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

Optimal. Leaf size=173 \[ \frac {1}{2} c^4 d^4 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 x-6 a c^2 d^4 \log (x)-\frac {1}{2} b c^3 d^4 x-4 i b c^3 d^4 x \tan ^{-1}(c x)-3 i b c^2 d^4 \text {Li}_2(-i c x)+3 i b c^2 d^4 \text {Li}_2(i c x)+4 i b c^2 d^4 \log (x)-\frac {b c d^4}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {4876, 4846, 260, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391, 321} \[ -3 i b c^2 d^4 \text {PolyLog}(2,-i c x)+3 i b c^2 d^4 \text {PolyLog}(2,i c x)+\frac {1}{2} c^4 d^4 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {d^4 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}-\frac {4 i c d^4 \left (a+b \tan ^{-1}(c x)\right )}{x}-4 i a c^3 d^4 x-6 a c^2 d^4 \log (x)-\frac {1}{2} b c^3 d^4 x+4 i b c^2 d^4 \log (x)-4 i b c^3 d^4 x \tan ^{-1}(c x)-\frac {b c d^4}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.15, size = 163, normalized size = 0.94 \[ \frac {d^4 \left (a c^4 x^4-8 i a c^3 x^3-12 a c^2 x^2 \log (x)-8 i a c x-a+b c^4 x^4 \tan ^{-1}(c x)-b c^3 x^3-8 i b c^3 x^3 \tan ^{-1}(c x)-6 i b c^2 x^2 \text {Li}_2(-i c x)+6 i b c^2 x^2 \text {Li}_2(i c x)+8 i b c^2 x^2 \log (c x)-b c x-8 i b c x \tan ^{-1}(c x)-b \tan ^{-1}(c x)\right )}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^4*(-a - (8*I)*a*c*x - b*c*x - (8*I)*a*c^3*x^3 - b*c^3*x^3 + a*c^4*x^4 - b*ArcTan[c*x] - (8*I)*b*c*x*ArcTan[
c*x] - (8*I)*b*c^3*x^3*ArcTan[c*x] + b*c^4*x^4*ArcTan[c*x] - 12*a*c^2*x^2*Log[x] + (8*I)*b*c^2*x^2*Log[c*x] -
(6*I)*b*c^2*x^2*PolyLog[2, (-I)*c*x] + (6*I)*b*c^2*x^2*PolyLog[2, I*c*x]))/(2*x^2)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {2 \, a c^{4} d^{4} x^{4} - 8 i \, a c^{3} d^{4} x^{3} - 12 \, a c^{2} d^{4} x^{2} + 8 i \, a c d^{4} x + 2 \, a d^{4} + {\left (i \, b c^{4} d^{4} x^{4} + 4 \, b c^{3} d^{4} x^{3} - 6 i \, b c^{2} d^{4} x^{2} - 4 \, b c d^{4} x + i \, b d^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.07, size = 248, normalized size = 1.43 \[ -4 i b \,c^{3} d^{4} x \arctan \left (c x \right )+\frac {c^{4} d^{4} a \,x^{2}}{2}-6 c^{2} d^{4} a \ln \left (c x \right )-4 i a \,c^{3} d^{4} x -\frac {d^{4} a}{2 x^{2}}-\frac {4 i c \,d^{4} b \arctan \left (c x \right )}{x}+\frac {c^{4} d^{4} b \arctan \left (c x \right ) x^{2}}{2}-6 c^{2} d^{4} b \ln \left (c x \right ) \arctan \left (c x \right )+4 i c^{2} d^{4} b \ln \left (c x \right )-\frac {d^{4} b \arctan \left (c x \right )}{2 x^{2}}-\frac {b \,c^{3} d^{4} x}{2}-3 i c^{2} d^{4} b \dilog \left (i c x +1\right )-\frac {b c \,d^{4}}{2 x}+3 i c^{2} d^{4} b \ln \left (c x \right ) \ln \left (-i c x +1\right )-3 i c^{2} d^{4} b \ln \left (c x \right ) \ln \left (i c x +1\right )-\frac {4 i c \,d^{4} a}{x}+3 i c^{2} d^{4} b \dilog \left (-i c x +1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 251, normalized size = 1.45 \[ \frac {1}{2} \, a c^{4} d^{4} x^{2} - 4 i \, a c^{3} d^{4} x - \frac {1}{2} \, b c^{3} d^{4} x + \frac {3}{2} \, \pi b c^{2} d^{4} \log \left (c^{2} x^{2} + 1\right ) - 6 \, b c^{2} d^{4} \arctan \left (c x\right ) \log \left (c x\right ) - 2 i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b c^{2} d^{4} + 3 i \, b c^{2} d^{4} {\rm Li}_2\left (i \, c x + 1\right ) - 3 i \, b c^{2} d^{4} {\rm Li}_2\left (-i \, c x + 1\right ) - 6 \, a c^{2} d^{4} \log \relax (x) - 2 i \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b c d^{4} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b d^{4} - \frac {4 i \, a c d^{4}}{x} - \frac {a d^{4}}{2 \, x^{2}} + \frac {1}{2} \, {\left (b c^{4} d^{4} x^{2} + b c^{2} d^{4}\right )} \arctan \left (c x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.89, size = 258, normalized size = 1.49 \[ \left \{\begin {array}{cl} -\frac {a\,d^4}{2\,x^2} & \text {\ if\ \ }c=0\\ \frac {a\,c^4\,d^4\,x^2}{2}-\frac {\frac {a\,d^4}{2}+a\,c\,d^4\,x\,4{}\mathrm {i}}{x^2}-6\,a\,c^2\,d^4\,\ln \relax (x)-\frac {b\,d^4\,\left (c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}\right )}{2\,c}-\frac {b\,c^3\,d^4\,x}{2}-\frac {b\,d^4\,\mathrm {atan}\left (c\,x\right )}{2\,x^2}+b\,c^4\,d^4\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )+b\,d^4\,\left (c^2\,\ln \relax (x)-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )\,4{}\mathrm {i}+b\,c^2\,d^4\,\ln \left (c^2\,x^2+1\right )\,2{}\mathrm {i}+b\,c^2\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}-b\,c^2\,d^4\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,3{}\mathrm {i}-a\,c^3\,d^4\,x\,4{}\mathrm {i}-\frac {b\,c\,d^4\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}}{x}-b\,c^3\,d^4\,x\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + c*d*x*1i)^4)/x^3,x)

[Out]

piecewise(c == 0, -(a*d^4)/(2*x^2), c ~= 0, - ((a*d^4)/2 + a*c*d^4*x*4i)/x^2 + b*d^4*(c^2*log(x) - (c^2*log(c^
2*x^2 + 1))/2)*4i + b*c^2*d^4*log(c^2*x^2 + 1)*2i + (a*c^4*d^4*x^2)/2 - 6*a*c^2*d^4*log(x) + b*c^2*d^4*dilog(-
c*x*1i + 1)*3i - b*c^2*d^4*dilog(c*x*1i + 1)*3i - (b*d^4*(c^3*atan(c*x) + c^2/x))/(2*c) - a*c^3*d^4*x*4i - (b
*c^3*d^4*x)/2 - (b*d^4*atan(c*x))/(2*x^2) - (b*c*d^4*atan(c*x)*4i)/x - b*c^3*d^4*x*atan(c*x)*4i + b*c^4*d^4*at
an(c*x)*(1/(2*c^2) + x^2/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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