3.1.98 \(\int \frac {(a+b \text {ArcTan}(c x))^2}{d+i c d x} \, dx\) [98]

Optimal. Leaf size=98 \[ \frac {i (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c d}+\frac {i b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x}\right )}{2 c d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4964, 5004, 5114, 6745} \begin {gather*} -\frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right ) (a+b \text {ArcTan}(c x))}{c d}+\frac {i \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))^2}{c d}+\frac {i b^2 \text {Li}_3\left (1-\frac {2}{i c x+1}\right )}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 5004

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 5114

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*(a + b*Ar
cTan[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 6745

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

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Mathematica [A]
time = 0.02, size = 95, normalized size = 0.97 \begin {gather*} \frac {i \left (2 (a+b \text {ArcTan}(c x))^2 \log \left (\frac {2 d}{d+i c d x}\right )+2 i b (a+b \text {ArcTan}(c x)) \text {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )+b^2 \text {PolyLog}\left (3,\frac {i+c x}{-i+c x}\right )\right )}{2 c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.50, size = 1003, normalized size = 10.23 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-2*I*b*a/d*ln(1+I*c*x)*arctan(c*x)+a^2/d*arctan(c*x)+1/2*I*b^2/d*polylog(3,-(1+I*c*x)^2/(c^2*x^2+1))-I*b^
2/d*ln(1+I*c*x)*arctan(c*x)^2-1/2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(
c*x)^2+1/2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((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)^2-1/2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1)/((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))*arctan(c*x)^2+1/2*b^2/d*Pi*csgn((1+I*c*x)^2/(c^2*x^2+1))*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^
2/(c^2*x^2+1)+1))*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*arctan(c*x)^2+1/2*b^2/d*arctan(c*x)^2*Pi*csgn((1+I*c*x)^
2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+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+1/2*b
^2/d*arctan(c*x)^2*Pi*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3-1/2*b^2/d*arctan(c*x)^2*Pi
*csgn((1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^
2+1)+1))-1/2*b^2/d*arctan(c*x)^2*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+1/2*b^2/d*Pi
*arctan(c*x)^2+b^2/d*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))-1/2*I*a^2/d*ln(c^2*x^2+1)+2/3*b^2/d*arcta
n(c*x)^3+I*b^2/d*arctan(c*x)^2*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-b*a/d*ln(1/2-1/2*I*c*x)*ln(1+I*c*x)+b*a/d*ln(1/
2-1/2*I*c*x)*ln(1/2+1/2*I*c*x)+b*a/d*dilog(1/2+1/2*I*c*x)+1/2*b*a/d*ln(1+I*c*x)^2)

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

[Out]

-I*a^2*log(I*c*d*x + d)/(c*d) + 1/96*(24*b^2*arctan(c*x)^3 + 12*I*b^2*arctan(c*x)^2*log(c^2*x^2 + 1) + 6*b^2*a
rctan(c*x)*log(c^2*x^2 + 1)^2 + 3*I*b^2*log(c^2*x^2 + 1)^3 - 8*(48*b^2*c*integrate(1/16*x*arctan(c*x)*log(c^2*
x^2 + 1)/(c^2*d*x^2 + d), x) - b^2*arctan(c*x)^3/(c*d) + 12*b^2*integrate(1/16*log(c^2*x^2 + 1)^2/(c^2*d*x^2 +
d), x) - 12*a*b*arctan(c*x)^2/(c*d))*c*d - 96*I*c*d*integrate(1/16*(20*b^2*c*x*arctan(c*x)^2 + 3*b^2*c*x*log(
c^2*x^2 + 1)^2 + 32*a*b*c*x*arctan(c*x) + 4*b^2*arctan(c*x)*log(c^2*x^2 + 1))/(c^2*d*x^2 + d), x))/(c*d)

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

[Out]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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