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

Optimal. Leaf size=88 \[ \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (1+i c x)^2}+\frac {3 b}{8 c^2 d^3 (-c x+i)}-\frac {i b}{8 c^2 d^3 (-c x+i)^2}+\frac {b \tan ^{-1}(c x)}{8 c^2 d^3} \]

[Out]

-1/8*I*b/c^2/d^3/(I-c*x)^2+3/8*b/c^2/d^3/(I-c*x)+1/8*b*arctan(c*x)/c^2/d^3+1/2*x^2*(a+b*arctan(c*x))/d^3/(1+I*
c*x)^2

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Rubi [A]  time = 0.08, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {37, 4872, 12, 88, 203} \[ \frac {x^2 \left (a+b \tan ^{-1}(c x)\right )}{2 d^3 (1+i c x)^2}+\frac {3 b}{8 c^2 d^3 (-c x+i)}-\frac {i b}{8 c^2 d^3 (-c x+i)^2}+\frac {b \tan ^{-1}(c x)}{8 c^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*b)/(c^2*d^3*(I - c*x)^2) + (3*b)/(8*c^2*d^3*(I - c*x)) + (b*ArcTan[c*x])/(8*c^2*d^3) + (x^2*(a + b*Arc
Tan[c*x]))/(2*d^3*(1 + I*c*x)^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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

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Mathematica [A]  time = 0.09, size = 63, normalized size = 0.72 \[ \frac {a (-4-8 i c x)-b \left (3 c^2 x^2+2 i c x+1\right ) \tan ^{-1}(c x)+b (-3 c x+2 i)}{8 c^2 d^3 (c x-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 83, normalized size = 0.94 \[ \frac {{\left (-16 i \, a - 6 \, b\right )} c x + {\left (-3 i \, b c^{2} x^{2} + 2 \, b c x - i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 8 \, a + 4 i \, b}{16 \, {\left (c^{4} d^{3} x^{2} - 2 i \, c^{3} d^{3} x - c^{2} d^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((-16*I*a - 6*b)*c*x + (-3*I*b*c^2*x^2 + 2*b*c*x - I*b)*log(-(c*x + I)/(c*x - I)) - 8*a + 4*I*b)/(c^4*d^3
*x^2 - 2*I*c^3*d^3*x - c^2*d^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*(a+b*arctan(c*x))/(d+I*c*d*x)^3,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.06, size = 128, normalized size = 1.45 \[ -\frac {i a}{c^{2} d^{3} \left (c x -i\right )}+\frac {a}{2 c^{2} d^{3} \left (c x -i\right )^{2}}-\frac {i b \arctan \left (c x \right )}{c^{2} d^{3} \left (c x -i\right )}+\frac {b \arctan \left (c x \right )}{2 c^{2} d^{3} \left (c x -i\right )^{2}}-\frac {3 b \arctan \left (c x \right )}{8 c^{2} d^{3}}-\frac {i b}{8 c^{2} d^{3} \left (c x -i\right )^{2}}-\frac {3 b}{8 c^{2} d^{3} \left (c x -i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I/c^2*a/d^3/(c*x-I)+1/2/c^2*a/d^3/(c*x-I)^2-I/c^2*b/d^3*arctan(c*x)/(c*x-I)+1/2/c^2*b/d^3*arctan(c*x)/(c*x-I)
^2-3/8*b*arctan(c*x)/c^2/d^3-1/8*I/c^2*b/d^3/(c*x-I)^2-3/8/c^2*b/d^3/(c*x-I)

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maxima [A]  time = 0.34, size = 71, normalized size = 0.81 \[ -\frac {{\left (8 i \, a + 3 \, b\right )} c x + {\left (3 \, b c^{2} x^{2} + 2 i \, b c x + b\right )} \arctan \left (c x\right ) + 4 \, a - 2 i \, b}{8 \, c^{4} d^{3} x^{2} - 16 i \, c^{3} d^{3} x - 8 \, c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((8*I*a + 3*b)*c*x + (3*b*c^2*x^2 + 2*I*b*c*x + b)*arctan(c*x) + 4*a - 2*I*b)/(8*c^4*d^3*x^2 - 16*I*c^3*d^3*x
- 8*c^2*d^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 10.22, size = 194, normalized size = 2.20 \[ \frac {b \left (\frac {3 i \log {\left (x - \frac {i}{c} \right )}}{16} - \frac {3 i \log {\left (x + \frac {i}{c} \right )}}{16}\right )}{c^{2} d^{3}} + \frac {\left (- 2 i b c x - b\right ) \log {\left (i c x + 1 \right )}}{4 i c^{4} d^{3} x^{2} + 8 c^{3} d^{3} x - 4 i c^{2} d^{3}} + \frac {\left (- 2 i b c x - b\right ) \log {\left (- i c x + 1 \right )}}{- 4 i c^{4} d^{3} x^{2} - 8 c^{3} d^{3} x + 4 i c^{2} d^{3}} + \frac {4 a - 2 i b + x \left (8 i a c + 3 b c\right )}{- 8 c^{4} d^{3} x^{2} + 16 i c^{3} d^{3} x + 8 c^{2} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(3*I*log(x - I/c)/16 - 3*I*log(x + I/c)/16)/(c**2*d**3) + (-2*I*b*c*x - b)*log(I*c*x + 1)/(4*I*c**4*d**3*x**
2 + 8*c**3*d**3*x - 4*I*c**2*d**3) + (-2*I*b*c*x - b)*log(-I*c*x + 1)/(-4*I*c**4*d**3*x**2 - 8*c**3*d**3*x + 4
*I*c**2*d**3) + (4*a - 2*I*b + x*(8*I*a*c + 3*b*c))/(-8*c**4*d**3*x**2 + 16*I*c**3*d**3*x + 8*c**2*d**3)

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