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

Optimal. Leaf size=129 \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+2 i a c d^2 x+a d^2 \log (x)-i b d^2 \log \left (c^2 x^2+1\right )+\frac{1}{2} b c d^2 x-\frac{1}{2} b d^2 \tan ^{-1}(c x)+2 i b c d^2 x \tan ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.12686, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4876, 4846, 260, 4848, 2391, 4852, 321, 203} \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)-\frac{1}{2} c^2 d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+2 i a c d^2 x+a d^2 \log (x)-i b d^2 \log \left (c^2 x^2+1\right )+\frac{1}{2} b c d^2 x-\frac{1}{2} b d^2 \tan ^{-1}(c x)+2 i b c d^2 x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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])

Rubi steps

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

Mathematica [A]  time = 0.0925627, size = 103, normalized size = 0.8 \[ -\frac{1}{2} d^2 \left (-i b \text{PolyLog}(2,-i c x)+i b \text{PolyLog}(2,i c x)+a c^2 x^2-4 i a c x-2 a \log (x)+2 i b \log \left (c^2 x^2+1\right )+b c^2 x^2 \tan ^{-1}(c x)-b c x-4 i b c x \tan ^{-1}(c x)+b \tan ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.043, size = 177, normalized size = 1.4 \begin{align*} 2\,iac{d}^{2}x-{\frac{{d}^{2}a{c}^{2}{x}^{2}}{2}}+{d}^{2}a\ln \left ( cx \right ) +2\,ibc{d}^{2}x\arctan \left ( cx \right ) -{\frac{{d}^{2}b\arctan \left ( cx \right ){c}^{2}{x}^{2}}{2}}+{d}^{2}b\arctan \left ( cx \right ) \ln \left ( cx \right ) +{\frac{bc{d}^{2}x}{2}}-ib{d}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) -{\frac{b{d}^{2}\arctan \left ( cx \right ) }{2}}+{\frac{i}{2}}{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}{d}^{2}b{\it dilog} \left ( 1+icx \right ) -{\frac{i}{2}}{d}^{2}b{\it dilog} \left ( 1-icx \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,x)

[Out]

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

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Maxima [A]  time = 2.14319, size = 204, normalized size = 1.58 \begin{align*} -\frac{1}{2} \, a c^{2} d^{2} x^{2} + 2 i \, a c d^{2} x + \frac{1}{2} \, b c d^{2} x - \frac{1}{4} \, \pi b d^{2} \log \left (c^{2} x^{2} + 1\right ) + b d^{2} \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + i \,{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{2} - \frac{1}{2} i \, b d^{2}{\rm Li}_2\left (i \, c x + 1\right ) + \frac{1}{2} i \, b d^{2}{\rm Li}_2\left (-i \, c x + 1\right ) + a d^{2} \log \left (x\right ) - \frac{1}{2} \,{\left (b c^{2} d^{2} x^{2} - b d^{2}{\left (2 i \, \arctan \left (0, c\right ) - 1\right )}\right )} \arctan \left (c 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,x, algorithm="maxima")

[Out]

-1/2*a*c^2*d^2*x^2 + 2*I*a*c*d^2*x + 1/2*b*c*d^2*x - 1/4*pi*b*d^2*log(c^2*x^2 + 1) + b*d^2*arctan(c*x)*log(x*a
bs(c)) + I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^2 - 1/2*I*b*d^2*dilog(I*c*x + 1) + 1/2*I*b*d^2*dilog(-I*
c*x + 1) + a*d^2*log(x) - 1/2*(b*c^2*d^2*x^2 - b*d^2*(2*I*arctan2(0, c) - 1))*arctan(c*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}, 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,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, x)

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

[Out]

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

[Out]

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