3.1.64 \(\int \frac {a+b \text {ArcTan}(c x)}{x^2 (d+i c d x)^3} \, dx\) [64]

Optimal. Leaf size=250 \[ \frac {b c}{8 d^3 (i-c x)^2}-\frac {9 i b c}{8 d^3 (i-c x)}+\frac {9 i b c \text {ArcTan}(c x)}{8 d^3}-\frac {a+b \text {ArcTan}(c x)}{d^3 x}+\frac {i c (a+b \text {ArcTan}(c x))}{2 d^3 (i-c x)^2}+\frac {2 c (a+b \text {ArcTan}(c x))}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}+\frac {b c \log (x)}{d^3}-\frac {3 i c (a+b \text {ArcTan}(c x)) \log \left (\frac {2}{1+i c x}\right )}{d^3}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {PolyLog}(2,-i c x)}{2 d^3}-\frac {3 b c \text {PolyLog}(2,i c x)}{2 d^3}+\frac {3 b c \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^3} \]

[Out]

1/8*b*c/d^3/(I-c*x)^2-9/8*I*b*c/d^3/(I-c*x)+9/8*I*b*c*arctan(c*x)/d^3+(-a-b*arctan(c*x))/d^3/x+1/2*I*c*(a+b*ar
ctan(c*x))/d^3/(I-c*x)^2+2*c*(a+b*arctan(c*x))/d^3/(I-c*x)-3*I*a*c*ln(x)/d^3+b*c*ln(x)/d^3-3*I*c*(a+b*arctan(c
*x))*ln(2/(1+I*c*x))/d^3-1/2*b*c*ln(c^2*x^2+1)/d^3+3/2*b*c*polylog(2,-I*c*x)/d^3-3/2*b*c*polylog(2,I*c*x)/d^3+
3/2*b*c*polylog(2,1-2/(1+I*c*x))/d^3

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Rubi [A]
time = 0.21, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 15, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {4996, 4946, 272, 36, 29, 31, 4940, 2438, 4972, 641, 46, 209, 4964, 2449, 2352} \begin {gather*} \frac {2 c (a+b \text {ArcTan}(c x))}{d^3 (-c x+i)}+\frac {i c (a+b \text {ArcTan}(c x))}{2 d^3 (-c x+i)^2}-\frac {a+b \text {ArcTan}(c x)}{d^3 x}-\frac {3 i c \log \left (\frac {2}{1+i c x}\right ) (a+b \text {ArcTan}(c x))}{d^3}-\frac {3 i a c \log (x)}{d^3}+\frac {9 i b c \text {ArcTan}(c x)}{8 d^3}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 d^3}-\frac {9 i b c}{8 d^3 (-c x+i)}+\frac {b c}{8 d^3 (-c x+i)^2}+\frac {b c \log (x)}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(8*d^3*(I - c*x)^2) - (((9*I)/8)*b*c)/(d^3*(I - c*x)) + (((9*I)/8)*b*c*ArcTan[c*x])/d^3 - (a + b*ArcTan[
c*x])/(d^3*x) + ((I/2)*c*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)^2) + (2*c*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) -
((3*I)*a*c*Log[x])/d^3 + (b*c*Log[x])/d^3 - ((3*I)*c*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/d^3 - (b*c*Log[1
+ c^2*x^2])/(2*d^3) + (3*b*c*PolyLog[2, (-I)*c*x])/(2*d^3) - (3*b*c*PolyLog[2, I*c*x])/(2*d^3) + (3*b*c*PolyLo
g[2, 1 - 2/(1 + I*c*x)])/(2*d^3)

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 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*} \int \frac {a+b \tan ^{-1}(c x)}{x^2 (d+i c d x)^3} \, dx &=\int \left (\frac {a+b \tan ^{-1}(c x)}{d^3 x^2}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac {i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^3}+\frac {2 c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^2}+\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d^3}-\frac {(3 i c) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}-\frac {\left (i c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{d^3}+\frac {\left (3 i c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{d^3}+\frac {\left (2 c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {(b c) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d^3}+\frac {(3 b c) \int \frac {\log (1-i c x)}{x} \, dx}{2 d^3}-\frac {(3 b c) \int \frac {\log (1+i c x)}{x} \, dx}{2 d^3}-\frac {\left (i b c^2\right ) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}+\frac {\left (3 i b c^2\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}+\frac {\left (2 b c^2\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}+\frac {(3 b c) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{d^3}-\frac {\left (i b c^2\right ) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{2 d^3}+\frac {\left (2 b c^2\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (i b c^2\right ) \int \left (-\frac {i}{2 (-i+c x)^3}+\frac {1}{4 (-i+c x)^2}-\frac {1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^3}+\frac {\left (2 b c^2\right ) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{d^3}-\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d^3}\\ &=\frac {b c}{8 d^3 (i-c x)^2}-\frac {9 i b c}{8 d^3 (i-c x)}-\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}+\frac {b c \log (x)}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {\left (i b c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3}+\frac {\left (i b c^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^3}\\ &=\frac {b c}{8 d^3 (i-c x)^2}-\frac {9 i b c}{8 d^3 (i-c x)}+\frac {9 i b c \tan ^{-1}(c x)}{8 d^3}-\frac {a+b \tan ^{-1}(c x)}{d^3 x}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {3 i a c \log (x)}{d^3}+\frac {b c \log (x)}{d^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}-\frac {b c \log \left (1+c^2 x^2\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 d^3}\\ \end {align*}

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Mathematica [A]
time = 1.03, size = 241, normalized size = 0.96 \begin {gather*} -\frac {\frac {32 a}{x}-\frac {16 i a c}{(-i+c x)^2}+\frac {64 a c}{-i+c x}+96 a c \text {ArcTan}(c x)+96 b c \text {ArcTan}(c x)^2+96 i a c \log (x)-48 i a c \log \left (1+c^2 x^2\right )+48 b c \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c x)}\right )+b c \left (20 \cos (2 \text {ArcTan}(c x))+\cos (4 \text {ArcTan}(c x))-32 \log \left (\frac {c x}{\sqrt {1+c^2 x^2}}\right )-20 i \sin (2 \text {ArcTan}(c x))-i \sin (4 \text {ArcTan}(c x))\right )+\frac {4 b \text {ArcTan}(c x) \left (8+10 i c x \cos (2 \text {ArcTan}(c x))+i c x \cos (4 \text {ArcTan}(c x))+24 i c x \log \left (1-e^{2 i \text {ArcTan}(c x)}\right )+10 c x \sin (2 \text {ArcTan}(c x))+c x \sin (4 \text {ArcTan}(c x))\right )}{x}}{32 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*((32*a)/x - ((16*I)*a*c)/(-I + c*x)^2 + (64*a*c)/(-I + c*x) + 96*a*c*ArcTan[c*x] + 96*b*c*ArcTan[c*x]^2
+ (96*I)*a*c*Log[x] - (48*I)*a*c*Log[1 + c^2*x^2] + 48*b*c*PolyLog[2, E^((2*I)*ArcTan[c*x])] + b*c*(20*Cos[2*A
rcTan[c*x]] + Cos[4*ArcTan[c*x]] - 32*Log[(c*x)/Sqrt[1 + c^2*x^2]] - (20*I)*Sin[2*ArcTan[c*x]] - I*Sin[4*ArcTa
n[c*x]]) + (4*b*ArcTan[c*x]*(8 + (10*I)*c*x*Cos[2*ArcTan[c*x]] + I*c*x*Cos[4*ArcTan[c*x]] + (24*I)*c*x*Log[1 -
E^((2*I)*ArcTan[c*x])] + 10*c*x*Sin[2*ArcTan[c*x]] + c*x*Sin[4*ArcTan[c*x]]))/x)/d^3

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Maple [A]
time = 0.14, size = 380, normalized size = 1.52

method	result	size
derivativedivides	\(c \left (\frac {i a}{2 d^{3} \left (c x -i\right )^{2}}-\frac {3 i b \arctan \left (c x \right ) \ln \left (c x \right )}{d^{3}}-\frac {3 a \arctan \left (c x \right )}{d^{3}}-\frac {2 a}{d^{3} \left (c x -i\right )}-\frac {a}{d^{3} c x}+\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {9 i b \arctan \left (c x \right )}{8 d^{3}}-\frac {3 i a \ln \left (c x \right )}{d^{3}}-\frac {2 b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {b \arctan \left (c x \right )}{d^{3} c x}+\frac {9 i b}{8 d^{3} \left (c x -i\right )}+\frac {3 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}+\frac {b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}+\frac {b \ln \left (c x \right )}{d^{3}}+\frac {3 b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (c x \right )}{2 d^{3}}-\frac {3 b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (-i c x \right )}{2 d^{3}}-\frac {3 b \dilog \left (-i c x \right )}{2 d^{3}}-\frac {3 b \dilog \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {3 b \ln \left (c x \right ) \ln \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {3 b \ln \left (c x -i\right )^{2}}{4 d^{3}}+\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}+\frac {3 b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}\right )\)	\(380\)
default	\(c \left (\frac {i a}{2 d^{3} \left (c x -i\right )^{2}}-\frac {3 i b \arctan \left (c x \right ) \ln \left (c x \right )}{d^{3}}-\frac {3 a \arctan \left (c x \right )}{d^{3}}-\frac {2 a}{d^{3} \left (c x -i\right )}-\frac {a}{d^{3} c x}+\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {9 i b \arctan \left (c x \right )}{8 d^{3}}-\frac {3 i a \ln \left (c x \right )}{d^{3}}-\frac {2 b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {b \arctan \left (c x \right )}{d^{3} c x}+\frac {9 i b}{8 d^{3} \left (c x -i\right )}+\frac {3 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}+\frac {b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}+\frac {b \ln \left (c x \right )}{d^{3}}+\frac {3 b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (c x \right )}{2 d^{3}}-\frac {3 b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (-i c x \right )}{2 d^{3}}-\frac {3 b \dilog \left (-i c x \right )}{2 d^{3}}-\frac {3 b \dilog \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {3 b \ln \left (c x \right ) \ln \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {3 b \ln \left (c x -i\right )^{2}}{4 d^{3}}+\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}+\frac {3 b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}\right )\)	\(380\)
risch	\(\frac {3 i c a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}-\frac {c b \ln \left (i c x +1\right )}{d^{3} \left (i c x +1\right )}-\frac {c b \ln \left (i c x +1\right )}{4 d^{3} \left (i c x +1\right )^{2}}+\frac {3 c b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{3}}-\frac {3 c b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{2 d^{3}}-\frac {c b \ln \left (-i c x +1\right )}{2 d^{3} \left (-i c x -1\right )}+\frac {3 c b \ln \left (-i c x +1\right )}{16 d^{3} \left (-i c x -1\right )^{2}}+\frac {9 i c b \arctan \left (c x \right )}{16 d^{3}}-\frac {i b \ln \left (-i c x +1\right )}{2 d^{3} x}+\frac {i b \ln \left (i c x +1\right )}{2 d^{3} x}-\frac {i c a}{2 d^{3} \left (-i c x -1\right )^{2}}+\frac {2 i c a}{d^{3} \left (-i c x -1\right )}-\frac {3 i c a \ln \left (-i c x \right )}{d^{3}}-\frac {7 b c \ln \left (c^{2} x^{2}+1\right )}{32 d^{3}}+\frac {c^{3} b \ln \left (-i c x +1\right ) x^{2}}{16 d^{3} \left (-i c x -1\right )^{2}}-\frac {a}{d^{3} x}+\frac {i c^{2} b \ln \left (-i c x +1\right ) x}{2 d^{3} \left (-i c x -1\right )}-\frac {i c^{2} b \ln \left (-i c x +1\right ) x}{8 d^{3} \left (-i c x -1\right )^{2}}-\frac {3 c a \arctan \left (c x \right )}{d^{3}}+\frac {c b \ln \left (-i c x \right )}{2 d^{3}}+\frac {c b}{8 d^{3} \left (-i c x -1\right )}-\frac {c b}{8 d^{3} \left (i c x +1\right )^{2}}+\frac {c b \ln \left (i c x \right )}{2 d^{3}}+\frac {3 c b \ln \left (i c x +1\right )^{2}}{4 d^{3}}-\frac {c b}{d^{3} \left (i c x +1\right )}+\frac {3 c b \dilog \left (\frac {1}{2}-\frac {i c x}{2}\right )}{2 d^{3}}-\frac {3 c b \dilog \left (-i c x +1\right )}{2 d^{3}}+\frac {3 c b \dilog \left (i c x +1\right )}{2 d^{3}}\)	\(516\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(1/2*I*a/d^3/(c*x-I)^2-3*I*b/d^3*arctan(c*x)*ln(c*x)-3*a/d^3*arctan(c*x)-2*a/d^3/(c*x-I)-a/d^3/c/x+3/2*I*a/d
^3*ln(c^2*x^2+1)+9/8*I*b/d^3*arctan(c*x)-3*I*a/d^3*ln(c*x)-2*b/d^3*arctan(c*x)/(c*x-I)-b/d^3*arctan(c*x)/c/x+9
/8*I*b/d^3/(c*x-I)+3*I*b/d^3*arctan(c*x)*ln(c*x-I)+1/8*b/d^3/(c*x-I)^2-1/2*b/d^3*ln(c^2*x^2+1)+1/2*I*b/d^3*arc
tan(c*x)/(c*x-I)^2+b/d^3*ln(c*x)+3/2*b/d^3*ln(-I*(-c*x+I))*ln(c*x)-3/2*b/d^3*ln(-I*(-c*x+I))*ln(-I*c*x)-3/2*b/
d^3*dilog(-I*c*x)-3/2*b/d^3*dilog(-I*(c*x+I))-3/2*b/d^3*ln(c*x)*ln(-I*(c*x+I))-3/4*b/d^3*ln(c*x-I)^2+3/2*b/d^3
*ln(c*x-I)*ln(-1/2*I*(c*x+I))+3/2*b/d^3*dilog(-1/2*I*(c*x+I)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 1.79, size = 263, normalized size = 1.05 \begin {gather*} -\frac {6 \, {\left (8 \, a - 3 i \, b\right )} c^{2} x^{2} + 4 \, {\left (-18 i \, a - 5 \, b\right )} c x + 24 \, {\left (b c^{3} x^{3} - 2 i \, b c^{2} x^{2} - b c x\right )} {\rm Li}_2\left (\frac {c x + i}{c x - i} + 1\right ) + 16 \, {\left ({\left (3 i \, a - b\right )} c^{3} x^{3} + 2 \, {\left (3 \, a + i \, b\right )} c^{2} x^{2} + {\left (-3 i \, a + b\right )} c x\right )} \log \left (x\right ) + 4 \, {\left (6 i \, b c^{2} x^{2} + 9 \, b c x - 2 i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) + 17 \, {\left (b c^{3} x^{3} - 2 i \, b c^{2} x^{2} - b c x\right )} \log \left (\frac {c x + i}{c}\right ) - {\left ({\left (48 i \, a + b\right )} c^{3} x^{3} + 2 \, {\left (48 \, a - i \, b\right )} c^{2} x^{2} + {\left (-48 i \, a - b\right )} c x\right )} \log \left (\frac {c x - i}{c}\right ) - 16 \, a}{16 \, {\left (c^{2} d^{3} x^{3} - 2 i \, c d^{3} x^{2} - d^{3} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(6*(8*a - 3*I*b)*c^2*x^2 + 4*(-18*I*a - 5*b)*c*x + 24*(b*c^3*x^3 - 2*I*b*c^2*x^2 - b*c*x)*dilog((c*x + I
)/(c*x - I) + 1) + 16*((3*I*a - b)*c^3*x^3 + 2*(3*a + I*b)*c^2*x^2 + (-3*I*a + b)*c*x)*log(x) + 4*(6*I*b*c^2*x
^2 + 9*b*c*x - 2*I*b)*log(-(c*x + I)/(c*x - I)) + 17*(b*c^3*x^3 - 2*I*b*c^2*x^2 - b*c*x)*log((c*x + I)/c) - ((
48*I*a + b)*c^3*x^3 + 2*(48*a - I*b)*c^2*x^2 + (-48*I*a - b)*c*x)*log((c*x - I)/c) - 16*a)/(c^2*d^3*x^3 - 2*I*
c*d^3*x^2 - d^3*x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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