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

Optimal. Leaf size=216 \[ -i b d \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 i b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2+b^2 (-d) \text {Li}_2\left (1-\frac {2}{i c x+1}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{i c x+1}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (\frac {2}{i c x+1}-1\right ) \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4850, 4988, 4884, 4994, 6610} \[ -i b d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+b^2 (-d) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 i b d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4850

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTan[c*x])^p*ArcTanh[1 - 2/(1 +
I*c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTan[c*x])^(p - 1)*ArcTanh[1 - 2/(1 + I*c*x)])/(1 + c^2*x^2), x], x
] /; FreeQ[{a, b, c}, x] && IGtQ[p, 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 4876

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 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 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 4988

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[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 \frac {(d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (i c d \left (a+b \tan ^{-1}(c x)\right )^2+\frac {d \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+(i c d) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )-(4 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b c^2 d\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+(2 i b c d) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx+(2 b c d) \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-(2 b c d) \int \frac {\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )+\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b^2 c d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )-\left (2 b^2 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )-b^2 d \text {Li}_2\left (1-\frac {2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x}\right )-\frac {1}{2} b^2 d \text {Li}_3\left (1-\frac {2}{1+i c x}\right )+\frac {1}{2} b^2 d \text {Li}_3\left (-1+\frac {2}{1+i c x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 272, normalized size = 1.26 \[ d \left (i a^2 c x+a^2 \log (c x)+i a b \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )+i a b (\text {Li}_2(-i c x)-\text {Li}_2(i c x))+b^2 \left (\text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left ((1+i c x) \tan ^{-1}(c x)+2 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+b^2 \left (i \tan ^{-1}(c x) \text {Li}_2\left (e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {1}{2} \text {Li}_3\left (e^{-2 i \tan ^{-1}(c x)}\right )-\frac {1}{2} \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )+\frac {2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac {i \pi ^3}{24}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

d*(I*a^2*c*x + a^2*Log[c*x] + I*a*b*(2*c*x*ArcTan[c*x] - Log[1 + c^2*x^2]) + b^2*(ArcTan[c*x]*((1 + I*c*x)*Arc
Tan[c*x] + (2*I)*Log[1 + E^((2*I)*ArcTan[c*x])]) + PolyLog[2, -E^((2*I)*ArcTan[c*x])]) + I*a*b*(PolyLog[2, (-I
)*c*x] - PolyLog[2, I*c*x]) + b^2*((-1/24*I)*Pi^3 + ((2*I)/3)*ArcTan[c*x]^3 + ArcTan[c*x]^2*Log[1 - E^((-2*I)*
ArcTan[c*x])] - ArcTan[c*x]^2*Log[1 + E^((2*I)*ArcTan[c*x])] + I*ArcTan[c*x]*PolyLog[2, E^((-2*I)*ArcTan[c*x])
] + I*ArcTan[c*x]*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + PolyLog[3, E^((-2*I)*ArcTan[c*x])]/2 - PolyLog[3, -E^((
2*I)*ArcTan[c*x])]/2))

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 i \, a^{2} c d x + 4 \, a^{2} d + {\left (-i \, b^{2} c d x - b^{2} d\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (4 \, a b c d x - 4 i \, a b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/4*(4*I*a^2*c*d*x + 4*a^2*d + (-I*b^2*c*d*x - b^2*d)*log(-(c*x + I)/(c*x - I))^2 - (4*a*b*c*d*x - 4*
I*a*b*d)*log(-(c*x + I)/(c*x - I)))/x, x)

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giac [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)*(a+b*arctan(c*x))^2/x,x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.79, size = 7034, normalized size = 32.56 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} i \, b^{2} c d x \arctan \left (c x\right )^{2} + 12 i \, b^{2} c^{3} d \int \frac {x^{3} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 4 \, b^{2} c^{3} d \int \frac {x^{3} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + i \, b^{2} c^{3} d \int \frac {x^{3} \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 8 \, b^{2} c^{3} d \int \frac {x^{3} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 4 i \, b^{2} c^{3} d \int \frac {x^{3} \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} - \frac {1}{4} \, b^{2} c d x \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right ) - \frac {1}{16} i \, b^{2} c d x \log \left (c^{2} x^{2} + 1\right )^{2} + \frac {1}{4} i \, b^{2} d \arctan \left (c x\right )^{3} + 12 \, b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} - 4 i \, b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b c^{2} d \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} - 8 i \, b^{2} c^{2} d \int \frac {x^{2} \arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {1}{96} \, b^{2} d \log \left (c^{2} x^{2} + 1\right )^{3} + i \, a^{2} c d x + 4 \, b^{2} c d \int \frac {x \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + i \, b^{2} c d \int \frac {x \log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + \frac {1}{16} \, b^{2} d \log \left (c^{2} x^{2} + 1\right )^{2} + i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} a b d + 12 \, b^{2} d \int \frac {\arctan \left (c x\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} - 4 i \, b^{2} d \int \frac {\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + b^{2} d \int \frac {\log \left (c^{2} x^{2} + 1\right )^{2}}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + 32 \, a b d \int \frac {\arctan \left (c x\right )}{16 \, {\left (c^{2} x^{3} + x\right )}}\,{d x} + a^{2} d \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*b^2*c*d*x*arctan(c*x)^2 + 12*I*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)^2/(c^2*x^3 + x), x) + 4*b^2*c^3*
d*integrate(1/16*x^3*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + I*b^2*c^3*d*integrate(1/16*x^3*log(c^2*x
^2 + 1)^2/(c^2*x^3 + x), x) + 8*b^2*c^3*d*integrate(1/16*x^3*arctan(c*x)/(c^2*x^3 + x), x) + 4*I*b^2*c^3*d*int
egrate(1/16*x^3*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) - 1/4*b^2*c*d*x*arctan(c*x)*log(c^2*x^2 + 1) - 1/16*I*b^2*c
*d*x*log(c^2*x^2 + 1)^2 + 1/4*I*b^2*d*arctan(c*x)^3 + 12*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)^2/(c^2*x^3 +
x), x) - 4*I*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + 32*a*b*c^2*d*integ
rate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) - 8*I*b^2*c^2*d*integrate(1/16*x^2*arctan(c*x)/(c^2*x^3 + x), x) +
1/96*b^2*d*log(c^2*x^2 + 1)^3 + I*a^2*c*d*x + 4*b^2*c*d*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^
3 + x), x) + I*b^2*c*d*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*x^3 + x), x) + 1/16*b^2*d*log(c^2*x^2 + 1)^2 +
I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*a*b*d + 12*b^2*d*integrate(1/16*arctan(c*x)^2/(c^2*x^3 + x), x) - 4*
I*b^2*d*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*x^3 + x), x) + b^2*d*integrate(1/16*log(c^2*x^2 + 1)^
2/(c^2*x^3 + x), x) + 32*a*b*d*integrate(1/16*arctan(c*x)/(c^2*x^3 + x), x) + a^2*d*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i d \left (\int a^{2} c\, dx + \int \left (- \frac {i a^{2}}{x}\right )\, dx + \int b^{2} c \operatorname {atan}^{2}{\left (c x \right )}\, dx + \int \left (- \frac {i b^{2} \operatorname {atan}^{2}{\left (c x \right )}}{x}\right )\, dx + \int 2 a b c \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- \frac {2 i a b \operatorname {atan}{\left (c x \right )}}{x}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*d*(Integral(a**2*c, x) + Integral(-I*a**2/x, x) + Integral(b**2*c*atan(c*x)**2, x) + Integral(-I*b**2*atan(c
*x)**2/x, x) + Integral(2*a*b*c*atan(c*x), x) + Integral(-2*I*a*b*atan(c*x)/x, x))

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