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

Optimal. Leaf size=89 \[ -b c d^2 \text{PolyLog}(2,-i c x)+b c d^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 x+2 i a c d^2 \log (x)-b c^2 d^2 x \tan ^{-1}(c x)+b c d^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.137576, antiderivative size = 89, 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, 4852, 266, 36, 29, 31, 4848, 2391} \[ -b c d^2 \text{PolyLog}(2,-i c x)+b c d^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 x+2 i a c d^2 \log (x)-b c^2 d^2 x \tan ^{-1}(c x)+b c d^2 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0918685, size = 79, normalized size = 0.89 \[ -\frac{d^2 \left (b c x \text{PolyLog}(2,-i c x)-b c x \text{PolyLog}(2,i c x)+a c^2 x^2-2 i a c x \log (x)+a+b c^2 x^2 \tan ^{-1}(c x)-b c x \log (c x)+b \tan ^{-1}(c x)\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 152, normalized size = 1.7 \begin{align*} -a{c}^{2}{d}^{2}x-{\frac{{d}^{2}a}{x}}+2\,ic{d}^{2}a\ln \left ( cx \right ) -b{c}^{2}{d}^{2}x\arctan \left ( cx \right ) -{\frac{b{d}^{2}\arctan \left ( cx \right ) }{x}}+2\,ic{d}^{2}b\arctan \left ( cx \right ) \ln \left ( cx \right ) -c{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) +c{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) -c{d}^{2}b{\it dilog} \left ( 1+icx \right ) +c{d}^{2}b{\it dilog} \left ( 1-icx \right ) +c{d}^{2}b\ln \left ( cx \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(-1/2*(2*a*c^2*d^2*x^2 - 4*I*a*c*d^2*x - 2*a*d^2 - (-I*b*c^2*d^2*x^2 - 2*b*c*d^2*x + I*b*d^2)*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^{2} \left (\int - a c^{2}\, dx + \int \frac{a}{x^{2}}\, dx + \int - b c^{2} \operatorname{atan}{\left (c x \right )}\, dx + \int \frac{b \operatorname{atan}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{2 i a c}{x}\, dx + \int \frac{2 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)**2*(a+b*atan(c*x))/x**2,x)

[Out]

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

[Out]

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