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

Optimal. Leaf size=250 \[ \frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac {a+b \tan ^{-1}(c x)}{d^3 x}-\frac {3 i c \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {3 i a c \log (x)}{d^3}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d^3}+\frac {3 b c \text {Li}_2(-i c x)}{2 d^3}-\frac {3 b c \text {Li}_2(i c x)}{2 d^3}+\frac {3 b c \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 d^3}-\frac {9 i b c}{8 d^3 (-c x+i)}+\frac {b c}{8 d^3 (-c x+i)^2}+\frac {b c \log (x)}{d^3}+\frac {9 i b c \tan ^{-1}(c x)}{8 d^3} \]

[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 = 0.29, 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 = {4876, 4852, 266, 36, 29, 31, 4848, 2391, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac {3 b c \text {PolyLog}(2,-i c x)}{2 d^3}-\frac {3 b c \text {PolyLog}(2,i c x)}{2 d^3}+\frac {3 b c \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 d^3}+\frac {2 c \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac {i c \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac {a+b \tan ^{-1}(c x)}{d^3 x}-\frac {3 i c \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {3 i a c \log (x)}{d^3}-\frac {b c \log \left (c^2 x^2+1\right )}{2 d^3}-\frac {9 i b c}{8 d^3 (-c x+i)}+\frac {b c}{8 d^3 (-c x+i)^2}+\frac {b c \log (x)}{d^3}+\frac {9 i b c \tan ^{-1}(c x)}{8 d^3} \]

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 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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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 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 4862

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 4876

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) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}+\frac {(3 b c) \operatorname {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) \operatorname {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 ) \operatorname {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 = 0.33, size = 227, normalized size = 0.91 \[ \frac {-\frac {8 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {16 c \left (a+b \tan ^{-1}(c x)\right )}{c x-i}+\frac {4 i c \left (a+b \tan ^{-1}(c x)\right )}{(c x-i)^2}-24 i c \log \left (\frac {2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )-24 i a c \log (x)+4 b c \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )+12 b c \text {Li}_2(-i c x)-12 b c \text {Li}_2(i c x)+12 b c \text {Li}_2\left (\frac {c x+i}{c x-i}\right )-8 i b c \left (-\tan ^{-1}(c x)+\frac {1}{-c x+i}\right )+\frac {b c \left (i c x+i (c x-i)^2 \tan ^{-1}(c x)+2\right )}{(c x-i)^2}}{8 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*I)*b*c*((I - c*x)^(-1) - ArcTan[c*x]) - (8*(a + b*ArcTan[c*x]))/x + ((4*I)*c*(a + b*ArcTan[c*x]))/(-I + c
*x)^2 - (16*c*(a + b*ArcTan[c*x]))/(-I + c*x) + (b*c*(2 + I*c*x + I*(-I + c*x)^2*ArcTan[c*x]))/(-I + c*x)^2 -
(24*I)*a*c*Log[x] - (24*I)*c*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + 4*b*c*(2*Log[x] - Log[1 + c^2*x^2]) +
12*b*c*PolyLog[2, (-I)*c*x] - 12*b*c*PolyLog[2, I*c*x] + 12*b*c*PolyLog[2, (I + c*x)/(-I + c*x)])/(8*d^3)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c^{3} d^{3} x^{5} - 6 i \, c^{2} d^{3} x^{4} - 6 \, c d^{3} x^{3} + 2 i \, d^{3} x^{2}}, x\right ) \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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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maple [A]  time = 0.08, size = 394, normalized size = 1.58 \[ -\frac {a}{d^{3} x}-\frac {3 i c b \arctan \left (c x \right ) \ln \left (c x \right )}{d^{3}}+\frac {i c b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}+\frac {9 i b c \arctan \left (c x \right )}{8 d^{3}}-\frac {3 c a \arctan \left (c x \right )}{d^{3}}-\frac {2 c a}{d^{3} \left (c x -i\right )}-\frac {b \arctan \left (c x \right )}{d^{3} x}+\frac {3 i c b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}+\frac {9 i c b}{8 d^{3} \left (c x -i\right )}+\frac {3 i c a \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}-\frac {2 c b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}+\frac {c b \ln \left (c x \right )}{d^{3}}-\frac {b c \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}+\frac {i c a}{2 d^{3} \left (c x -i\right )^{2}}-\frac {3 i c a \ln \left (c x \right )}{d^{3}}+\frac {c b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {3 c b \dilog \left (-i \left (c x +i\right )\right )}{2 d^{3}}-\frac {3 c b \ln \left (c x \right ) \ln \left (-i \left (c x +i\right )\right )}{2 d^{3}}+\frac {3 c b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (c x \right )}{2 d^{3}}-\frac {3 c b \ln \left (-i \left (-c x +i\right )\right ) \ln \left (-i c x \right )}{2 d^{3}}-\frac {3 c b \dilog \left (-i c x \right )}{2 d^{3}}+\frac {3 c b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}+\frac {3 c b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 d^{3}}-\frac {3 c b \ln \left (c x -i\right )^{2}}{4 d^{3}} \]

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)

[Out]

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

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maxima [B]  time = 0.43, size = 519, normalized size = 2.08 \[ -\frac {17 i \, b c^{3} x^{3} \arctan \left (1, c x\right ) + {\left (b {\left (34 \, \arctan \left (1, c x\right ) - 18 i\right )} + 48 \, a\right )} c^{2} x^{2} + {\left (b {\left (-17 i \, \arctan \left (1, c x\right ) - 20\right )} - 72 i \, a\right )} c x + {\left (12 \, b c^{3} x^{3} - 24 i \, b c^{2} x^{2} - 12 \, b c x\right )} \arctan \left (c x\right )^{2} + {\left (3 \, b c^{3} x^{3} - 6 i \, b c^{2} x^{2} - 3 \, b c x\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + {\left (-12 i \, b c^{3} x^{3} - 24 \, b c^{2} x^{2} + 12 i \, b c x\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + {\left (48 i \, b c^{3} x^{3} + 96 \, b c^{2} x^{2} - 48 i \, b c x\right )} \arctan \left (c x\right ) \log \left (c x\right ) + {\left ({\left (48 \, a - i \, b\right )} c^{3} x^{3} + {\left (-96 i \, a + 46 \, b\right )} c^{2} x^{2} - {\left (48 \, a + 71 i \, b\right )} c x - 16 \, b\right )} \arctan \left (c x\right ) + {\left (24 \, b c^{3} x^{3} - 48 i \, b c^{2} x^{2} - 24 \, b c x\right )} {\rm Li}_2\left (i \, c x + 1\right ) - {\left (24 \, b c^{3} x^{3} - 48 i \, b c^{2} x^{2} - 24 \, b c x\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) - {\left (24 \, b c^{3} x^{3} - 48 i \, b c^{2} x^{2} - 24 \, b c x\right )} {\rm Li}_2\left (-i \, c x + 1\right ) - {\left (4 \, {\left ({\left (3 i \, \pi - 2\right )} b + 6 i \, a\right )} c^{3} x^{3} + {\left ({\left (24 \, \pi + 16 i\right )} b + 48 \, a\right )} c^{2} x^{2} + 4 \, {\left ({\left (-3 i \, \pi + 2\right )} b - 6 i \, a\right )} c x + {\left (6 \, b c^{3} x^{3} - 12 i \, b c^{2} x^{2} - 6 \, b c x\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right )\right )} \log \left (c^{2} x^{2} + 1\right ) + {\left ({\left (48 i \, a - 16 \, b\right )} c^{3} x^{3} + 32 \, {\left (3 \, a + i \, b\right )} c^{2} x^{2} + {\left (-48 i \, a + 16 \, b\right )} c x\right )} \log \relax (x) - 16 \, a}{16 \, {\left (c^{2} d^{3} x^{3} - 2 i \, c d^{3} x^{2} - d^{3} x\right )}} \]

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]

-1/16*(17*I*b*c^3*x^3*arctan2(1, c*x) + (b*(34*arctan2(1, c*x) - 18*I) + 48*a)*c^2*x^2 + (b*(-17*I*arctan2(1,
c*x) - 20) - 72*I*a)*c*x + (12*b*c^3*x^3 - 24*I*b*c^2*x^2 - 12*b*c*x)*arctan(c*x)^2 + (3*b*c^3*x^3 - 6*I*b*c^2
*x^2 - 3*b*c*x)*log(c^2*x^2 + 1)^2 + (-12*I*b*c^3*x^3 - 24*b*c^2*x^2 + 12*I*b*c*x)*arctan(c*x)*log(1/4*c^2*x^2
+ 1/4) + (48*I*b*c^3*x^3 + 96*b*c^2*x^2 - 48*I*b*c*x)*arctan(c*x)*log(c*x) + ((48*a - I*b)*c^3*x^3 + (-96*I*a
+ 46*b)*c^2*x^2 - (48*a + 71*I*b)*c*x - 16*b)*arctan(c*x) + (24*b*c^3*x^3 - 48*I*b*c^2*x^2 - 24*b*c*x)*dilog(
I*c*x + 1) - (24*b*c^3*x^3 - 48*I*b*c^2*x^2 - 24*b*c*x)*dilog(1/2*I*c*x + 1/2) - (24*b*c^3*x^3 - 48*I*b*c^2*x^
2 - 24*b*c*x)*dilog(-I*c*x + 1) - (4*((3*I*pi - 2)*b + 6*I*a)*c^3*x^3 + ((24*pi + 16*I)*b + 48*a)*c^2*x^2 + 4*
((-3*I*pi + 2)*b - 6*I*a)*c*x + (6*b*c^3*x^3 - 12*I*b*c^2*x^2 - 6*b*c*x)*log(1/4*c^2*x^2 + 1/4))*log(c^2*x^2 +
1) + ((48*I*a - 16*b)*c^3*x^3 + 32*(3*a + I*b)*c^2*x^2 + (-48*I*a + 16*b)*c*x)*log(x) - 16*a)/(c^2*d^3*x^3 -
2*I*c*d^3*x^2 - d^3*x)

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

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

[Out]

Timed out

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