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

Optimal. Leaf size=180 \[ \frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c d^3 (-c x+i)}-\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c d^3 (-c x+i)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^3 (1+i c x)^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3}+\frac{3 b^2}{16 c d^3 (-c x+i)}+\frac{i b^2}{16 c d^3 (-c x+i)^2}-\frac{3 b^2 \tan ^{-1}(c x)}{16 c d^3} \]

[Out]

((I/16)*b^2)/(c*d^3*(I - c*x)^2) + (3*b^2)/(16*c*d^3*(I - c*x)) - (3*b^2*ArcTan[c*x])/(16*c*d^3) - (b*(a + b*A
rcTan[c*x]))/(4*c*d^3*(I - c*x)^2) + ((I/4)*b*(a + b*ArcTan[c*x]))/(c*d^3*(I - c*x)) - ((I/8)*(a + b*ArcTan[c*
x])^2)/(c*d^3) + ((I/2)*(a + b*ArcTan[c*x])^2)/(c*d^3*(1 + I*c*x)^2)

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Rubi [A]  time = 0.17955, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 13, 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} \[ \frac{i b \left (a+b \tan ^{-1}(c x)\right )}{4 c d^3 (-c x+i)}-\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c d^3 (-c x+i)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^3 (1+i c x)^2}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c d^3}+\frac{3 b^2}{16 c d^3 (-c x+i)}+\frac{i b^2}{16 c d^3 (-c x+i)^2}-\frac{3 b^2 \tan ^{-1}(c x)}{16 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/16)*b^2)/(c*d^3*(I - c*x)^2) + (3*b^2)/(16*c*d^3*(I - c*x)) - (3*b^2*ArcTan[c*x])/(16*c*d^3) - (b*(a + b*A
rcTan[c*x]))/(4*c*d^3*(I - c*x)^2) + ((I/4)*b*(a + b*ArcTan[c*x]))/(c*d^3*(I - c*x)) - ((I/8)*(a + b*ArcTan[c*
x])^2)/(c*d^3) + ((I/2)*(a + b*ArcTan[c*x])^2)/(c*d^3*(1 + I*c*x)^2)

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

Mathematica [A]  time = 0.178431, size = 110, normalized size = 0.61 \[ -\frac{i \left (8 a^2+4 a b (c x-2 i)+b (c x+i) \tan ^{-1}(c x) (4 a (c x-3 i)+b (-5-3 i c x))+2 b^2 \left (c^2 x^2-2 i c x+3\right ) \tan ^{-1}(c x)^2+b^2 (-4-3 i c x)\right )}{16 c d^3 (c x-i)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/16)*(8*a^2 + b^2*(-4 - (3*I)*c*x) + 4*a*b*(-2*I + c*x) + b*(I + c*x)*(b*(-5 - (3*I)*c*x) + 4*a*(-3*I + c*
x))*ArcTan[c*x] + 2*b^2*(3 - (2*I)*c*x + c^2*x^2)*ArcTan[c*x]^2))/(c*d^3*(-I + c*x)^2)

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Maple [B]  time = 0.072, size = 405, normalized size = 2.3 \begin{align*}{\frac{{\frac{i}{16}}{b}^{2}\ln \left ( -{\frac{i}{2}} \left ( -cx+i \right ) \right ) \ln \left ( cx+i \right ) }{c{d}^{3}}}-{\frac{{\frac{i}{32}}{b}^{2} \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{c{d}^{3}}}-{\frac{{b}^{2}\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{8\,c{d}^{3}}}-{\frac{{b}^{2}\arctan \left ( cx \right ) }{4\,c{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{2}}{a}^{2}}{c{d}^{3} \left ( 1+icx \right ) ^{2}}}+{\frac{{b}^{2}\arctan \left ( cx \right ) \ln \left ( cx+i \right ) }{8\,c{d}^{3}}}+{\frac{iab\arctan \left ( cx \right ) }{c{d}^{3} \left ( 1+icx \right ) ^{2}}}-{\frac{{\frac{i}{16}}{b}^{2}\ln \left ( -{\frac{i}{2}} \left ( -cx+i \right ) \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c{d}^{3}}}-{\frac{{\frac{i}{4}}ab}{c{d}^{3} \left ( cx-i \right ) }}-{\frac{3\,{b}^{2}}{16\,c{d}^{3} \left ( cx-i \right ) }}-{\frac{{\frac{i}{4}}ab\arctan \left ( cx \right ) }{c{d}^{3}}}-{\frac{3\,{b}^{2}\arctan \left ( cx \right ) }{16\,c{d}^{3}}}+{\frac{{\frac{i}{16}}{b}^{2}}{c{d}^{3} \left ( cx-i \right ) ^{2}}}-{\frac{{\frac{i}{32}}{b}^{2} \left ( \ln \left ( cx+i \right ) \right ) ^{2}}{c{d}^{3}}}+{\frac{{\frac{i}{16}}{b}^{2}\ln \left ( cx-i \right ) \ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{c{d}^{3}}}-{\frac{{\frac{i}{4}}{b}^{2}\arctan \left ( cx \right ) }{c{d}^{3} \left ( cx-i \right ) }}-{\frac{ab}{4\,c{d}^{3} \left ( cx-i \right ) ^{2}}}+{\frac{{\frac{i}{2}}{b}^{2} \left ( \arctan \left ( cx \right ) \right ) ^{2}}{c{d}^{3} \left ( 1+icx \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*I/c*b^2/d^3*ln(-1/2*I*(-c*x+I))*ln(c*x+I)-1/32*I/c*b^2/d^3*ln(c*x-I)^2-1/8/c*b^2/d^3*arctan(c*x)*ln(c*x-I
)-1/4/c*b^2/d^3*arctan(c*x)/(c*x-I)^2+1/2*I/c*a^2/d^3/(1+I*c*x)^2+1/8/c*b^2/d^3*arctan(c*x)*ln(c*x+I)+I/c*a*b/
d^3/(1+I*c*x)^2*arctan(c*x)-1/16*I/c*b^2/d^3*ln(-1/2*I*(-c*x+I))*ln(-1/2*I*(c*x+I))-1/4*I/c*a*b/d^3/(c*x-I)-3/
16/c*b^2/d^3/(c*x-I)-1/4*I/c*a*b/d^3*arctan(c*x)-3/16*b^2*arctan(c*x)/c/d^3+1/16*I/c*b^2/d^3/(c*x-I)^2-1/32*I/
c*b^2/d^3*ln(c*x+I)^2+1/16*I/c*b^2/d^3*ln(c*x-I)*ln(-1/2*I*(c*x+I))-1/4*I/c*b^2/d^3*arctan(c*x)/(c*x-I)-1/4/c*
a*b/d^3/(c*x-I)^2+1/2*I/c*b^2/d^3/(1+I*c*x)^2*arctan(c*x)^2

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Maxima [A]  time = 1.17536, size = 182, normalized size = 1.01 \begin{align*} -\frac{{\left (4 i \, a b + 3 \, b^{2}\right )} c x +{\left (2 i \, b^{2} c^{2} x^{2} + 4 \, b^{2} c x + 6 i \, b^{2}\right )} \arctan \left (c x\right )^{2} + 8 i \, a^{2} + 8 \, a b - 4 i \, b^{2} +{\left ({\left (4 i \, a b + 3 \, b^{2}\right )} c^{2} x^{2} +{\left (8 \, a b - 2 i \, b^{2}\right )} c x + 12 i \, a b + 5 \, b^{2}\right )} \arctan \left (c x\right )}{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))^2/(d+I*c*d*x)^3,x, algorithm="maxima")

[Out]

-((4*I*a*b + 3*b^2)*c*x + (2*I*b^2*c^2*x^2 + 4*b^2*c*x + 6*I*b^2)*arctan(c*x)^2 + 8*I*a^2 + 8*a*b - 4*I*b^2 +
((4*I*a*b + 3*b^2)*c^2*x^2 + (8*a*b - 2*I*b^2)*c*x + 12*I*a*b + 5*b^2)*arctan(c*x))/(16*c^3*d^3*x^2 - 32*I*c^2
*d^3*x - 16*c*d^3)

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Fricas [A]  time = 2.35746, size = 363, normalized size = 2.02 \begin{align*} \frac{{\left (-8 i \, a b - 6 \, b^{2}\right )} c x +{\left (i \, b^{2} c^{2} x^{2} + 2 \, b^{2} c x + 3 i \, b^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 16 i \, a^{2} - 16 \, a b + 8 i \, b^{2} +{\left ({\left (4 \, a b - 3 i \, b^{2}\right )} c^{2} x^{2} +{\left (-8 i \, a b - 2 \, b^{2}\right )} c x + 12 \, a b - 5 i \, b^{2}\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{32 \, c^{3} d^{3} x^{2} - 64 i \, c^{2} d^{3} x - 32 \, c d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-8*I*a*b - 6*b^2)*c*x + (I*b^2*c^2*x^2 + 2*b^2*c*x + 3*I*b^2)*log(-(c*x + I)/(c*x - I))^2 - 16*I*a^2 - 16*a*
b + 8*I*b^2 + ((4*a*b - 3*I*b^2)*c^2*x^2 + (-8*I*a*b - 2*b^2)*c*x + 12*a*b - 5*I*b^2)*log(-(c*x + I)/(c*x - I)
))/(32*c^3*d^3*x^2 - 64*I*c^2*d^3*x - 32*c*d^3)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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