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

Optimal. Leaf size=89 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 x+2 i a c d^2 \log (x)-b c^2 d^2 x \tan ^{-1}(c x)-b c d^2 \text {Li}_2(-i c x)+b c d^2 \text {Li}_2(i c x)+b c d^2 \log (x) \]

[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*ln(x)+b*c*d^2*ln(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 = 0.14, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 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 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 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 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 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)^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*}

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Mathematica [A]  time = 0.10, size = 79, normalized size = 0.89 \[ -\frac {d^2 \left (a c^2 x^2-2 i a c x \log (x)+a+b c^2 x^2 \tan ^{-1}(c x)+b c x \text {Li}_2(-i c x)-b c x \text {Li}_2(i c x)-b c x \log (c x)+b \tan ^{-1}(c x)\right )}{x} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {2 \, a c^{2} d^{2} x^{2} - 4 i \, a c d^{2} x - 2 \, a d^{2} - {\left (-i \, b c^{2} d^{2} x^{2} - 2 \, b c d^{2} x + i \, b d^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{2}}, x\right ) \]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 152, normalized size = 1.71 \[ -a \,c^{2} d^{2} x +2 i c \,d^{2} a \ln \left (c x \right )-\frac {d^{2} a}{x}-b \,c^{2} d^{2} x \arctan \left (c x \right )+2 i c \,d^{2} b \arctan \left (c x \right ) \ln \left (c x \right )-\frac {d^{2} b \arctan \left (c x \right )}{x}-c \,d^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )+c \,d^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )-c \,d^{2} b \dilog \left (i c x +1\right )+c \,d^{2} b \dilog \left (-i c x +1\right )+c \,d^{2} b \ln \left (c x \right ) \]

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

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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mupad [B]  time = 0.60, size = 141, normalized size = 1.58 \[ \left \{\begin {array}{cl} -\frac {a\,d^2}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d^2\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+b\,c\,d^2\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )+\frac {b\,c\,d^2\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {a\,d^2\,\left (c^2\,x^2+1-c\,x\,\ln \relax (x)\,2{}\mathrm {i}\right )}{x}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{x}-b\,c^2\,d^2\,x\,\mathrm {atan}\left (c\,x\right ) & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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