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

Optimal. Leaf size=192 \[ -\frac{4 i b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{3 c}+\frac{1}{3} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c}+\frac{8 b d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c}-2 i a b d^2 x+\frac{i b^2 d^2 \log \left (c^2 x^2+1\right )}{c}+\frac{b^2 d^2 \tan ^{-1}(c x)}{3 c}-2 i b^2 d^2 x \tan ^{-1}(c x)-\frac{1}{3} b^2 d^2 x \]

[Out]

(-2*I)*a*b*d^2*x - (b^2*d^2*x)/3 + (b^2*d^2*ArcTan[c*x])/(3*c) - (2*I)*b^2*d^2*x*ArcTan[c*x] + (b*c*d^2*x^2*(a
+ b*ArcTan[c*x]))/3 - ((I/3)*d^2*(1 + I*c*x)^3*(a + b*ArcTan[c*x])^2)/c + (8*b*d^2*(a + b*ArcTan[c*x])*Log[2/
(1 - I*c*x)])/(3*c) + (I*b^2*d^2*Log[1 + c^2*x^2])/c - (((4*I)/3)*b^2*d^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/c

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Rubi [A]  time = 0.196765, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4864, 4846, 260, 4852, 321, 203, 1586, 4854, 2402, 2315} \[ -\frac{4 i b^2 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{3 c}+\frac{1}{3} b c d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{i d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 c}+\frac{8 b d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{3 c}-2 i a b d^2 x+\frac{i b^2 d^2 \log \left (c^2 x^2+1\right )}{c}+\frac{b^2 d^2 \tan ^{-1}(c x)}{3 c}-2 i b^2 d^2 x \tan ^{-1}(c x)-\frac{1}{3} b^2 d^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*I)*a*b*d^2*x - (b^2*d^2*x)/3 + (b^2*d^2*ArcTan[c*x])/(3*c) - (2*I)*b^2*d^2*x*ArcTan[c*x] + (b*c*d^2*x^2*(a
+ b*ArcTan[c*x]))/3 - ((I/3)*d^2*(1 + I*c*x)^3*(a + b*ArcTan[c*x])^2)/c + (8*b*d^2*(a + b*ArcTan[c*x])*Log[2/
(1 - I*c*x)])/(3*c) + (I*b^2*d^2*Log[1 + c^2*x^2])/c - (((4*I)/3)*b^2*d^2*PolyLog[2, 1 - 2/(1 - I*c*x)])/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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.636959, size = 205, normalized size = 1.07 \[ -\frac{d^2 \left (4 i b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+a^2 c^3 x^3-3 i a^2 c^2 x^2-3 a^2 c x-a b c^2 x^2+4 a b \log \left (c^2 x^2+1\right )-b \tan ^{-1}(c x) \left (a \left (-2 c^3 x^3+6 i c^2 x^2+6 c x+6 i\right )+b \left (c^2 x^2-6 i c x+1\right )+8 b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+6 i a b c x-3 i b^2 \log \left (c^2 x^2+1\right )+b^2 c x+b^2 (c x-i)^3 \tan ^{-1}(c x)^2\right )}{3 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(d^2*(-3*a^2*c*x + (6*I)*a*b*c*x + b^2*c*x - (3*I)*a^2*c^2*x^2 - a*b*c^2*x^2 + a^2*c^3*x^3 + b^2*(-I + c*x)^3
*ArcTan[c*x]^2 - b*ArcTan[c*x]*(b*(1 - (6*I)*c*x + c^2*x^2) + a*(6*I + 6*c*x + (6*I)*c^2*x^2 - 2*c^3*x^3) + 8*
b*Log[1 + E^((2*I)*ArcTan[c*x])]) + 4*a*b*Log[1 + c^2*x^2] - (3*I)*b^2*Log[1 + c^2*x^2] + (4*I)*b^2*PolyLog[2,
-E^((2*I)*ArcTan[c*x])]))/(3*c)

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Maple [B]  time = 0.089, size = 523, normalized size = 2.7 \begin{align*}{\frac{{d}^{2}{b}^{2}\arctan \left ( cx \right ) }{3\,c}}-{\frac{{\frac{i}{3}}{d}^{2}{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{c}}+{\frac{c{d}^{2}ab{x}^{2}}{3}}-{\frac{4\,{d}^{2}ab\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,c}}-{\frac{4\,{d}^{2}{b}^{2}\arctan \left ( cx \right ) \ln \left ({c}^{2}{x}^{2}+1 \right ) }{3\,c}}+{\frac{i{d}^{2}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{c}}+ic{x}^{2}{a}^{2}{d}^{2}-{\frac{{c}^{2}{d}^{2}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{3}}{3}}+2\,{d}^{2}ab\arctan \left ( cx \right ) x-{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}{\it dilog} \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}+{\frac{{\frac{i}{3}}{d}^{2}{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{c}}+{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}+2\,ic{d}^{2}ab\arctan \left ( cx \right ){x}^{2}-{\frac{{c}^{2}{x}^{3}{a}^{2}{d}^{2}}{3}}+{d}^{2}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}x-{\frac{{\frac{i}{3}}{d}^{2}{a}^{2}}{c}}+{\frac{c{d}^{2}{b}^{2}\arctan \left ( cx \right ){x}^{2}}{3}}-{\frac{{b}^{2}{d}^{2}x}{3}}+ic{d}^{2}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}{x}^{2}-{\frac{2\,{c}^{2}{d}^{2}ab\arctan \left ( cx \right ){x}^{3}}{3}}+{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c}}-{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}\ln \left ( cx+i \right ) \ln \left ({\frac{i}{2}} \left ( cx-i \right ) \right ) }{c}}-{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx-i \right ) }{c}}+{\frac{{\frac{2\,i}{3}}{d}^{2}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) \ln \left ( cx+i \right ) }{c}}+{\frac{2\,i{d}^{2}ab\arctan \left ( cx \right ) }{c}}+x{a}^{2}{d}^{2}-2\,iab{d}^{2}x-2\,i{b}^{2}{d}^{2}x\arctan \left ( cx \right ) +{\frac{i{d}^{2}{b}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b^2*d^2*arctan(c*x)/c-1/3*I/c*d^2*b^2*ln(c*x+I)^2+1/3*c*d^2*a*b*x^2-4/3/c*d^2*a*b*ln(c^2*x^2+1)-4/3/c*d^2*
b^2*arctan(c*x)*ln(c^2*x^2+1)+I/c*d^2*b^2*arctan(c*x)^2+I*c*x^2*a^2*d^2-1/3*c^2*d^2*b^2*arctan(c*x)^2*x^3+2*d^
2*a*b*arctan(c*x)*x-2/3*I/c*d^2*b^2*dilog(1/2*I*(c*x-I))+1/3*I/c*d^2*b^2*ln(c*x-I)^2+2/3*I/c*d^2*b^2*dilog(-1/
2*I*(c*x+I))+2*I*c*d^2*a*b*arctan(c*x)*x^2-1/3*c^2*x^3*a^2*d^2+d^2*b^2*arctan(c*x)^2*x-1/3*I/c*d^2*a^2+1/3*c*d
^2*b^2*arctan(c*x)*x^2-1/3*b^2*d^2*x+I*c*d^2*b^2*arctan(c*x)^2*x^2-2/3*c^2*d^2*a*b*arctan(c*x)*x^3+2/3*I/c*d^2
*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-2/3*I/c*d^2*b^2*ln(c*x+I)*ln(1/2*I*(c*x-I))-2/3*I/c*d^2*b^2*ln(c^2*x^2+1)*ln
(c*x-I)+2/3*I/c*d^2*b^2*ln(c^2*x^2+1)*ln(c*x+I)+2*I/c*d^2*a*b*arctan(c*x)+x*a^2*d^2-2*I*a*b*d^2*x-2*I*b^2*d^2*
x*arctan(c*x)+I*b^2*d^2*ln(c^2*x^2+1)/c

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

[Out]

-1/3*a^2*c^2*d^2*x^3 - 36*b^2*c^4*d^2*integrate(1/48*x^4*arctan(c*x)^2/(c^2*x^2 + 1), x) - 3*b^2*c^4*d^2*integ
rate(1/48*x^4*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) - 4*b^2*c^4*d^2*integrate(1/48*x^4*log(c^2*x^2 + 1)/(c^2*x^
2 + 1), x) + 24*b^2*c^3*d^2*integrate(1/48*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 32*b^2*c^3*d^2
*integrate(1/48*x^3*arctan(c*x)/(c^2*x^2 + 1), x) - 1/3*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4
))*a*b*c^2*d^2 + I*a^2*c*d^2*x^2 + 24*b^2*c^2*d^2*integrate(1/48*x^2*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) + 2*I*
(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*a*b*c*d^2 + 1/4*b^2*d^2*arctan(c*x)^3/c + 24*b^2*c*d^2*integra
te(1/48*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^2 + 1), x) - 24*b^2*c*d^2*integrate(1/48*x*arctan(c*x)/(c^2*x^2
+ 1), x) + a^2*d^2*x + 3*b^2*d^2*integrate(1/48*log(c^2*x^2 + 1)^2/(c^2*x^2 + 1), x) + (2*c*x*arctan(c*x) - lo
g(c^2*x^2 + 1))*a*b*d^2/c - 1/48*(4*b^2*c^2*d^2*x^3 - 12*I*b^2*c*d^2*x^2 - 12*b^2*d^2*x)*arctan(c*x)^2 - 1/48*
(4*I*b^2*c^2*d^2*x^3 + 12*b^2*c*d^2*x^2 - 12*I*b^2*d^2*x)*arctan(c*x)*log(c^2*x^2 + 1) + 1/48*(b^2*c^2*d^2*x^3
- 3*I*b^2*c*d^2*x^2 - 3*b^2*d^2*x)*log(c^2*x^2 + 1)^2 + I*integrate(1/24*(36*(b^2*c^3*d^2*x^3 + b^2*c*d^2*x)*
arctan(c*x)^2 + 3*(b^2*c^3*d^2*x^3 + b^2*c*d^2*x)*log(c^2*x^2 + 1)^2 + 4*(b^2*c^4*d^2*x^4 - 6*b^2*c^2*d^2*x^2)
*arctan(c*x) + 2*(4*b^2*c^3*d^2*x^3 - 3*b^2*c*d^2*x + 3*(b^2*c^4*d^2*x^4 - b^2*d^2)*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}{12} \,{\left (b^{2} c^{2} d^{2} x^{3} - 3 i \, b^{2} c d^{2} x^{2} - 3 \, b^{2} d^{2} x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} +{\rm integral}\left (-\frac{3 \, a^{2} c^{4} d^{2} x^{4} - 6 i \, a^{2} c^{3} d^{2} x^{3} - 6 i \, a^{2} c d^{2} x - 3 \, a^{2} d^{2} -{\left (-3 i \, a b c^{4} d^{2} x^{4} -{\left (6 \, a b - i \, b^{2}\right )} c^{3} d^{2} x^{3} + 3 \, b^{2} c^{2} d^{2} x^{2} - 3 \,{\left (2 \, a b + i \, b^{2}\right )} c d^{2} x + 3 i \, a b d^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{3 \,{\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)^2*(a+b*arctan(c*x))^2,x, algorithm="fricas")

[Out]

1/12*(b^2*c^2*d^2*x^3 - 3*I*b^2*c*d^2*x^2 - 3*b^2*d^2*x)*log(-(c*x + I)/(c*x - I))^2 + integral(-1/3*(3*a^2*c^
4*d^2*x^4 - 6*I*a^2*c^3*d^2*x^3 - 6*I*a^2*c*d^2*x - 3*a^2*d^2 - (-3*I*a*b*c^4*d^2*x^4 - (6*a*b - I*b^2)*c^3*d^
2*x^3 + 3*b^2*c^2*d^2*x^2 - 3*(2*a*b + I*b^2)*c*d^2*x + 3*I*a*b*d^2)*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)**2*(a+b*atan(c*x))**2,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 )}^{2}{\left (b \arctan \left (c x\right ) + a\right )}^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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