3.27 \(\int \frac{(d+i c d x)^3 (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=189 \[ \frac{1}{2} b c^3 d^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} b c^3 d^3 \text{PolyLog}(2,i c x)+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-i a c^3 d^3 \log (x)+\frac{5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac{3 i b c^2 d^3}{2 x}-\frac{10}{3} b c^3 d^3 \log (x)-\frac{3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac{b c d^3}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.203295, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {4876, 4852, 266, 44, 325, 203, 36, 29, 31, 4848, 2391} \[ \frac{1}{2} b c^3 d^3 \text{PolyLog}(2,-i c x)-\frac{1}{2} b c^3 d^3 \text{PolyLog}(2,i c x)+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-i a c^3 d^3 \log (x)+\frac{5}{3} b c^3 d^3 \log \left (c^2 x^2+1\right )-\frac{3 i b c^2 d^3}{2 x}-\frac{10}{3} b c^3 d^3 \log (x)-\frac{3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac{b c d^3}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

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

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 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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

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 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)^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=\int \left (\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4}+\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx+\left (3 i c d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx-\left (3 c^2 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (i c^3 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac{1}{3} \left (b c d^3\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx+\frac{1}{2} \left (3 i b c^2 d^3\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac{1}{2} \left (b c^3 d^3\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} \left (b c^3 d^3\right ) \int \frac{\log (1+i c x)}{x} \, dx-\left (3 b c^3 d^3\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{3 i b c^2 d^3}{2 x}-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac{1}{2} b c^3 d^3 \text{Li}_2(-i c x)-\frac{1}{2} b c^3 d^3 \text{Li}_2(i c x)+\frac{1}{6} \left (b c d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{2} \left (3 b c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{2} \left (3 i b c^4 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{3 i b c^2 d^3}{2 x}-\frac{3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)+\frac{1}{2} b c^3 d^3 \text{Li}_2(-i c x)-\frac{1}{2} b c^3 d^3 \text{Li}_2(i c x)+\frac{1}{6} \left (b c d^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{x^2}-\frac{c^2}{x}+\frac{c^4}{1+c^2 x}\right ) \, dx,x,x^2\right )-\frac{1}{2} \left (3 b c^3 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} \left (3 b c^5 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d^3}{6 x^2}-\frac{3 i b c^2 d^3}{2 x}-\frac{3}{2} i b c^3 d^3 \tan ^{-1}(c x)-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+\frac{3 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-i a c^3 d^3 \log (x)-\frac{10}{3} b c^3 d^3 \log (x)+\frac{5}{3} b c^3 d^3 \log \left (1+c^2 x^2\right )+\frac{1}{2} b c^3 d^3 \text{Li}_2(-i c x)-\frac{1}{2} b c^3 d^3 \text{Li}_2(i c x)\\ \end{align*}

Mathematica [C]  time = 0.094224, size = 170, normalized size = 0.9 \[ \frac{d^3 \left (-9 i b c^2 x^2 \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},-c^2 x^2\right )+3 b c^3 x^3 \text{PolyLog}(2,-i c x)-3 b c^3 x^3 \text{PolyLog}(2,i c x)+18 a c^2 x^2-6 i a c^3 x^3 \log (x)-9 i a c x-2 a-20 b c^3 x^3 \log (x)+10 b c^3 x^3 \log \left (c^2 x^2+1\right )+18 b c^2 x^2 \tan ^{-1}(c x)-b c x-9 i b c x \tan ^{-1}(c x)-2 b \tan ^{-1}(c x)\right )}{6 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

(d^3*(-2*a - (9*I)*a*c*x - b*c*x + 18*a*c^2*x^2 - 2*b*ArcTan[c*x] - (9*I)*b*c*x*ArcTan[c*x] + 18*b*c^2*x^2*Arc
Tan[c*x] - (9*I)*b*c^2*x^2*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)] - (6*I)*a*c^3*x^3*Log[x] - 20*b*c^3*x^3
*Log[x] + 10*b*c^3*x^3*Log[1 + c^2*x^2] + 3*b*c^3*x^3*PolyLog[2, (-I)*c*x] - 3*b*c^3*x^3*PolyLog[2, I*c*x]))/(
6*x^3)

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Maple [A]  time = 0.051, size = 255, normalized size = 1.4 \begin{align*}{\frac{-{\frac{3\,i}{2}}c{d}^{3}b\arctan \left ( cx \right ) }{{x}^{2}}}+3\,{\frac{{c}^{2}{d}^{3}a}{x}}-{\frac{{d}^{3}a}{3\,{x}^{3}}}-i{c}^{3}{d}^{3}b\arctan \left ( cx \right ) \ln \left ( cx \right ) -{\frac{{\frac{3\,i}{2}}b{c}^{2}{d}^{3}}{x}}+3\,{\frac{b{c}^{2}{d}^{3}\arctan \left ( cx \right ) }{x}}-{\frac{b{d}^{3}\arctan \left ( cx \right ) }{3\,{x}^{3}}}-{\frac{{\frac{3\,i}{2}}c{d}^{3}a}{{x}^{2}}}+{\frac{{c}^{3}{d}^{3}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) }{2}}-{\frac{{c}^{3}{d}^{3}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) }{2}}+{\frac{{c}^{3}{d}^{3}b{\it dilog} \left ( 1+icx \right ) }{2}}-{\frac{{c}^{3}{d}^{3}b{\it dilog} \left ( 1-icx \right ) }{2}}+{\frac{5\,b{c}^{3}{d}^{3}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3}}-i{c}^{3}{d}^{3}a\ln \left ( cx \right ) -{\frac{3\,i}{2}}b{c}^{3}{d}^{3}\arctan \left ( cx \right ) -{\frac{bc{d}^{3}}{6\,{x}^{2}}}-{\frac{10\,{c}^{3}{d}^{3}b\ln \left ( cx \right ) }{3}} \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^4,x)

[Out]

-3/2*I*c*d^3*b*arctan(c*x)/x^2+3*c^2*d^3*a/x-1/3*d^3*a/x^3-I*c^3*d^3*b*arctan(c*x)*ln(c*x)-3/2*I*b*c^2*d^3/x+3
*c^2*d^3*b*arctan(c*x)/x-1/3*d^3*b*arctan(c*x)/x^3-3/2*I*c*d^3*a/x^2+1/2*c^3*d^3*b*ln(c*x)*ln(1+I*c*x)-1/2*c^3
*d^3*b*ln(c*x)*ln(1-I*c*x)+1/2*c^3*d^3*b*dilog(1+I*c*x)-1/2*c^3*d^3*b*dilog(1-I*c*x)+5/3*b*c^3*d^3*ln(c^2*x^2+
1)-I*c^3*d^3*a*ln(c*x)-3/2*I*b*c^3*d^3*arctan(c*x)-1/6*b*c*d^3/x^2-10/3*c^3*d^3*b*ln(c*x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -i \, b c^{3} d^{3} \int \frac{\arctan \left (c x\right )}{x}\,{d x} - i \, a c^{3} d^{3} \log \left (x\right ) + \frac{3}{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 c^{2} d^{3} - \frac{3}{2} i \,{\left ({\left (c \arctan \left (c x\right ) + \frac{1}{x}\right )} c + \frac{\arctan \left (c x\right )}{x^{2}}\right )} b c d^{3} + \frac{1}{6} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac{1}{x^{2}}\right )} c - \frac{2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{3} + \frac{3 \, a c^{2} d^{3}}{x} - \frac{3 i \, a c d^{3}}{2 \, x^{2}} - \frac{a d^{3}}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

-I*b*c^3*d^3*integrate(arctan(c*x)/x, x) - I*a*c^3*d^3*log(x) + 3/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arcta
n(c*x)/x)*b*c^2*d^3 - 3/2*I*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^2)*b*c*d^3 + 1/6*((c^2*log(c^2*x^2 + 1) -
c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*d^3 + 3*a*c^2*d^3/x - 3/2*I*a*c*d^3/x^2 - 1/3*a*d^3/x^3

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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^{4}}, 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^4,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^4, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{3} \left (\int \frac{a}{x^{4}}\, dx + \int - \frac{3 a c^{2}}{x^{2}}\, dx + \int \frac{b \operatorname{atan}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{3 i a c}{x^{3}}\, dx + \int - \frac{i a c^{3}}{x}\, dx + \int - \frac{3 b c^{2} \operatorname{atan}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{3 i b c \operatorname{atan}{\left (c x \right )}}{x^{3}}\, dx + \int - \frac{i b c^{3} \operatorname{atan}{\left (c x \right )}}{x}\, 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**4,x)

[Out]

d**3*(Integral(a/x**4, x) + Integral(-3*a*c**2/x**2, x) + Integral(b*atan(c*x)/x**4, x) + Integral(3*I*a*c/x**
3, x) + Integral(-I*a*c**3/x, x) + Integral(-3*b*c**2*atan(c*x)/x**2, x) + Integral(3*I*b*c*atan(c*x)/x**3, x)
+ Integral(-I*b*c**3*atan(c*x)/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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((I*c*d*x + d)^3*(b*arctan(c*x) + a)/x^4, x)