\(\int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx\) [58]

Optimal result

Integrand size = 23, antiderivative size = 256 \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=-\frac {3 a x}{c^4 d^3}-\frac {i b x}{2 c^4 d^3}-\frac {b}{8 c^5 d^3 (i-c x)^2}-\frac {15 i b}{8 c^5 d^3 (i-c x)}+\frac {19 i b \arctan (c x)}{8 c^5 d^3}-\frac {3 b x \arctan (c x)}{c^4 d^3}+\frac {i x^2 (a+b \arctan (c x))}{2 c^3 d^3}-\frac {i (a+b \arctan (c x))}{2 c^5 d^3 (i-c x)^2}+\frac {4 (a+b \arctan (c x))}{c^5 d^3 (i-c x)}+\frac {6 i (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c^5 d^3}+\frac {3 b \log \left (1+c^2 x^2\right )}{2 c^5 d^3}-\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c^5 d^3} \]

[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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4996, 4930, 266, 4946, 327, 209, 4972, 641, 46, 4964, 2449, 2352} \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=\frac {4 (a+b \arctan (c x))}{c^5 d^3 (-c x+i)}-\frac {i (a+b \arctan (c x))}{2 c^5 d^3 (-c x+i)^2}+\frac {6 i \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c^5 d^3}+\frac {i x^2 (a+b \arctan (c x))}{2 c^3 d^3}-\frac {3 a x}{c^4 d^3}+\frac {19 i b \arctan (c x)}{8 c^5 d^3}-\frac {3 b x \arctan (c x)}{c^4 d^3}-\frac {3 b \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c^5 d^3}-\frac {15 i b}{8 c^5 d^3 (-c x+i)}-\frac {b}{8 c^5 d^3 (-c x+i)^2}-\frac {i b x}{2 c^4 d^3}+\frac {3 b \log \left (c^2 x^2+1\right )}{2 c^5 d^3} \]

[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 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 266

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 327

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 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 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 4930

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

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

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=\frac {-96 a c x+16 i a c^2 x^2-\frac {16 i a}{(-i+c x)^2}-\frac {128 a}{-i+c x}+192 a \arctan (c x)-96 i a \log \left (1+c^2 x^2\right )+b \left (-16 i c x+192 \arctan (c x)^2-28 \cos (2 \arctan (c x))+\cos (4 \arctan (c x))+48 \log \left (1+c^2 x^2\right )+96 \operatorname {PolyLog}\left (2,-e^{2 i \arctan (c x)}\right )+28 i \sin (2 \arctan (c x))+4 i \arctan (c x) \left (4+24 i c x+4 c^2 x^2-14 \cos (2 \arctan (c x))+\cos (4 \arctan (c x))+48 \log \left (1+e^{2 i \arctan (c x)}\right )+14 i \sin (2 \arctan (c x))-i \sin (4 \arctan (c x))\right )-i \sin (4 \arctan (c x))\right )}{32 c^5 d^3} \]

[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)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.42

method	result	size
derivativedivides	\(\frac {-\frac {3 a c x}{d^{3}}-\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{d^{3}}-\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}-\frac {4 a}{d^{3} \left (c x -i\right )}+\frac {6 a \arctan \left (c x \right )}{d^{3}}+\frac {43 i b \arctan \left (c x \right )}{16 d^{3}}-\frac {3 b \arctan \left (c x \right ) c x}{d^{3}}-\frac {i a}{2 d^{3} \left (c x -i\right )^{2}}+\frac {i b \arctan \left (c x \right ) c^{2} x^{2}}{2 d^{3}}-\frac {4 b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {6 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}-\frac {b}{2 d^{3}}-\frac {i b c x}{2 d^{3}}+\frac {5 b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{64 d^{3}}-\frac {5 i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{32 d^{3}}+\frac {i a \,c^{2} x^{2}}{2 d^{3}}+\frac {15 i b}{8 d^{3} \left (c x -i\right )}-\frac {5 i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{16 d^{3}}-\frac {b}{8 d^{3} \left (c x -i\right )^{2}}+\frac {43 b \ln \left (c^{2} x^{2}+1\right )}{32 d^{3}}+\frac {5 i b \arctan \left (\frac {c x}{2}\right )}{32 d^{3}}-\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}+\frac {3 b \ln \left (c x -i\right )^{2}}{2 d^{3}}-\frac {3 b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}}{c^{5}}\)	\(364\)
default	\(\frac {-\frac {3 a c x}{d^{3}}-\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{d^{3}}-\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (c x -i\right )^{2}}-\frac {4 a}{d^{3} \left (c x -i\right )}+\frac {6 a \arctan \left (c x \right )}{d^{3}}+\frac {43 i b \arctan \left (c x \right )}{16 d^{3}}-\frac {3 b \arctan \left (c x \right ) c x}{d^{3}}-\frac {i a}{2 d^{3} \left (c x -i\right )^{2}}+\frac {i b \arctan \left (c x \right ) c^{2} x^{2}}{2 d^{3}}-\frac {4 b \arctan \left (c x \right )}{d^{3} \left (c x -i\right )}-\frac {6 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{d^{3}}-\frac {b}{2 d^{3}}-\frac {i b c x}{2 d^{3}}+\frac {5 b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{64 d^{3}}-\frac {5 i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{32 d^{3}}+\frac {i a \,c^{2} x^{2}}{2 d^{3}}+\frac {15 i b}{8 d^{3} \left (c x -i\right )}-\frac {5 i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{16 d^{3}}-\frac {b}{8 d^{3} \left (c x -i\right )^{2}}+\frac {43 b \ln \left (c^{2} x^{2}+1\right )}{32 d^{3}}+\frac {5 i b \arctan \left (\frac {c x}{2}\right )}{32 d^{3}}-\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}+\frac {3 b \ln \left (c x -i\right )^{2}}{2 d^{3}}-\frac {3 b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{d^{3}}}{c^{5}}\)	\(364\)
parts	\(-\frac {5 i b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{32 c^{5} d^{3}}-\frac {3 a x}{c^{4} d^{3}}+\frac {4 a}{d^{3} c^{5} \left (-c x +i\right )}+\frac {5 i b \arctan \left (\frac {c x}{2}\right )}{32 c^{5} d^{3}}+\frac {6 a \arctan \left (c x \right )}{c^{5} d^{3}}+\frac {43 i b \arctan \left (c x \right )}{16 c^{5} d^{3}}-\frac {3 b x \arctan \left (c x \right )}{c^{4} d^{3}}+\frac {15 i b}{8 c^{5} d^{3} \left (c x -i\right )}-\frac {4 b \arctan \left (c x \right )}{c^{5} d^{3} \left (c x -i\right )}-\frac {5 i b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{16 c^{5} d^{3}}-\frac {i b x}{2 c^{4} d^{3}}+\frac {i a \,x^{2}}{2 c^{3} d^{3}}-\frac {b}{2 c^{5} d^{3}}+\frac {5 b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{64 c^{5} d^{3}}+\frac {i b \arctan \left (c x \right ) x^{2}}{2 c^{3} d^{3}}-\frac {6 i b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{5} d^{3}}-\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{c^{5} d^{3}}-\frac {i a}{2 d^{3} c^{5} \left (-c x +i\right )^{2}}-\frac {b}{8 c^{5} d^{3} \left (c x -i\right )^{2}}+\frac {43 b \ln \left (c^{2} x^{2}+1\right )}{32 c^{5} d^{3}}-\frac {i b \arctan \left (c x \right )}{2 c^{5} d^{3} \left (c x -i\right )^{2}}+\frac {3 b \ln \left (c x -i\right )^{2}}{2 c^{5} d^{3}}-\frac {3 b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{5} d^{3}}-\frac {3 b \operatorname {dilog}\left (-\frac {i \left (c x +i\right )}{2}\right )}{c^{5} d^{3}}\)	\(425\)
risch	\(-\frac {3 a x}{c^{4} d^{3}}+\frac {43 b \ln \left (c^{2} x^{2}+1\right )}{32 c^{5} d^{3}}-\frac {9 b}{8 c^{5} d^{3}}-\frac {b}{8 c^{5} d^{3} \left (-c x +i\right )^{2}}+\frac {5 b \ln \left (-i c x +1\right )}{4 c^{5} d^{3}}-\frac {3 b \operatorname {dilog}\left (\frac {1}{2}-\frac {i c x}{2}\right )}{c^{5} d^{3}}-\frac {b}{8 c^{5} d^{3} \left (-i c x -1\right )}-\frac {5 i a}{2 c^{5} d^{3}}-\frac {3 b \ln \left (i c x +1\right )^{2}}{2 c^{5} d^{3}}+\frac {6 a \arctan \left (c x \right )}{c^{5} d^{3}}-\frac {b \ln \left (-i c x +1\right ) x^{2}}{16 c^{3} d^{3} \left (-i c x -1\right )^{2}}-\frac {3 i b \ln \left (-i c x +1\right ) x}{2 c^{4} d^{3}}+\frac {i b \ln \left (-i c x +1\right ) x}{8 c^{4} d^{3} \left (-i c x -1\right )^{2}}+\frac {i b \ln \left (-i c x +1\right ) x}{c^{4} d^{3} \left (-i c x -1\right )}+\left (\frac {b \left (\frac {1}{2} c \,x^{2}+3 i x \right )}{2 c^{4} d^{3}}+\frac {4 i b \,d^{3} x +\frac {7 b \,d^{3}}{2 c}}{2 c^{4} d^{6} \left (c x -i\right )^{2}}\right ) \ln \left (i c x +1\right )+\frac {i a}{2 c^{5} d^{3} \left (-i c x -1\right )^{2}}-\frac {2 i b}{c^{5} d^{3} \left (-c x +i\right )}-\frac {x^{2} b \ln \left (-i c x +1\right )}{4 c^{3} d^{3}}-\frac {3 \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) b}{c^{5} d^{3}}+\frac {3 b \ln \left (\frac {1}{2}+\frac {i c x}{2}\right ) \ln \left (-i c x +1\right )}{c^{5} d^{3}}-\frac {b \ln \left (-i c x +1\right )}{c^{5} d^{3} \left (-i c x -1\right )}-\frac {3 b \ln \left (-i c x +1\right )}{16 c^{5} d^{3} \left (-i c x -1\right )^{2}}+\frac {4 i a}{c^{5} d^{3} \left (-i c x -1\right )}+\frac {43 i b \arctan \left (c x \right )}{16 c^{5} d^{3}}-\frac {i b x}{2 c^{4} d^{3}}+\frac {i a \,x^{2}}{2 c^{3} d^{3}}-\frac {3 i a \ln \left (c^{2} x^{2}+1\right )}{c^{5} d^{3}}\)	\(557\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(-1/2*(b*x^4*log(-(c*x + I)/(c*x - I)) - 2*I*a*x^4)/(c^3*d^3*x^3 - 3*I*c^2*d^3*x^2 - 3*c*d^3*x + I*d^3
), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.40 \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=\frac {8 i \, a c^{4} x^{4} - 8 \, {\left (4 \, a + i \, b\right )} c^{3} x^{3} + {\left (b {\left (5 i \, \arctan \left (1, c x\right ) - 16\right )} + 88 i \, a\right )} c^{2} x^{2} + 2 \, {\left (b {\left (5 \, \arctan \left (1, c x\right ) + 19 i\right )} - 8 \, a\right )} c x + 24 \, {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \arctan \left (c x\right )^{2} + 6 \, {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} - 24 \, {\left (i \, b c^{2} x^{2} + 2 \, b c x - i \, b\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + b {\left (-5 i \, \arctan \left (1, c x\right ) + 28\right )} + {\left (8 i \, b c^{4} x^{4} - 32 \, b c^{3} x^{3} + {\left (96 \, a + 131 i \, b\right )} c^{2} x^{2} - 2 \, {\left (96 i \, a - 35 \, b\right )} c x - 96 \, a + 13 i \, b\right )} \arctan \left (c x\right ) - 48 \, {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) - 12 \, {\left (2 \, {\left (2 i \, a - b\right )} c^{2} x^{2} + 4 \, {\left (2 \, a + i \, b\right )} c x + {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) - 4 i \, a + 2 \, b\right )} \log \left (c^{2} x^{2} + 1\right ) + 56 i \, a}{16 \, {\left (c^{7} d^{3} x^{2} - 2 i \, c^{6} d^{3} x - c^{5} d^{3}\right )}} \]

[In]

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

[Out]

1/16*(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 + 2*(b*(5*arctan
2(1, c*x) + 19*I) - 8*a)*c*x + 24*(b*c^2*x^2 - 2*I*b*c*x - b)*arctan(c*x)^2 + 6*(b*c^2*x^2 - 2*I*b*c*x - b)*lo
g(c^2*x^2 + 1)^2 - 24*(I*b*c^2*x^2 + 2*b*c*x - I*b)*arctan(c*x)*log(1/4*c^2*x^2 + 1/4) + b*(-5*I*arctan2(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 - 2*(96*I*a - 35*b)*c*x - 96*a + 13*I*b)*a
rctan(c*x) - 48*(b*c^2*x^2 - 2*I*b*c*x - b)*dilog(1/2*I*c*x + 1/2) - 12*(2*(2*I*a - b)*c^2*x^2 + 4*(2*a + I*b)
*c*x + (b*c^2*x^2 - 2*I*b*c*x - b)*log(1/4*c^2*x^2 + 1/4) - 4*I*a + 2*b)*log(c^2*x^2 + 1) + 56*I*a)/(c^7*d^3*x
^2 - 2*I*c^6*d^3*x - c^5*d^3)

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \arctan (c x))}{(d+i c d x)^3} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

[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)