3.3 \(\int x (d+i c d x) (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=91 \[ \frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d \log \left (c^2 x^2+1\right )}{6 c^2}+\frac{b d \tan ^{-1}(c x)}{2 c^2}-\frac{b d x}{2 c}-\frac{1}{6} i b d x^2 \]

[Out]

-(b*d*x)/(2*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

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Rubi [A]  time = 0.0773123, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {43, 4872, 12, 801, 635, 203, 260} \[ \frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d \log \left (c^2 x^2+1\right )}{6 c^2}+\frac{b d \tan ^{-1}(c x)}{2 c^2}-\frac{b d x}{2 c}-\frac{1}{6} i b d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d*x)/(2*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 43

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 4872

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]))

Rule 12

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

Rule 801

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 635

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 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]

Rubi steps

\begin{align*} \int x (d+i c d x) \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )-(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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )-(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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )-(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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )-\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 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )+(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 \tan ^{-1}(c x)}{2 c^2}+\frac{1}{2} d x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} i c d x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d \log \left (1+c^2 x^2\right )}{6 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0468314, size = 76, normalized size = 0.84 \[ \frac{d \left (c x (a c x (3+2 i c x)+b (-3-i c x))+i b \log \left (c^2 x^2+1\right )+b \left (2 i c^3 x^3+3 c^2 x^2+3\right ) \tan ^{-1}(c x)\right )}{6 c^2} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.028, size = 87, normalized size = 1. \begin{align*}{\frac{i}{3}}cda{x}^{3}+{\frac{da{x}^{2}}{2}}+{\frac{i}{3}}cdb\arctan \left ( cx \right ){x}^{3}+{\frac{db\arctan \left ( cx \right ){x}^{2}}{2}}-{\frac{i}{6}}bd{x}^{2}-{\frac{dbx}{2\,c}}+{\frac{{\frac{i}{6}}db\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{2}}}+{\frac{db\arctan \left ( cx \right ) }{2\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.46907, size = 119, normalized size = 1.31 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.80572, size = 243, normalized size = 2.67 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 1.7801, size = 128, normalized size = 1.41 \begin{align*} \frac{i a c d x^{3}}{3} - \frac{b d x}{2 c} - \frac{i b d \log{\left (x - \frac{i}{c} \right )}}{12 c^{2}} + \frac{5 i b d \log{\left (x + \frac{i}{c} \right )}}{12 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 )} + \left (\frac{b c d x^{3}}{6} - \frac{i b d x^{2}}{4}\right ) \log{\left (i c x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Giac [A]  time = 1.1604, size = 134, normalized size = 1.47 \begin{align*} -\frac{4 \, b c^{3} d x^{3} \arctan \left (c x\right ) + 4 \, a c^{3} d x^{3} - 6 \, b c^{2} d i x^{2} \arctan \left (c x\right ) - 6 \, a c^{2} d i x^{2} - 2 \, b c^{2} d x^{2} + 6 \, b c d i x + 5 \, b d \log \left (c i x - 1\right ) - b d \log \left (-c i x - 1\right )}{12 \, c^{2} i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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