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

Optimal. Leaf size=77 \[ -\frac{1}{2} b c d \text{PolyLog}(2,-i c x)+\frac{1}{2} b c d \text{PolyLog}(2,i c x)-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{x}+i a c d \log (x)-\frac{1}{2} b c d \log \left (c^2 x^2+1\right )+b c d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.100228, antiderivative size = 77, 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, 4852, 266, 36, 29, 31, 4848, 2391} \[ -\frac{1}{2} b c d \text{PolyLog}(2,-i c x)+\frac{1}{2} b c d \text{PolyLog}(2,i c x)-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{x}+i a c d \log (x)-\frac{1}{2} b c d \log \left (c^2 x^2+1\right )+b c d \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(a + b*ArcTan[c*x]))/x) + I*a*c*d*Log[x] + b*c*d*Log[x] - (b*c*d*Log[1 + c^2*x^2])/2 - (b*c*d*PolyLog[2,
(-I)*c*x])/2 + (b*c*d*PolyLog[2, I*c*x])/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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.048796, size = 75, normalized size = 0.97 \[ \frac{d \left (-b c x \text{PolyLog}(2,-i c x)+b c x \text{PolyLog}(2,i c x)+2 i a c x \log (x)-2 a-b c x \log \left (c^2 x^2+1\right )+2 b c x \log (x)-2 b \tan ^{-1}(c x)\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.044, size = 127, normalized size = 1.7 \begin{align*} -{\frac{da}{x}}+icda\ln \left ( cx \right ) -{\frac{db\arctan \left ( cx \right ) }{x}}+icdb\arctan \left ( cx \right ) \ln \left ( cx \right ) -{\frac{cdb\ln \left ( cx \right ) \ln \left ( 1+icx \right ) }{2}}+{\frac{cdb\ln \left ( cx \right ) \ln \left ( 1-icx \right ) }{2}}-{\frac{cdb{\it dilog} \left ( 1+icx \right ) }{2}}+{\frac{cdb{\it dilog} \left ( 1-icx \right ) }{2}}-{\frac{bcd\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2}}+cdb\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-d*a/x+I*c*d*a*ln(c*x)-d*b*arctan(c*x)/x+I*c*d*b*arctan(c*x)*ln(c*x)-1/2*c*d*b*ln(c*x)*ln(1+I*c*x)+1/2*c*d*b*l
n(c*x)*ln(1-I*c*x)-1/2*c*d*b*dilog(1+I*c*x)+1/2*c*d*b*dilog(1-I*c*x)-1/2*b*c*d*ln(c^2*x^2+1)+c*d*b*ln(c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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