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

Optimal. Leaf size=152 \[ -\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {i b d^2 \tan ^{-1}(c x)}{2 c^3}+\frac {i b d^2 x}{2 c^2}+\frac {4 b d^2 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac {1}{20} b c d^2 x^4-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3 \]

[Out]

1/2*I*b*d^2*x/c^2-4/15*b*d^2*x^2/c-1/6*I*b*d^2*x^3+1/20*b*c*d^2*x^4-1/2*I*b*d^2*arctan(c*x)/c^3+1/3*d^2*x^3*(a
+b*arctan(c*x))+1/2*I*c*d^2*x^4*(a+b*arctan(c*x))-1/5*c^2*d^2*x^5*(a+b*arctan(c*x))+4/15*b*d^2*ln(c^2*x^2+1)/c
^3

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Rubi [A]  time = 0.15, antiderivative size = 152, 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} \[ -\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {4 b d^2 \log \left (c^2 x^2+1\right )}{15 c^3}+\frac {i b d^2 x}{2 c^2}-\frac {i b d^2 \tan ^{-1}(c x)}{2 c^3}+\frac {1}{20} b c d^2 x^4-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*b*d^2*x)/c^2 - (4*b*d^2*x^2)/(15*c) - (I/6)*b*d^2*x^3 + (b*c*d^2*x^4)/20 - ((I/2)*b*d^2*ArcTan[c*x])/c^
3 + (d^2*x^3*(a + b*ArcTan[c*x]))/3 + (I/2)*c*d^2*x^4*(a + b*ArcTan[c*x]) - (c^2*d^2*x^5*(a + b*ArcTan[c*x]))/
5 + (4*b*d^2*Log[1 + c^2*x^2])/(15*c^3)

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^2 (d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {d^2 x^3 \left (10+15 i c x-6 c^2 x^2\right )}{30 \left (1+c^2 x^2\right )} \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{30} \left (b c d^2\right ) \int \frac {x^3 \left (10+15 i c x-6 c^2 x^2\right )}{1+c^2 x^2} \, dx\\ &=\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{30} \left (b c d^2\right ) \int \left (-\frac {15 i}{c^3}+\frac {16 x}{c^2}+\frac {15 i x^2}{c}-6 x^3+\frac {15 i-16 c x}{c^3 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=\frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4+\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {\left (b d^2\right ) \int \frac {15 i-16 c x}{1+c^2 x^2} \, dx}{30 c^2}\\ &=\frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4+\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {\left (i b d^2\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 c^2}+\frac {\left (8 b d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{15 c}\\ &=\frac {i b d^2 x}{2 c^2}-\frac {4 b d^2 x^2}{15 c}-\frac {1}{6} i b d^2 x^3+\frac {1}{20} b c d^2 x^4-\frac {i b d^2 \tan ^{-1}(c x)}{2 c^3}+\frac {1}{3} d^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{2} i c d^2 x^4 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{5} c^2 d^2 x^5 \left (a+b \tan ^{-1}(c x)\right )+\frac {4 b d^2 \log \left (1+c^2 x^2\right )}{15 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 116, normalized size = 0.76 \[ \frac {d^2 \left (2 a c^3 x^3 \left (-6 c^2 x^2+15 i c x+10\right )+16 b \log \left (c^2 x^2+1\right )+b c x \left (3 c^3 x^3-10 i c^2 x^2-16 c x+30 i\right )+2 b \left (-6 c^5 x^5+15 i c^4 x^4+10 c^3 x^3-15 i\right ) \tan ^{-1}(c x)\right )}{60 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(2*a*c^3*x^3*(10 + (15*I)*c*x - 6*c^2*x^2) + b*c*x*(30*I - 16*c*x - (10*I)*c^2*x^2 + 3*c^3*x^3) + 2*b*(-1
5*I + 10*c^3*x^3 + (15*I)*c^4*x^4 - 6*c^5*x^5)*ArcTan[c*x] + 16*b*Log[1 + c^2*x^2]))/(60*c^3)

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fricas [A]  time = 0.44, size = 160, normalized size = 1.05 \[ -\frac {12 \, a c^{5} d^{2} x^{5} - {\left (30 i \, a + 3 \, b\right )} c^{4} d^{2} x^{4} - 10 \, {\left (2 \, a - i \, b\right )} c^{3} d^{2} x^{3} + 16 \, b c^{2} d^{2} x^{2} - 30 i \, b c d^{2} x - 31 \, b d^{2} \log \left (\frac {c x + i}{c}\right ) - b d^{2} \log \left (\frac {c x - i}{c}\right ) - {\left (-6 i \, b c^{5} d^{2} x^{5} - 15 \, b c^{4} d^{2} x^{4} + 10 i \, b c^{3} d^{2} x^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{60 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*a*c^5*d^2*x^5 - (30*I*a + 3*b)*c^4*d^2*x^4 - 10*(2*a - I*b)*c^3*d^2*x^3 + 16*b*c^2*d^2*x^2 - 30*I*b*
c*d^2*x - 31*b*d^2*log((c*x + I)/c) - b*d^2*log((c*x - I)/c) - (-6*I*b*c^5*d^2*x^5 - 15*b*c^4*d^2*x^4 + 10*I*b
*c^3*d^2*x^3)*log(-(c*x + I)/(c*x - I)))/c^3

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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^2*(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 = 154, normalized size = 1.01 \[ -\frac {c^{2} d^{2} a \,x^{5}}{5}+\frac {i c \,d^{2} a \,x^{4}}{2}+\frac {d^{2} a \,x^{3}}{3}-\frac {c^{2} d^{2} b \arctan \left (c x \right ) x^{5}}{5}+\frac {i c \,d^{2} b \arctan \left (c x \right ) x^{4}}{2}+\frac {d^{2} b \arctan \left (c x \right ) x^{3}}{3}+\frac {i b \,d^{2} x}{2 c^{2}}+\frac {b c \,d^{2} x^{4}}{20}-\frac {i b \,d^{2} x^{3}}{6}-\frac {4 b \,d^{2} x^{2}}{15 c}+\frac {4 b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{15 c^{3}}-\frac {i b \,d^{2} \arctan \left (c x \right )}{2 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*c^2*d^2*a*x^5+1/2*I*c*d^2*a*x^4+1/3*d^2*a*x^3-1/5*c^2*d^2*b*arctan(c*x)*x^5+1/2*I*c*d^2*b*arctan(c*x)*x^4
+1/3*d^2*b*arctan(c*x)*x^3+1/2*I*b*d^2*x/c^2+1/20*b*c*d^2*x^4-1/6*I*b*d^2*x^3-4/15*b*d^2*x^2/c+4/15*b*d^2*ln(c
^2*x^2+1)/c^3-1/2*I*b*d^2*arctan(c*x)/c^3

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maxima [A]  time = 0.41, size = 174, normalized size = 1.14 \[ -\frac {1}{5} \, a c^{2} d^{2} x^{5} + \frac {1}{2} i \, a c d^{2} x^{4} - \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} \, a d^{2} x^{3} + \frac {1}{6} i \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c d^{2} + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*a*c^2*d^2*x^5 + 1/2*I*a*c*d^2*x^4 - 1/20*(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^2*d^2 + 1/3*a*d^2*x^3 + 1/6*I*(3*x^4*arctan(c*x) - c*((c^2*x^3 - 3*x)/c^4 + 3*arctan(c*x)/c^5))*b
*c*d^2 + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d^2

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mupad [B]  time = 0.74, size = 140, normalized size = 0.92 \[ -\frac {\frac {d^2\,\left (-16\,b\,\ln \left (c^2\,x^2+1\right )+b\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{60}+\frac {4\,b\,c^2\,d^2\,x^2}{15}-\frac {b\,c\,d^2\,x\,1{}\mathrm {i}}{2}}{c^3}+\frac {d^2\,\left (20\,a\,x^3+20\,b\,x^3\,\mathrm {atan}\left (c\,x\right )-b\,x^3\,10{}\mathrm {i}\right )}{60}-\frac {c^2\,d^2\,\left (12\,a\,x^5+12\,b\,x^5\,\mathrm {atan}\left (c\,x\right )\right )}{60}+\frac {c\,d^2\,\left (a\,x^4\,30{}\mathrm {i}+3\,b\,x^4+b\,x^4\,\mathrm {atan}\left (c\,x\right )\,30{}\mathrm {i}\right )}{60} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d^2*(20*a*x^3 - b*x^3*10i + 20*b*x^3*atan(c*x)))/60 - ((d^2*(b*atan(c*x)*30i - 16*b*log(c^2*x^2 + 1)))/60 + (
4*b*c^2*d^2*x^2)/15 - (b*c*d^2*x*1i)/2)/c^3 - (c^2*d^2*(12*a*x^5 + 12*b*x^5*atan(c*x)))/60 + (c*d^2*(a*x^4*30i
+ 3*b*x^4 + b*x^4*atan(c*x)*30i))/60

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sympy [A]  time = 4.13, size = 250, normalized size = 1.64 \[ - \frac {a c^{2} d^{2} x^{5}}{5} - \frac {4 b d^{2} x^{2}}{15 c} + \frac {i b d^{2} x}{2 c^{2}} - \frac {b d^{2} \left (- \frac {\log {\left (47 b c d^{2} x - 47 i b d^{2} \right )}}{60} - \frac {49 \log {\left (47 b c d^{2} x + 47 i b d^{2} \right )}}{120}\right )}{c^{3}} - x^{4} \left (- \frac {i a c d^{2}}{2} - \frac {b c d^{2}}{20}\right ) - x^{3} \left (- \frac {a d^{2}}{3} + \frac {i b d^{2}}{6}\right ) + \left (\frac {i b c^{2} d^{2} x^{5}}{10} + \frac {b c d^{2} x^{4}}{4} - \frac {i b d^{2} x^{3}}{6}\right ) \log {\left (i c x + 1 \right )} - \frac {\left (12 i b c^{5} d^{2} x^{5} + 30 b c^{4} d^{2} x^{4} - 20 i b c^{3} d^{2} x^{3} - 13 b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{120 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c**2*d**2*x**5/5 - 4*b*d**2*x**2/(15*c) + I*b*d**2*x/(2*c**2) - b*d**2*(-log(47*b*c*d**2*x - 47*I*b*d**2)/6
0 - 49*log(47*b*c*d**2*x + 47*I*b*d**2)/120)/c**3 - x**4*(-I*a*c*d**2/2 - b*c*d**2/20) - x**3*(-a*d**2/3 + I*b
*d**2/6) + (I*b*c**2*d**2*x**5/10 + b*c*d**2*x**4/4 - I*b*d**2*x**3/6)*log(I*c*x + 1) - (12*I*b*c**5*d**2*x**5
+ 30*b*c**4*d**2*x**4 - 20*I*b*c**3*d**2*x**3 - 13*b*d**2)*log(-I*c*x + 1)/(120*c**3)

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