\(\int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx\) [1]

Optimal result

Integrand size = 21, antiderivative size = 117 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4-\frac {b d \arctan (c x)}{4 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {i b d \log \left (1+c^2 x^2\right )}{10 c^4} \]

[Out]

1/4*b*d*x/c^3+1/10*I*b*d*x^2/c^2-1/12*b*d*x^3/c-1/20*I*b*d*x^4-1/4*b*d*arctan(c*x)/c^4+1/4*d*x^4*(a+b*arctan(c
*x))+1/5*I*c*d*x^5*(a+b*arctan(c*x))-1/10*I*b*d*ln(c^2*x^2+1)/c^4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {45, 4992, 12, 815, 649, 209, 266} \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{5} i c d x^5 (a+b \arctan (c x))+\frac {1}{4} d x^4 (a+b \arctan (c x))-\frac {b d \arctan (c x)}{4 c^4}+\frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {i b d \log \left (c^2 x^2+1\right )}{10 c^4}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4 \]

[In]

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

[Out]

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

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 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 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 815

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x)^m*((f + g*x)/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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 {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-(b c) \int \frac {d x^4 (5+4 i c x)}{20 \left (1+c^2 x^2\right )} \, dx \\ & = \frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {1}{20} (b c d) \int \frac {x^4 (5+4 i c x)}{1+c^2 x^2} \, dx \\ & = \frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {1}{20} (b c d) \int \left (-\frac {5}{c^4}-\frac {4 i x}{c^3}+\frac {5 x^2}{c^2}+\frac {4 i x^3}{c}+\frac {5+4 i c x}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx \\ & = \frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {(b d) \int \frac {5+4 i c x}{1+c^2 x^2} \, dx}{20 c^3} \\ & = \frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {(b d) \int \frac {1}{1+c^2 x^2} \, dx}{4 c^3}-\frac {(i b d) \int \frac {x}{1+c^2 x^2} \, dx}{5 c^2} \\ & = \frac {b d x}{4 c^3}+\frac {i b d x^2}{10 c^2}-\frac {b d x^3}{12 c}-\frac {1}{20} i b d x^4-\frac {b d \arctan (c x)}{4 c^4}+\frac {1}{4} d x^4 (a+b \arctan (c x))+\frac {1}{5} i c d x^5 (a+b \arctan (c x))-\frac {i b d \log \left (1+c^2 x^2\right )}{10 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {d \left (3 a c^4 x^4 (5+4 i c x)+b c x \left (15+6 i c x-5 c^2 x^2-3 i c^3 x^3\right )+3 b \left (-5+5 c^4 x^4+4 i c^5 x^5\right ) \arctan (c x)-6 i b \log \left (1+c^2 x^2\right )\right )}{60 c^4} \]

[In]

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

[Out]

(d*(3*a*c^4*x^4*(5 + (4*I)*c*x) + b*c*x*(15 + (6*I)*c*x - 5*c^2*x^2 - (3*I)*c^3*x^3) + 3*b*(-5 + 5*c^4*x^4 + (
4*I)*c^5*x^5)*ArcTan[c*x] - (6*I)*b*Log[1 + c^2*x^2]))/(60*c^4)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85

method	result	size
parts	\(a d \left (\frac {1}{5} i c \,x^{5}+\frac {1}{4} x^{4}\right )+\frac {b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(99\)
derivativedivides	\(\frac {a d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(105\)
default	\(\frac {a d \left (\frac {1}{5} i c^{5} x^{5}+\frac {1}{4} c^{4} x^{4}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{5} x^{5}}{5}+\frac {c^{4} x^{4} \arctan \left (c x \right )}{4}+\frac {c x}{4}-\frac {i c^{4} x^{4}}{20}-\frac {c^{3} x^{3}}{12}+\frac {i c^{2} x^{2}}{10}-\frac {i \ln \left (c^{2} x^{2}+1\right )}{10}-\frac {\arctan \left (c x \right )}{4}\right )}{c^{4}}\)	\(105\)
parallelrisch	\(-\frac {-12 i c^{5} b d \arctan \left (c x \right ) x^{5}-12 i x^{5} a \,c^{5} d +3 i x^{4} b \,c^{4} d -15 x^{4} \arctan \left (c x \right ) b \,c^{4} d -15 a \,c^{4} d \,x^{4}+5 c^{3} x^{3} d b -6 i x^{2} b \,c^{2} d +6 i b d \ln \left (c^{2} x^{2}+1\right )-15 b c d x +15 b d \arctan \left (c x \right )}{60 c^{4}}\)	\(118\)
risch	\(\frac {d b \left (4 x^{5} c -5 i x^{4}\right ) \ln \left (i c x +1\right )}{40}-\frac {d c b \,x^{5} \ln \left (-i c x +1\right )}{10}+\frac {i d c a \,x^{5}}{5}+\frac {a d \,x^{4}}{4}+\frac {i d \,x^{4} b \ln \left (-i c x +1\right )}{8}-\frac {i b d \,x^{4}}{20}-\frac {b d \,x^{3}}{12 c}+\frac {i b d \,x^{2}}{10 c^{2}}+\frac {b d x}{4 c^{3}}-\frac {b d \arctan \left (c x \right )}{4 c^{4}}-\frac {i d b \ln \left (25 c^{2} x^{2}+25\right )}{10 c^{4}}\)	\(142\)

[In]

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

[Out]

a*d*(1/5*I*c*x^5+1/4*x^4)+b*d/c^4*(1/5*I*arctan(c*x)*c^5*x^5+1/4*c^4*x^4*arctan(c*x)+1/4*c*x-1/20*I*c^4*x^4-1/
12*c^3*x^3+1/10*I*c^2*x^2-1/10*I*ln(c^2*x^2+1)-1/4*arctan(c*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.06 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {24 i \, a c^{5} d x^{5} + 6 \, {\left (5 \, a - i \, b\right )} c^{4} d x^{4} - 10 \, b c^{3} d x^{3} + 12 i \, b c^{2} d x^{2} + 30 \, b c d x - 27 i \, b d \log \left (\frac {c x + i}{c}\right ) + 3 i \, b d \log \left (\frac {c x - i}{c}\right ) - 3 \, {\left (4 \, b c^{5} d x^{5} - 5 i \, b c^{4} d x^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{120 \, c^{4}} \]

[In]

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

[Out]

1/120*(24*I*a*c^5*d*x^5 + 6*(5*a - I*b)*c^4*d*x^4 - 10*b*c^3*d*x^3 + 12*I*b*c^2*d*x^2 + 30*b*c*d*x - 27*I*b*d*
log((c*x + I)/c) + 3*I*b*d*log((c*x - I)/c) - 3*(4*b*c^5*d*x^5 - 5*I*b*c^4*d*x^4)*log(-(c*x + I)/(c*x - I)))/c
^4

Sympy [A] (verification not implemented)

Time = 1.88 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.57 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {i a c d x^{5}}{5} - \frac {b d x^{3}}{12 c} + \frac {i b d x^{2}}{10 c^{2}} + \frac {b d x}{4 c^{3}} + \frac {b d \left (\frac {i \log {\left (25 b c d x - 25 i b d \right )}}{40} - \frac {11 i \log {\left (25 b c d x + 25 i b d \right )}}{60}\right )}{c^{4}} + x^{4} \left (\frac {a d}{4} - \frac {i b d}{20}\right ) + \left (\frac {b c d x^{5}}{10} - \frac {i b d x^{4}}{8}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 12 b c^{5} d x^{5} + 15 i b c^{4} d x^{4} - 5 i b d\right ) \log {\left (- i c x + 1 \right )}}{120 c^{4}} \]

[In]

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

[Out]

I*a*c*d*x**5/5 - b*d*x**3/(12*c) + I*b*d*x**2/(10*c**2) + b*d*x/(4*c**3) + b*d*(I*log(25*b*c*d*x - 25*I*b*d)/4
0 - 11*I*log(25*b*c*d*x + 25*I*b*d)/60)/c**4 + x**4*(a*d/4 - I*b*d/20) + (b*c*d*x**5/10 - I*b*d*x**4/8)*log(I*
c*x + 1) + (-12*b*c**5*d*x**5 + 15*I*b*c**4*d*x**4 - 5*I*b*d)*log(-I*c*x + 1)/(120*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{5} i \, a c d x^{5} + \frac {1}{4} \, a d x^{4} + \frac {1}{20} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c d + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d \]

[In]

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

[Out]

1/5*I*a*c*d*x^5 + 1/4*a*d*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*d + 1/12*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b*d

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int x^3 (d+i c d x) (a+b \arctan (c x)) \, dx=-\frac {\frac {d\,\left (15\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,6{}\mathrm {i}\right )}{60}-\frac {b\,c\,d\,x}{4}+\frac {b\,c^3\,d\,x^3}{12}-\frac {b\,c^2\,d\,x^2\,1{}\mathrm {i}}{10}}{c^4}+\frac {d\,\left (15\,a\,x^4+15\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,3{}\mathrm {i}\right )}{60}+\frac {c\,d\,\left (a\,x^5\,12{}\mathrm {i}+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,12{}\mathrm {i}\right )}{60} \]

[In]

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

[Out]

(d*(15*a*x^4 - b*x^4*3i + 15*b*x^4*atan(c*x)))/60 - ((d*(15*b*atan(c*x) + b*log(c^2*x^2 + 1)*6i))/60 - (b*c*d*
x)/4 - (b*c^2*d*x^2*1i)/10 + (b*c^3*d*x^3)/12)/c^4 + (c*d*(a*x^5*12i + b*x^5*atan(c*x)*12i))/60