\(\int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx\) [22]

Optimal result

Integrand size = 21, antiderivative size = 157 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {3 b d^3 x}{5 c}-\frac {3 i b d^3 (i-c x)^2}{20 c^2}-\frac {b d^3 (i-c x)^3}{20 c^2}+\frac {i b d^3 (i-c x)^4}{20 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {6 i b d^3 \log (i+c x)}{5 c^2} \]

[Out]

-3/5*b*d^3*x/c-3/20*I*b*d^3*(I-c*x)^2/c^2-1/20*b*d^3*(I-c*x)^3/c^2+1/20*I*b*d^3*(I-c*x)^4/c^2+1/4*d^3*(1+I*c*x
)^4*(a+b*arctan(c*x))/c^2-1/5*d^3*(1+I*c*x)^5*(a+b*arctan(c*x))/c^2+6/5*I*b*d^3*ln(I+c*x)/c^2

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {45, 4992, 12, 78} \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}+\frac {i b d^3 (-c x+i)^4}{20 c^2}-\frac {b d^3 (-c x+i)^3}{20 c^2}-\frac {3 i b d^3 (-c x+i)^2}{20 c^2}+\frac {6 i b d^3 \log (c x+i)}{5 c^2}-\frac {3 b d^3 x}{5 c} \]

[In]

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

[Out]

(-3*b*d^3*x)/(5*c) - (((3*I)/20)*b*d^3*(I - c*x)^2)/c^2 - (b*d^3*(I - c*x)^3)/(20*c^2) + ((I/20)*b*d^3*(I - c*
x)^4)/c^2 + (d^3*(1 + I*c*x)^4*(a + b*ArcTan[c*x]))/(4*c^2) - (d^3*(1 + I*c*x)^5*(a + b*ArcTan[c*x]))/(5*c^2)
+ (((6*I)/5)*b*d^3*Log[I + c*x])/c^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 4992

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_))^(q_.), x_Symbol] :> With[{u = I
ntHide[(f*x)^m*(d + e*x)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^2*x^
2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && NeQ[q, -1] && IntegerQ[2*m] && ((IGtQ[m, 0] && IGtQ[q, 0
]) || (ILtQ[m + q + 1, 0] && LtQ[m*q, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-(b c) \int \frac {d^3 (i-c x)^3 (-1+4 i c x)}{20 c^2 (i+c x)} \, dx \\ & = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-\frac {\left (b d^3\right ) \int \frac {(i-c x)^3 (-1+4 i c x)}{i+c x} \, dx}{20 c} \\ & = \frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}-\frac {\left (b d^3\right ) \int \left (12+4 i (i-c x)^3+6 i (-i+c x)-3 (-i+c x)^2-\frac {24 i}{i+c x}\right ) \, dx}{20 c} \\ & = -\frac {3 b d^3 x}{5 c}-\frac {3 i b d^3 (i-c x)^2}{20 c^2}-\frac {b d^3 (i-c x)^3}{20 c^2}+\frac {i b d^3 (i-c x)^4}{20 c^2}+\frac {d^3 (1+i c x)^4 (a+b \arctan (c x))}{4 c^2}-\frac {d^3 (1+i c x)^5 (a+b \arctan (c x))}{5 c^2}+\frac {6 i b d^3 \log (i+c x)}{5 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.84 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {d^3 \left (c x \left (b \left (-25-12 i c x+5 c^2 x^2+i c^3 x^3\right )+a c x \left (10+20 i c x-15 c^2 x^2-4 i c^3 x^3\right )\right )+b \left (25+10 c^2 x^2+20 i c^3 x^3-15 c^4 x^4-4 i c^5 x^5\right ) \arctan (c x)+12 i b \log \left (1+c^2 x^2\right )\right )}{20 c^2} \]

[In]

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

[Out]

(d^3*(c*x*(b*(-25 - (12*I)*c*x + 5*c^2*x^2 + I*c^3*x^3) + a*c*x*(10 + (20*I)*c*x - 15*c^2*x^2 - (4*I)*c^3*x^3)
) + b*(25 + 10*c^2*x^2 + (20*I)*c^3*x^3 - 15*c^4*x^4 - (4*I)*c^5*x^5)*ArcTan[c*x] + (12*I)*b*Log[1 + c^2*x^2])
)/(20*c^2)

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.92

method	result	size
parts	\(a \,d^{3} \left (-\frac {1}{5} i c^{3} x^{5}-\frac {3}{4} c^{2} x^{4}+i c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\)	\(145\)
derivativedivides	\(\frac {a \,d^{3} \left (-\frac {1}{5} i c^{5} x^{5}-\frac {3}{4} c^{4} x^{4}+i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\)	\(151\)
default	\(\frac {a \,d^{3} \left (-\frac {1}{5} i c^{5} x^{5}-\frac {3}{4} c^{4} x^{4}+i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b \,d^{3} \left (-\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}-\frac {3 c^{4} x^{4} \arctan \left (c x \right )}{4}+i \arctan \left (c x \right ) c^{3} x^{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {5 c x}{4}+\frac {i c^{4} x^{4}}{20}+\frac {c^{3} x^{3}}{4}-\frac {3 i c^{2} x^{2}}{5}+\frac {3 i \ln \left (c^{2} x^{2}+1\right )}{5}+\frac {5 \arctan \left (c x \right )}{4}\right )}{c^{2}}\)	\(151\)
parallelrisch	\(\frac {-4 i c^{5} b \,d^{3} \arctan \left (c x \right ) x^{5}-4 i x^{5} a \,c^{5} d^{3}+i x^{4} b \,c^{4} d^{3}-15 x^{4} \arctan \left (c x \right ) b \,c^{4} d^{3}+20 i x^{3} \arctan \left (c x \right ) b \,c^{3} d^{3}-15 a \,c^{4} d^{3} x^{4}+20 i x^{3} a \,c^{3} d^{3}+5 b \,c^{3} d^{3} x^{3}-12 i x^{2} b \,c^{2} d^{3}+10 x^{2} \arctan \left (c x \right ) b \,c^{2} d^{3}+10 x^{2} d^{3} c^{2} a +12 i b \,d^{3} \ln \left (c^{2} x^{2}+1\right )-25 b c \,d^{3} x +25 b \,d^{3} \arctan \left (c x \right )}{20 c^{2}}\)	\(196\)
risch	\(-\frac {d^{3} b \left (4 c^{3} x^{5}-15 i c^{2} x^{4}-20 c \,x^{3}+10 i x^{2}\right ) \ln \left (i c x +1\right )}{40}+\frac {d^{3} c^{3} b \,x^{5} \ln \left (-i c x +1\right )}{10}-\frac {i a \,c^{3} d^{3} x^{5}}{5}-\frac {3 a \,c^{2} d^{3} x^{4}}{4}-\frac {3 i d^{3} c^{2} x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {d^{3} c b \,x^{3} \ln \left (-i c x +1\right )}{2}+\frac {i b \,c^{2} d^{3} x^{4}}{20}+\frac {b c \,d^{3} x^{3}}{4}+i a c \,d^{3} x^{3}+\frac {a \,d^{3} x^{2}}{2}+\frac {i d^{3} x^{2} b \ln \left (-i c x +1\right )}{4}-\frac {3 i b \,d^{3} x^{2}}{5}-\frac {5 b \,d^{3} x}{4 c}+\frac {5 d^{3} b \arctan \left (c x \right )}{4 c^{2}}+\frac {3 i d^{3} b \ln \left (625 c^{2} x^{2}+625\right )}{5 c^{2}}\)	\(245\)

[In]

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

[Out]

a*d^3*(-1/5*I*c^3*x^5-3/4*c^2*x^4+I*c*x^3+1/2*x^2)+b*d^3/c^2*(-1/5*I*arctan(c*x)*c^5*x^5-3/4*c^4*x^4*arctan(c*
x)+I*arctan(c*x)*c^3*x^3+1/2*c^2*x^2*arctan(c*x)-5/4*c*x+1/20*I*c^4*x^4+1/4*c^3*x^3-3/5*I*c^2*x^2+3/5*I*ln(c^2
*x^2+1)+5/4*arctan(c*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.13 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {-8 i \, a c^{5} d^{3} x^{5} - 2 \, {\left (15 \, a - i \, b\right )} c^{4} d^{3} x^{4} - 10 \, {\left (-4 i \, a - b\right )} c^{3} d^{3} x^{3} + 4 \, {\left (5 \, a - 6 i \, b\right )} c^{2} d^{3} x^{2} - 50 \, b c d^{3} x + 49 i \, b d^{3} \log \left (\frac {c x + i}{c}\right ) - i \, b d^{3} \log \left (\frac {c x - i}{c}\right ) + {\left (4 \, b c^{5} d^{3} x^{5} - 15 i \, b c^{4} d^{3} x^{4} - 20 \, b c^{3} d^{3} x^{3} + 10 i \, b c^{2} d^{3} x^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{40 \, c^{2}} \]

[In]

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

[Out]

1/40*(-8*I*a*c^5*d^3*x^5 - 2*(15*a - I*b)*c^4*d^3*x^4 - 10*(-4*I*a - b)*c^3*d^3*x^3 + 4*(5*a - 6*I*b)*c^2*d^3*
x^2 - 50*b*c*d^3*x + 49*I*b*d^3*log((c*x + I)/c) - I*b*d^3*log((c*x - I)/c) + (4*b*c^5*d^3*x^5 - 15*I*b*c^4*d^
3*x^4 - 20*b*c^3*d^3*x^3 + 10*I*b*c^2*d^3*x^2)*log(-(c*x + I)/(c*x - I)))/c^2

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (138) = 276\).

Time = 2.16 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.89 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=- \frac {i a c^{3} d^{3} x^{5}}{5} - \frac {5 b d^{3} x}{4 c} - \frac {b d^{3} \left (\frac {i \log {\left (19 b c d^{3} x - 19 i b d^{3} \right )}}{40} - \frac {37 i \log {\left (19 b c d^{3} x + 19 i b d^{3} \right )}}{40}\right )}{c^{2}} - x^{4} \cdot \left (\frac {3 a c^{2} d^{3}}{4} - \frac {i b c^{2} d^{3}}{20}\right ) - x^{3} \left (- i a c d^{3} - \frac {b c d^{3}}{4}\right ) - x^{2} \left (- \frac {a d^{3}}{2} + \frac {3 i b d^{3}}{5}\right ) + \left (- \frac {b c^{3} d^{3} x^{5}}{10} + \frac {3 i b c^{2} d^{3} x^{4}}{8} + \frac {b c d^{3} x^{3}}{2} - \frac {i b d^{3} x^{2}}{4}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (4 b c^{5} d^{3} x^{5} - 15 i b c^{4} d^{3} x^{4} - 20 b c^{3} d^{3} x^{3} + 10 i b c^{2} d^{3} x^{2} + 12 i b d^{3}\right ) \log {\left (- i c x + 1 \right )}}{40 c^{2}} \]

[In]

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

[Out]

-I*a*c**3*d**3*x**5/5 - 5*b*d**3*x/(4*c) - b*d**3*(I*log(19*b*c*d**3*x - 19*I*b*d**3)/40 - 37*I*log(19*b*c*d**
3*x + 19*I*b*d**3)/40)/c**2 - x**4*(3*a*c**2*d**3/4 - I*b*c**2*d**3/20) - x**3*(-I*a*c*d**3 - b*c*d**3/4) - x*
*2*(-a*d**3/2 + 3*I*b*d**3/5) + (-b*c**3*d**3*x**5/10 + 3*I*b*c**2*d**3*x**4/8 + b*c*d**3*x**3/2 - I*b*d**3*x*
*2/4)*log(I*c*x + 1) + (4*b*c**5*d**3*x**5 - 15*I*b*c**4*d**3*x**4 - 20*b*c**3*d**3*x**3 + 10*I*b*c**2*d**3*x*
*2 + 12*I*b*d**3)*log(-I*c*x + 1)/(40*c**2)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.41 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=-\frac {1}{5} i \, a c^{3} d^{3} x^{5} - \frac {3}{4} \, a c^{2} d^{3} x^{4} - \frac {1}{20} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{3} d^{3} + i \, a c d^{3} x^{3} - \frac {1}{4} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{2} d^{3} + \frac {1}{2} i \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c d^{3} + \frac {1}{2} \, a d^{3} x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{3} \]

[In]

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

[Out]

-1/5*I*a*c^3*d^3*x^5 - 3/4*a*c^2*d^3*x^4 - 1/20*I*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*log(c^2*x^
2 + 1)/c^6))*b*c^3*d^3 + I*a*c*d^3*x^3 - 1/4*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))
*b*c^2*d^3 + 1/2*I*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*c*d^3 + 1/2*a*d^3*x^2 + 1/2*(x^2
*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b*d^3

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.79 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.02 \[ \int x (d+i c d x)^3 (a+b \arctan (c x)) \, dx=\frac {\frac {d^3\,\left (25\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,12{}\mathrm {i}\right )}{20}-\frac {5\,b\,c\,d^3\,x}{4}}{c^2}+\frac {d^3\,\left (10\,a\,x^2+10\,b\,x^2\,\mathrm {atan}\left (c\,x\right )-b\,x^2\,12{}\mathrm {i}\right )}{20}-\frac {c^3\,d^3\,\left (a\,x^5\,4{}\mathrm {i}+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,4{}\mathrm {i}\right )}{20}+\frac {c\,d^3\,\left (a\,x^3\,20{}\mathrm {i}+5\,b\,x^3+b\,x^3\,\mathrm {atan}\left (c\,x\right )\,20{}\mathrm {i}\right )}{20}-\frac {c^2\,d^3\,\left (15\,a\,x^4+15\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,1{}\mathrm {i}\right )}{20} \]

[In]

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

[Out]

((d^3*(25*b*atan(c*x) + b*log(c^2*x^2 + 1)*12i))/20 - (5*b*c*d^3*x)/4)/c^2 + (d^3*(10*a*x^2 - b*x^2*12i + 10*b
*x^2*atan(c*x)))/20 - (c^3*d^3*(a*x^5*4i + b*x^5*atan(c*x)*4i))/20 + (c*d^3*(a*x^3*20i + 5*b*x^3 + b*x^3*atan(
c*x)*20i))/20 - (c^2*d^3*(15*a*x^4 - b*x^4*1i + 15*b*x^4*atan(c*x)))/20