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

Optimal. Leaf size=306 \[ -\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac {6 c^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac {6 a c^2 \log (x)}{d^3}-\frac {3 i b c^2 \text {Li}_2(-i c x)}{d^3}+\frac {3 i b c^2 \text {Li}_2(i c x)}{d^3}-\frac {3 i b c^2 \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{d^3}+\frac {3 i b c^2 \log \left (c^2 x^2+1\right )}{2 d^3}-\frac {13 b c^2}{8 d^3 (-c x+i)}-\frac {i b c^2}{8 d^3 (-c x+i)^2}-\frac {3 i b c^2 \log (x)}{d^3}+\frac {9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac {b c}{2 d^3 x} \]

[Out]

-1/2*b*c/d^3/x-1/8*I*b*c^2/d^3/(I-c*x)^2-13/8*b*c^2/d^3/(I-c*x)+9/8*b*c^2*arctan(c*x)/d^3+1/2*(-a-b*arctan(c*x
))/d^3/x^2+3*I*c*(a+b*arctan(c*x))/d^3/x+1/2*c^2*(a+b*arctan(c*x))/d^3/(I-c*x)^2-3*I*c^2*(a+b*arctan(c*x))/d^3
/(I-c*x)-6*a*c^2*ln(x)/d^3-3*I*b*c^2*ln(x)/d^3-6*c^2*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/d^3+3/2*I*b*c^2*ln(c^2*
x^2+1)/d^3-3*I*b*c^2*polylog(2,-I*c*x)/d^3+3*I*b*c^2*polylog(2,I*c*x)/d^3-3*I*b*c^2*polylog(2,1-2/(1+I*c*x))/d
^3

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Rubi [A]  time = 0.32, antiderivative size = 306, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 16, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.696, Rules used = {4876, 4852, 325, 203, 266, 36, 29, 31, 4848, 2391, 4862, 627, 44, 4854, 2402, 2315} \[ -\frac {3 i b c^2 \text {PolyLog}(2,-i c x)}{d^3}+\frac {3 i b c^2 \text {PolyLog}(2,i c x)}{d^3}-\frac {3 i b c^2 \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-c x+i)}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (-c x+i)^2}-\frac {6 c^2 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac {6 a c^2 \log (x)}{d^3}+\frac {3 i b c^2 \log \left (c^2 x^2+1\right )}{2 d^3}-\frac {13 b c^2}{8 d^3 (-c x+i)}-\frac {i b c^2}{8 d^3 (-c x+i)^2}-\frac {3 i b c^2 \log (x)}{d^3}+\frac {9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac {b c}{2 d^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c)/(2*d^3*x) - ((I/8)*b*c^2)/(d^3*(I - c*x)^2) - (13*b*c^2)/(8*d^3*(I - c*x)) + (9*b*c^2*ArcTan[c*x])/(8*d
^3) - (a + b*ArcTan[c*x])/(2*d^3*x^2) + ((3*I)*c*(a + b*ArcTan[c*x]))/(d^3*x) + (c^2*(a + b*ArcTan[c*x]))/(2*d
^3*(I - c*x)^2) - ((3*I)*c^2*(a + b*ArcTan[c*x]))/(d^3*(I - c*x)) - (6*a*c^2*Log[x])/d^3 - ((3*I)*b*c^2*Log[x]
)/d^3 - (6*c^2*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/d^3 + (((3*I)/2)*b*c^2*Log[1 + c^2*x^2])/d^3 - ((3*I)*b
*c^2*PolyLog[2, (-I)*c*x])/d^3 + ((3*I)*b*c^2*PolyLog[2, I*c*x])/d^3 - ((3*I)*b*c^2*PolyLog[2, 1 - 2/(1 + I*c*
x)])/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 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 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^3 (d+i c d x)^3} \, dx &=\int \left (\frac {a+b \tan ^{-1}(c x)}{d^3 x^3}-\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x^2}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}-\frac {c^3 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^3}-\frac {3 i c^3 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (-i+c x)^2}+\frac {6 c^3 \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^3} \, dx}{d^3}-\frac {(3 i c) \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx}{d^3}-\frac {\left (6 c^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx}{d^3}-\frac {\left (3 i c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{d^3}-\frac {c^3 \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{d^3}+\frac {\left (6 c^3\right ) \int \frac {a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {6 a c^2 \log (x)}{d^3}-\frac {6 c^2 \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^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}-\frac {\left (3 i b c^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac {\left (3 i b c^2\right ) \int \frac {\log (1-i c x)}{x} \, dx}{d^3}+\frac {\left (3 i b c^2\right ) \int \frac {\log (1+i c x)}{x} \, dx}{d^3}-\frac {\left (3 i b c^3\right ) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{d^3}-\frac {\left (b c^3\right ) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}+\frac {\left (6 b c^3\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {b c}{2 d^3 x}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {6 a c^2 \log (x)}{d^3}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b c^2 \text {Li}_2(-i c x)}{d^3}+\frac {3 i b c^2 \text {Li}_2(i c x)}{d^3}-\frac {\left (3 i b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (6 i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{d^3}-\frac {\left (3 i b c^3\right ) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{d^3}-\frac {\left (b c^3\right ) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{2 d^3}-\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=-\frac {b c}{2 d^3 x}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {6 a c^2 \log (x)}{d^3}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}-\frac {3 i b c^2 \text {Li}_2(-i c x)}{d^3}+\frac {3 i b c^2 \text {Li}_2(i c x)}{d^3}-\frac {3 i b c^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}-\frac {\left (3 i b c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^3}-\frac {\left (3 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^3}-\frac {\left (b c^3\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 (3 i b c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )}{2 d^3}\\ &=-\frac {b c}{2 d^3 x}-\frac {i b c^2}{8 d^3 (i-c x)^2}-\frac {13 b c^2}{8 d^3 (i-c x)}-\frac {b c^2 \tan ^{-1}(c x)}{2 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {6 a c^2 \log (x)}{d^3}-\frac {3 i b c^2 \log (x)}{d^3}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {3 i b c^2 \log \left (1+c^2 x^2\right )}{2 d^3}-\frac {3 i b c^2 \text {Li}_2(-i c x)}{d^3}+\frac {3 i b c^2 \text {Li}_2(i c x)}{d^3}-\frac {3 i b c^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}+\frac {\left (b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 d^3}+\frac {\left (3 b c^3\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^3}\\ &=-\frac {b c}{2 d^3 x}-\frac {i b c^2}{8 d^3 (i-c x)^2}-\frac {13 b c^2}{8 d^3 (i-c x)}+\frac {9 b c^2 \tan ^{-1}(c x)}{8 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 d^3 x^2}+\frac {3 i c \left (a+b \tan ^{-1}(c x)\right )}{d^3 x}+\frac {c^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (i-c x)^2}-\frac {3 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{d^3 (i-c x)}-\frac {6 a c^2 \log (x)}{d^3}-\frac {3 i b c^2 \log (x)}{d^3}-\frac {6 c^2 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{d^3}+\frac {3 i b c^2 \log \left (1+c^2 x^2\right )}{2 d^3}-\frac {3 i b c^2 \text {Li}_2(-i c x)}{d^3}+\frac {3 i b c^2 \text {Li}_2(i c x)}{d^3}-\frac {3 i b c^2 \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{d^3}\\ \end {align*}

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Mathematica [C]  time = 0.59, size = 285, normalized size = 0.93 \[ -\frac {-\frac {24 i c^2 \left (a+b \tan ^{-1}(c x)\right )}{c x-i}-\frac {4 c^2 \left (a+b \tan ^{-1}(c x)\right )}{(c x-i)^2}+48 c^2 \log \left (\frac {2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{x^2}-\frac {24 i c \left (a+b \tan ^{-1}(c x)\right )}{x}+48 a c^2 \log (x)+\frac {4 b c \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )}{x}+24 i b c^2 \text {Li}_2(-i c x)-24 i b c^2 \text {Li}_2(i c x)+24 i b c^2 \text {Li}_2\left (\frac {c x+i}{c x-i}\right )+12 i b c^2 \left (2 \log (x)-\log \left (c^2 x^2+1\right )\right )+12 b c^2 \left (-\tan ^{-1}(c x)+\frac {1}{-c x+i}\right )-\frac {b c^2 \left (c x+(c x-i)^2 \tan ^{-1}(c x)-2 i\right )}{(c x-i)^2}}{8 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/8*(12*b*c^2*((I - c*x)^(-1) - ArcTan[c*x]) + (4*(a + b*ArcTan[c*x]))/x^2 - ((24*I)*c*(a + b*ArcTan[c*x]))/x
- (4*c^2*(a + b*ArcTan[c*x]))/(-I + c*x)^2 - ((24*I)*c^2*(a + b*ArcTan[c*x]))/(-I + c*x) - (b*c^2*(-2*I + c*x
+ (-I + c*x)^2*ArcTan[c*x]))/(-I + c*x)^2 + (4*b*c*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)])/x + 48*a*c^2*
Log[x] + 48*c^2*(a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)] + (12*I)*b*c^2*(2*Log[x] - Log[1 + c^2*x^2]) + (24*I)
*b*c^2*PolyLog[2, (-I)*c*x] - (24*I)*b*c^2*PolyLog[2, I*c*x] + (24*I)*b*c^2*PolyLog[2, (I + c*x)/(-I + c*x)])/
d^3

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fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c^{3} d^{3} x^{6} - 6 i \, c^{2} d^{3} x^{5} - 6 \, c d^{3} x^{4} + 2 i \, d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.10, size = 481, normalized size = 1.57 \[ -\frac {a}{2 d^{3} x^{2}}+\frac {13 c^{2} b}{8 d^{3} \left (c x -i\right )}+\frac {c^{2} a}{2 d^{3} \left (c x -i\right )^{2}}+\frac {3 c^{2} a \ln \left (c^{2} x^{2}+1\right )}{d^{3}}-\frac {6 c^{2} a \ln \left (c x \right )}{d^{3}}-\frac {b \arctan \left (c x \right )}{2 d^{3} x^{2}}-\frac {3 i c^{2} b \ln \left (c x \right )}{d^{3}}+\frac {3 i c^{2} a}{d^{3} \left (c x -i\right )}+\frac {3 i c a}{d^{3} x}+\frac {6 c^{2} b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}+\frac {c^{2} b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}+\frac {6 i c^{2} a \arctan \left (c x \right )}{d^{3}}-\frac {3 i c^{2} b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}+\frac {3 i c^{2} b \dilog \left (-i \left (c x +i\right )\right )}{d^{3}}+\frac {3 i c^{2} b \dilog \left (-i c x \right )}{d^{3}}-\frac {6 c^{2} b \ln \left (c x \right ) \arctan \left (c x \right )}{d^{3}}+\frac {3 i c^{2} b \ln \left (-i c x \right ) \ln \left (-i \left (-c x +i\right )\right )}{d^{3}}-\frac {b c}{2 d^{3} x}+\frac {9 b \,c^{2} \arctan \left (c x \right )}{8 d^{3}}+\frac {3 i c^{2} b \ln \left (c x \right ) \ln \left (-i \left (c x +i\right )\right )}{d^{3}}-\frac {3 i c^{2} b \ln \left (c x \right ) \ln \left (-i \left (-c x +i\right )\right )}{d^{3}}-\frac {3 i c^{2} b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}+\frac {3 i c b \arctan \left (c x \right )}{d^{3} x}+\frac {3 i c^{2} b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}+\frac {3 i b \,c^{2} \ln \left (c^{2} x^{2}+1\right )}{2 d^{3}}-\frac {i c^{2} b}{8 d^{3} \left (c x -i\right )^{2}}+\frac {3 i c^{2} b \ln \left (c x -i\right )^{2}}{2 d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d^3/x^2+3/2*I*b*c^2*ln(c^2*x^2+1)/d^3+3*c^2*a/d^3*ln(c^2*x^2+1)+13/8*c^2*b/d^3/(c*x-I)-6*c^2*a/d^3*ln(c
*x)+1/2*c^2*a/d^3/(c*x-I)^2-1/2*b/d^3*arctan(c*x)/x^2-1/8*I*c^2*b/d^3/(c*x-I)^2-3*I*c^2*b/d^3*dilog(-1/2*I*(I+
c*x))+3/2*I*c^2*b/d^3*ln(c*x-I)^2-3*I*c^2*b/d^3*ln(c*x)+3*I*c^2*a/d^3/(c*x-I)+3*I*c*a/d^3/x+3*I*c^2*b/d^3*dilo
g(-I*(I+c*x))-6*c^2*b/d^3*ln(c*x)*arctan(c*x)+6*c^2*b/d^3*arctan(c*x)*ln(c*x-I)+1/2*c^2*b/d^3*arctan(c*x)/(c*x
-I)^2+3*I*c^2*b/d^3*arctan(c*x)/(c*x-I)+3*I*c*b/d^3*arctan(c*x)/x-3*I*c^2*b/d^3*ln(c*x)*ln(-I*(-c*x+I))-3*I*c^
2*b/d^3*ln(c*x-I)*ln(-1/2*I*(I+c*x))-1/2*b*c/d^3/x+9/8*b*c^2*arctan(c*x)/d^3+3*I*c^2*b/d^3*ln(-I*c*x)*ln(-I*(-
c*x+I))+3*I*c^2*b/d^3*ln(c*x)*ln(-I*(I+c*x))+6*I*c^2*a/d^3*arctan(c*x)+3*I*c^2*b/d^3*dilog(-I*c*x)

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maxima [B]  time = 0.44, size = 594, normalized size = 1.94 \[ -\frac {33 \, b c^{4} x^{4} \arctan \left (1, c x\right ) + 6 \, {\left (b {\left (-11 i \, \arctan \left (1, c x\right ) - 3\right )} - 16 i \, a\right )} c^{3} x^{3} - {\left (b {\left (33 \, \arctan \left (1, c x\right ) - 12 i\right )} + 144 \, a\right )} c^{2} x^{2} - {\left (-32 i \, a + 8 \, b\right )} c x - {\left (24 i \, b c^{4} x^{4} + 48 \, b c^{3} x^{3} - 24 i \, b c^{2} x^{2}\right )} \arctan \left (c x\right )^{2} - {\left (6 i \, b c^{4} x^{4} + 12 \, b c^{3} x^{3} - 6 i \, b c^{2} x^{2}\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - {\left (24 \, b c^{4} x^{4} - 48 i \, b c^{3} x^{3} - 24 \, b c^{2} x^{2}\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + {\left (96 \, b c^{4} x^{4} - 192 i \, b c^{3} x^{3} - 96 \, b c^{2} x^{2}\right )} \arctan \left (c x\right ) \log \left (c x\right ) - {\left ({\left (96 i \, a - 15 \, b\right )} c^{4} x^{4} + 6 \, {\left (32 \, a + 21 i \, b\right )} c^{3} x^{3} + {\left (-96 i \, a + 159 \, b\right )} c^{2} x^{2} - 32 i \, b c x + 8 \, b\right )} \arctan \left (c x\right ) - {\left (48 i \, b c^{4} x^{4} + 96 \, b c^{3} x^{3} - 48 i \, b c^{2} x^{2}\right )} {\rm Li}_2\left (i \, c x + 1\right ) - {\left (-48 i \, b c^{4} x^{4} - 96 \, b c^{3} x^{3} + 48 i \, b c^{2} x^{2}\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) - {\left (-48 i \, b c^{4} x^{4} - 96 \, b c^{3} x^{3} + 48 i \, b c^{2} x^{2}\right )} {\rm Li}_2\left (-i \, c x + 1\right ) - {\left ({\left ({\left (24 \, \pi + 24 i\right )} b + 48 \, a\right )} c^{4} x^{4} - 48 \, {\left ({\left (i \, \pi - 1\right )} b + 2 i \, a\right )} c^{3} x^{3} - {\left ({\left (24 \, \pi + 24 i\right )} b + 48 \, a\right )} c^{2} x^{2} + {\left (-12 i \, b c^{4} x^{4} - 24 \, b c^{3} x^{3} + 12 i \, b c^{2} x^{2}\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right )\right )} \log \left (c^{2} x^{2} + 1\right ) + {\left (48 \, {\left (2 \, a + i \, b\right )} c^{4} x^{4} - {\left (192 i \, a - 96 \, b\right )} c^{3} x^{3} - 48 \, {\left (2 \, a + i \, b\right )} c^{2} x^{2}\right )} \log \relax (x) - 8 \, a}{16 \, {\left (c^{2} d^{3} x^{4} - 2 i \, c d^{3} x^{3} - d^{3} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(33*b*c^4*x^4*arctan2(1, c*x) + 6*(b*(-11*I*arctan2(1, c*x) - 3) - 16*I*a)*c^3*x^3 - (b*(33*arctan2(1, c
*x) - 12*I) + 144*a)*c^2*x^2 - (-32*I*a + 8*b)*c*x - (24*I*b*c^4*x^4 + 48*b*c^3*x^3 - 24*I*b*c^2*x^2)*arctan(c
*x)^2 - (6*I*b*c^4*x^4 + 12*b*c^3*x^3 - 6*I*b*c^2*x^2)*log(c^2*x^2 + 1)^2 - (24*b*c^4*x^4 - 48*I*b*c^3*x^3 - 2
4*b*c^2*x^2)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) + (96*b*c^4*x^4 - 192*I*b*c^3*x^3 - 96*b*c^2*x^2)*arctan(c*x)*
log(c*x) - ((96*I*a - 15*b)*c^4*x^4 + 6*(32*a + 21*I*b)*c^3*x^3 + (-96*I*a + 159*b)*c^2*x^2 - 32*I*b*c*x + 8*b
)*arctan(c*x) - (48*I*b*c^4*x^4 + 96*b*c^3*x^3 - 48*I*b*c^2*x^2)*dilog(I*c*x + 1) - (-48*I*b*c^4*x^4 - 96*b*c^
3*x^3 + 48*I*b*c^2*x^2)*dilog(1/2*I*c*x + 1/2) - (-48*I*b*c^4*x^4 - 96*b*c^3*x^3 + 48*I*b*c^2*x^2)*dilog(-I*c*
x + 1) - (((24*pi + 24*I)*b + 48*a)*c^4*x^4 - 48*((I*pi - 1)*b + 2*I*a)*c^3*x^3 - ((24*pi + 24*I)*b + 48*a)*c^
2*x^2 + (-12*I*b*c^4*x^4 - 24*b*c^3*x^3 + 12*I*b*c^2*x^2)*log(1/4*c^2*x^2 + 1/4))*log(c^2*x^2 + 1) + (48*(2*a
+ I*b)*c^4*x^4 - (192*I*a - 96*b)*c^3*x^3 - 48*(2*a + I*b)*c^2*x^2)*log(x) - 8*a)/(c^2*d^3*x^4 - 2*I*c*d^3*x^3
- d^3*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \left (\int \frac {a}{c^{3} x^{6} - 3 i c^{2} x^{5} - 3 c x^{4} + i x^{3}}\, dx + \int \frac {b \operatorname {atan}{\left (c x \right )}}{c^{3} x^{6} - 3 i c^{2} x^{5} - 3 c x^{4} + i x^{3}}\, dx\right )}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*(Integral(a/(c**3*x**6 - 3*I*c**2*x**5 - 3*c*x**4 + I*x**3), x) + Integral(b*atan(c*x)/(c**3*x**6 - 3*I*c**2
*x**5 - 3*c*x**4 + I*x**3), x))/d**3

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