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

Optimal. Leaf size=77 \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{x}+i a c d \log (x)-\frac {1}{2} b c d \log \left (c^2 x^2+1\right )-\frac {1}{2} b c d \text {Li}_2(-i c x)+\frac {1}{2} b c d \text {Li}_2(i c x)+b c d \log (x) \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 77, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.05, size = 75, normalized size = 0.97 \[ \frac {d \left (2 i a c x \log (x)-2 a-b c x \log \left (c^2 x^2+1\right )-b c x \text {Li}_2(-i c x)+b c x \text {Li}_2(i c x)+2 b c x \log (x)-2 b \tan ^{-1}(c x)\right )}{2 x} \]

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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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 i \, a c d x + 2 \, a d - {\left (b c d x - i \, b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{2}}, x\right ) \]

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 127, normalized size = 1.65 \[ i c d a \ln \left (c x \right )-\frac {d a}{x}+i c d b \arctan \left (c x \right ) \ln \left (c x \right )-\frac {d b \arctan \left (c x \right )}{x}-\frac {c d b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {c d b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {c d b \dilog \left (i c x +1\right )}{2}+\frac {c d b \dilog \left (-i c x +1\right )}{2}+c d b \ln \left (c x \right )-\frac {b c d \ln \left (c^{2} x^{2}+1\right )}{2} \]

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]

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

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

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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mupad [B]  time = 0.86, size = 93, normalized size = 1.21 \[ \left \{\begin {array}{cl} -\frac {a\,d}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+\frac {b\,c\,d\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )}{2}+\frac {a\,d\,\left (-1+c\,x\,\ln \relax (x)\,1{}\mathrm {i}\right )}{x}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{x} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i))/x^2,x)

[Out]

piecewise(c == 0, -(a*d)/x, c ~= 0, (b*d*(c^2*log(x) - (c^2*log(c^2*x^2 + 1))/2))/c + (b*c*d*(dilog(- c*x*1i +
1) - dilog(c*x*1i + 1)))/2 + (a*d*(c*x*log(x)*1i - 1))/x - (b*d*atan(c*x))/x)

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

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]

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

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