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

Optimal. Leaf size=122 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2 d^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (-c x+i)}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2}-\frac{b}{2 c^2 d^2 (-c x+i)}+\frac{b \tan ^{-1}(c x)}{2 c^2 d^2} \]

[Out]

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

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Rubi [A]  time = 0.146809, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {4876, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )}{2 c^2 d^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (-c x+i)}+\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2}-\frac{b}{2 c^2 d^2 (-c x+i)}+\frac{b \tan ^{-1}(c x)}{2 c^2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-b/(2*c^2*d^2*(I - c*x)) + (b*ArcTan[c*x])/(2*c^2*d^2) - (I*(a + b*ArcTan[c*x]))/(c^2*d^2*(I - c*x)) + ((a + b
*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^2*d^2) + ((I/2)*b*PolyLog[2, 1 - 2/(1 + I*c*x)])/(c^2*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 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 \left (a+b \tan ^{-1}(c x)\right )}{(d+i c d x)^2} \, dx &=\int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c d^2 (-i+c x)^2}-\frac{a+b \tan ^{-1}(c x)}{c d^2 (-i+c x)}\right ) \, dx\\ &=-\frac{i \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c d^2}-\frac{\int \frac{a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c d^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (i-c x)}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^2 d^2}-\frac{(i b) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c d^2}-\frac{b \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c d^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (i-c x)}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^2 d^2}+\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )}{c^2 d^2}-\frac{(i b) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c d^2}\\ &=-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (i-c x)}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^2 d^2}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d^2}-\frac{(i b) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c d^2}\\ &=-\frac{b}{2 c^2 d^2 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (i-c x)}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^2 d^2}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d^2}+\frac{b \int \frac{1}{1+c^2 x^2} \, dx}{2 c d^2}\\ &=-\frac{b}{2 c^2 d^2 (i-c x)}+\frac{b \tan ^{-1}(c x)}{2 c^2 d^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (i-c x)}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{c^2 d^2}+\frac{i b \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d^2}\\ \end{align*}

Mathematica [A]  time = 0.0897185, size = 128, normalized size = 1.05 \[ \frac{i b \text{PolyLog}\left (2,-\frac{c x+i}{-c x+i}\right )}{2 c^2 d^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2 (-c x+i)}+\frac{\log \left (\frac{2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d^2}-\frac{b \left (-\frac{\tan ^{-1}(c x)}{c}+\frac{1}{c (-c x+i)}\right )}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.055, size = 293, normalized size = 2.4 \begin{align*} -{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,{c}^{2}{d}^{2}}}-{\frac{ia\arctan \left ( cx \right ) }{{c}^{2}{d}^{2}}}+{\frac{ia}{{c}^{2}{d}^{2} \left ( cx-i \right ) }}-{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{{c}^{2}{d}^{2}}}+{\frac{ib\arctan \left ( cx \right ) }{{c}^{2}{d}^{2} \left ( cx-i \right ) }}+{\frac{{\frac{i}{16}}b\ln \left ({c}^{4}{x}^{4}+10\,{c}^{2}{x}^{2}+9 \right ) }{{c}^{2}{d}^{2}}}+{\frac{b}{8\,{c}^{2}{d}^{2}}\arctan \left ({\frac{{c}^{3}{x}^{3}}{6}}+{\frac{7\,cx}{6}} \right ) }-{\frac{b}{8\,{c}^{2}{d}^{2}}\arctan \left ({\frac{cx}{2}} \right ) }+{\frac{b}{4\,{c}^{2}{d}^{2}}\arctan \left ({\frac{cx}{2}}-{\frac{i}{2}} \right ) }+{\frac{b}{2\,{c}^{2}{d}^{2} \left ( cx-i \right ) }}-{\frac{{\frac{i}{8}}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{2}{d}^{2}}}+{\frac{b\arctan \left ( cx \right ) }{4\,{c}^{2}{d}^{2}}}+{\frac{{\frac{i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{{c}^{2}{d}^{2}}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{{c}^{2}{d}^{2}}}-{\frac{{\frac{i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{{c}^{2}{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

a*(I/(c^3*d^2*x - I*c^2*d^2) - log(c*x - I)/(c^2*d^2)) - 1/8*(32*(I*c*x + 1)*arctan(c*x)^2 + 32*c*x*arctan2(1,
c*x) - 8*(-I*c*x - 1)*log(c^2*x^2 + 1)^2 - (8*c^3*d^2*x - 8*I*c^2*d^2)*((c*(x/(c^5*d^2*x^2 + c^3*d^2) + arcta
n(c*x)/(c^4*d^2)) - 2*arctan(c*x)/(c^5*d^2*x^2 + c^3*d^2))*c + 8*integrate(1/4*log(c^2*x^2 + 1)/(c^5*d^2*x^4 +
2*c^3*d^2*x^2 + c*d^2), x)) - 8*(I*c^3*d^2*x + c^2*d^2)*(c*(c^2/(c^7*d^2*x^2 + c^5*d^2) + log(c^2*x^2 + 1)/(c
^5*d^2*x^2 + c^3*d^2)) + 16*integrate(1/4*arctan(c*x)/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x)) + (8*c^4*d^2*
x - 8*I*c^3*d^2)*(c*(x/(c^5*d^2*x^2 + c^3*d^2) + arctan(c*x)/(c^4*d^2)) - 8*c*integrate(1/4*x^2*log(c^2*x^2 +
1)/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - 2*arctan(c*x)/(c^5*d^2*x^2 + c^3*d^2)) + 8*(-I*c^4*d^2*x - c^3*
d^2)*(16*c*integrate(1/4*x^2*arctan(c*x)/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - c^2/(c^7*d^2*x^2 + c^5*d^
2) - log(c^2*x^2 + 1)/(c^5*d^2*x^2 + c^3*d^2)) + 8*(16*c^5*d^2*x - 16*I*c^4*d^2)*integrate(1/4*(2*c*x^3*arctan
(c*x) + x^2*log(c^2*x^2 + 1))/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) + 128*(-I*c^5*d^2*x - c^4*d^2)*integra
te(1/4*(c*x^3*log(c^2*x^2 + 1) - 2*x^2*arctan(c*x))/(c^5*d^2*x^4 + 2*c^3*d^2*x^2 + c*d^2), x) - 32*I*arctan(c*
x) - 32*I*arctan2(1, c*x) + 16*log(c^2*x^2 + 1))*b/(8*c^3*d^2*x - 8*I*c^2*d^2)

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

[Out]

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

[Out]

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