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

Optimal. Leaf size=167 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (-c x+i)}+\frac{2 i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2}-\frac{a x}{c^2 d^2}+\frac{b \log \left (c^2 x^2+1\right )}{2 c^3 d^2}-\frac{i b}{2 c^3 d^2 (-c x+i)}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{i b \tan ^{-1}(c x)}{2 c^3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.190293, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4862, 627, 44, 203, 4854, 2402, 2315} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (-c x+i)}+\frac{2 i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^2}-\frac{a x}{c^2 d^2}+\frac{b \log \left (c^2 x^2+1\right )}{2 c^3 d^2}-\frac{i b}{2 c^3 d^2 (-c x+i)}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{i b \tan ^{-1}(c x)}{2 c^3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 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 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]

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )}{(d+i c d x)^2} \, dx &=\int \left (-\frac{a+b \tan ^{-1}(c x)}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^2 d^2 (-i+c x)^2}-\frac{2 i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (-i+c x)}\right ) \, dx\\ &=-\frac{(2 i) \int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c^2 d^2}-\frac{\int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^2 d^2}+\frac{\int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^2 d^2}\\ &=-\frac{a x}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (i-c x)}+\frac{2 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^2}-\frac{(2 i b) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^2}+\frac{b \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^2 d^2}-\frac{b \int \tan ^{-1}(c x) \, dx}{c^2 d^2}\\ &=-\frac{a x}{c^2 d^2}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (i-c x)}+\frac{2 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^2}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^3 d^2}+\frac{b \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c^2 d^2}+\frac{b \int \frac{x}{1+c^2 x^2} \, dx}{c d^2}\\ &=-\frac{a x}{c^2 d^2}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (i-c x)}+\frac{2 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{b \log \left (1+c^2 x^2\right )}{2 c^3 d^2}-\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{b \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^2}\\ &=-\frac{a x}{c^2 d^2}-\frac{i b}{2 c^3 d^2 (i-c x)}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (i-c x)}+\frac{2 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{b \log \left (1+c^2 x^2\right )}{2 c^3 d^2}-\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{(i b) \int \frac{1}{1+c^2 x^2} \, dx}{2 c^2 d^2}\\ &=-\frac{a x}{c^2 d^2}-\frac{i b}{2 c^3 d^2 (i-c x)}+\frac{i b \tan ^{-1}(c x)}{2 c^3 d^2}-\frac{b x \tan ^{-1}(c x)}{c^2 d^2}+\frac{a+b \tan ^{-1}(c x)}{c^3 d^2 (i-c x)}+\frac{2 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^3 d^2}+\frac{b \log \left (1+c^2 x^2\right )}{2 c^3 d^2}-\frac{b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^2}\\ \end{align*}

Mathematica [A]  time = 0.75328, size = 153, normalized size = 0.92 \[ -\frac{b \left (-4 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )-2 \log \left (c^2 x^2+1\right )-8 \tan ^{-1}(c x)^2-i \sin \left (2 \tan ^{-1}(c x)\right )+\cos \left (2 \tan ^{-1}(c x)\right )+2 \tan ^{-1}(c x) \left (2 c x-4 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+\sin \left (2 \tan ^{-1}(c x)\right )+i \cos \left (2 \tan ^{-1}(c x)\right )\right )\right )+4 i a \log \left (c^2 x^2+1\right )+4 a c x+\frac{4 a}{c x-i}-8 a \tan ^{-1}(c x)}{4 c^3 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(4*a*c*x + (4*a)/(-I + c*x) - 8*a*ArcTan[c*x] + (4*I)*a*Log[1 + c^2*x^2] + b*(-8*ArcTan[c*x]^2 + Cos[2*ArcTan
[c*x]] - 2*Log[1 + c^2*x^2] - 4*PolyLog[2, -E^((2*I)*ArcTan[c*x])] - I*Sin[2*ArcTan[c*x]] + 2*ArcTan[c*x]*(2*c
*x + I*Cos[2*ArcTan[c*x]] - (4*I)*Log[1 + E^((2*I)*ArcTan[c*x])] + Sin[2*ArcTan[c*x]])))/(4*c^3*d^2)

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Maple [B]  time = 0.056, size = 316, normalized size = 1.9 \begin{align*} -{\frac{ax}{{c}^{2}{d}^{2}}}+2\,{\frac{a\arctan \left ( cx \right ) }{{c}^{3}{d}^{2}}}-{\frac{ia\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{3}{d}^{2}}}-{\frac{a}{{c}^{3}{d}^{2} \left ( cx-i \right ) }}-{\frac{bx\arctan \left ( cx \right ) }{{c}^{2}{d}^{2}}}-{\frac{{\frac{i}{4}}b}{{c}^{3}{d}^{2}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }-{\frac{b\arctan \left ( cx \right ) }{{c}^{3}{d}^{2} \left ( cx-i \right ) }}-{\frac{b\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{3}{d}^{2}}}-{\frac{b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{3}{d}^{2}}}+{\frac{b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{2\,{c}^{3}{d}^{2}}}+{\frac{b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{16\,{c}^{3}{d}^{2}}}+{\frac{{\frac{i}{2}}b}{{c}^{3}{d}^{2} \left ( cx-i \right ) }}+{\frac{{\frac{i}{8}}b}{{c}^{3}{d}^{2}}\arctan \left ({\frac{cx}{2}} \right ) }-{\frac{{\frac{i}{8}}b}{{c}^{3}{d}^{2}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }-{\frac{2\,ib\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{c}^{3}{d}^{2}}}+{\frac{3\,b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{8\,{c}^{3}{d}^{2}}}+{\frac{{\frac{3\,i}{4}}b\arctan \left ( cx \right ) }{{c}^{3}{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*x/c^2/d^2+2/c^3*a/d^2*arctan(c*x)-I/c^3*a/d^2*ln(c^2*x^2+1)-1/c^3*a/d^2/(c*x-I)-b*x*arctan(c*x)/c^2/d^2-1/4
*I/c^3*b/d^2*arctan(1/2*c*x-1/2*I)-1/c^3*b/d^2*arctan(c*x)/(c*x-I)-1/c^3*b/d^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/
c^3*b/d^2*dilog(-1/2*I*(c*x+I))+1/2/c^3*b/d^2*ln(c*x-I)^2+1/16/c^3*b/d^2*ln(c^4*x^4+10*c^2*x^2+9)+1/2*I/c^3*b/
d^2/(c*x-I)+1/8*I/c^3*b/d^2*arctan(1/2*c*x)-1/8*I/c^3*b/d^2*arctan(1/6*c^3*x^3+7/6*c*x)-2*I/c^3*b/d^2*arctan(c
*x)*ln(c*x-I)+3/8*b*ln(c^2*x^2+1)/c^3/d^2+3/4*I/c^3*b/d^2*arctan(c*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(1/(c^4*d^2*x - I*c^3*d^2) + x/(c^2*d^2) + 2*I*log(c*x - I)/(c^3*d^2)) - 1/8*(-16*I*c^2*x^2 - 8*(4*c*x - 4*
I)*arctan(c*x)^2 - 8*(c*x - I)*log(c^2*x^2 + 1)^2 + 8*(-I*c^4*d^2*x - c^3*d^2)*((c*(x/(c^6*d^2*x^2 + c^4*d^2)
+ arctan(c*x)/(c^5*d^2)) - 2*arctan(c*x)/(c^6*d^2*x^2 + c^4*d^2))*c + 8*integrate(1/4*log(c^2*x^2 + 1)/(c^6*d^
2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x)) + (8*c^4*d^2*x - 8*I*c^3*d^2)*(c*(c^2/(c^8*d^2*x^2 + c^6*d^2) + log(c^2*
x^2 + 1)/(c^6*d^2*x^2 + c^4*d^2)) + 16*integrate(1/4*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x))
- 16*(-I*c^5*d^2*x - c^4*d^2)*(c*(x/(c^6*d^2*x^2 + c^4*d^2) + arctan(c*x)/(c^5*d^2)) - 8*c*integrate(1/4*x^2*l
og(c^2*x^2 + 1)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 2*arctan(c*x)/(c^6*d^2*x^2 + c^4*d^2)) + (16*c^5
*d^2*x - 16*I*c^4*d^2)*(16*c*integrate(1/4*x^2*arctan(c*x)/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - c^2/(
c^8*d^2*x^2 + c^6*d^2) - log(c^2*x^2 + 1)/(c^6*d^2*x^2 + c^4*d^2)) - 16*c*x + 16*(c^2*x^2 - I*c*x + 1)*arctan(
c*x) + 8*(4*c^7*d^2*x - 4*I*c^6*d^2)*integrate(1/4*(2*c*x^4*arctan(c*x) + x^3*log(c^2*x^2 + 1))/(c^6*d^2*x^4 +
2*c^4*d^2*x^2 + c^2*d^2), x) + 32*(-I*c^7*d^2*x - c^6*d^2)*integrate(1/4*(c*x^4*log(c^2*x^2 + 1) - 2*x^3*arct
an(c*x))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 96*(I*c^6*d^2*x + c^5*d^2)*integrate(1/4*(2*c*x^3*arcta
n(c*x) + x^2*log(c^2*x^2 + 1))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) + 8*(12*c^6*d^2*x - 12*I*c^5*d^2)*i
ntegrate(1/4*(c*x^3*log(c^2*x^2 + 1) - 2*x^2*arctan(c*x))/(c^6*d^2*x^4 + 2*c^4*d^2*x^2 + c^2*d^2), x) - 8*(-I*
c^2*x^2 - 2*I)*log(c^2*x^2 + 1))*b/(4*c^4*d^2*x - 4*I*c^3*d^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{-i \, b x^{2} \log \left (-\frac{c x + i}{c x - i}\right ) - 2 \, a x^{2}}{2 \, c^{2} d^{2} x^{2} - 4 i \, c d^{2} x - 2 \, d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-I*b*x^2*log(-(c*x + I)/(c*x - I)) - 2*a*x^2)/(2*c^2*d^2*x^2 - 4*I*c*d^2*x - 2*d^2), 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(x**2*(a+b*atan(c*x))/(d+I*c*d*x)**2,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 )} x^{2}}{{\left (i \, c d x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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