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

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 166, normalized size of antiderivative = 1.00, 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 12

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

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 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]

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 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 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]))

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*}

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Mathematica [A]  time = 0.13, size = 124, normalized size = 0.75 \[ \frac {d^2 \left (3 a c^4 x^4 \left (-10 c^2 x^2+24 i c x+15\right )-36 i b \log \left (c^2 x^2+1\right )+3 b \left (-10 c^6 x^6+24 i c^5 x^5+15 c^4 x^4-25\right ) \tan ^{-1}(c x)+b c x \left (6 c^4 x^4-18 i c^3 x^3-25 c^2 x^2+36 i c x+75\right )\right )}{180 c^4} \]

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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fricas [A]  time = 0.46, size = 172, normalized size = 1.04 \[ -\frac {60 \, a c^{6} d^{2} x^{6} - {\left (144 i \, a + 12 \, b\right )} c^{5} d^{2} x^{5} - 18 \, {\left (5 \, a - 2 i \, b\right )} c^{4} d^{2} x^{4} + 50 \, b c^{3} d^{2} x^{3} - 72 i \, b c^{2} d^{2} x^{2} - 150 \, b c d^{2} x + 147 i \, b d^{2} \log \left (\frac {c x + i}{c}\right ) - 3 i \, b d^{2} \log \left (\frac {c x - i}{c}\right ) - {\left (-30 i \, b c^{6} d^{2} x^{6} - 72 \, b c^{5} d^{2} x^{5} + 45 i \, b c^{4} d^{2} x^{4}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{360 \, c^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [A]  time = 0.03, size = 166, normalized size = 1.00 \[ -\frac {c^{2} d^{2} a \,x^{6}}{6}+\frac {2 i c \,d^{2} a \,x^{5}}{5}+\frac {d^{2} a \,x^{4}}{4}-\frac {c^{2} d^{2} b \arctan \left (c x \right ) x^{6}}{6}+\frac {2 i c \,d^{2} b \arctan \left (c x \right ) x^{5}}{5}+\frac {d^{2} b \arctan \left (c x \right ) x^{4}}{4}+\frac {5 b \,d^{2} x}{12 c^{3}}+\frac {b c \,d^{2} x^{5}}{30}-\frac {i b \,d^{2} x^{4}}{10}-\frac {5 b \,d^{2} x^{3}}{36 c}+\frac {i b \,d^{2} x^{2}}{5 c^{2}}-\frac {i b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{5 c^{4}}-\frac {5 b \,d^{2} \arctan \left (c x \right )}{12 c^{4}} \]

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 = 0.42, size = 185, normalized size = 1.11 \[ -\frac {1}{6} \, a c^{2} d^{2} x^{6} + \frac {2}{5} i \, a c d^{2} x^{5} + \frac {1}{4} \, a d^{2} x^{4} - \frac {1}{90} \, {\left (15 \, x^{6} \arctan \left (c x\right ) - c {\left (\frac {3 \, c^{4} x^{5} - 5 \, c^{2} x^{3} + 15 \, x}{c^{6}} - \frac {15 \, \arctan \left (c x\right )}{c^{7}}\right )}\right )} b c^{2} d^{2} + \frac {1}{10} i \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c d^{2} + \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b d^{2} \]

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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mupad [B]  time = 0.76, size = 152, normalized size = 0.92 \[ -\frac {\frac {d^2\,\left (75\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,36{}\mathrm {i}\right )}{180}+\frac {5\,b\,c^3\,d^2\,x^3}{36}-\frac {5\,b\,c\,d^2\,x}{12}-\frac {b\,c^2\,d^2\,x^2\,1{}\mathrm {i}}{5}}{c^4}+\frac {d^2\,\left (45\,a\,x^4+45\,b\,x^4\,\mathrm {atan}\left (c\,x\right )-b\,x^4\,18{}\mathrm {i}\right )}{180}-\frac {c^2\,d^2\,\left (30\,a\,x^6+30\,b\,x^6\,\mathrm {atan}\left (c\,x\right )\right )}{180}+\frac {c\,d^2\,\left (a\,x^5\,72{}\mathrm {i}+6\,b\,x^5+b\,x^5\,\mathrm {atan}\left (c\,x\right )\,72{}\mathrm {i}\right )}{180} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atan(c*x))*(d + c*d*x*1i)^2,x)

[Out]

(d^2*(45*a*x^4 - b*x^4*18i + 45*b*x^4*atan(c*x)))/180 - ((d^2*(75*b*atan(c*x) + b*log(c^2*x^2 + 1)*36i))/180 -
(b*c^2*d^2*x^2*1i)/5 + (5*b*c^3*d^2*x^3)/36 - (5*b*c*d^2*x)/12)/c^4 - (c^2*d^2*(30*a*x^6 + 30*b*x^6*atan(c*x)
))/180 + (c*d^2*(a*x^5*72i + 6*b*x^5 + b*x^5*atan(c*x)*72i))/180

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sympy [A]  time = 4.53, size = 270, normalized size = 1.63 \[ - \frac {a c^{2} d^{2} x^{6}}{6} - \frac {5 b d^{2} x^{3}}{36 c} + \frac {i b d^{2} x^{2}}{5 c^{2}} + \frac {5 b d^{2} x}{12 c^{3}} - \frac {b d^{2} \left (- \frac {i \log {\left (291 b c d^{2} x - 291 i b d^{2} \right )}}{120} + \frac {71 i \log {\left (291 b c d^{2} x + 291 i b d^{2} \right )}}{210}\right )}{c^{4}} - x^{5} \left (- \frac {2 i a c d^{2}}{5} - \frac {b c d^{2}}{30}\right ) - x^{4} \left (- \frac {a d^{2}}{4} + \frac {i b d^{2}}{10}\right ) + \left (\frac {i b c^{2} d^{2} x^{6}}{12} + \frac {b c d^{2} x^{5}}{5} - \frac {i b d^{2} x^{4}}{8}\right ) \log {\left (i c x + 1 \right )} - \frac {\left (70 i b c^{6} d^{2} x^{6} + 168 b c^{5} d^{2} x^{5} - 105 i b c^{4} d^{2} x^{4} + 59 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{840 c^{4}} \]

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) - b*d**2*(-I*log(29
1*b*c*d**2*x - 291*I*b*d**2)/120 + 71*I*log(291*b*c*d**2*x + 291*I*b*d**2)/210)/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)*l
og(I*c*x + 1) - (70*I*b*c**6*d**2*x**6 + 168*b*c**5*d**2*x**5 - 105*I*b*c**4*d**2*x**4 + 59*I*b*d**2)*log(-I*c
*x + 1)/(840*c**4)

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