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

Optimal. Leaf size=170 \[ \frac{1}{2} i b d^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^3 \text{PolyLog}(2,i c x)-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+3 i a c d^3 x+a d^3 \log (x)+\frac{1}{6} i b c^2 d^3 x^2-\frac{5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac{3}{2} b c d^3 x-\frac{3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x) \]

[Out]

(3*I)*a*c*d^3*x + (3*b*c*d^3*x)/2 + (I/6)*b*c^2*d^3*x^2 - (3*b*d^3*ArcTan[c*x])/2 + (3*I)*b*c*d^3*x*ArcTan[c*x
] - (3*c^2*d^3*x^2*(a + b*ArcTan[c*x]))/2 - (I/3)*c^3*d^3*x^3*(a + b*ArcTan[c*x]) + a*d^3*Log[x] - ((5*I)/3)*b
*d^3*Log[1 + c^2*x^2] + (I/2)*b*d^3*PolyLog[2, (-I)*c*x] - (I/2)*b*d^3*PolyLog[2, I*c*x]

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Rubi [A]  time = 0.17451, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4848, 2391, 4852, 321, 203, 266, 43} \[ \frac{1}{2} i b d^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^3 \text{PolyLog}(2,i c x)-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+3 i a c d^3 x+a d^3 \log (x)+\frac{1}{6} i b c^2 d^3 x^2-\frac{5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac{3}{2} b c d^3 x-\frac{3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*I)*a*c*d^3*x + (3*b*c*d^3*x)/2 + (I/6)*b*c^2*d^3*x^2 - (3*b*d^3*ArcTan[c*x])/2 + (3*I)*b*c*d^3*x*ArcTan[c*x
] - (3*c^2*d^3*x^2*(a + b*ArcTan[c*x]))/2 - (I/3)*c^3*d^3*x^3*(a + b*ArcTan[c*x]) + a*d^3*Log[x] - ((5*I)/3)*b
*d^3*Log[1 + c^2*x^2] + (I/2)*b*d^3*PolyLog[2, (-I)*c*x] - (I/2)*b*d^3*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 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])

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x} \, dx &=\int \left (3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )-i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )\right ) \, dx\\ &=d^3 \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+\left (3 i c d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (3 c^2 d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (i c^3 d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=3 i a c d^3 x-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac{1}{2} \left (i b d^3\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} \left (i b d^3\right ) \int \frac{\log (1+i c x)}{x} \, dx+\left (3 i b c d^3\right ) \int \tan ^{-1}(c x) \, dx+\frac{1}{2} \left (3 b c^3 d^3\right ) \int \frac{x^2}{1+c^2 x^2} \, dx+\frac{1}{3} \left (i b c^4 d^3\right ) \int \frac{x^3}{1+c^2 x^2} \, dx\\ &=3 i a c d^3 x+\frac{3}{2} b c d^3 x+3 i b c d^3 x \tan ^{-1}(c x)-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac{1}{2} i b d^3 \text{Li}_2(-i c x)-\frac{1}{2} i b d^3 \text{Li}_2(i c x)-\frac{1}{2} \left (3 b c d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx-\left (3 i b c^2 d^3\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{1}{6} \left (i b c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{x}{1+c^2 x} \, dx,x,x^2\right )\\ &=3 i a c d^3 x+\frac{3}{2} b c d^3 x-\frac{3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x)-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)-\frac{3}{2} i b d^3 \log \left (1+c^2 x^2\right )+\frac{1}{2} i b d^3 \text{Li}_2(-i c x)-\frac{1}{2} i b d^3 \text{Li}_2(i c x)+\frac{1}{6} \left (i b c^4 d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{c^2}-\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=3 i a c d^3 x+\frac{3}{2} b c d^3 x+\frac{1}{6} i b c^2 d^3 x^2-\frac{3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x)-\frac{3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)-\frac{5}{3} i b d^3 \log \left (1+c^2 x^2\right )+\frac{1}{2} i b d^3 \text{Li}_2(-i c x)-\frac{1}{2} i b d^3 \text{Li}_2(i c x)\\ \end{align*}

Mathematica [A]  time = 0.13178, size = 139, normalized size = 0.82 \[ -\frac{1}{6} i d^3 \left (-3 b \text{PolyLog}(2,-i c x)+3 b \text{PolyLog}(2,i c x)+2 a c^3 x^3-9 i a c^2 x^2-18 a c x+6 i a \log (x)-b c^2 x^2+10 b \log \left (c^2 x^2+1\right )+2 b c^3 x^3 \tan ^{-1}(c x)-9 i b c^2 x^2 \tan ^{-1}(c x)+9 i b c x-18 b c x \tan ^{-1}(c x)-9 i b \tan ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-I/6)*d^3*(-18*a*c*x + (9*I)*b*c*x - (9*I)*a*c^2*x^2 - b*c^2*x^2 + 2*a*c^3*x^3 - (9*I)*b*ArcTan[c*x] - 18*b*c
*x*ArcTan[c*x] - (9*I)*b*c^2*x^2*ArcTan[c*x] + 2*b*c^3*x^3*ArcTan[c*x] + (6*I)*a*Log[x] + 10*b*Log[1 + c^2*x^2
] - 3*b*PolyLog[2, (-I)*c*x] + 3*b*PolyLog[2, I*c*x])

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Maple [A]  time = 0.041, size = 220, normalized size = 1.3 \begin{align*} 3\,iac{d}^{3}x-{\frac{i}{2}}{d}^{3}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) -{\frac{3\,{d}^{3}a{c}^{2}{x}^{2}}{2}}+{d}^{3}a\ln \left ( cx \right ) +{\frac{i}{2}}{d}^{3}b{\it dilog} \left ( 1+icx \right ) +{\frac{i}{2}}{d}^{3}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{3\,{d}^{3}b\arctan \left ( cx \right ){c}^{2}{x}^{2}}{2}}+{d}^{3}b\arctan \left ( cx \right ) \ln \left ( cx \right ) +3\,ibc{d}^{3}x\arctan \left ( cx \right ) -{\frac{i}{3}}{d}^{3}a{c}^{3}{x}^{3}+{\frac{i}{6}}b{c}^{2}{d}^{3}{x}^{2}-{\frac{i}{3}}{d}^{3}b\arctan \left ( cx \right ){c}^{3}{x}^{3}+{\frac{3\,bc{d}^{3}x}{2}}-{\frac{i}{2}}{d}^{3}b{\it dilog} \left ( 1-icx \right ) -{\frac{5\,i}{3}}b{d}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) -{\frac{3\,b{d}^{3}\arctan \left ( cx \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*I*a*c*d^3*x-1/2*I*d^3*b*ln(c*x)*ln(1-I*c*x)-3/2*d^3*a*c^2*x^2+d^3*a*ln(c*x)+1/2*I*d^3*b*dilog(1+I*c*x)+1/2*I
*d^3*b*ln(c*x)*ln(1+I*c*x)-3/2*d^3*b*arctan(c*x)*c^2*x^2+d^3*b*arctan(c*x)*ln(c*x)+3*I*b*c*d^3*x*arctan(c*x)-1
/3*I*d^3*a*c^3*x^3+1/6*I*b*c^2*d^3*x^2-1/3*I*d^3*b*arctan(c*x)*c^3*x^3+3/2*b*c*d^3*x-1/2*I*d^3*b*dilog(1-I*c*x
)-5/3*I*b*d^3*ln(c^2*x^2+1)-3/2*b*d^3*arctan(c*x)

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Maxima [A]  time = 2.16782, size = 259, normalized size = 1.52 \begin{align*} -\frac{1}{3} i \, a c^{3} d^{3} x^{3} - \frac{3}{2} \, a c^{2} d^{3} x^{2} + \frac{1}{6} i \, b c^{2} d^{3} x^{2} + 3 i \, a c d^{3} x + \frac{3}{2} \, b c d^{3} x - \frac{1}{12} \,{\left (3 \, \pi + 2 i\right )} b d^{3} \log \left (c^{2} x^{2} + 1\right ) + b d^{3} \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + \frac{3}{2} i \,{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3} - \frac{1}{2} i \, b d^{3}{\rm Li}_2\left (i \, c x + 1\right ) + \frac{1}{2} i \, b d^{3}{\rm Li}_2\left (-i \, c x + 1\right ) + a d^{3} \log \left (x\right ) - \frac{1}{12} \,{\left (4 i \, b c^{3} d^{3} x^{3} + 18 \, b c^{2} d^{3} x^{2} - 6 \, b d^{3}{\left (2 i \, \arctan \left (0, c\right ) - 3\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)^3*(a+b*arctan(c*x))/x,x, algorithm="maxima")

[Out]

-1/3*I*a*c^3*d^3*x^3 - 3/2*a*c^2*d^3*x^2 + 1/6*I*b*c^2*d^3*x^2 + 3*I*a*c*d^3*x + 3/2*b*c*d^3*x - 1/12*(3*pi +
2*I)*b*d^3*log(c^2*x^2 + 1) + b*d^3*arctan(c*x)*log(x*abs(c)) + 3/2*I*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b
*d^3 - 1/2*I*b*d^3*dilog(I*c*x + 1) + 1/2*I*b*d^3*dilog(-I*c*x + 1) + a*d^3*log(x) - 1/12*(4*I*b*c^3*d^3*x^3 +
18*b*c^2*d^3*x^2 - 6*b*d^3*(2*I*arctan2(0, c) - 3))*arctan(c*x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{-2 i \, a c^{3} d^{3} x^{3} - 6 \, a c^{2} d^{3} x^{2} + 6 i \, a c d^{3} x + 2 \, a d^{3} +{\left (b c^{3} d^{3} x^{3} - 3 i \, b c^{2} d^{3} x^{2} - 3 \, b c d^{3} x + i \, b d^{3}\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)^3*(a+b*arctan(c*x))/x,x, algorithm="fricas")

[Out]

integral(1/2*(-2*I*a*c^3*d^3*x^3 - 6*a*c^2*d^3*x^2 + 6*I*a*c*d^3*x + 2*a*d^3 + (b*c^3*d^3*x^3 - 3*I*b*c^2*d^3*
x^2 - 3*b*c*d^3*x + I*b*d^3)*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^{3} \left (\int \frac{a}{x}\, dx + \int 3 i a c\, dx + \int - 3 a c^{2} x\, dx + \int \frac{b \operatorname{atan}{\left (c x \right )}}{x}\, dx + \int - i a c^{3} x^{2}\, dx + \int 3 i b c \operatorname{atan}{\left (c x \right )}\, dx + \int - 3 b c^{2} x \operatorname{atan}{\left (c x \right )}\, dx + \int - i b c^{3} x^{2} \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)**3*(a+b*atan(c*x))/x,x)

[Out]

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

[Out]

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