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

Optimal. Leaf size=201 \[ 2 b c^3 d^4 \text{PolyLog}(2,-i c x)-2 b c^3 d^4 \text{PolyLog}(2,i c x)+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+a c^4 d^4 x-4 i a c^3 d^4 \log (x)+\frac{8}{3} b c^3 d^4 \log \left (c^2 x^2+1\right )-\frac{2 i b c^2 d^4}{x}-\frac{19}{3} b c^3 d^4 \log (x)-2 i b c^3 d^4 \tan ^{-1}(c x)+b c^4 d^4 x \tan ^{-1}(c x)-\frac{b c d^4}{6 x^2} \]

[Out]

-(b*c*d^4)/(6*x^2) - ((2*I)*b*c^2*d^4)/x + a*c^4*d^4*x - (2*I)*b*c^3*d^4*ArcTan[c*x] + b*c^4*d^4*x*ArcTan[c*x]
- (d^4*(a + b*ArcTan[c*x]))/(3*x^3) - ((2*I)*c*d^4*(a + b*ArcTan[c*x]))/x^2 + (6*c^2*d^4*(a + b*ArcTan[c*x]))
/x - (4*I)*a*c^3*d^4*Log[x] - (19*b*c^3*d^4*Log[x])/3 + (8*b*c^3*d^4*Log[1 + c^2*x^2])/3 + 2*b*c^3*d^4*PolyLog
[2, (-I)*c*x] - 2*b*c^3*d^4*PolyLog[2, I*c*x]

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Rubi [A]  time = 0.217667, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {4876, 4846, 260, 4852, 266, 44, 325, 203, 36, 29, 31, 4848, 2391} \[ 2 b c^3 d^4 \text{PolyLog}(2,-i c x)-2 b c^3 d^4 \text{PolyLog}(2,i c x)+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+a c^4 d^4 x-4 i a c^3 d^4 \log (x)+\frac{8}{3} b c^3 d^4 \log \left (c^2 x^2+1\right )-\frac{2 i b c^2 d^4}{x}-\frac{19}{3} b c^3 d^4 \log (x)-2 i b c^3 d^4 \tan ^{-1}(c x)+b c^4 d^4 x \tan ^{-1}(c x)-\frac{b c d^4}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^4)/(6*x^2) - ((2*I)*b*c^2*d^4)/x + a*c^4*d^4*x - (2*I)*b*c^3*d^4*ArcTan[c*x] + b*c^4*d^4*x*ArcTan[c*x]
- (d^4*(a + b*ArcTan[c*x]))/(3*x^3) - ((2*I)*c*d^4*(a + b*ArcTan[c*x]))/x^2 + (6*c^2*d^4*(a + b*ArcTan[c*x]))
/x - (4*I)*a*c^3*d^4*Log[x] - (19*b*c^3*d^4*Log[x])/3 + (8*b*c^3*d^4*Log[1 + c^2*x^2])/3 + 2*b*c^3*d^4*PolyLog
[2, (-I)*c*x] - 2*b*c^3*d^4*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 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)^4 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=\int \left (c^4 d^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^4}+\frac{4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac{4 i c^3 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^4 \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx+\left (4 i c d^4\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx-\left (6 c^2 d^4\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (4 i c^3 d^4\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+\left (c^4 d^4\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=a c^4 d^4 x-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 \log (x)+\frac{1}{3} \left (b c d^4\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx+\left (2 i b c^2 d^4\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\left (2 b c^3 d^4\right ) \int \frac{\log (1-i c x)}{x} \, dx-\left (2 b c^3 d^4\right ) \int \frac{\log (1+i c x)}{x} \, dx-\left (6 b c^3 d^4\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx+\left (b c^4 d^4\right ) \int \tan ^{-1}(c x) \, dx\\ &=-\frac{2 i b c^2 d^4}{x}+a c^4 d^4 x+b c^4 d^4 x \tan ^{-1}(c x)-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 \log (x)+2 b c^3 d^4 \text{Li}_2(-i c x)-2 b c^3 d^4 \text{Li}_2(i c x)+\frac{1}{6} \left (b c d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\left (3 b c^3 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\left (2 i b c^4 d^4\right ) \int \frac{1}{1+c^2 x^2} \, dx-\left (b c^5 d^4\right ) \int \frac{x}{1+c^2 x^2} \, dx\\ &=-\frac{2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \tan ^{-1}(c x)+b c^4 d^4 x \tan ^{-1}(c x)-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 \log (x)-\frac{1}{2} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \text{Li}_2(-i c x)-2 b c^3 d^4 \text{Li}_2(i c x)+\frac{1}{6} \left (b c d^4\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 )-\left (3 b c^3 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\left (3 b c^5 d^4\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b c d^4}{6 x^2}-\frac{2 i b c^2 d^4}{x}+a c^4 d^4 x-2 i b c^3 d^4 \tan ^{-1}(c x)+b c^4 d^4 x \tan ^{-1}(c x)-\frac{d^4 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{2 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{6 c^2 d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 \log (x)-\frac{19}{3} b c^3 d^4 \log (x)+\frac{8}{3} b c^3 d^4 \log \left (1+c^2 x^2\right )+2 b c^3 d^4 \text{Li}_2(-i c x)-2 b c^3 d^4 \text{Li}_2(i c x)\\ \end{align*}

Mathematica [A]  time = 0.147416, size = 193, normalized size = 0.96 \[ \frac{d^4 \left (12 b c^3 x^3 \text{PolyLog}(2,-i c x)-12 b c^3 x^3 \text{PolyLog}(2,i c x)+6 a c^4 x^4+36 a c^2 x^2-24 i a c^3 x^3 \log (x)-12 i a c x-2 a-12 i b c^2 x^2-38 b c^3 x^3 \log (c x)+16 b c^3 x^3 \log \left (c^2 x^2+1\right )+6 b c^4 x^4 \tan ^{-1}(c x)-12 i b c^3 x^3 \tan ^{-1}(c x)+36 b c^2 x^2 \tan ^{-1}(c x)-b c x-12 i b c x \tan ^{-1}(c x)-2 b \tan ^{-1}(c x)\right )}{6 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^4*(-2*a - (12*I)*a*c*x - b*c*x + 36*a*c^2*x^2 - (12*I)*b*c^2*x^2 + 6*a*c^4*x^4 - 2*b*ArcTan[c*x] - (12*I)*b
*c*x*ArcTan[c*x] + 36*b*c^2*x^2*ArcTan[c*x] - (12*I)*b*c^3*x^3*ArcTan[c*x] + 6*b*c^4*x^4*ArcTan[c*x] - (24*I)*
a*c^3*x^3*Log[x] - 38*b*c^3*x^3*Log[c*x] + 16*b*c^3*x^3*Log[1 + c^2*x^2] + 12*b*c^3*x^3*PolyLog[2, (-I)*c*x] -
12*b*c^3*x^3*PolyLog[2, I*c*x]))/(6*x^3)

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Maple [A]  time = 0.047, size = 277, normalized size = 1.4 \begin{align*} a{c}^{4}{d}^{4}x-4\,i{c}^{3}{d}^{4}a\ln \left ( cx \right ) +6\,{\frac{{c}^{2}{d}^{4}a}{x}}-{\frac{{d}^{4}a}{3\,{x}^{3}}}-{\frac{2\,ic{d}^{4}b\arctan \left ( cx \right ) }{{x}^{2}}}+b{c}^{4}{d}^{4}x\arctan \left ( cx \right ) -{\frac{2\,ib{c}^{2}{d}^{4}}{x}}+6\,{\frac{{c}^{2}{d}^{4}b\arctan \left ( cx \right ) }{x}}-{\frac{b{d}^{4}\arctan \left ( cx \right ) }{3\,{x}^{3}}}-4\,i{c}^{3}{d}^{4}b\arctan \left ( cx \right ) \ln \left ( cx \right ) +2\,{c}^{3}{d}^{4}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -2\,{c}^{3}{d}^{4}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +2\,{c}^{3}{d}^{4}b{\it dilog} \left ( 1+icx \right ) -2\,{c}^{3}{d}^{4}b{\it dilog} \left ( 1-icx \right ) +{\frac{8\,b{c}^{3}{d}^{4}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{3}}-{\frac{2\,ic{d}^{4}a}{{x}^{2}}}-{\frac{{d}^{4}bc}{6\,{x}^{2}}}-2\,ib{c}^{3}{d}^{4}\arctan \left ( cx \right ) -{\frac{19\,{c}^{3}{d}^{4}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)^4*(a+b*arctan(c*x))/x^4,x)

[Out]

a*c^4*d^4*x-4*I*c^3*d^4*a*ln(c*x)+6*c^2*d^4*a/x-1/3*d^4*a/x^3-2*I*c*d^4*b*arctan(c*x)/x^2+b*c^4*d^4*x*arctan(c
*x)-2*I*b*c^2*d^4/x+6*c^2*d^4*b*arctan(c*x)/x-1/3*d^4*b*arctan(c*x)/x^3-4*I*c^3*d^4*b*arctan(c*x)*ln(c*x)+2*c^
3*d^4*b*ln(c*x)*ln(1+I*c*x)-2*c^3*d^4*b*ln(c*x)*ln(1-I*c*x)+2*c^3*d^4*b*dilog(1+I*c*x)-2*c^3*d^4*b*dilog(1-I*c
*x)+8/3*b*c^3*d^4*ln(c^2*x^2+1)-2*I*c*d^4*a/x^2-1/6*b*c*d^4/x^2-2*I*b*c^3*d^4*arctan(c*x)-19/3*c^3*d^4*b*ln(c*
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^4*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

a*c^4*d^4*x + 1/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*c^3*d^4 - 4*I*b*c^3*d^4*integrate(arctan(c*x)/x, x)
- 4*I*a*c^3*d^4*log(x) + 3*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*c^2*d^4 - 2*I*((c*arctan(c*x
) + 1/x)*c + arctan(c*x)/x^2)*b*c*d^4 + 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^4 + 6*a*c^2*d^4/x - 2*I*a*c*d^4/x^2 - 1/3*a*d^4/x^3

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{2 \, a c^{4} d^{4} x^{4} - 8 i \, a c^{3} d^{4} x^{3} - 12 \, a c^{2} d^{4} x^{2} + 8 i \, a c d^{4} x + 2 \, a d^{4} +{\left (i \, b c^{4} d^{4} x^{4} + 4 \, b c^{3} d^{4} x^{3} - 6 i \, b c^{2} d^{4} x^{2} - 4 \, b c d^{4} x + i \, b d^{4}\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)^4*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

integral(1/2*(2*a*c^4*d^4*x^4 - 8*I*a*c^3*d^4*x^3 - 12*a*c^2*d^4*x^2 + 8*I*a*c*d^4*x + 2*a*d^4 + (I*b*c^4*d^4*
x^4 + 4*b*c^3*d^4*x^3 - 6*I*b*c^2*d^4*x^2 - 4*b*c*d^4*x + I*b*d^4)*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^{4} \left (\int a c^{4}\, dx + \int \frac{a}{x^{4}}\, dx + \int - \frac{6 a c^{2}}{x^{2}}\, dx + \int b c^{4} \operatorname{atan}{\left (c x \right )}\, dx + \int \frac{b \operatorname{atan}{\left (c x \right )}}{x^{4}}\, dx + \int \frac{4 i a c}{x^{3}}\, dx + \int - \frac{4 i a c^{3}}{x}\, dx + \int - \frac{6 b c^{2} \operatorname{atan}{\left (c x \right )}}{x^{2}}\, dx + \int \frac{4 i b c \operatorname{atan}{\left (c x \right )}}{x^{3}}\, dx + \int - \frac{4 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)**4*(a+b*atan(c*x))/x**4,x)

[Out]

d**4*(Integral(a*c**4, x) + Integral(a/x**4, x) + Integral(-6*a*c**2/x**2, x) + Integral(b*c**4*atan(c*x), x)
+ Integral(b*atan(c*x)/x**4, x) + Integral(4*I*a*c/x**3, x) + Integral(-4*I*a*c**3/x, x) + Integral(-6*b*c**2*
atan(c*x)/x**2, x) + Integral(4*I*b*c*atan(c*x)/x**3, x) + Integral(-4*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 )}^{4}{\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)^4*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

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