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

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

[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)

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Rubi [A]  time = 0.133452, antiderivative size = 98, normalized size of antiderivative = 1., 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 = {4854, 4884, 4994, 6610} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c d}+\frac{i b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c d}+\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c d} \]

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 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 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}{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*}

Mathematica [A]  time = 0.0337886, size = 95, normalized size = 0.97 \[ \frac{i \left (2 i b \text{PolyLog}\left (2,\frac{c x+i}{c x-i}\right ) \left (a+b \tan ^{-1}(c x)\right )+b^2 \text{PolyLog}\left (3,\frac{c x+i}{c x-i}\right )+2 \log \left (\frac{2 d}{d+i c d x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2\right )}{2 c d} \]

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]  time = 0.305, size = 1062, normalized size = 10.8 \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/(d+I*c*d*x),x)

[Out]

-I/c*b^2/d*ln(1+I*c*x)*arctan(c*x)^2+1/c*a^2/d*arctan(c*x)-1/2*I/c*a^2/d*ln(c^2*x^2+1)+I/c*b^2/d*arctan(c*x)^2
*ln(2*I*(1+I*c*x)^2/(c^2*x^2+1))-1/2/c*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*ar
ctan(c*x)^2-1/2/c*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/c*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/c*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/c*b^2/d*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))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1
))^2*Pi+1/2/c*b^2/d*arctan(c*x)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*Pi-1/2/c*b^2/d
*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))*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*
c*x)^2/(c^2*x^2+1)+1))*Pi-1/2/c*b^2/d*arctan(c*x)^2*csgn(I*(1+I*c*x)^2/(c^2*x^2+1)/((1+I*c*x)^2/(c^2*x^2+1)+1)
)^2*Pi+1/2/c*b^2/d*Pi*arctan(c*x)^2+1/c*b^2/d*arctan(c*x)*polylog(2,-(1+I*c*x)^2/(c^2*x^2+1))+1/2*I/c*b^2/d*po
lylog(3,-(1+I*c*x)^2/(c^2*x^2+1))+2/3/c*b^2/d*arctan(c*x)^3-2*I/c*a*b/d*ln(1+I*c*x)*arctan(c*x)-1/c*a*b/d*ln(1
/2-1/2*I*c*x)*ln(1+I*c*x)+1/c*a*b/d*ln(1/2-1/2*I*c*x)*ln(1/2*I*c*x+1/2)+1/c*a*b/d*dilog(1/2*I*c*x+1/2)+1/2/c*a
*b/d*ln(1+I*c*x)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{i \, a^{2} \log \left (i \, c d x + d\right )}{c d} + \frac{24 \, b^{2} \arctan \left (c x\right )^{3} + 6 \, b^{2} \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )^{2} - 2 \,{\left (12 \, b^{2} c \int \frac{x \arctan \left (c x\right ) \log \left (c^{2} x^{2} + 1\right )}{c^{2} d x^{2} + d}\,{d x} - \frac{4 \, b^{2} \arctan \left (c x\right )^{3}}{c d} + 3 \, b^{2} \int \frac{\log \left (c^{2} x^{2} + 1\right )^{2}}{c^{2} d x^{2} + d}\,{d x} - \frac{48 \, a b \arctan \left (c x\right )^{2}}{c d}\right )} c d - 6 i \, c d \int \frac{16 \,{\left (b^{2} c x \arctan \left (c x\right )^{2} + 2 \, a b c x \arctan \left (c x\right )\right )}}{c^{2} d x^{2} + d}\,{d x}}{96 \, c d} \end{align*}

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., 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 \, c d x - 4 i \, d}, x\right ) \end{align*}

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((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 - 4*I*d), 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/(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}}{i \, c d x + d}\,{d x} \end{align*}

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]

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