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

Optimal. Leaf size=110 \[ -\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac {i a x}{c d}-\frac {i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^2 d}+\frac {i b \log \left (c^2 x^2+1\right )}{2 c^2 d}-\frac {i b x \tan ^{-1}(c x)}{c d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {4866, 4846, 260, 4854, 2402, 2315} \[ -\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^2 d}-\frac {\log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac {i a x}{c d}+\frac {i b \log \left (c^2 x^2+1\right )}{2 c^2 d}-\frac {i b x \tan ^{-1}(c x)}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

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 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 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 4866

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f)/e, Int[((f*x)^(m - 1)*(a + b*ArcTan[c*x])^p)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 108, normalized size = 0.98 \[ \frac {2 i \tan ^{-1}(c x) \left (a-b c x+i b \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )+a \log \left (c^2 x^2+1\right )-2 i a c x+i b \log \left (c^2 x^2+1\right )+i b \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )+2 i b \tan ^{-1}(c x)^2}{2 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a x}{2 \, c d x - 2 i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x*log(-(c*x + I)/(c*x - I)) - 2*I*a*x)/(2*c*d*x - 2*I*d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.07, size = 244, normalized size = 2.22 \[ -\frac {i a x}{d c}+\frac {a \ln \left (c^{2} x^{2}+1\right )}{2 c^{2} d}+\frac {i a \arctan \left (c x \right )}{c^{2} d}-\frac {i b x \arctan \left (c x \right )}{d c}+\frac {b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{2} d}-\frac {i b \ln \left (-\frac {i \left (c x +i\right )}{2}\right ) \ln \left (c x -i\right )}{2 c^{2} d}-\frac {i b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c^{2} d}+\frac {i b \ln \left (c x -i\right )^{2}}{4 c^{2} d}+\frac {i b \ln \left (c^{8} x^{8}+12 c^{6} x^{6}+30 c^{4} x^{4}+28 c^{2} x^{2}+9\right )}{8 c^{2} d}-\frac {b \arctan \left (\frac {1}{12} c^{3} x^{3}+\frac {13}{12} c x \right )}{4 c^{2} d}-\frac {b \arctan \left (\frac {c x}{4}\right )}{4 c^{2} d}+\frac {b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a*x/d/c+1/2/c^2*a/d*ln(c^2*x^2+1)+I/c^2*a/d*arctan(c*x)-I*b*x*arctan(c*x)/d/c+1/c^2*b/d*arctan(c*x)*ln(c*x-
I)-1/2*I/c^2*b/d*ln(-1/2*I*(I+c*x))*ln(c*x-I)-1/2*I/c^2*b/d*dilog(-1/2*I*(I+c*x))+1/4*I/c^2*b/d*ln(c*x-I)^2+1/
8*I/c^2*b/d*ln(c^8*x^8+12*c^6*x^6+30*c^4*x^4+28*c^2*x^2+9)-1/4/c^2*b/d*arctan(1/12*c^3*x^3+13/12*c*x)-1/4/c^2*
b/d*arctan(1/4*c*x)+1/2/c^2*b/d*arctan(1/2*c*x-1/2*I)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (-\frac {i \, x}{c d} + \frac {\log \left (i \, c x + 1\right )}{c^{2} d}\right )} - \frac {{\left (2 i \, {\left (2 \, {\left (\frac {x}{c^{3} d} - \frac {\arctan \left (c x\right )}{c^{4} d}\right )} \arctan \left (c x\right ) + \frac {\arctan \left (c x\right )^{2} - \log \left (c^{2} x^{2} + 1\right )}{c^{4} d}\right )} c^{4} d + 2 \, c^{4} d \int \frac {x^{2} \log \left (c^{2} x^{2} + 1\right )}{c^{3} d x^{2} + c d}\,{d x} - 8 \, c^{3} d \int \frac {x \arctan \left (c x\right )}{c^{3} d x^{2} + c d}\,{d x} + 2 \, c^{2} d \int \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{3} d x^{2} + c d}\,{d x} - 2 \, c x \log \left (c^{2} x^{2} + 1\right ) + 4 \, c x - 4 \, {\left (-i \, c x + 1\right )} \arctan \left (c x\right ) - 2 i \, \arctan \left (c x\right )^{2} - 2 i \, \log \left (2 \, c^{3} d x^{2} + 2 \, c d\right )\right )} b}{8 \, c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-I*x/(c*d) + log(I*c*x + 1)/(c^2*d)) - 1/8*(8*I*c^4*d*integrate(1/2*x^2*arctan(c*x)/(c^3*d*x^2 + c*d), x) +
4*c^4*d*integrate(1/2*x^2*log(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) - 16*c^3*d*integrate(1/2*x*arctan(c*x)/(c^3*
d*x^2 + c*d), x) + 8*I*c^3*d*integrate(1/2*x*log(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) + 4*c^2*d*integrate(1/2*lo
g(c^2*x^2 + 1)/(c^3*d*x^2 + c*d), x) - 2*c*x*log(c^2*x^2 + 1) + 4*c*x - 4*(-I*c*x + 1)*arctan(c*x) - 2*I*arcta
n(c*x)^2 - I*log(c^2*x^2 + 1)^2 - 2*I*log(2*c^3*d*x^2 + 2*c*d))*b/(c^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {i \left (\int \left (- \frac {i b \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 a c^{2} x^{2}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {b c x}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 i a c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {i b c^{2} x^{2}}{c^{2} x^{2} + 1}\right )\, dx + \int \frac {2 b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {i b c^{2} x^{2} \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{2 c d} + \frac {\left (b c x + i b \log {\left (i c x + 1 \right )}\right ) \log {\left (- i c x + 1 \right )}}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*(Integral(-I*b*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(2*a*c**2*x**2/(c**2*x**2 + 1), x) + Integral(-
b*c*x/(c**2*x**2 + 1), x) + Integral(2*I*a*c*x/(c**2*x**2 + 1), x) + Integral(-I*b*c**2*x**2/(c**2*x**2 + 1),
x) + Integral(2*b*c*x*log(I*c*x + 1)/(c**2*x**2 + 1), x) + Integral(-I*b*c**2*x**2*log(I*c*x + 1)/(c**2*x**2 +
1), x))/(2*c*d) + (b*c*x + I*b*log(I*c*x + 1))*log(-I*c*x + 1)/(2*c**2*d)

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