\(\int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx\) [57]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 23, antiderivative size = 244 \[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=-\frac {b c}{2 d^2 x}-\frac {b c^2}{2 d^2 (i-c x)}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {2 i b c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {i b c^2 \log \left (1+c^2 x^2\right )}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {3 i b c^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {4996, 4946, 331, 209, 272, 36, 29, 31, 4940, 2438, 4972, 641, 46, 4964, 2449, 2352} \[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=-\frac {i c^2 (a+b \arctan (c x))}{d^2 (-c x+i)}-\frac {3 c^2 \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{d^2}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {3 a c^2 \log (x)}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {3 i b c^2 \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{2 d^2}+\frac {i b c^2 \log \left (c^2 x^2+1\right )}{d^2}-\frac {b c^2}{2 d^2 (-c x+i)}-\frac {2 i b c^2 \log (x)}{d^2}-\frac {b c}{2 d^2 x} \]

[In]

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

[Out]

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

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 272

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 331

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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2352

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

Rule 2438

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 2449

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 4940

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 4946

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

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[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 4972

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b*
ArcTan[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 4996

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*} \text {integral}& = \int \left (\frac {a+b \arctan (c x)}{d^2 x^3}-\frac {2 i c (a+b \arctan (c x))}{d^2 x^2}-\frac {3 c^2 (a+b \arctan (c x))}{d^2 x}-\frac {i c^3 (a+b \arctan (c x))}{d^2 (-i+c x)^2}+\frac {3 c^3 (a+b \arctan (c x))}{d^2 (-i+c x)}\right ) \, dx \\ & = \frac {\int \frac {a+b \arctan (c x)}{x^3} \, dx}{d^2}-\frac {(2 i c) \int \frac {a+b \arctan (c x)}{x^2} \, dx}{d^2}-\frac {\left (3 c^2\right ) \int \frac {a+b \arctan (c x)}{x} \, dx}{d^2}-\frac {\left (i c^3\right ) \int \frac {a+b \arctan (c x)}{(-i+c x)^2} \, dx}{d^2}+\frac {\left (3 c^3\right ) \int \frac {a+b \arctan (c x)}{-i+c x} \, dx}{d^2} \\ & = -\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {(b c) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d^2}-\frac {\left (3 i b c^2\right ) \int \frac {\log (1-i c x)}{x} \, dx}{2 d^2}+\frac {\left (3 i b c^2\right ) \int \frac {\log (1+i c x)}{x} \, dx}{2 d^2}-\frac {\left (2 i b c^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d^2}-\frac {\left (i b c^3\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^2}+\frac {\left (3 b c^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^2} \\ & = -\frac {b c}{2 d^2 x}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {\left (i b c^2\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{d^2}-\frac {\left (3 i b c^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{d^2}-\frac {\left (i b c^3\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^2}-\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2} \\ & = -\frac {b c}{2 d^2 x}-\frac {b c^2 \arctan (c x)}{2 d^2}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {3 i b c^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^2}-\frac {\left (i b c^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{d^2}-\frac {\left (i b c^3\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^2}+\frac {\left (i b c^4\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{d^2} \\ & = -\frac {b c}{2 d^2 x}-\frac {b c^2}{2 d^2 (i-c x)}-\frac {b c^2 \arctan (c x)}{2 d^2}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {2 i b c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {i b c^2 \log \left (1+c^2 x^2\right )}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {3 i b c^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^2}+\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^2} \\ & = -\frac {b c}{2 d^2 x}-\frac {b c^2}{2 d^2 (i-c x)}-\frac {a+b \arctan (c x)}{2 d^2 x^2}+\frac {2 i c (a+b \arctan (c x))}{d^2 x}-\frac {i c^2 (a+b \arctan (c x))}{d^2 (i-c x)}-\frac {3 a c^2 \log (x)}{d^2}-\frac {2 i b c^2 \log (x)}{d^2}-\frac {3 c^2 (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{d^2}+\frac {i b c^2 \log \left (1+c^2 x^2\right )}{d^2}-\frac {3 i b c^2 \operatorname {PolyLog}(2,-i c x)}{2 d^2}+\frac {3 i b c^2 \operatorname {PolyLog}(2,i c x)}{2 d^2}-\frac {3 i b c^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.25 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=-\frac {-b c^2 \left (\frac {1}{-i+c x}+\arctan (c x)\right )+\frac {a+b \arctan (c x)}{x^2}-\frac {4 i c (a+b \arctan (c x))}{x}-\frac {2 i c^2 (a+b \arctan (c x))}{-i+c x}+\frac {b c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-c^2 x^2\right )}{x}+6 a c^2 \log (x)+6 c^2 (a+b \arctan (c x)) \log \left (\frac {2 i}{i-c x}\right )+2 i b c^2 \left (2 \log (x)-\log \left (1+c^2 x^2\right )\right )+3 i b c^2 \operatorname {PolyLog}(2,-i c x)-3 i b c^2 \operatorname {PolyLog}(2,i c x)+3 i b c^2 \operatorname {PolyLog}\left (2,\frac {i+c x}{-i+c x}\right )}{2 d^2} \]

[In]

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

[Out]

-1/2*(-(b*c^2*((-I + c*x)^(-1) + ArcTan[c*x])) + (a + b*ArcTan[c*x])/x^2 - ((4*I)*c*(a + b*ArcTan[c*x]))/x - (
(2*I)*c^2*(a + b*ArcTan[c*x]))/(-I + c*x) + (b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 6*a*c^2*Log[
x] + 6*c^2*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + (2*I)*b*c^2*(2*Log[x] - Log[1 + c^2*x^2]) + (3*I)*b*c^2*
PolyLog[2, (-I)*c*x] - (3*I)*b*c^2*PolyLog[2, I*c*x] + (3*I)*b*c^2*PolyLog[2, (I + c*x)/(-I + c*x)])/d^2

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.17

method result size
derivativedivides \(c^{2} \left (-\frac {a}{2 d^{2} c^{2} x^{2}}+\frac {2 i a}{d^{2} c x}-\frac {3 a \ln \left (c x \right )}{d^{2}}+\frac {i a}{d^{2} \left (c x -i\right )}+\frac {3 a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {3 i a \arctan \left (c x \right )}{d^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {2 i \arctan \left (c x \right )}{c x}-3 \arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \arctan \left (c x \right )}{c x -i}+3 \arctan \left (c x \right ) \ln \left (c x -i\right )-\frac {3 i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}+\frac {3 i \ln \left (c x -i\right )^{2}}{4}+i \ln \left (c^{2} x^{2}+1\right )-\frac {1}{2 c x}-2 i \ln \left (c x \right )+\frac {1}{2 c x -2 i}\right )}{d^{2}}\right )\) \(286\)
default \(c^{2} \left (-\frac {a}{2 d^{2} c^{2} x^{2}}+\frac {2 i a}{d^{2} c x}-\frac {3 a \ln \left (c x \right )}{d^{2}}+\frac {i a}{d^{2} \left (c x -i\right )}+\frac {3 a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {3 i a \arctan \left (c x \right )}{d^{2}}+\frac {b \left (-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {2 i \arctan \left (c x \right )}{c x}-3 \arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \arctan \left (c x \right )}{c x -i}+3 \arctan \left (c x \right ) \ln \left (c x -i\right )-\frac {3 i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}+\frac {3 i \ln \left (c x -i\right )^{2}}{4}+i \ln \left (c^{2} x^{2}+1\right )-\frac {1}{2 c x}-2 i \ln \left (c x \right )+\frac {1}{2 c x -2 i}\right )}{d^{2}}\right )\) \(286\)
parts \(-\frac {a}{2 d^{2} x^{2}}+\frac {2 i a c}{x \,d^{2}}-\frac {3 a \,c^{2} \ln \left (x \right )}{d^{2}}-\frac {i a \,c^{2}}{d^{2} \left (-c x +i\right )}+\frac {3 c^{2} a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {3 i c^{2} a \arctan \left (c x \right )}{d^{2}}+\frac {b \,c^{2} \left (-\frac {\arctan \left (c x \right )}{2 c^{2} x^{2}}+\frac {2 i \arctan \left (c x \right )}{c x}-3 \arctan \left (c x \right ) \ln \left (c x \right )+\frac {i \arctan \left (c x \right )}{c x -i}+3 \arctan \left (c x \right ) \ln \left (c x -i\right )-\frac {3 i \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {3 i \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}-\frac {3 i \operatorname {dilog}\left (i c x +1\right )}{2}+\frac {3 i \operatorname {dilog}\left (-i c x +1\right )}{2}-\frac {3 i \left (\operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )+\ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )\right )}{2}+\frac {3 i \ln \left (c x -i\right )^{2}}{4}+i \ln \left (c^{2} x^{2}+1\right )-\frac {1}{2 c x}-2 i \ln \left (c x \right )+\frac {1}{2 c x -2 i}\right )}{d^{2}}\) \(291\)
risch \(-\frac {a}{2 d^{2} x^{2}}+\frac {b \,c^{2} \arctan \left (c x \right )}{4 d^{2}}-\frac {b c}{2 d^{2} x}+\frac {c^{3} b \ln \left (-i c x +1\right ) x}{4 d^{2} \left (-i c x -1\right )}+\frac {i c^{2} b \ln \left (-i c x +1\right )}{4 d^{2} \left (-i c x -1\right )}+\frac {3 i c^{2} b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d^{2}}-\frac {3 i c^{2} b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{2}}+\frac {i b \,c^{2} \ln \left (i c x +1\right )}{2 d^{2} \left (i c x +1\right )}-\frac {3 c^{2} a \ln \left (-i c x \right )}{d^{2}}+\frac {c^{2} a}{d^{2} \left (-i c x -1\right )}+\frac {3 c^{2} a \ln \left (c^{2} x^{2}+1\right )}{2 d^{2}}+\frac {b c \ln \left (i c x +1\right )}{d^{2} x}-\frac {3 i b \,c^{2} \ln \left (i c x +1\right )^{2}}{4 d^{2}}-\frac {3 i b \,c^{2} \operatorname {dilog}\left (i c x +1\right )}{2 d^{2}}-\frac {5 i b \,c^{2} \ln \left (i c x \right )}{4 d^{2}}+\frac {5 i b \,c^{2} \ln \left (i c x +1\right )}{4 d^{2}}+\frac {i b \ln \left (i c x +1\right )}{4 d^{2} x^{2}}+\frac {i b \,c^{2}}{2 d^{2} \left (i c x +1\right )}-\frac {c b \ln \left (-i c x +1\right )}{d^{2} x}-\frac {3 i c^{2} b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{2}}-\frac {i b \,c^{2} \ln \left (c^{2} x^{2}+1\right )}{8 d^{2}}+\frac {3 i c^{2} b \operatorname {dilog}\left (-i c x +1\right )}{2 d^{2}}-\frac {3 i c^{2} b \ln \left (-i c x \right )}{4 d^{2}}+\frac {3 i c^{2} \ln \left (-i c x +1\right ) b}{4 d^{2}}-\frac {i b \ln \left (-i c x +1\right )}{4 d^{2} x^{2}}+\frac {2 i a c}{x \,d^{2}}+\frac {3 i c^{2} a \arctan \left (c x \right )}{d^{2}}\) \(495\)

[In]

int((a+b*arctan(c*x))/x^3/(d+I*c*d*x)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.91 \[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=\frac {12 i \, a c^{2} x^{2} + 2 \, {\left (3 \, a + i \, b\right )} c x - 6 \, {\left (-i \, b c^{3} x^{3} - b c^{2} x^{2}\right )} {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) - 4 \, {\left ({\left (3 \, a + 2 i \, b\right )} c^{3} x^{3} + {\left (-3 i \, a + 2 \, b\right )} c^{2} x^{2}\right )} \log \left (x\right ) - {\left (6 \, b c^{2} x^{2} - 3 i \, b c x + b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 4 \, {\left (-i \, b c^{3} x^{3} - b c^{2} x^{2}\right )} \log \left (\frac {c x + i}{c}\right ) + 4 \, {\left ({\left (3 \, a + i \, b\right )} c^{3} x^{3} - {\left (3 i \, a - b\right )} c^{2} x^{2}\right )} \log \left (\frac {c x - i}{c}\right ) + 2 i \, a}{4 \, {\left (c d^{2} x^{3} - i \, d^{2} x^{2}\right )}} \]

[In]

integrate((a+b*arctan(c*x))/x^3/(d+I*c*d*x)^2,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=- \frac {\int \frac {a}{c^{2} x^{5} - 2 i c x^{4} - x^{3}}\, dx + \int \frac {b \operatorname {atan}{\left (c x \right )}}{c^{2} x^{5} - 2 i c x^{4} - x^{3}}\, dx}{d^{2}} \]

[In]

integrate((a+b*atan(c*x))/x**3/(d+I*c*d*x)**2,x)

[Out]

-(Integral(a/(c**2*x**5 - 2*I*c*x**4 - x**3), x) + Integral(b*atan(c*x)/(c**2*x**5 - 2*I*c*x**4 - x**3), x))/d
**2

Maxima [F]

\[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{2} x^{3}} \,d x } \]

[In]

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

[Out]

(-2*I*c*integrate(arctan(c*x)/(c^4*d^2*x^6 + 2*c^2*d^2*x^4 + d^2*x^2), x) - integrate((c^2*x^2 - 1)*arctan(c*x
)/(c^4*d^2*x^7 + 2*c^2*d^2*x^5 + d^2*x^3), x))*b - 1/2*a*(-2*I*c^2/(c*d^2*x - I*d^2) - 6*c^2*log(c*x - I)/d^2
+ 6*c^2*log(x)/d^2 - (4*I*c*x - 1)/(d^2*x^2))

Giac [F]

\[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=\int { \frac {b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{2} x^{3}} \,d x } \]

[In]

integrate((a+b*arctan(c*x))/x^3/(d+I*c*d*x)^2,x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \arctan (c x)}{x^3 (d+i c d x)^2} \, dx=\int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^3\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

[In]

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

[Out]

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