3.12 \(\int x (d+i c d x)^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=136 \[ -\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d^2 \log \left (c^2 x^2+1\right )}{3 c^2}+\frac{3 b d^2 \tan ^{-1}(c x)}{4 c^2}+\frac{1}{12} b c d^2 x^3-\frac{3 b d^2 x}{4 c}-\frac{1}{3} i b d^2 x^2 \]

[Out]

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

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Rubi [A]  time = 0.127064, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {43, 4872, 12, 1802, 635, 203, 260} \[ -\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d^2 \log \left (c^2 x^2+1\right )}{3 c^2}+\frac{3 b d^2 \tan ^{-1}(c x)}{4 c^2}+\frac{1}{12} b c d^2 x^3-\frac{3 b d^2 x}{4 c}-\frac{1}{3} i b d^2 x^2 \]

Antiderivative was successfully verified.

[In]

Int[x*(d + I*c*d*x)^2*(a + b*ArcTan[c*x]),x]

[Out]

(-3*b*d^2*x)/(4*c) - (I/3)*b*d^2*x^2 + (b*c*d^2*x^3)/12 + (3*b*d^2*ArcTan[c*x])/(4*c^2) + (d^2*x^2*(a + b*ArcT
an[c*x]))/2 + ((2*I)/3)*c*d^2*x^3*(a + b*ArcTan[c*x]) - (c^2*d^2*x^4*(a + b*ArcTan[c*x]))/4 + ((I/3)*b*d^2*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 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

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)^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^2 \left (6+8 i c x-3 c^2 x^2\right )}{12 \left (1+c^2 x^2\right )} \, dx\\ &=\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{12} \left (b c d^2\right ) \int \frac{x^2 \left (6+8 i c x-3 c^2 x^2\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{12} \left (b c d^2\right ) \int \left (\frac{9}{c^2}+\frac{8 i x}{c}-3 x^2+\frac{i (9 i-8 c x)}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{3 b d^2 x}{4 c}-\frac{1}{3} i b d^2 x^2+\frac{1}{12} b c d^2 x^3+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac{\left (i b d^2\right ) \int \frac{9 i-8 c x}{1+c^2 x^2} \, dx}{12 c}\\ &=-\frac{3 b d^2 x}{4 c}-\frac{1}{3} i b d^2 x^2+\frac{1}{12} b c d^2 x^3+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{3} \left (2 i b d^2\right ) \int \frac{x}{1+c^2 x^2} \, dx+\frac{\left (3 b d^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{4 c}\\ &=-\frac{3 b d^2 x}{4 c}-\frac{1}{3} i b d^2 x^2+\frac{1}{12} b c d^2 x^3+\frac{3 b d^2 \tan ^{-1}(c x)}{4 c^2}+\frac{1}{2} d^2 x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{3} i c d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{4} c^2 d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d^2 \log \left (1+c^2 x^2\right )}{3 c^2}\\ \end{align*}

Mathematica [A]  time = 0.0910559, size = 101, normalized size = 0.74 \[ \frac{d^2 \left (c x \left (a c x \left (-3 c^2 x^2+8 i c x+6\right )+b \left (c^2 x^2-4 i c x-9\right )\right )+4 i b \log \left (c^2 x^2+1\right )+b \left (-3 c^4 x^4+8 i c^3 x^3+6 c^2 x^2+9\right ) \tan ^{-1}(c x)\right )}{12 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(d + I*c*d*x)^2*(a + b*ArcTan[c*x]),x]

[Out]

(d^2*(c*x*(a*c*x*(6 + (8*I)*c*x - 3*c^2*x^2) + b*(-9 - (4*I)*c*x + c^2*x^2)) + b*(9 + 6*c^2*x^2 + (8*I)*c^3*x^
3 - 3*c^4*x^4)*ArcTan[c*x] + (4*I)*b*Log[1 + c^2*x^2]))/(12*c^2)

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Maple [A]  time = 0.027, size = 141, normalized size = 1. \begin{align*} -{\frac{{c}^{2}{d}^{2}a{x}^{4}}{4}}+{\frac{2\,i}{3}}c{d}^{2}a{x}^{3}+{\frac{{d}^{2}a{x}^{2}}{2}}-{\frac{{c}^{2}{d}^{2}b\arctan \left ( cx \right ){x}^{4}}{4}}+{\frac{2\,i}{3}}c{d}^{2}b\arctan \left ( cx \right ){x}^{3}+{\frac{{d}^{2}b\arctan \left ( cx \right ){x}^{2}}{2}}-{\frac{3\,{d}^{2}bx}{4\,c}}+{\frac{bc{d}^{2}{x}^{3}}{12}}-{\frac{i}{3}}b{d}^{2}{x}^{2}+{\frac{{\frac{i}{3}}{d}^{2}b\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{2}}}+{\frac{3\,b{d}^{2}\arctan \left ( cx \right ) }{4\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x)

[Out]

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

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Maxima [A]  time = 1.47229, size = 209, normalized size = 1.54 \begin{align*} -\frac{1}{4} \, a c^{2} d^{2} x^{4} + \frac{2}{3} i \, a c d^{2} x^{3} - \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 c^{2} d^{2} + \frac{1}{3} 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^{2} + \frac{1}{2} \, a d^{2} 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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.79827, size = 333, normalized size = 2.45 \begin{align*} -\frac{6 \, a c^{4} d^{2} x^{4} -{\left (16 i \, a + 2 \, b\right )} c^{3} d^{2} x^{3} - 4 \,{\left (3 \, a - 2 i \, b\right )} c^{2} d^{2} x^{2} + 18 \, b c d^{2} x - 17 i \, b d^{2} \log \left (\frac{c x + i}{c}\right ) + i \, b d^{2} \log \left (\frac{c x - i}{c}\right ) -{\left (-3 i \, b c^{4} d^{2} x^{4} - 8 \, b c^{3} d^{2} x^{3} + 6 i \, b c^{2} d^{2} x^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{24 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

-1/24*(6*a*c^4*d^2*x^4 - (16*I*a + 2*b)*c^3*d^2*x^3 - 4*(3*a - 2*I*b)*c^2*d^2*x^2 + 18*b*c*d^2*x - 17*I*b*d^2*
log((c*x + I)/c) + I*b*d^2*log((c*x - I)/c) - (-3*I*b*c^4*d^2*x^4 - 8*b*c^3*d^2*x^3 + 6*I*b*c^2*d^2*x^2)*log(-
(c*x + I)/(c*x - I)))/c^2

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Sympy [A]  time = 2.71854, size = 202, normalized size = 1.49 \begin{align*} - \frac{a c^{2} d^{2} x^{4}}{4} - \frac{3 b d^{2} x}{4 c} - \frac{i b d^{2} \log{\left (x - \frac{i}{c} \right )}}{24 c^{2}} + \frac{17 i b d^{2} \log{\left (x + \frac{i}{c} \right )}}{24 c^{2}} - x^{3} \left (- \frac{2 i a c d^{2}}{3} - \frac{b c d^{2}}{12}\right ) - x^{2} \left (- \frac{a d^{2}}{2} + \frac{i b d^{2}}{3}\right ) + \left (- \frac{i b c^{2} d^{2} x^{4}}{8} - \frac{b c d^{2} x^{3}}{3} + \frac{i b d^{2} x^{2}}{4}\right ) \log{\left (- i c x + 1 \right )} + \left (\frac{i b c^{2} d^{2} x^{4}}{8} + \frac{b c d^{2} x^{3}}{3} - \frac{i b d^{2} 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)**2*(a+b*atan(c*x)),x)

[Out]

-a*c**2*d**2*x**4/4 - 3*b*d**2*x/(4*c) - I*b*d**2*log(x - I/c)/(24*c**2) + 17*I*b*d**2*log(x + I/c)/(24*c**2)
- x**3*(-2*I*a*c*d**2/3 - b*c*d**2/12) - x**2*(-a*d**2/2 + I*b*d**2/3) + (-I*b*c**2*d**2*x**4/8 - b*c*d**2*x**
3/3 + I*b*d**2*x**2/4)*log(-I*c*x + 1) + (I*b*c**2*d**2*x**4/8 + b*c*d**2*x**3/3 - I*b*d**2*x**2/4)*log(I*c*x
+ 1)

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Giac [A]  time = 1.15196, size = 207, normalized size = 1.52 \begin{align*} -\frac{6 \, b c^{4} d^{2} x^{4} \arctan \left (c x\right ) + 6 \, a c^{4} d^{2} x^{4} - 16 \, b c^{3} d^{2} i x^{3} \arctan \left (c x\right ) - 16 \, a c^{3} d^{2} i x^{3} - 2 \, b c^{3} d^{2} x^{3} + 8 \, b c^{2} d^{2} i x^{2} - 12 \, b c^{2} d^{2} x^{2} \arctan \left (c x\right ) - 12 \, a c^{2} d^{2} x^{2} + 18 \, b c d^{2} x - 17 \, b d^{2} i \log \left (c i x - 1\right ) + b d^{2} i \log \left (-c i x - 1\right )}{24 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

-1/24*(6*b*c^4*d^2*x^4*arctan(c*x) + 6*a*c^4*d^2*x^4 - 16*b*c^3*d^2*i*x^3*arctan(c*x) - 16*a*c^3*d^2*i*x^3 - 2
*b*c^3*d^2*x^3 + 8*b*c^2*d^2*i*x^2 - 12*b*c^2*d^2*x^2*arctan(c*x) - 12*a*c^2*d^2*x^2 + 18*b*c*d^2*x - 17*b*d^2
*i*log(c*i*x - 1) + b*d^2*i*log(-c*i*x - 1))/c^2