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

Optimal. Leaf size=88 \[ \frac {i b \text {Li}_2\left (\frac {2}{i c x+1}-1\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {\log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}+\frac {b^2 \text {Li}_3\left (\frac {2}{i c x+1}-1\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {4868, 4884, 4994, 6610} \[ \frac {i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x}\right )}{2 d}+\frac {\log \left (2-\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4868

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

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Mathematica [A]  time = 0.12, size = 113, normalized size = 1.28 \[ \frac {2 i b \text {Li}_2\left (\frac {c x+i}{i-c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 \left (\log \left (\frac {2 i}{-c x+i}\right )+2 \tanh ^{-1}\left (\frac {c x+i}{c x-i}\right )\right ) \left (a+b \tan ^{-1}(c x)\right )^2+b^2 \text {Li}_3\left (\frac {c x+i}{i-c x}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {i \, b^{2} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 4 \, a b \log \left (-\frac {c x + i}{c x - i}\right ) - 4 i \, a^{2}}{4 \, c d x^{2} - 4 i \, d x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [C]  time = 0.50, size = 1741, normalized size = 19.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a^{2} {\left (\frac {\log \left (i \, c x + 1\right )}{d} - \frac {\log \relax (x)}{d}\right )} + \frac {-24 i \, b^{2} \arctan \left (c x\right )^{3} + 12 \, b^{2} \arctan \left (c x\right )^{2} \log \left (c^{2} x^{2} + 1\right ) - 6 i \, b^{2} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} + 3 \, b^{2} \log \left (c^{2} x^{2} + 1\right )^{3} - {\left (48 \, b^{2} c^{2} \int \frac {x^{2} \arctan \left (c x\right )^{2}}{c^{2} d x^{3} + d x}\,{d x} + \frac {2 \, b^{2} \log \left (c^{2} x^{2} + 1\right )^{3}}{d} + {\left (\frac {\log \left (c^{2} x^{2} + 1\right )^{3}}{d} - \frac {3 \, {\left (\log \left (c^{2} x^{2} + 1\right )^{2} \log \left (-c^{2} x^{2}\right ) + 2 \, {\rm Li}_2\left (c^{2} x^{2} + 1\right ) \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\rm Li}_{3}(c^{2} x^{2} + 1)\right )}}{d}\right )} b^{2} - 72 \, b^{2} \int \frac {\arctan \left (c x\right )^{2}}{c^{2} d x^{3} + d x}\,{d x} - 192 \, a b \int \frac {\arctan \left (c x\right )}{c^{2} d x^{3} + d x}\,{d x} - \frac {12 \, {\left (2 \, c^{2} d \mathit {sage}_{0} x - \arctan \left (c x\right )^{2} \log \left (c^{2} x^{2} + 1\right )\right )} b^{2}}{d}\right )} d - 2 i \, {\left (\frac {4 \, b^{2} \arctan \left (c x\right )^{3}}{d} - 3 \, b^{2} c \int \frac {x \log \left (c^{2} x^{2} + 1\right )^{2}}{c^{2} d x^{3} + d x}\,{d x} + \frac {48 \, a b \arctan \left (c x\right )^{2}}{d} + 12 \, b^{2} \int \frac {\arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{c^{2} d x^{3} + d x}\,{d x}\right )} d}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*(log(I*c*x + 1)/d - log(x)/d) + 1/96*(-24*I*b^2*arctan(c*x)^3 + 12*b^2*arctan(c*x)^2*log(c^2*x^2 + 1) - 6
*I*b^2*arctan(c*x)*log(c^2*x^2 + 1)^2 + 3*b^2*log(c^2*x^2 + 1)^3 - 2*(384*b^2*c^2*integrate(1/16*x^2*arctan(c*
x)^2/(c^2*d*x^3 + d*x), x) + 192*b^2*c*integrate(1/16*x*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^3 + d*x), x) + b
^2*log(c^2*x^2 + 1)^3/d - 576*b^2*integrate(1/16*arctan(c*x)^2/(c^2*d*x^3 + d*x), x) - 48*b^2*integrate(1/16*l
og(c^2*x^2 + 1)^2/(c^2*d*x^3 + d*x), x) - 1536*a*b*integrate(1/16*arctan(c*x)/(c^2*d*x^3 + d*x), x))*d - 8*I*(
b^2*arctan(c*x)^3/d - 12*b^2*c*integrate(1/16*x*log(c^2*x^2 + 1)^2/(c^2*d*x^3 + d*x), x) + 12*a*b*arctan(c*x)^
2/d + 48*b^2*integrate(1/16*arctan(c*x)*log(c^2*x^2 + 1)/(c^2*d*x^3 + d*x), x))*d)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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