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]

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

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Rubi [A]  time = 0.13, antiderivative size = 136, normalized size of antiderivative = 1.00, 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 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 (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*}

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Mathematica [A]  time = 0.10, 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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fricas [A]  time = 0.46, size = 148, normalized size = 1.09 \[ -\frac {6 \, a c^{4} d^{2} x^{4} - {\left (16 i \, a + 2 \, b\right )} c^{3} d^{2} x^{3} - 4 \, {\left (3 \, a - 2 i \, b\right )} c^{2} d^{2} x^{2} + 18 \, b c d^{2} x - 17 i \, b d^{2} \log \left (\frac {c x + i}{c}\right ) + i \, b d^{2} \log \left (\frac {c x - i}{c}\right ) - {\left (-3 i \, b c^{4} d^{2} x^{4} - 8 \, b c^{3} d^{2} x^{3} + 6 i \, b c^{2} d^{2} x^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{24 \, c^{2}} \]

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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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*(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 = 141, normalized size = 1.04 \[ -\frac {c^{2} d^{2} a \,x^{4}}{4}+\frac {2 i c \,d^{2} a \,x^{3}}{3}+\frac {d^{2} a \,x^{2}}{2}-\frac {c^{2} d^{2} b \arctan \left (c x \right ) x^{4}}{4}+\frac {2 i c \,d^{2} b \arctan \left (c x \right ) x^{3}}{3}+\frac {d^{2} b \arctan \left (c x \right ) x^{2}}{2}-\frac {3 b \,d^{2} x}{4 c}+\frac {b c \,d^{2} x^{3}}{12}-\frac {i b \,d^{2} x^{2}}{3}+\frac {i b \,d^{2} \ln \left (c^{2} x^{2}+1\right )}{3 c^{2}}+\frac {3 b \,d^{2} \arctan \left (c x \right )}{4 c^{2}} \]

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 = 0.41, size = 155, normalized size = 1.14 \[ -\frac {1}{4} \, a c^{2} d^{2} x^{4} + \frac {2}{3} i \, a c d^{2} x^{3} - \frac {1}{12} \, {\left (3 \, x^{4} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{3} - 3 \, x}{c^{4}} + \frac {3 \, \arctan \left (c x\right )}{c^{5}}\right )}\right )} b c^{2} d^{2} + \frac {1}{3} i \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b c d^{2} + \frac {1}{2} \, a d^{2} x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b d^{2} \]

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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mupad [B]  time = 0.68, size = 125, normalized size = 0.92 \[ \frac {\frac {d^2\,\left (9\,b\,\mathrm {atan}\left (c\,x\right )+b\,\ln \left (c^2\,x^2+1\right )\,4{}\mathrm {i}\right )}{12}-\frac {3\,b\,c\,d^2\,x}{4}}{c^2}+\frac {d^2\,\left (6\,a\,x^2+6\,b\,x^2\,\mathrm {atan}\left (c\,x\right )-b\,x^2\,4{}\mathrm {i}\right )}{12}-\frac {c^2\,d^2\,\left (3\,a\,x^4+3\,b\,x^4\,\mathrm {atan}\left (c\,x\right )\right )}{12}+\frac {c\,d^2\,\left (a\,x^3\,8{}\mathrm {i}+b\,x^3+b\,x^3\,\mathrm {atan}\left (c\,x\right )\,8{}\mathrm {i}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((d^2*(9*b*atan(c*x) + b*log(c^2*x^2 + 1)*4i))/12 - (3*b*c*d^2*x)/4)/c^2 + (d^2*(6*a*x^2 - b*x^2*4i + 6*b*x^2*
atan(c*x)))/12 - (c^2*d^2*(3*a*x^4 + 3*b*x^4*atan(c*x)))/12 + (c*d^2*(a*x^3*8i + b*x^3 + b*x^3*atan(c*x)*8i))/
12

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sympy [A]  time = 3.84, size = 240, normalized size = 1.76 \[ - \frac {a c^{2} d^{2} x^{4}}{4} - \frac {3 b d^{2} x}{4 c} - \frac {b d^{2} \left (\frac {i \log {\left (67 b c d^{2} x - 67 i b d^{2} \right )}}{24} - \frac {31 i \log {\left (67 b c d^{2} x + 67 i b d^{2} \right )}}{60}\right )}{c^{2}} - x^{3} \left (- \frac {2 i a c d^{2}}{3} - \frac {b c d^{2}}{12}\right ) - x^{2} \left (- \frac {a d^{2}}{2} + \frac {i b d^{2}}{3}\right ) + \left (\frac {i b c^{2} d^{2} x^{4}}{8} + \frac {b c d^{2} x^{3}}{3} - \frac {i b d^{2} x^{2}}{4}\right ) \log {\left (i c x + 1 \right )} - \frac {\left (15 i b c^{4} d^{2} x^{4} + 40 b c^{3} d^{2} x^{3} - 30 i b c^{2} d^{2} x^{2} - 23 i b d^{2}\right ) \log {\left (- i c x + 1 \right )}}{120 c^{2}} \]

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) - b*d**2*(I*log(67*b*c*d**2*x - 67*I*b*d**2)/24 - 31*I*log(67*b*c*d**2*
x + 67*I*b*d**2)/60)/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) - (15*I*b*c**4*d**2*x**4 + 40*b*c**3*d**2*x**3
- 30*I*b*c**2*d**2*x**2 - 23*I*b*d**2)*log(-I*c*x + 1)/(120*c**2)

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