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

Optimal result

Integrand size = 19, antiderivative size = 91 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=-\frac {b d x}{2 c}-\frac {1}{6} i b d x^2+\frac {b d \arctan (c x)}{2 c^2}+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))+\frac {i b d \log \left (1+c^2 x^2\right )}{6 c^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {45, 4992, 12, 815, 649, 209, 266} \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{3} i c d x^3 (a+b \arctan (c x))+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {b d \arctan (c x)}{2 c^2}+\frac {i b d \log \left (c^2 x^2+1\right )}{6 c^2}-\frac {b d x}{2 c}-\frac {1}{6} i b d x^2 \]

[In]

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

[Out]

-1/2*(b*d*x)/c - (I/6)*b*d*x^2 + (b*d*ArcTan[c*x])/(2*c^2) + (d*x^2*(a + b*ArcTan[c*x]))/2 + (I/3)*c*d*x^3*(a
+ b*ArcTan[c*x]) + ((I/6)*b*d*Log[1 + c^2*x^2])/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 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}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))-(b c) \int \frac {d x^2 (3+2 i c x)}{6+6 c^2 x^2} \, dx \\ & = \frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))-(b c d) \int \frac {x^2 (3+2 i c x)}{6+6 c^2 x^2} \, dx \\ & = \frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))-(b c d) \int \left (\frac {1}{2 c^2}+\frac {i x}{3 c}+\frac {i (3 i-2 c x)}{c^2 \left (6+6 c^2 x^2\right )}\right ) \, dx \\ & = -\frac {b d x}{2 c}-\frac {1}{6} i b d x^2+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))-\frac {(i b d) \int \frac {3 i-2 c x}{6+6 c^2 x^2} \, dx}{c} \\ & = -\frac {b d x}{2 c}-\frac {1}{6} i b d x^2+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))+(2 i b d) \int \frac {x}{6+6 c^2 x^2} \, dx+\frac {(3 b d) \int \frac {1}{6+6 c^2 x^2} \, dx}{c} \\ & = -\frac {b d x}{2 c}-\frac {1}{6} i b d x^2+\frac {b d \arctan (c x)}{2 c^2}+\frac {1}{2} d x^2 (a+b \arctan (c x))+\frac {1}{3} i c d x^3 (a+b \arctan (c x))+\frac {i b d \log \left (1+c^2 x^2\right )}{6 c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {d \left (c x (b (-3-i c x)+a c x (3+2 i c x))+b \left (3+3 c^2 x^2+2 i c^3 x^3\right ) \arctan (c x)+i b \log \left (1+c^2 x^2\right )\right )}{6 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.90

method	result	size
parts	\(a d \left (\frac {1}{3} i c \,x^{3}+\frac {1}{2} x^{2}\right )+\frac {b d \left (\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {i c^{2} x^{2}}{6}-\frac {c x}{2}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{6}+\frac {\arctan \left (c x \right )}{2}\right )}{c^{2}}\)	\(82\)
derivativedivides	\(\frac {a d \left (\frac {1}{3} i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {i c^{2} x^{2}}{6}-\frac {c x}{2}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{6}+\frac {\arctan \left (c x \right )}{2}\right )}{c^{2}}\)	\(88\)
default	\(\frac {a d \left (\frac {1}{3} i c^{3} x^{3}+\frac {1}{2} c^{2} x^{2}\right )+b d \left (\frac {i \arctan \left (c x \right ) c^{3} x^{3}}{3}+\frac {c^{2} x^{2} \arctan \left (c x \right )}{2}-\frac {i c^{2} x^{2}}{6}-\frac {c x}{2}+\frac {i \ln \left (c^{2} x^{2}+1\right )}{6}+\frac {\arctan \left (c x \right )}{2}\right )}{c^{2}}\)	\(88\)
parallelrisch	\(\frac {2 i x^{3} \arctan \left (c x \right ) b \,c^{3} d +2 i a \,c^{3} d \,x^{3}-i x^{2} b \,c^{2} d +3 x^{2} \arctan \left (c x \right ) b \,c^{2} d +3 a \,c^{2} d \,x^{2}+i b d \ln \left (c^{2} x^{2}+1\right )-3 b c d x +3 b d \arctan \left (c x \right )}{6 c^{2}}\)	\(97\)
risch	\(\frac {d b \left (2 c \,x^{3}-3 i x^{2}\right ) \ln \left (i c x +1\right )}{12}-\frac {d c b \,x^{3} \ln \left (-i c x +1\right )}{6}+\frac {i a c d \,x^{3}}{3}+\frac {x^{2} d a}{2}+\frac {i d \,x^{2} b \ln \left (-i c x +1\right )}{4}-\frac {i b d \,x^{2}}{6}-\frac {b d x}{2 c}+\frac {b d \arctan \left (c x \right )}{2 c^{2}}+\frac {i d b \ln \left (9 c^{2} x^{2}+9\right )}{6 c^{2}}\)	\(121\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {4 i \, a c^{3} d x^{3} + 2 \, {\left (3 \, a - i \, b\right )} c^{2} d x^{2} - 6 \, b c d x + 5 i \, b d \log \left (\frac {c x + i}{c}\right ) - i \, b d \log \left (\frac {c x - i}{c}\right ) - {\left (2 \, b c^{3} d x^{3} - 3 i \, b c^{2} d x^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{12 \, c^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.55 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.74 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {i a c d x^{3}}{3} - \frac {b d x}{2 c} + \frac {b d \left (- \frac {i \log {\left (9 b c d x - 9 i b d \right )}}{12} + \frac {7 i \log {\left (9 b c d x + 9 i b d \right )}}{24}\right )}{c^{2}} + x^{2} \left (\frac {a d}{2} - \frac {i b d}{6}\right ) + \left (\frac {b c d x^{3}}{6} - \frac {i b d x^{2}}{4}\right ) \log {\left (i c x + 1 \right )} + \frac {\left (- 4 b c^{3} d x^{3} + 6 i b c^{2} d x^{2} + 3 i b d\right ) \log {\left (- i c x + 1 \right )}}{24 c^{2}} \]

[In]

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

[Out]

I*a*c*d*x**3/3 - b*d*x/(2*c) + b*d*(-I*log(9*b*c*d*x - 9*I*b*d)/12 + 7*I*log(9*b*c*d*x + 9*I*b*d)/24)/c**2 + x
**2*(a*d/2 - I*b*d/6) + (b*c*d*x**3/6 - I*b*d*x**2/4)*log(I*c*x + 1) + (-4*b*c**3*d*x**3 + 6*I*b*c**2*d*x**2 +
3*I*b*d)*log(-I*c*x + 1)/(24*c**2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {1}{3} i \, a c d x^{3} + \frac {1}{6} i \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c d + \frac {1}{2} \, a d x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.96 \[ \int x (d+i c d x) (a+b \arctan (c x)) \, dx=\frac {d\,\left (3\,a\,x^2+3\,b\,x^2\,\mathrm {atan}\left (c\,x\right )-b\,x^2\,1{}\mathrm {i}\right )}{6}+\frac {\frac {d\,\left (3\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,1{}\mathrm {i}\right )}{6}-\frac {b\,c\,d\,x}{2}}{c^2}+\frac {c\,d\,\left (a\,x^3\,2{}\mathrm {i}+b\,x^3\,\mathrm {atan}\left (c\,x\right )\,2{}\mathrm {i}\right )}{6} \]

[In]

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

[Out]

(d*(3*a*x^2 - b*x^2*1i + 3*b*x^2*atan(c*x)))/6 + ((d*(3*b*atan(c*x) + b*log(c^2*x^2 + 1)*1i))/6 - (b*c*d*x)/2)
/c^2 + (c*d*(a*x^3*2i + b*x^3*atan(c*x)*2i))/6