3.83 \(\int \frac{(d+i c d x)^2 (a+b \tan ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=267 \[ -\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}(2,-i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}(2,i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )-\frac{8}{3} a b c^3 d^2 \log (x)-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{8}{3} b c^3 d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-i b^2 c^3 d^2 \log \left (c^2 x^2+1\right )-\frac{b^2 c^2 d^2}{3 x}+2 i b^2 c^3 d^2 \log (x)-\frac{1}{3} b^2 c^3 d^2 \tan ^{-1}(c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.268309, antiderivative size = 267, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 14, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.56, Rules used = {37, 4874, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391, 4854, 2402, 2315} \[ -\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}(2,-i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}(2,i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )-\frac{8}{3} a b c^3 d^2 \log (x)-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{8}{3} b c^3 d^2 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-i b^2 c^3 d^2 \log \left (c^2 x^2+1\right )-\frac{b^2 c^2 d^2}{3 x}+2 i b^2 c^3 d^2 \log (x)-\frac{1}{3} b^2 c^3 d^2 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^2*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 4874

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_)*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[(a + b*ArcTan[c*x])^p, u, x] - Dist[b*c*p, Int[ExpandIntegrand[(a +
b*ArcTan[c*x])^(p - 1), u/(1 + c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && IGtQ[p, 1] && EqQ[c
^2*d^2 + e^2, 0] && IntegersQ[m, q] && NeQ[m, -1] && NeQ[q, -1] && ILtQ[m + q + 1, 0] && LtQ[m*q, 0]

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]

Rule 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2/(1 + (e*x)/d)])/(1 + c^2*x^
2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

\begin{align*} \int \frac{(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )^2}{x^4} \, dx &=-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-(2 b c) \int \left (-\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac{i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{4 c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x}-\frac{4 c^3 d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 (i+c x)}\right ) \, dx\\ &=-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}+\frac{1}{3} \left (2 b c d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx+\left (2 i b c^2 d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\frac{1}{3} \left (8 b c^3 d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+\frac{1}{3} \left (8 b c^4 d^2\right ) \int \frac{a+b \tan ^{-1}(c x)}{i+c x} \, dx\\ &=-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{8}{3} a b c^3 d^2 \log (x)-\frac{8}{3} b c^3 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{3} \left (b^2 c^2 d^2\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac{1}{3} \left (4 i b^2 c^3 d^2\right ) \int \frac{\log (1-i c x)}{x} \, dx+\frac{1}{3} \left (4 i b^2 c^3 d^2\right ) \int \frac{\log (1+i c x)}{x} \, dx+\left (2 i b^2 c^3 d^2\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx+\frac{1}{3} \left (8 b^2 c^4 d^2\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac{b^2 c^2 d^2}{3 x}-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{8}{3} a b c^3 d^2 \log (x)-\frac{8}{3} b c^3 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )-\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(-i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(i c x)+\left (i b^2 c^3 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\frac{1}{3} \left (8 i b^2 c^3 d^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )-\frac{1}{3} \left (b^2 c^4 d^2\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{b^2 c^2 d^2}{3 x}-\frac{1}{3} b^2 c^3 d^2 \tan ^{-1}(c x)-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{8}{3} a b c^3 d^2 \log (x)-\frac{8}{3} b c^3 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )-\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(-i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )+\left (i b^2 c^3 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\left (i b^2 c^5 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2 d^2}{3 x}-\frac{1}{3} b^2 c^3 d^2 \tan ^{-1}(c x)-\frac{b c d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^2}-\frac{2 i b c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{d^2 (1+i c x)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{3 x^3}-\frac{8}{3} a b c^3 d^2 \log (x)+2 i b^2 c^3 d^2 \log (x)-\frac{8}{3} b c^3 d^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )-i b^2 c^3 d^2 \log \left (1+c^2 x^2\right )-\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(-i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2(i c x)+\frac{4}{3} i b^2 c^3 d^2 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.65824, size = 253, normalized size = 0.95 \[ \frac{d^2 \left (4 i b^2 c^3 x^3 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+3 a^2 c^2 x^2-3 i a^2 c x-a^2-6 i a b c^2 x^2-8 a b c^3 x^3 \log (c x)+4 a b c^3 x^3 \log \left (c^2 x^2+1\right )-b \tan ^{-1}(c x) \left (a \left (6 i c^3 x^3-6 c^2 x^2+6 i c x+2\right )+b c x \left (c^2 x^2+6 i c x+1\right )+8 b c^3 x^3 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-a b c x-b^2 c^2 x^2+6 i b^2 c^3 x^3 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+b^2 (-1-i c x)^3 \tan ^{-1}(c x)^2\right )}{3 x^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^2*(a + b*ArcTan[c*x])^2)/x^4,x]

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.116, size = 669, normalized size = 2.5 \begin{align*} \text{result too large to display} \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))^2/x^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))^2/x^4,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

1/12*(12*x^3*integral(-1/3*(3*a^2*c^4*d^2*x^4 - 6*I*a^2*c^3*d^2*x^3 - 6*I*a^2*c*d^2*x - 3*a^2*d^2 - (-3*I*a*b*
c^4*d^2*x^4 - 3*(2*a*b + I*b^2)*c^3*d^2*x^3 - 3*b^2*c^2*d^2*x^2 - (6*a*b - I*b^2)*c*d^2*x + 3*I*a*b*d^2)*log(-
(c*x + I)/(c*x - I)))/(c^2*x^6 + x^4), x) - (3*b^2*c^2*d^2*x^2 - 3*I*b^2*c*d^2*x - b^2*d^2)*log(-(c*x + I)/(c*
x - I))^2)/x^3

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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))**2/x**4,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

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 )}^{2}}{x^{4}}\,{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))^2/x^4,x, algorithm="giac")

[Out]

integrate((I*c*d*x + d)^2*(b*arctan(c*x) + a)^2/x^4, x)