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

Optimal. Leaf size=293 \[ -4 i a b c^4 d^3 \log (x)-4 i b c^4 d^3 \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {7 b c^3 d^3 \left (a+b \tan ^{-1}(c x)\right )}{2 x}-\frac {i b c^2 d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {d^3 (1+i c x)^4 \left (a+b \tan ^{-1}(c x)\right )^2}{4 x^4}-\frac {b c d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^3}+2 b^2 c^4 d^3 \text {Li}_2(-i c x)-2 b^2 c^4 d^3 \text {Li}_2(i c x)-2 b^2 c^4 d^3 \text {Li}_2\left (1-\frac {2}{1-i c x}\right )-\frac {11}{3} b^2 c^4 d^3 \log (x)-i b^2 c^4 d^3 \tan ^{-1}(c x)-\frac {i b^2 c^3 d^3}{x}-\frac {b^2 c^2 d^3}{12 x^2}+\frac {11}{6} b^2 c^4 d^3 \log \left (c^2 x^2+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.32, antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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 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 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 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 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 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]

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.90, size = 322, normalized size = 1.10 \[ \frac {d^3 \left (12 i a^2 c^3 x^3+18 a^2 c^2 x^2-12 i a^2 c x-3 a^2-48 i a b c^4 x^4 \log (c x)+42 a b c^3 x^3-12 i a b c^2 x^2+24 i a b c^4 x^4 \log \left (c^2 x^2+1\right )+2 b \tan ^{-1}(c x) \left (3 a \left (7 c^4 x^4+4 i c^3 x^3+6 c^2 x^2-4 i c x-1\right )-24 i b c^4 x^4 \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )+b c x \left (-6 i c^3 x^3+21 c^2 x^2-6 i c x-1\right )\right )-2 a b c x-24 b^2 c^4 x^4 \text {Li}_2\left (e^{2 i \tan ^{-1}(c x)}\right )-b^2 c^4 x^4-12 i b^2 c^3 x^3-b^2 c^2 x^2-44 b^2 c^4 x^4 \log \left (\frac {c x}{\sqrt {c^2 x^2+1}}\right )-3 b^2 (c x-i)^4 \tan ^{-1}(c x)^2\right )}{12 x^4} \]

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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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \frac {16 \, x^{4} {\rm integral}\left (\frac {-4 i \, a^{2} c^{5} d^{3} x^{5} - 12 \, a^{2} c^{4} d^{3} x^{4} + 8 i \, a^{2} c^{3} d^{3} x^{3} - 8 \, a^{2} c^{2} d^{3} x^{2} + 12 i \, a^{2} c d^{3} x + 4 \, a^{2} d^{3} + {\left (4 \, a b c^{5} d^{3} x^{5} - 4 \, {\left (3 i \, a b - b^{2}\right )} c^{4} d^{3} x^{4} - {\left (8 \, a b + 6 i \, b^{2}\right )} c^{3} d^{3} x^{3} - 4 \, {\left (2 i \, a b + b^{2}\right )} c^{2} d^{3} x^{2} - {\left (12 \, a b - i \, b^{2}\right )} c d^{3} x + 4 i \, a b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{4 \, {\left (c^{2} x^{7} + x^{5}\right )}}, x\right ) + {\left (-4 i \, b^{2} c^{3} d^{3} x^{3} - 6 \, b^{2} c^{2} d^{3} x^{2} + 4 i \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2}}{16 \, x^{4}} \]

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 - 4*(3*I*a*b - b^2)*c^4*d^3*x^4 - (8*a*b + 6*I*b^2)*c^3*d
^3*x^3 - 4*(2*I*a*b + 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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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)^3*(a+b*arctan(c*x))^2/x^5,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.12, size = 757, normalized size = 2.58 \[ -\frac {d^{3} a^{2}}{4 x^{4}}-\frac {2 i c \,d^{3} a b \arctan \left (c x \right )}{x^{3}}+\frac {2 i c^{3} d^{3} a b \arctan \left (c x \right )}{x}-\frac {b^{2} c^{2} d^{3}}{12 x^{2}}-\frac {i c^{2} d^{3} a b}{x^{2}}+\frac {11 b^{2} c^{4} d^{3} \ln \left (c^{2} x^{2}+1\right )}{6}+c^{4} d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (c^{2} x^{2}+1\right )-c^{4} d^{3} b^{2} \ln \left (c x +i\right ) \ln \left (\frac {i \left (c x -i\right )}{2}\right )-\frac {i b^{2} c^{3} d^{3}}{x}-i b^{2} c^{4} d^{3} \arctan \left (c x \right )-c^{4} d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (c^{2} x^{2}+1\right )+c^{4} d^{3} b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )-\frac {i c \,d^{3} a^{2}}{x^{3}}+\frac {i c^{3} d^{3} a^{2}}{x}+2 c^{4} d^{3} b^{2} \ln \left (c x \right ) \ln \left (i c x +1\right )-2 c^{4} d^{3} b^{2} \ln \left (c x \right ) \ln \left (-i c x +1\right )+\frac {7 c^{3} d^{3} a b}{2 x}-\frac {c \,d^{3} a b}{6 x^{3}}+\frac {7 c^{3} d^{3} b^{2} \arctan \left (c x \right )}{2 x}-\frac {d^{3} a b \arctan \left (c x \right )}{2 x^{4}}+\frac {3 c^{2} d^{3} b^{2} \arctan \left (c x \right )^{2}}{2 x^{2}}+\frac {7 c^{4} d^{3} a b \arctan \left (c x \right )}{2}-\frac {i c \,d^{3} b^{2} \arctan \left (c x \right )^{2}}{x^{3}}-\frac {i c^{2} d^{3} b^{2} \arctan \left (c x \right )}{x^{2}}+2 i c^{4} d^{3} a b \ln \left (c^{2} x^{2}+1\right )-4 i c^{4} d^{3} a b \ln \left (c x \right )-4 i c^{4} d^{3} b^{2} \arctan \left (c x \right ) \ln \left (c x \right )+\frac {i c^{3} d^{3} b^{2} \arctan \left (c x \right )^{2}}{x}+2 i c^{4} d^{3} b^{2} \arctan \left (c x \right ) \ln \left (c^{2} x^{2}+1\right )-\frac {c \,d^{3} b^{2} \arctan \left (c x \right )}{6 x^{3}}+\frac {3 c^{2} d^{3} a b \arctan \left (c x \right )}{x^{2}}-\frac {c^{4} d^{3} b^{2} \ln \left (c x +i\right )^{2}}{2}+2 c^{4} d^{3} b^{2} \dilog \left (i c x +1\right )-2 c^{4} d^{3} b^{2} \dilog \left (-i c x +1\right )+c^{4} d^{3} b^{2} \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )-c^{4} d^{3} b^{2} \dilog \left (\frac {i \left (c x -i\right )}{2}\right )+\frac {c^{4} d^{3} b^{2} \ln \left (c x -i\right )^{2}}{2}-\frac {11 c^{4} d^{3} b^{2} \ln \left (c x \right )}{3}+\frac {7 c^{4} d^{3} b^{2} \arctan \left (c x \right )^{2}}{4}+\frac {3 c^{2} d^{3} a^{2}}{2 x^{2}}-\frac {d^{3} b^{2} \arctan \left (c x \right )^{2}}{4 x^{4}} \]

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]

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

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i)^3)/x^5,x)

[Out]

int(((a + b*atan(c*x))^2*(d + c*d*x*1i)^3)/x^5, x)

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

[Out]

Timed out

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