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

Optimal. Leaf size=182 \[ \frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (-c x+i)}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (-c x+i)}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}-\frac{3 i b^3}{4 c d^2 (-c x+i)}+\frac{3 i b^3 \tan ^{-1}(c x)}{4 c d^2} \]

[Out]

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

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Rubi [A]  time = 0.219803, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4864, 4862, 627, 44, 203, 4884} \[ \frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right )}{2 c d^2 (-c x+i)}+\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2 (-c x+i)}-\frac{3 b \left (a+b \tan ^{-1}(c x)\right )^2}{4 c d^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{c d^2 (1+i c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^3}{2 c d^2}-\frac{3 i b^3}{4 c d^2 (-c x+i)}+\frac{3 i b^3 \tan ^{-1}(c x)}{4 c d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])^3/(d + I*c*d*x)^2,x]

[Out]

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

Rule 4864

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

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])

Rule 4884

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

Rubi steps

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

Mathematica [A]  time = 0.222033, size = 121, normalized size = 0.66 \[ \frac{3 i b \left (-2 a^2+2 i a b+b^2\right ) (c x+i) \tan ^{-1}(c x)-6 i a^2 b+4 a^3-3 b^2 (b+2 i a) (c x+i) \tan ^{-1}(c x)^2-6 a b^2+2 b^3 (1-i c x) \tan ^{-1}(c x)^3+3 i b^3}{4 c d^2 (c x-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])^3/(d + I*c*d*x)^2,x]

[Out]

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

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Maple [B]  time = 0.319, size = 551, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))^3/(d+I*c*d*x)^2,x)

[Out]

3/4*I/c*a*b^2/d^2*ln(-1/2*I*(-c*x+I))*ln(c*x+I)+3*I/c*a^2*b/d^2/(1+I*c*x)*arctan(c*x)-1/2/c*b^3/d^2/(c*x-I)*ar
ctan(c*x)^3-3/8*I/c*a*b^2/d^2*ln(c*x-I)^2-3/4*b^3/d^2/(c*x-I)*arctan(c*x)^2*x-3/4*I/c*b^3/d^2/(c*x-I)*arctan(c
*x)^2-3/4/c*b^3/d^2/(c*x-I)*arctan(c*x)+I/c*b^3/d^2/(1+I*c*x)*arctan(c*x)^3-3*I/c*a*b^2/d^2*arctan(c*x)/(c*x-I
)+3/4*I/c*a*b^2/d^2*ln(c*x-I)*ln(-1/2*I*(c*x+I))-3/2/c*a*b^2/d^2*arctan(c*x)*ln(c*x-I)-1/2*I*b^3/d^2/(c*x-I)*a
rctan(c*x)^3*x+3/2/c*a*b^2/d^2*arctan(c*x)*ln(c*x+I)-3/4*I/c*a*b^2/d^2*ln(-1/2*I*(-c*x+I))*ln(-1/2*I*(c*x+I))-
3/2*I/c*a^2*b/d^2*arctan(c*x)-3/2/c*a*b^2/d^2/(c*x-I)-3/2/c*a*b^2/d^2*arctan(c*x)+3/4*I/c*b^3/d^2/(c*x-I)-3/2*
I/c*a^2*b/d^2/(c*x-I)-3/8*I/c*a*b^2/d^2*ln(c*x+I)^2+3*I/c*a*b^2/d^2/(1+I*c*x)*arctan(c*x)^2+3/4*I*b^3/d^2/(c*x
-I)*arctan(c*x)*x+I/c*a^3/d^2/(1+I*c*x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.88012, size = 401, normalized size = 2.2 \begin{align*} -\frac{{\left (b^{3} c x + i \, b^{3}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{3} - 16 \, a^{3} + 24 i \, a^{2} b + 24 \, a b^{2} - 12 i \, b^{3} +{\left (6 \, a b^{2} - 3 i \, b^{3} -{\left (6 i \, a b^{2} + 3 \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} -{\left (12 i \, a^{2} b + 12 \, a b^{2} - 6 i \, b^{3} +{\left (12 \, a^{2} b - 12 i \, a b^{2} - 6 \, b^{3}\right )} c x\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{16 \,{\left (c^{2} d^{2} x - i \, c d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^2,x, algorithm="fricas")

[Out]

-1/16*((b^3*c*x + I*b^3)*log(-(c*x + I)/(c*x - I))^3 - 16*a^3 + 24*I*a^2*b + 24*a*b^2 - 12*I*b^3 + (6*a*b^2 -
3*I*b^3 - (6*I*a*b^2 + 3*b^3)*c*x)*log(-(c*x + I)/(c*x - I))^2 - (12*I*a^2*b + 12*a*b^2 - 6*I*b^3 + (12*a^2*b
- 12*I*a*b^2 - 6*b^3)*c*x)*log(-(c*x + I)/(c*x - I)))/(c^2*d^2*x - I*c*d^2)

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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))**3/(d+I*c*d*x)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.16034, size = 782, normalized size = 4.3 \begin{align*} -\frac{\frac{6 \, b^{3} d i^{2} \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{2}}{c d i x + d} - 2 \, b^{3} i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{3} + \frac{4 \, b^{3} d i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{3}}{c d i x + d} - \frac{12 \, a b^{2} d i^{2} \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )}{c d i x + d} + 6 \, a b^{2} i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{2} - \frac{12 \, a b^{2} d i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{2}}{c d i x + d} + \frac{6 \, a^{2} b d i^{2}}{c d i x + d} - \frac{3 \, b^{3} d i^{2}}{c d i x + d} - 6 \, a^{2} b i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right ) + 3 \, b^{3} i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right ) + \frac{12 \, a^{2} b d i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )}{c d i x + d} - \frac{6 \, b^{3} d i \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )}{c d i x + d} + 3 \, b^{3} \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )^{2} - \frac{4 \, a^{3} d i}{c d i x + d} + \frac{6 \, a b^{2} d i}{c d i x + d} - 6 \, a b^{2} \arctan \left (\frac{{\left (c d i x + d\right )}{\left (\frac{d i^{2}}{c d i x + d} + 1\right )} i}{d}\right )}{4 \, c d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))^3/(d+I*c*d*x)^2,x, algorithm="giac")

[Out]

-1/4*(6*b^3*d*i^2*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)^2/(c*d*i*x + d) - 2*b^3*i*arctan((c*d*i*
x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)^3 + 4*b^3*d*i*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)^3/(c*d
*i*x + d) - 12*a*b^2*d*i^2*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)/(c*d*i*x + d) + 6*a*b^2*i*arcta
n((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)^2 - 12*a*b^2*d*i*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)
*i/d)^2/(c*d*i*x + d) + 6*a^2*b*d*i^2/(c*d*i*x + d) - 3*b^3*d*i^2/(c*d*i*x + d) - 6*a^2*b*i*arctan((c*d*i*x +
d)*(d*i^2/(c*d*i*x + d) + 1)*i/d) + 3*b^3*i*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d) + 12*a^2*b*d*i
*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)/(c*d*i*x + d) - 6*b^3*d*i*arctan((c*d*i*x + d)*(d*i^2/(c*
d*i*x + d) + 1)*i/d)/(c*d*i*x + d) + 3*b^3*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d)^2 - 4*a^3*d*i/(
c*d*i*x + d) + 6*a*b^2*d*i/(c*d*i*x + d) - 6*a*b^2*arctan((c*d*i*x + d)*(d*i^2/(c*d*i*x + d) + 1)*i/d))/(c*d^2
)