3.58 \(\int \frac {x^4 (a+b \tan ^{-1}(c x))}{(d+i c d x)^3} \, dx\)

Optimal. Leaf size=256 \[ \frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (-c x+i)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (-c x+i)^2}+\frac {6 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {3 a x}{c^4 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{c^5 d^3}-\frac {15 i b}{8 c^5 d^3 (-c x+i)}-\frac {b}{8 c^5 d^3 (-c x+i)^2}+\frac {19 i b \tan ^{-1}(c x)}{8 c^5 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {3 b \log \left (c^2 x^2+1\right )}{2 c^5 d^3} \]

[Out]

-3*a*x/c^4/d^3-1/2*I*b*x/c^4/d^3-1/8*b/c^5/d^3/(I-c*x)^2-15/8*I*b/c^5/d^3/(I-c*x)+19/8*I*b*arctan(c*x)/c^5/d^3
-3*b*x*arctan(c*x)/c^4/d^3+1/2*I*x^2*(a+b*arctan(c*x))/c^3/d^3-1/2*I*(a+b*arctan(c*x))/c^5/d^3/(I-c*x)^2+4*(a+
b*arctan(c*x))/c^5/d^3/(I-c*x)+6*I*(a+b*arctan(c*x))*ln(2/(1+I*c*x))/c^5/d^3+3/2*b*ln(c^2*x^2+1)/c^5/d^3-3*b*p
olylog(2,1-2/(1+I*c*x))/c^5/d^3

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Rubi [A]  time = 0.28, antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4876, 4846, 260, 4852, 321, 203, 4862, 627, 44, 4854, 2402, 2315} \[ -\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (-c x+i)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (-c x+i)^2}+\frac {6 i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3}-\frac {3 a x}{c^4 d^3}+\frac {3 b \log \left (c^2 x^2+1\right )}{2 c^5 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {15 i b}{8 c^5 d^3 (-c x+i)}-\frac {b}{8 c^5 d^3 (-c x+i)^2}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {19 i b \tan ^{-1}(c x)}{8 c^5 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcTan[c*x]))/(d + I*c*d*x)^3,x]

[Out]

(-3*a*x)/(c^4*d^3) - ((I/2)*b*x)/(c^4*d^3) - b/(8*c^5*d^3*(I - c*x)^2) - (((15*I)/8)*b)/(c^5*d^3*(I - c*x)) +
(((19*I)/8)*b*ArcTan[c*x])/(c^5*d^3) - (3*b*x*ArcTan[c*x])/(c^4*d^3) + ((I/2)*x^2*(a + b*ArcTan[c*x]))/(c^3*d^
3) - ((I/2)*(a + b*ArcTan[c*x]))/(c^5*d^3*(I - c*x)^2) + (4*(a + b*ArcTan[c*x]))/(c^5*d^3*(I - c*x)) + ((6*I)*
(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c^5*d^3) + (3*b*Log[1 + c^2*x^2])/(2*c^5*d^3) - (3*b*PolyLog[2, 1 - 2
/(1 + I*c*x)])/(c^5*d^3)

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 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 4846

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

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 {x^4 \left (a+b \tan ^{-1}(c x)\right )}{(d+i c d x)^3} \, dx &=\int \left (-\frac {3 \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac {i x \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-i+c x)^3}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-i+c x)^2}-\frac {6 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-i+c x)}\right ) \, dx\\ &=\frac {i \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{c^4 d^3}-\frac {(6 i) \int \frac {a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c^4 d^3}-\frac {3 \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^4 d^3}+\frac {4 \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^4 d^3}+\frac {i \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^3 d^3}\\ &=-\frac {3 a x}{c^4 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (i-c x)^2}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (i-c x)}+\frac {6 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {(i b) \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 c^4 d^3}-\frac {(6 i b) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^4 d^3}-\frac {(3 b) \int \tan ^{-1}(c x) \, dx}{c^4 d^3}+\frac {(4 b) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^4 d^3}-\frac {(i b) \int \frac {x^2}{1+c^2 x^2} \, dx}{2 c^2 d^3}\\ &=-\frac {3 a x}{c^4 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (i-c x)^2}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (i-c x)}+\frac {6 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}-\frac {(6 b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^5 d^3}+\frac {(i b) \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{2 c^4 d^3}+\frac {(i b) \int \frac {1}{1+c^2 x^2} \, dx}{2 c^4 d^3}+\frac {(4 b) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{c^4 d^3}+\frac {(3 b) \int \frac {x}{1+c^2 x^2} \, dx}{c^3 d^3}\\ &=-\frac {3 a x}{c^4 d^3}-\frac {i b x}{2 c^4 d^3}+\frac {i b \tan ^{-1}(c x)}{2 c^5 d^3}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (i-c x)^2}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (i-c x)}+\frac {6 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {3 b \log \left (1+c^2 x^2\right )}{2 c^5 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {(i b) \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 c^4 d^3}+\frac {(4 b) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^4 d^3}\\ &=-\frac {3 a x}{c^4 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (i-c x)^2}-\frac {15 i b}{8 c^5 d^3 (i-c x)}+\frac {i b \tan ^{-1}(c x)}{2 c^5 d^3}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (i-c x)^2}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (i-c x)}+\frac {6 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {3 b \log \left (1+c^2 x^2\right )}{2 c^5 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^3}-\frac {(i b) \int \frac {1}{1+c^2 x^2} \, dx}{8 c^4 d^3}+\frac {(2 i b) \int \frac {1}{1+c^2 x^2} \, dx}{c^4 d^3}\\ &=-\frac {3 a x}{c^4 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (i-c x)^2}-\frac {15 i b}{8 c^5 d^3 (i-c x)}+\frac {19 i b \tan ^{-1}(c x)}{8 c^5 d^3}-\frac {3 b x \tan ^{-1}(c x)}{c^4 d^3}+\frac {i x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c^3 d^3}-\frac {i \left (a+b \tan ^{-1}(c x)\right )}{2 c^5 d^3 (i-c x)^2}+\frac {4 \left (a+b \tan ^{-1}(c x)\right )}{c^5 d^3 (i-c x)}+\frac {6 i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {3 b \log \left (1+c^2 x^2\right )}{2 c^5 d^3}-\frac {3 b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{c^5 d^3}\\ \end {align*}

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Mathematica [A]  time = 1.06, size = 235, normalized size = 0.92 \[ \frac {16 i a c^2 x^2-96 i a \log \left (c^2 x^2+1\right )-96 a c x-\frac {128 a}{c x-i}-\frac {16 i a}{(c x-i)^2}+192 a \tan ^{-1}(c x)+b \left (48 \log \left (c^2 x^2+1\right )+4 i \tan ^{-1}(c x) \left (4 c^2 x^2+24 i c x+48 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+14 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )-14 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )+4\right )+96 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-16 i c x+192 \tan ^{-1}(c x)^2+28 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )-28 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )\right )}{32 c^5 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcTan[c*x]))/(d + I*c*d*x)^3,x]

[Out]

(-96*a*c*x + (16*I)*a*c^2*x^2 - ((16*I)*a)/(-I + c*x)^2 - (128*a)/(-I + c*x) + 192*a*ArcTan[c*x] - (96*I)*a*Lo
g[1 + c^2*x^2] + b*((-16*I)*c*x + 192*ArcTan[c*x]^2 - 28*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*x]] + 48*Log[1 +
c^2*x^2] + 96*PolyLog[2, -E^((2*I)*ArcTan[c*x])] + (28*I)*Sin[2*ArcTan[c*x]] + (4*I)*ArcTan[c*x]*(4 + (24*I)*c
*x + 4*c^2*x^2 - 14*Cos[2*ArcTan[c*x]] + Cos[4*ArcTan[c*x]] + 48*Log[1 + E^((2*I)*ArcTan[c*x])] + (14*I)*Sin[2
*ArcTan[c*x]] - I*Sin[4*ArcTan[c*x]]) - I*Sin[4*ArcTan[c*x]]))/(32*c^5*d^3)

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{4} \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a x^{4}}{2 \, c^{3} d^{3} x^{3} - 6 i \, c^{2} d^{3} x^{2} - 6 \, c d^{3} x + 2 i \, d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))/(d+I*c*d*x)^3,x, algorithm="fricas")

[Out]

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 423, normalized size = 1.65 \[ -\frac {3 a x}{c^{4} d^{3}}-\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{c^{5} d^{3}}-\frac {4 a}{c^{5} d^{3} \left (c x -i\right )}-\frac {i b \arctan \left (c x \right )}{2 c^{5} d^{3} \left (c x -i\right )^{2}}+\frac {6 a \arctan \left (c x \right )}{c^{5} d^{3}}+\frac {i a \,x^{2}}{2 c^{3} d^{3}}-\frac {3 b x \arctan \left (c x \right )}{c^{4} d^{3}}+\frac {5 i b \arctan \left (\frac {c x}{2}\right )}{32 c^{5} d^{3}}-\frac {4 b \arctan \left (c x \right )}{c^{5} d^{3} \left (c x -i\right )}+\frac {15 i b}{8 c^{5} d^{3} \left (c x -i\right )}-\frac {5 i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{32 c^{5} d^{3}}-\frac {b}{2 c^{5} d^{3}}-\frac {5 i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{16 c^{5} d^{3}}+\frac {5 b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{64 c^{5} d^{3}}-\frac {i b x}{2 c^{4} d^{3}}-\frac {i a}{2 c^{5} d^{3} \left (c x -i\right )^{2}}+\frac {i b \arctan \left (c x \right ) x^{2}}{2 c^{3} d^{3}}-\frac {6 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{5} d^{3}}-\frac {b}{8 c^{5} d^{3} \left (c x -i\right )^{2}}+\frac {43 b \ln \left (c^{2} x^{2}+1\right )}{32 c^{5} d^{3}}+\frac {43 i b \arctan \left (c x \right )}{16 c^{5} d^{3}}+\frac {3 b \ln \left (c x -i\right )^{2}}{2 c^{5} d^{3}}-\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{5} d^{3}}-\frac {3 b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{5} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctan(c*x))/(d+I*c*d*x)^3,x)

[Out]

-3*a*x/c^4/d^3-3*I/c^5*a/d^3*ln(c^2*x^2+1)-4/c^5*a/d^3/(c*x-I)-1/2*I/c^5*b/d^3*arctan(c*x)/(c*x-I)^2+6/c^5*a/d
^3*arctan(c*x)+1/2*I/c^3*a/d^3*x^2-3*b*x*arctan(c*x)/c^4/d^3+5/32*I/c^5*b/d^3*arctan(1/2*c*x)-4/c^5*b/d^3*arct
an(c*x)/(c*x-I)+15/8*I/c^5*b/d^3/(c*x-I)-5/32*I/c^5*b/d^3*arctan(1/6*c^3*x^3+7/6*c*x)-1/2/c^5*b/d^3-5/16*I/c^5
*b/d^3*arctan(1/2*c*x-1/2*I)+5/64/c^5*b/d^3*ln(c^4*x^4+10*c^2*x^2+9)-1/2*I*b*x/c^4/d^3-1/2*I/c^5*a/d^3/(c*x-I)
^2+1/2*I/c^3*b/d^3*arctan(c*x)*x^2-6*I/c^5*b/d^3*arctan(c*x)*ln(c*x-I)-1/8/c^5*b/d^3/(c*x-I)^2+43/32*b*ln(c^2*
x^2+1)/c^5/d^3+43/16*I/c^5*b/d^3*arctan(c*x)+3/2/c^5*b/d^3*ln(c*x-I)^2-3/c^5*b/d^3*ln(c*x-I)*ln(-1/2*I*(I+c*x)
)-3/c^5*b/d^3*dilog(-1/2*I*(I+c*x))

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maxima [A]  time = 0.45, size = 356, normalized size = 1.39 \[ \frac {8 i \, a c^{4} x^{4} - 8 \, {\left (4 \, a + i \, b\right )} c^{3} x^{3} + {\left (b {\left (5 i \, \arctan \left (1, c x\right ) - 16\right )} + 88 i \, a\right )} c^{2} x^{2} + {\left (b {\left (10 \, \arctan \left (1, c x\right ) + 38 i\right )} - 16 \, a\right )} c x + {\left (24 \, b c^{2} x^{2} - 48 i \, b c x - 24 \, b\right )} \arctan \left (c x\right )^{2} + {\left (6 \, b c^{2} x^{2} - 12 i \, b c x - 6 \, b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + {\left (-24 i \, b c^{2} x^{2} - 48 \, b c x + 24 i \, b\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + b {\left (-5 i \, \arctan \left (1, c x\right ) + 28\right )} + {\left (8 i \, b c^{4} x^{4} - 32 \, b c^{3} x^{3} + {\left (96 \, a + 131 i \, b\right )} c^{2} x^{2} + {\left (-192 i \, a + 70 \, b\right )} c x - 96 \, a + 13 i \, b\right )} \arctan \left (c x\right ) - {\left (48 \, b c^{2} x^{2} - 96 i \, b c x - 48 \, b\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) + {\left ({\left (-48 i \, a + 24 \, b\right )} c^{2} x^{2} - 48 \, {\left (2 \, a + i \, b\right )} c x - {\left (12 \, b c^{2} x^{2} - 24 i \, b c x - 12 \, b\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + 48 i \, a - 24 \, b\right )} \log \left (c^{2} x^{2} + 1\right ) + 56 i \, a}{16 \, c^{7} d^{3} x^{2} - 32 i \, c^{6} d^{3} x - 16 \, c^{5} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctan(c*x))/(d+I*c*d*x)^3,x, algorithm="maxima")

[Out]

(8*I*a*c^4*x^4 - 8*(4*a + I*b)*c^3*x^3 + (b*(5*I*arctan2(1, c*x) - 16) + 88*I*a)*c^2*x^2 + (b*(10*arctan2(1, c
*x) + 38*I) - 16*a)*c*x + (24*b*c^2*x^2 - 48*I*b*c*x - 24*b)*arctan(c*x)^2 + (6*b*c^2*x^2 - 12*I*b*c*x - 6*b)*
log(c^2*x^2 + 1)^2 + (-24*I*b*c^2*x^2 - 48*b*c*x + 24*I*b)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) + b*(-5*I*arctan
2(1, c*x) + 28) + (8*I*b*c^4*x^4 - 32*b*c^3*x^3 + (96*a + 131*I*b)*c^2*x^2 + (-192*I*a + 70*b)*c*x - 96*a + 13
*I*b)*arctan(c*x) - (48*b*c^2*x^2 - 96*I*b*c*x - 48*b)*dilog(1/2*I*c*x + 1/2) + ((-48*I*a + 24*b)*c^2*x^2 - 48
*(2*a + I*b)*c*x - (12*b*c^2*x^2 - 24*I*b*c*x - 12*b)*log(1/4*c^2*x^2 + 1/4) + 48*I*a - 24*b)*log(c^2*x^2 + 1)
+ 56*I*a)/(16*c^7*d^3*x^2 - 32*I*c^6*d^3*x - 16*c^5*d^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*atan(c*x)))/(d + c*d*x*1i)^3,x)

[Out]

int((x^4*(a + b*atan(c*x)))/(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(x**4*(a+b*atan(c*x))/(d+I*c*d*x)**3,x)

[Out]

Timed out

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