3.10 \(\int x^3 (d+i c d x)^2 (a+b \tan ^{-1}(c x)) \, dx\)

Optimal. Leaf size=166 \[ -\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d^2 x^2}{5 c^2}-\frac{i b d^2 \log \left (c^2 x^2+1\right )}{5 c^4}+\frac{5 b d^2 x}{12 c^3}-\frac{5 b d^2 \tan ^{-1}(c x)}{12 c^4}+\frac{1}{30} b c d^2 x^5-\frac{5 b d^2 x^3}{36 c}-\frac{1}{10} i b d^2 x^4 \]

[Out]

(5*b*d^2*x)/(12*c^3) + ((I/5)*b*d^2*x^2)/c^2 - (5*b*d^2*x^3)/(36*c) - (I/10)*b*d^2*x^4 + (b*c*d^2*x^5)/30 - (5
*b*d^2*ArcTan[c*x])/(12*c^4) + (d^2*x^4*(a + b*ArcTan[c*x]))/4 + ((2*I)/5)*c*d^2*x^5*(a + b*ArcTan[c*x]) - (c^
2*d^2*x^6*(a + b*ArcTan[c*x]))/6 - ((I/5)*b*d^2*Log[1 + c^2*x^2])/c^4

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Rubi [A]  time = 0.16337, antiderivative size = 166, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {43, 4872, 12, 1802, 635, 203, 260} \[ -\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{i b d^2 x^2}{5 c^2}-\frac{i b d^2 \log \left (c^2 x^2+1\right )}{5 c^4}+\frac{5 b d^2 x}{12 c^3}-\frac{5 b d^2 \tan ^{-1}(c x)}{12 c^4}+\frac{1}{30} b c d^2 x^5-\frac{5 b d^2 x^3}{36 c}-\frac{1}{10} i b d^2 x^4 \]

Antiderivative was successfully verified.

[In]

Int[x^3*(d + I*c*d*x)^2*(a + b*ArcTan[c*x]),x]

[Out]

(5*b*d^2*x)/(12*c^3) + ((I/5)*b*d^2*x^2)/c^2 - (5*b*d^2*x^3)/(36*c) - (I/10)*b*d^2*x^4 + (b*c*d^2*x^5)/30 - (5
*b*d^2*ArcTan[c*x])/(12*c^4) + (d^2*x^4*(a + b*ArcTan[c*x]))/4 + ((2*I)/5)*c*d^2*x^5*(a + b*ArcTan[c*x]) - (c^
2*d^2*x^6*(a + b*ArcTan[c*x]))/6 - ((I/5)*b*d^2*Log[1 + c^2*x^2])/c^4

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^3 (d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac{d^2 x^4 \left (15+24 i c x-10 c^2 x^2\right )}{60 \left (1+c^2 x^2\right )} \, dx\\ &=\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{60} \left (b c d^2\right ) \int \frac{x^4 \left (15+24 i c x-10 c^2 x^2\right )}{1+c^2 x^2} \, dx\\ &=\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{60} \left (b c d^2\right ) \int \left (-\frac{25}{c^4}-\frac{24 i x}{c^3}+\frac{25 x^2}{c^2}+\frac{24 i x^3}{c}-10 x^4+\frac{25+24 i c x}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=\frac{5 b d^2 x}{12 c^3}+\frac{i b d^2 x^2}{5 c^2}-\frac{5 b d^2 x^3}{36 c}-\frac{1}{10} i b d^2 x^4+\frac{1}{30} b c d^2 x^5+\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{\left (b d^2\right ) \int \frac{25+24 i c x}{1+c^2 x^2} \, dx}{60 c^3}\\ &=\frac{5 b d^2 x}{12 c^3}+\frac{i b d^2 x^2}{5 c^2}-\frac{5 b d^2 x^3}{36 c}-\frac{1}{10} i b d^2 x^4+\frac{1}{30} b c d^2 x^5+\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{\left (5 b d^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{12 c^3}-\frac{\left (2 i b d^2\right ) \int \frac{x}{1+c^2 x^2} \, dx}{5 c^2}\\ &=\frac{5 b d^2 x}{12 c^3}+\frac{i b d^2 x^2}{5 c^2}-\frac{5 b d^2 x^3}{36 c}-\frac{1}{10} i b d^2 x^4+\frac{1}{30} b c d^2 x^5-\frac{5 b d^2 \tan ^{-1}(c x)}{12 c^4}+\frac{1}{4} d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac{2}{5} i c d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac{1}{6} c^2 d^2 x^6 \left (a+b \tan ^{-1}(c x)\right )-\frac{i b d^2 \log \left (1+c^2 x^2\right )}{5 c^4}\\ \end{align*}

Mathematica [A]  time = 0.128977, size = 124, normalized size = 0.75 \[ \frac{d^2 \left (3 a c^4 x^4 \left (-10 c^2 x^2+24 i c x+15\right )+b c x \left (6 c^4 x^4-18 i c^3 x^3-25 c^2 x^2+36 i c x+75\right )-36 i b \log \left (c^2 x^2+1\right )+3 b \left (-10 c^6 x^6+24 i c^5 x^5+15 c^4 x^4-25\right ) \tan ^{-1}(c x)\right )}{180 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(d + I*c*d*x)^2*(a + b*ArcTan[c*x]),x]

[Out]

(d^2*(3*a*c^4*x^4*(15 + (24*I)*c*x - 10*c^2*x^2) + b*c*x*(75 + (36*I)*c*x - 25*c^2*x^2 - (18*I)*c^3*x^3 + 6*c^
4*x^4) + 3*b*(-25 + 15*c^4*x^4 + (24*I)*c^5*x^5 - 10*c^6*x^6)*ArcTan[c*x] - (36*I)*b*Log[1 + c^2*x^2]))/(180*c
^4)

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Maple [A]  time = 0.027, size = 166, normalized size = 1. \begin{align*} -{\frac{{c}^{2}{d}^{2}a{x}^{6}}{6}}+{\frac{2\,i}{5}}c{d}^{2}a{x}^{5}+{\frac{{d}^{2}a{x}^{4}}{4}}-{\frac{{c}^{2}{d}^{2}b\arctan \left ( cx \right ){x}^{6}}{6}}+{\frac{2\,i}{5}}c{d}^{2}b\arctan \left ( cx \right ){x}^{5}+{\frac{{d}^{2}b\arctan \left ( cx \right ){x}^{4}}{4}}+{\frac{5\,{d}^{2}bx}{12\,{c}^{3}}}+{\frac{bc{d}^{2}{x}^{5}}{30}}-{\frac{i}{10}}b{d}^{2}{x}^{4}-{\frac{5\,{d}^{2}b{x}^{3}}{36\,c}}+{\frac{{\frac{i}{5}}b{d}^{2}{x}^{2}}{{c}^{2}}}-{\frac{{\frac{i}{5}}b{d}^{2}\ln \left ({c}^{2}{x}^{2}+1 \right ) }{{c}^{4}}}-{\frac{5\,b{d}^{2}\arctan \left ( cx \right ) }{12\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x)

[Out]

-1/6*c^2*d^2*a*x^6+2/5*I*c*d^2*a*x^5+1/4*d^2*a*x^4-1/6*c^2*d^2*b*arctan(c*x)*x^6+2/5*I*c*d^2*b*arctan(c*x)*x^5
+1/4*d^2*b*arctan(c*x)*x^4+5/12*b*d^2*x/c^3+1/30*b*c*d^2*x^5-1/10*I*b*d^2*x^4-5/36*b*d^2*x^3/c+1/5*I*b*d^2*x^2
/c^2-1/5*I*b*d^2*ln(c^2*x^2+1)/c^4-5/12*b*d^2*arctan(c*x)/c^4

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Maxima [A]  time = 1.50962, size = 250, normalized size = 1.51 \begin{align*} -\frac{1}{6} \, a c^{2} d^{2} x^{6} + \frac{2}{5} i \, a c d^{2} x^{5} + \frac{1}{4} \, a d^{2} x^{4} - \frac{1}{90} \,{\left (15 \, x^{6} \arctan \left (c x\right ) - c{\left (\frac{3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac{15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{2} d^{2} + \frac{1}{10} 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^{2} + \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^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

-1/6*a*c^2*d^2*x^6 + 2/5*I*a*c*d^2*x^5 + 1/4*a*d^2*x^4 - 1/90*(15*x^6*arctan(c*x) - c*((3*c^4*x^5 - 5*c^2*x^3
+ 15*x)/c^6 - 15*arctan(c*x)/c^7))*b*c^2*d^2 + 1/10*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^2 + 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^2

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Fricas [A]  time = 2.83353, size = 404, normalized size = 2.43 \begin{align*} -\frac{60 \, a c^{6} d^{2} x^{6} -{\left (144 i \, a + 12 \, b\right )} c^{5} d^{2} x^{5} - 18 \,{\left (5 \, a - 2 i \, b\right )} c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} - 72 i \, b c^{2} d^{2} x^{2} - 150 \, b c d^{2} x + 147 i \, b d^{2} \log \left (\frac{c x + i}{c}\right ) - 3 i \, b d^{2} \log \left (\frac{c x - i}{c}\right ) -{\left (-30 i \, b c^{6} d^{2} x^{6} - 72 \, b c^{5} d^{2} x^{5} + 45 i \, b c^{4} d^{2} x^{4}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{360 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

-1/360*(60*a*c^6*d^2*x^6 - (144*I*a + 12*b)*c^5*d^2*x^5 - 18*(5*a - 2*I*b)*c^4*d^2*x^4 + 50*b*c^3*d^2*x^3 - 72
*I*b*c^2*d^2*x^2 - 150*b*c*d^2*x + 147*I*b*d^2*log((c*x + I)/c) - 3*I*b*d^2*log((c*x - I)/c) - (-30*I*b*c^6*d^
2*x^6 - 72*b*c^5*d^2*x^5 + 45*I*b*c^4*d^2*x^4)*log(-(c*x + I)/(c*x - I)))/c^4

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Sympy [A]  time = 2.7756, size = 233, normalized size = 1.4 \begin{align*} - \frac{a c^{2} d^{2} x^{6}}{6} - \frac{5 b d^{2} x^{3}}{36 c} + \frac{i b d^{2} x^{2}}{5 c^{2}} + \frac{5 b d^{2} x}{12 c^{3}} + \frac{i b d^{2} \log{\left (x - \frac{i}{c} \right )}}{120 c^{4}} - \frac{49 i b d^{2} \log{\left (x + \frac{i}{c} \right )}}{120 c^{4}} - x^{5} \left (- \frac{2 i a c d^{2}}{5} - \frac{b c d^{2}}{30}\right ) - x^{4} \left (- \frac{a d^{2}}{4} + \frac{i b d^{2}}{10}\right ) + \left (- \frac{i b c^{2} d^{2} x^{6}}{12} - \frac{b c d^{2} x^{5}}{5} + \frac{i b d^{2} x^{4}}{8}\right ) \log{\left (- i c x + 1 \right )} + \left (\frac{i b c^{2} d^{2} x^{6}}{12} + \frac{b c d^{2} x^{5}}{5} - \frac{i b d^{2} x^{4}}{8}\right ) \log{\left (i c x + 1 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(d+I*c*d*x)**2*(a+b*atan(c*x)),x)

[Out]

-a*c**2*d**2*x**6/6 - 5*b*d**2*x**3/(36*c) + I*b*d**2*x**2/(5*c**2) + 5*b*d**2*x/(12*c**3) + I*b*d**2*log(x -
I/c)/(120*c**4) - 49*I*b*d**2*log(x + I/c)/(120*c**4) - x**5*(-2*I*a*c*d**2/5 - b*c*d**2/30) - x**4*(-a*d**2/4
+ I*b*d**2/10) + (-I*b*c**2*d**2*x**6/12 - b*c*d**2*x**5/5 + I*b*d**2*x**4/8)*log(-I*c*x + 1) + (I*b*c**2*d**
2*x**6/12 + b*c*d**2*x**5/5 - I*b*d**2*x**4/8)*log(I*c*x + 1)

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Giac [A]  time = 1.20699, size = 247, normalized size = 1.49 \begin{align*} -\frac{60 \, b c^{6} d^{2} i x^{6} \arctan \left (c x\right ) + 60 \, a c^{6} d^{2} i x^{6} - 12 \, b c^{5} d^{2} i x^{5} + 144 \, b c^{5} d^{2} x^{5} \arctan \left (c x\right ) + 144 \, a c^{5} d^{2} x^{5} - 90 \, b c^{4} d^{2} i x^{4} \arctan \left (c x\right ) - 90 \, a c^{4} d^{2} i x^{4} - 36 \, b c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} i x^{3} + 72 \, b c^{2} d^{2} x^{2} - 150 \, b c d^{2} i x + 3 \, b d^{2} \log \left (c i x + 1\right ) - 147 \, b d^{2} \log \left (-c i x + 1\right )}{360 \, c^{4} i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(d+I*c*d*x)^2*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

-1/360*(60*b*c^6*d^2*i*x^6*arctan(c*x) + 60*a*c^6*d^2*i*x^6 - 12*b*c^5*d^2*i*x^5 + 144*b*c^5*d^2*x^5*arctan(c*
x) + 144*a*c^5*d^2*x^5 - 90*b*c^4*d^2*i*x^4*arctan(c*x) - 90*a*c^4*d^2*i*x^4 - 36*b*c^4*d^2*x^4 + 50*b*c^3*d^2
*i*x^3 + 72*b*c^2*d^2*x^2 - 150*b*c*d^2*i*x + 3*b*d^2*log(c*i*x + 1) - 147*b*d^2*log(-c*i*x + 1))/(c^4*i)