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

Optimal. Leaf size=92 \[ \frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c d^3 (1+i c x)^2}+\frac{i b}{8 c d^3 (-c x+i)}-\frac{b}{8 c d^3 (-c x+i)^2}-\frac{i b \tan ^{-1}(c x)}{8 c d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0549023, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4862, 627, 44, 203} \[ \frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 c d^3 (1+i c x)^2}+\frac{i b}{8 c d^3 (-c x+i)}-\frac{b}{8 c d^3 (-c x+i)^2}-\frac{i b \tan ^{-1}(c x)}{8 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4862

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(a + b*
ArcTan[c*x]))/(e*(q + 1)), x] - Dist[(b*c)/(e*(q + 1)), Int[(d + e*x)^(q + 1)/(1 + c^2*x^2), x], x] /; FreeQ[{
a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[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])

Rubi steps

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

Mathematica [A]  time = 0.0397565, size = 55, normalized size = 0.6 \[ -\frac{i \left (4 a+b \left (c^2 x^2-2 i c x+3\right ) \tan ^{-1}(c x)+b (c x-2 i)\right )}{8 c d^3 (c x-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.036, size = 93, normalized size = 1. \begin{align*}{\frac{{\frac{i}{2}}a}{c{d}^{3} \left ( 1+icx \right ) ^{2}}}+{\frac{{\frac{i}{2}}b\arctan \left ( cx \right ) }{c{d}^{3} \left ( 1+icx \right ) ^{2}}}-{\frac{{\frac{i}{8}}b\arctan \left ( cx \right ) }{c{d}^{3}}}-{\frac{b}{8\,c{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{{\frac{i}{8}}b}{c{d}^{3} \left ( cx-i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.03283, size = 89, normalized size = 0.97 \begin{align*} -\frac{i \, b c x +{\left (i \, b c^{2} x^{2} + 2 \, b c x + 3 i \, b\right )} \arctan \left (c x\right ) + 4 i \, a + 2 \, b}{8 \, c^{3} d^{3} x^{2} - 16 i \, c^{2} d^{3} x - 8 \, c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.24684, size = 177, normalized size = 1.92 \begin{align*} \frac{-2 i \, b c x +{\left (b c^{2} x^{2} - 2 i \, b c x + 3 \, b\right )} \log \left (-\frac{c x + i}{c x - i}\right ) - 8 i \, a - 4 \, b}{16 \, c^{3} d^{3} x^{2} - 32 i \, c^{2} d^{3} x - 16 \, c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*I*b*c*x + (b*c^2*x^2 - 2*I*b*c*x + 3*b)*log(-(c*x + I)/(c*x - I)) - 8*I*a - 4*b)/(16*c^3*d^3*x^2 - 32*I*c^
2*d^3*x - 16*c*d^3)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.18408, size = 174, normalized size = 1.89 \begin{align*} \frac{b c^{2} x^{2} \log \left (c x + i\right ) - b c^{2} x^{2} \log \left (c x - i\right ) - 2 \, b c i x \log \left (c x + i\right ) + 2 \, b c i x \log \left (c x - i\right ) - 2 \, b c i x - 8 \, b i \arctan \left (c x\right ) - 8 \, a i - b \log \left (c x + i\right ) + b \log \left (c x - i\right ) - 4 \, b}{16 \,{\left (c^{3} d^{3} x^{2} - 2 \, c^{2} d^{3} i x - c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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