3.16 \(\int \frac{(d+i c d x)^2 (a+b \tan ^{-1}(c x))}{x^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac{1}{2} i b c^2 d^2 \text{PolyLog}(2,-i c x)+\frac{1}{2} i b c^2 d^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac{b c d^2}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.153578, antiderivative size = 152, 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, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391} \[ -\frac{1}{2} i b c^2 d^2 \text{PolyLog}(2,-i c x)+\frac{1}{2} i b c^2 d^2 \text{PolyLog}(2,i c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-i b c^2 d^2 \log \left (c^2 x^2+1\right )+2 i b c^2 d^2 \log (x)-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac{b c d^2}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

-(b*c*d^2)/(2*x) - (b*c^2*d^2*ArcTan[c*x])/2 - (d^2*(a + b*ArcTan[c*x]))/(2*x^2) - ((2*I)*c*d^2*(a + b*ArcTan[
c*x]))/x - a*c^2*d^2*Log[x] + (2*I)*b*c^2*d^2*Log[x] - I*b*c^2*d^2*Log[1 + c^2*x^2] - (I/2)*b*c^2*d^2*PolyLog[
2, (-I)*c*x] + (I/2)*b*c^2*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 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 325

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

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^3} \, dx &=\int \left (\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^2 \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (2 i c d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (c^2 d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+\frac{1}{2} \left (b c d^2\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac{1}{2} \left (i b c^2 d^2\right ) \int \frac{\log (1-i c x)}{x} \, dx+\frac{1}{2} \left (i b c^2 d^2\right ) \int \frac{\log (1+i c x)}{x} \, dx+\left (2 i b c^2 d^2\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{b c d^2}{2 x}-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac{1}{2} i b c^2 d^2 \text{Li}_2(-i c x)+\frac{1}{2} i b c^2 d^2 \text{Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{2} \left (b c^3 d^2\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b c d^2}{2 x}-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)-\frac{1}{2} i b c^2 d^2 \text{Li}_2(-i c x)+\frac{1}{2} i b c^2 d^2 \text{Li}_2(i c x)+\left (i b c^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\left (i b c^4 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d^2}{2 x}-\frac{1}{2} b c^2 d^2 \tan ^{-1}(c x)-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-a c^2 d^2 \log (x)+2 i b c^2 d^2 \log (x)-i b c^2 d^2 \log \left (1+c^2 x^2\right )-\frac{1}{2} i b c^2 d^2 \text{Li}_2(-i c x)+\frac{1}{2} i b c^2 d^2 \text{Li}_2(i c x)\\ \end{align*}

Mathematica [C]  time = 0.0710135, size = 139, normalized size = 0.91 \[ -\frac{d^2 \left (b c x \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )+i b c^2 x^2 \text{PolyLog}(2,-i c x)-i b c^2 x^2 \text{PolyLog}(2,i c x)+2 a c^2 x^2 \log (x)+4 i a c x+a-4 i b c^2 x^2 \log (x)+2 i b c^2 x^2 \log \left (c^2 x^2+1\right )+4 i b c x \tan ^{-1}(c x)+b \tan ^{-1}(c x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x]))/x^3,x]

[Out]

-(d^2*(a + (4*I)*a*c*x + b*ArcTan[c*x] + (4*I)*b*c*x*ArcTan[c*x] + b*c*x*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2
*x^2)] + 2*a*c^2*x^2*Log[x] - (4*I)*b*c^2*x^2*Log[x] + (2*I)*b*c^2*x^2*Log[1 + c^2*x^2] + I*b*c^2*x^2*PolyLog[
2, (-I)*c*x] - I*b*c^2*x^2*PolyLog[2, I*c*x]))/(2*x^2)

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Maple [A]  time = 0.05, size = 217, normalized size = 1.4 \begin{align*} -{\frac{{d}^{2}a}{2\,{x}^{2}}}-{\frac{2\,ic{d}^{2}a}{x}}-{c}^{2}{d}^{2}a\ln \left ( cx \right ) -{\frac{b{d}^{2}\arctan \left ( cx \right ) }{2\,{x}^{2}}}-{\frac{2\,ic{d}^{2}b\arctan \left ( cx \right ) }{x}}-{c}^{2}{d}^{2}b\arctan \left ( cx \right ) \ln \left ( cx \right ) -{\frac{i}{2}}{c}^{2}{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) +{\frac{i}{2}}{c}^{2}{d}^{2}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) -{\frac{i}{2}}{c}^{2}{d}^{2}b{\it dilog} \left ( 1+icx \right ) +{\frac{i}{2}}{c}^{2}{d}^{2}b{\it dilog} \left ( 1-icx \right ) -ib{c}^{2}{d}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) -{\frac{b{c}^{2}{d}^{2}\arctan \left ( cx \right ) }{2}}-{\frac{bc{d}^{2}}{2\,x}}+2\,i{c}^{2}{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^3,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -b c^{2} d^{2} \int \frac{\arctan \left (c x\right )}{x}\,{d x} - a c^{2} d^{2} \log \left (x\right ) - i \,{\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 c d^{2} - \frac{1}{2} \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b d^{2} - \frac{2 i \, a c d^{2}}{x} - \frac{a d^{2}}{2 \, x^{2}} \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^3,x, algorithm="maxima")

[Out]

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

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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^{3}}, 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^3,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^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{a}{x^{3}}\, dx + \int - \frac{a c^{2}}{x}\, dx + \int \frac{b \operatorname{atan}{\left (c x \right )}}{x^{3}}\, dx + \int \frac{2 i a c}{x^{2}}\, dx + \int - \frac{b c^{2} \operatorname{atan}{\left (c x \right )}}{x}\, dx + \int \frac{2 i b c \operatorname{atan}{\left (c x \right )}}{x^{2}}\, 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**3,x)

[Out]

d**2*(Integral(a/x**3, x) + Integral(-a*c**2/x, x) + Integral(b*atan(c*x)/x**3, x) + Integral(2*I*a*c/x**2, x)
+ Integral(-b*c**2*atan(c*x)/x, x) + Integral(2*I*b*c*atan(c*x)/x**2, 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^{3}}\,{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^3,x, algorithm="giac")

[Out]

integrate((I*c*d*x + d)^2*(b*arctan(c*x) + a)/x^3, x)