 c x \tan ^{-1}(c x)^2-6 a b^2 c x \tan ^{-1}(c x)-12 i a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+b^3 c^2 x^2 \tan ^{-1}(c x)^3-b^3 \tan ^{-1}(c x)^3-2 i b^3 c x \tan ^{-1}(c x)^3+3 i b^3 \tan ^{-1}(c x)^2-3 b^3 c x \tan ^{-1}(c x)^2-6 i b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 b^3 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.787, size = 7451, normalized size = 33.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

12*b^3*c^3*d*integrate(1/64*x^3*arctan(c*x)^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c^3*d*integrate(1/64*x^
3*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) + 12*b^3*c^3*d*integrate(1/64*x^3*arctan(c*x)^2/(c^2*x^2 + 1), x) - 3*b
^3*c^3*d*integrate(1/64*x^3*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 1/2*I*a^3*c*d*x^2 + 7/32*b^3*d*arctan(c*x)^
4/c + 56*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)^3/(c^2*x^2 + 1), x) + 6*b^3*c^2*d*integrate(1/64*x^2*arctan(
c*x)*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + 192*a*b^2*c^2*d*integrate(1/64*x^2*arctan(c*x)^2/(c^2*x^2 + 1), x)
+ 36*b^3*c^2*d*integrate(1/64*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 3/2*I*(x^2*arctan(c*x) - c
*(x/c^2 - arctan(c*x)/c^3))*a^2*b*c*d + a*b^2*d*arctan(c*x)^3/c + 12*b^3*c*d*integrate(1/64*x*arctan(c*x)^2*lo
g(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - b^3*c*d*integrate(1/64*x*log(c^2*x^2 + 1)^3/(c^2*x^2 + 1), x) - 24*b^3*c*d*
integrate(1/64*x*arctan(c*x)^2/(c^2*x^2 + 1), x) + 6*b^3*c*d*integrate(1/64*x*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1)
, x) + a^3*d*x + 6*b^3*d*integrate(1/64*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/c - 1/128*(-8*I*b^3*c*d*x^2 - 16*b^3*d*x)*arctan(c*x)^3 - 3/32*(b^3*c*d*x^2 -
2*I*b^3*d*x)*arctan(c*x)^2*log(c^2*x^2 + 1) - 1/128*(6*I*b^3*c*d*x^2 + 12*b^3*d*x)*arctan(c*x)*log(c^2*x^2 + 1
)^2 + 1/128*(b^3*c*d*x^2 - 2*I*b^3*d*x)*log(c^2*x^2 + 1)^3 + I*integrate(1/64*(56*(b^3*c^3*d*x^3 + b^3*c*d*x)*
arctan(c*x)^3 + (b^3*c^2*d*x^2 + b^3*d)*log(c^2*x^2 + 1)^3 + 12*(16*a*b^2*c^3*d*x^3 - 3*b^3*c^2*d*x^2 + 16*a*b
^2*c*d*x)*arctan(c*x)^2 + 3*(3*b^3*c^2*d*x^2 + 2*(b^3*c^3*d*x^3 + b^3*c*d*x)*arctan(c*x))*log(c^2*x^2 + 1)^2 -
12*((b^3*c^2*d*x^2 + b^3*d)*arctan(c*x)^2 - (b^3*c^3*d*x^3 - 2*b^3*c*d*x)*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}{16} \,{\left (b^{3} c d x^{2} - 2 i \, b^{3} d x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{3} +{\rm integral}\left (\frac{8 i \, a^{3} c^{3} d x^{3} + 8 \, a^{3} c^{2} d x^{2} + 8 i \, a^{3} c d x + 8 \, a^{3} d +{\left (-6 i \, a b^{2} c^{3} d x^{3} - 3 \,{\left (2 \, a b^{2} - i \, b^{3}\right )} c^{2} d x^{2} - 6 \, a b^{2} d +{\left (-6 i \, a b^{2} + 6 \, b^{3}\right )} c d x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} -{\left (12 \, a^{2} b c^{3} d x^{3} - 12 i \, a^{2} b c^{2} d x^{2} + 12 \, a^{2} b c d x - 12 i \, a^{2} b d\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{8 \,{\left (c^{2} x^{2} + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(b^3*c*d*x^2 - 2*I*b^3*d*x)*log(-(c*x + I)/(c*x - I))^3 + integral(1/8*(8*I*a^3*c^3*d*x^3 + 8*a^3*c^2*d*x
^2 + 8*I*a^3*c*d*x + 8*a^3*d + (-6*I*a*b^2*c^3*d*x^3 - 3*(2*a*b^2 - I*b^3)*c^2*d*x^2 - 6*a*b^2*d + (-6*I*a*b^2
+ 6*b^3)*c*d*x)*log(-(c*x + I)/(c*x - I))^2 - (12*a^2*b*c^3*d*x^3 - 12*I*a^2*b*c^2*d*x^2 + 12*a^2*b*c*d*x - 1
2*I*a^2*b*d)*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((d+I*c*d*x)*(a+b*atan(c*x))**3,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 )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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