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

Optimal. Leaf size=88 \[ \frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{\log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d} \]

[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)

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Rubi [A]  time = 0.160682, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, 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*}

Mathematica [A]  time = 0.113741, size = 113, normalized size = 1.28 \[ \frac{2 i b \text{PolyLog}\left (2,\frac{c x+i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+b^2 \text{PolyLog}\left (3,\frac{c x+i}{-c x+i}\right )+2 \left (\log \left (\frac{2 i}{-c x+i}\right )+2 \tanh ^{-1}\left (\frac{c x+i}{c x-i}\right )\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d} \]

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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Maple [C]  time = 0.349, size = 1741, normalized size = 19.8 \begin{align*} \text{result too large to display} \end{align*}

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]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -a^{2}{\left (\frac{\log \left (i \, c x + 1\right )}{d} - \frac{\log \left (x\right )}{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} \int \frac{x \arctan \left (c x\right )^{2}}{c^{2} x^{2} + 1}\,{d 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} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 \,{\left (c d x^{2} - i \, d x\right )}}, x\right ) \end{align*}

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

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )} x}\,{d x} \end{align*}

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]

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