3.92 \(\int \frac{(d+i c d x)^3 (a+b \tan ^{-1}(c x))^2}{x^5} \, dx\)

Optimal. Leaf size=293 \[ 2 b^2 c^4 d^3 \text{PolyLog}(2,-i c x)-2 b^2 c^4 d^3 \text{PolyLog}(2,i c x)-2 b^2 c^4 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-4 i a b c^4 d^3 \log (x)+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-4 i b c^4 d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2 d^3}{12 x^2}+\frac{11}{6} b^2 c^4 d^3 \log \left (c^2 x^2+1\right )-\frac{i b^2 c^3 d^3}{x}-\frac{11}{3} b^2 c^4 d^3 \log (x)-i b^2 c^4 d^3 \tan ^{-1}(c x) \]

[Out]

-(b^2*c^2*d^3)/(12*x^2) - (I*b^2*c^3*d^3)/x - I*b^2*c^4*d^3*ArcTan[c*x] - (b*c*d^3*(a + b*ArcTan[c*x]))/(6*x^3
) - (I*b*c^2*d^3*(a + b*ArcTan[c*x]))/x^2 + (7*b*c^3*d^3*(a + b*ArcTan[c*x]))/(2*x) - (d^3*(1 + I*c*x)^4*(a +
b*ArcTan[c*x])^2)/(4*x^4) - (4*I)*a*b*c^4*d^3*Log[x] - (11*b^2*c^4*d^3*Log[x])/3 - (4*I)*b*c^4*d^3*(a + b*ArcT
an[c*x])*Log[2/(1 - I*c*x)] + (11*b^2*c^4*d^3*Log[1 + c^2*x^2])/6 + 2*b^2*c^4*d^3*PolyLog[2, (-I)*c*x] - 2*b^2
*c^4*d^3*PolyLog[2, I*c*x] - 2*b^2*c^4*d^3*PolyLog[2, 1 - 2/(1 - I*c*x)]

________________________________________________________________________________________

Rubi [A]  time = 0.321967, antiderivative size = 293, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 15, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {37, 4874, 4852, 266, 44, 325, 203, 36, 29, 31, 4848, 2391, 4854, 2402, 2315} \[ 2 b^2 c^4 d^3 \text{PolyLog}(2,-i c x)-2 b^2 c^4 d^3 \text{PolyLog}(2,i c x)-2 b^2 c^4 d^3 \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-4 i a b c^4 d^3 \log (x)+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-4 i b c^4 d^3 \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac{b^2 c^2 d^3}{12 x^2}+\frac{11}{6} b^2 c^4 d^3 \log \left (c^2 x^2+1\right )-\frac{i b^2 c^3 d^3}{x}-\frac{11}{3} b^2 c^4 d^3 \log (x)-i b^2 c^4 d^3 \tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^5,x]

[Out]

-(b^2*c^2*d^3)/(12*x^2) - (I*b^2*c^3*d^3)/x - I*b^2*c^4*d^3*ArcTan[c*x] - (b*c*d^3*(a + b*ArcTan[c*x]))/(6*x^3
) - (I*b*c^2*d^3*(a + b*ArcTan[c*x]))/x^2 + (7*b*c^3*d^3*(a + b*ArcTan[c*x]))/(2*x) - (d^3*(1 + I*c*x)^4*(a +
b*ArcTan[c*x])^2)/(4*x^4) - (4*I)*a*b*c^4*d^3*Log[x] - (11*b^2*c^4*d^3*Log[x])/3 - (4*I)*b*c^4*d^3*(a + b*ArcT
an[c*x])*Log[2/(1 - I*c*x)] + (11*b^2*c^4*d^3*Log[1 + c^2*x^2])/6 + 2*b^2*c^4*d^3*PolyLog[2, (-I)*c*x] - 2*b^2
*c^4*d^3*PolyLog[2, I*c*x] - 2*b^2*c^4*d^3*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 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]

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)^3 \left (a+b \tan ^{-1}(c x)\right )^2}{x^5} \, dx &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-(2 b c) \int \left (-\frac{d^3 \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac{i c d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac{7 c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{4 x^2}+\frac{2 i c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac{2 i c^4 d^3 \left (a+b \tan ^{-1}(c x)\right )}{i+c x}\right ) \, dx\\ &=-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}+\frac{1}{2} \left (b c d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^4} \, dx+\left (2 i b c^2 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^3} \, dx-\frac{1}{2} \left (7 b c^3 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x^2} \, dx-\left (4 i b c^4 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+\left (4 i b c^5 d^3\right ) \int \frac{a+b \tan ^{-1}(c x)}{i+c x} \, dx\\ &=-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-4 i a b c^4 d^3 \log (x)-4 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{1}{6} \left (b^2 c^2 d^3\right ) \int \frac{1}{x^3 \left (1+c^2 x^2\right )} \, dx+\left (i b^2 c^3 d^3\right ) \int \frac{1}{x^2 \left (1+c^2 x^2\right )} \, dx+\left (2 b^2 c^4 d^3\right ) \int \frac{\log (1-i c x)}{x} \, dx-\left (2 b^2 c^4 d^3\right ) \int \frac{\log (1+i c x)}{x} \, dx-\frac{1}{2} \left (7 b^2 c^4 d^3\right ) \int \frac{1}{x \left (1+c^2 x^2\right )} \, dx+\left (4 i b^2 c^5 d^3\right ) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=-\frac{i b^2 c^3 d^3}{x}-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-4 i a b c^4 d^3 \log (x)-4 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+2 b^2 c^4 d^3 \text{Li}_2(-i c x)-2 b^2 c^4 d^3 \text{Li}_2(i c x)+\frac{1}{12} \left (b^2 c^2 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )-\frac{1}{4} \left (7 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )-\left (4 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )-\left (i b^2 c^5 d^3\right ) \int \frac{1}{1+c^2 x^2} \, dx\\ &=-\frac{i b^2 c^3 d^3}{x}-i b^2 c^4 d^3 \tan ^{-1}(c x)-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-4 i a b c^4 d^3 \log (x)-4 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+2 b^2 c^4 d^3 \text{Li}_2(-i c x)-2 b^2 c^4 d^3 \text{Li}_2(i c x)-2 b^2 c^4 d^3 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )+\frac{1}{12} \left (b^2 c^2 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}{4} \left (7 b^2 c^4 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{4} \left (7 b^2 c^6 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-\frac{b^2 c^2 d^3}{12 x^2}-\frac{i b^2 c^3 d^3}{x}-i b^2 c^4 d^3 \tan ^{-1}(c x)-\frac{b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}-\frac{i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac{7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac{d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-4 i a b c^4 d^3 \log (x)-\frac{11}{3} b^2 c^4 d^3 \log (x)-4 i b c^4 d^3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )+\frac{11}{6} b^2 c^4 d^3 \log \left (1+c^2 x^2\right )+2 b^2 c^4 d^3 \text{Li}_2(-i c x)-2 b^2 c^4 d^3 \text{Li}_2(i c x)-2 b^2 c^4 d^3 \text{Li}_2\left (1-\frac{2}{1-i c x}\right )\\ \end{align*}

Mathematica [A]  time = 0.834639, size = 322, normalized size = 1.1 \[ \frac{d^3 \left (-24 b^2 c^4 x^4 \text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+12 i a^2 c^3 x^3+18 a^2 c^2 x^2-12 i a^2 c x-3 a^2+42 a b c^3 x^3-12 i a b c^2 x^2-48 i a b c^4 x^4 \log (c x)+24 i a b c^4 x^4 \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (3 a \left (7 c^4 x^4+4 i c^3 x^3+6 c^2 x^2-4 i c x-1\right )+b c x \left (-6 i c^3 x^3+21 c^2 x^2-6 i c x-1\right )-24 i b c^4 x^4 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )-2 a b c x-b^2 c^4 x^4-12 i b^2 c^3 x^3-b^2 c^2 x^2-44 b^2 c^4 x^4 \log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )-3 b^2 (c x-i)^4 \tan ^{-1}(c x)^2\right )}{12 x^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((d + I*c*d*x)^3*(a + b*ArcTan[c*x])^2)/x^5,x]

[Out]

(d^3*(-3*a^2 - (12*I)*a^2*c*x - 2*a*b*c*x + 18*a^2*c^2*x^2 - (12*I)*a*b*c^2*x^2 - b^2*c^2*x^2 + (12*I)*a^2*c^3
*x^3 + 42*a*b*c^3*x^3 - (12*I)*b^2*c^3*x^3 - b^2*c^4*x^4 - 3*b^2*(-I + c*x)^4*ArcTan[c*x]^2 + 2*b*ArcTan[c*x]*
(b*c*x*(-1 - (6*I)*c*x + 21*c^2*x^2 - (6*I)*c^3*x^3) + 3*a*(-1 - (4*I)*c*x + 6*c^2*x^2 + (4*I)*c^3*x^3 + 7*c^4
*x^4) - (24*I)*b*c^4*x^4*Log[1 - E^((2*I)*ArcTan[c*x])]) - (48*I)*a*b*c^4*x^4*Log[c*x] - 44*b^2*c^4*x^4*Log[(c
*x)/Sqrt[1 + c^2*x^2]] + (24*I)*a*b*c^4*x^4*Log[1 + c^2*x^2] - 24*b^2*c^4*x^4*PolyLog[2, E^((2*I)*ArcTan[c*x])
]))/(12*x^4)

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Maple [B]  time = 0.111, size = 757, normalized size = 2.6 \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)^3*(a+b*arctan(c*x))^2/x^5,x)

[Out]

c^4*d^3*b^2*dilog(-1/2*I*(c*x+I))-1/2*c^4*d^3*b^2*ln(c*x+I)^2+1/2*c^4*d^3*b^2*ln(c*x-I)^2+7/4*c^4*d^3*b^2*arct
an(c*x)^2-11/3*c^4*d^3*b^2*ln(c*x)+2*c^4*d^3*b^2*dilog(1+I*c*x)-2*c^4*d^3*b^2*dilog(1-I*c*x)-c^4*d^3*b^2*dilog
(1/2*I*(c*x-I))-1/4*d^3*b^2*arctan(c*x)^2/x^4+3/2*c^2*d^3*a^2/x^2-1/12*b^2*c^2*d^3/x^2+2*c^4*d^3*b^2*ln(c*x)*l
n(1+I*c*x)+7/2*c^3*d^3*b^2*arctan(c*x)/x-1/2*d^3*a*b*arctan(c*x)/x^4-I*c*d^3*a^2/x^3-1/6*c*d^3*a*b/x^3+7/2*c^3
*d^3*a*b/x-2*c^4*d^3*b^2*ln(c*x)*ln(1-I*c*x)+c^4*d^3*b^2*ln(c*x+I)*ln(c^2*x^2+1)-c^4*d^3*b^2*ln(c*x+I)*ln(1/2*
I*(c*x-I))-c^4*d^3*b^2*ln(c*x-I)*ln(c^2*x^2+1)+c^4*d^3*b^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))+7/2*c^4*d^3*a*b*arctan
(c*x)+I*c^3*d^3*a^2/x+3/2*c^2*d^3*b^2*arctan(c*x)^2/x^2-1/6*c*d^3*b^2*arctan(c*x)/x^3+2*I*c^4*d^3*a*b*ln(c^2*x
^2+1)-I*c^2*d^3*b^2*arctan(c*x)/x^2-I*c*d^3*b^2*arctan(c*x)^2/x^3-I*c^2*d^3*a*b/x^2+2*I*c^4*d^3*b^2*arctan(c*x
)*ln(c^2*x^2+1)-4*I*c^4*d^3*a*b*ln(c*x)+I*c^3*d^3*b^2*arctan(c*x)^2/x+3*c^2*d^3*a*b*arctan(c*x)/x^2-4*I*c^4*d^
3*b^2*arctan(c*x)*ln(c*x)-1/4*d^3*a^2/x^4-I*b^2*c^3*d^3/x-I*b^2*c^4*d^3*arctan(c*x)-2*I*c*d^3*a*b*arctan(c*x)/
x^3+2*I*c^3*d^3*a*b*arctan(c*x)/x+11/6*b^2*c^4*d^3*ln(c^2*x^2+1)

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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)^3*(a+b*arctan(c*x))^2/x^5,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{16 \, x^{4}{\rm integral}\left (\frac{-4 i \, a^{2} c^{5} d^{3} x^{5} - 12 \, a^{2} c^{4} d^{3} x^{4} + 8 i \, a^{2} c^{3} d^{3} x^{3} - 8 \, a^{2} c^{2} d^{3} x^{2} + 12 i \, a^{2} c d^{3} x + 4 \, a^{2} d^{3} +{\left (4 \, a b c^{5} d^{3} x^{5} +{\left (-12 i \, a b + 4 \, b^{2}\right )} c^{4} d^{3} x^{4} - 2 \,{\left (4 \, a b + 3 i \, b^{2}\right )} c^{3} d^{3} x^{3} +{\left (-8 i \, a b - 4 \, b^{2}\right )} c^{2} d^{3} x^{2} -{\left (12 \, a b - i \, b^{2}\right )} c d^{3} x + 4 i \, a b d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{4 \,{\left (c^{2} x^{7} + x^{5}\right )}}, x\right ) +{\left (-4 i \, b^{2} c^{3} d^{3} x^{3} - 6 \, b^{2} c^{2} d^{3} x^{2} + 4 i \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2}}{16 \, x^{4}} \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))^2/x^5,x, algorithm="fricas")

[Out]

1/16*(16*x^4*integral(1/4*(-4*I*a^2*c^5*d^3*x^5 - 12*a^2*c^4*d^3*x^4 + 8*I*a^2*c^3*d^3*x^3 - 8*a^2*c^2*d^3*x^2
+ 12*I*a^2*c*d^3*x + 4*a^2*d^3 + (4*a*b*c^5*d^3*x^5 + (-12*I*a*b + 4*b^2)*c^4*d^3*x^4 - 2*(4*a*b + 3*I*b^2)*c
^3*d^3*x^3 + (-8*I*a*b - 4*b^2)*c^2*d^3*x^2 - (12*a*b - I*b^2)*c*d^3*x + 4*I*a*b*d^3)*log(-(c*x + I)/(c*x - I)
))/(c^2*x^7 + x^5), x) + (-4*I*b^2*c^3*d^3*x^3 - 6*b^2*c^2*d^3*x^2 + 4*I*b^2*c*d^3*x + b^2*d^3)*log(-(c*x + I)
/(c*x - I))^2)/x^4

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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)**3*(a+b*atan(c*x))**2/x**5,x)

[Out]

Exception raised: AttributeError

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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 )}^{2}}{x^{5}}\,{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))^2/x^5,x, algorithm="giac")

[Out]

integrate((I*c*d*x + d)^3*(b*arctan(c*x) + a)^2/x^5, x)