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]

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

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Rubi [A]  time = 0.18, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.19, 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]

((-1/16*I)*(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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fricas [A]  time = 0.62, size = 158, normalized size = 0.88 \[ -\frac {2 \, {\left (4 i \, a b + 3 \, b^{2}\right )} c x - {\left (i \, b^{2} c^{2} x^{2} + 2 \, b^{2} c x + 3 i \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 16 i \, a^{2} + 16 \, a b - 8 i \, b^{2} - {\left ({\left (4 \, a b - 3 i \, b^{2}\right )} c^{2} x^{2} - 2 \, {\left (4 i \, a b + b^{2}\right )} c x + 12 \, a b - 5 i \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{32 \, {\left (c^{3} d^{3} x^{2} - 2 i \, c^{2} d^{3} x - c d^{3}\right )}} \]

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]

-1/32*(2*(4*I*a*b + 3*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 - 2*(4*I*a*b + b^2)*c*x + 12*a*b - 5*I*b^2)*log(-(c*x + I)/(c*
x - I)))/(c^3*d^3*x^2 - 2*I*c^2*d^3*x - c*d^3)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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maple [B]  time = 0.07, size = 405, normalized size = 2.25 \[ -\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16 c \,d^{3}}+\frac {i a^{2}}{2 c \,d^{3} \left (i c x +1\right )^{2}}+\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x +i\right )}{8 c \,d^{3}}-\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x -i\right )}{8 c \,d^{3}}-\frac {b^{2} \arctan \left (c x \right )}{4 c \,d^{3} \left (c x -i\right )^{2}}-\frac {i a b}{4 c \,d^{3} \left (c x -i\right )}+\frac {i b^{2}}{16 c \,d^{3} \left (c x -i\right )^{2}}+\frac {i b^{2} \arctan \left (c x \right )^{2}}{2 c \,d^{3} \left (i c x +1\right )^{2}}-\frac {i b^{2} \ln \left (c x -i\right )^{2}}{32 c \,d^{3}}-\frac {3 b^{2} \arctan \left (c x \right )}{16 c \,d^{3}}-\frac {3 b^{2}}{16 c \,d^{3} \left (c x -i\right )}-\frac {i b^{2} \arctan \left (c x \right )}{4 c \,d^{3} \left (c x -i\right )}+\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{16 c \,d^{3}}-\frac {i b^{2} \ln \left (c x +i\right )^{2}}{32 c \,d^{3}}+\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (c x +i\right )}{16 c \,d^{3}}+\frac {i a b \arctan \left (c x \right )}{c \,d^{3} \left (i c x +1\right )^{2}}-\frac {a b}{4 c \,d^{3} \left (c x -i\right )^{2}}-\frac {i a b \arctan \left (c x \right )}{4 c \,d^{3}} \]

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

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maxima [A]  time = 0.67, size = 136, normalized size = 0.76 \[ -\frac {{\left (4 i \, a b + 3 \, b^{2}\right )} c x + {\left (2 i \, b^{2} c^{2} x^{2} + 4 \, b^{2} c x + 6 i \, b^{2}\right )} \arctan \left (c x\right )^{2} + 8 i \, a^{2} + 8 \, a b - 4 i \, b^{2} + {\left ({\left (4 i \, a b + 3 \, b^{2}\right )} c^{2} x^{2} + 2 \, {\left (4 \, a b - i \, b^{2}\right )} c x + 12 i \, a b + 5 \, b^{2}\right )} \arctan \left (c x\right )}{16 \, c^{3} d^{3} x^{2} - 32 i \, c^{2} d^{3} x - 16 \, c d^{3}} \]

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 + 2*(4*a*b - 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))^2/(d + c*d*x*1i)^3,x)

[Out]

int((a + b*atan(c*x))^2/(d + c*d*x*1i)^3, x)

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sympy [B]  time = 65.27, size = 464, normalized size = 2.58 \[ \frac {b \left (4 a - 3 i b\right ) \log {\left (- \frac {b \left (4 a - 3 i b\right )}{c} + x \left (4 i a b + 3 b^{2}\right ) \right )}}{32 c d^{3}} - \frac {b \left (4 a - 3 i b\right ) \log {\left (\frac {b \left (4 a - 3 i b\right )}{c} + x \left (4 i a b + 3 b^{2}\right ) \right )}}{32 c d^{3}} + \frac {\left (4 i a b + i b^{2} c x + 2 b^{2}\right ) \log {\left (i c x + 1 \right )}}{- 8 i c^{3} d^{3} x^{2} - 16 c^{2} d^{3} x + 8 i c d^{3}} + \frac {\left (- b^{2} c^{2} x^{2} + 2 i b^{2} c x - 3 b^{2}\right ) \log {\left (i c x + 1 \right )}^{2}}{32 i c^{3} d^{3} x^{2} + 64 c^{2} d^{3} x - 32 i c d^{3}} + \frac {8 i a^{2} + 8 a b - 4 i b^{2} + x \left (4 i a b c + 3 b^{2} c\right )}{- 16 c^{3} d^{3} x^{2} + 32 i c^{2} d^{3} x + 16 c d^{3}} + \frac {\left (- 8 a b + i b^{2} c^{2} x^{2} \log {\left (i c x + 1 \right )} + 2 b^{2} c x \log {\left (i c x + 1 \right )} - 2 b^{2} c x + 3 i b^{2} \log {\left (i c x + 1 \right )} + 4 i b^{2}\right ) \log {\left (- i c x + 1 \right )}}{- 16 c^{3} d^{3} x^{2} + 32 i c^{2} d^{3} x + 16 c d^{3}} + \frac {\left (- i b^{2} c^{2} x^{2} - 2 b^{2} c x - 3 i b^{2}\right ) \log {\left (- i c x + 1 \right )}^{2}}{- 32 c^{3} d^{3} x^{2} + 64 i c^{2} d^{3} x + 32 c d^{3}} \]

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]

b*(4*a - 3*I*b)*log(-b*(4*a - 3*I*b)/c + x*(4*I*a*b + 3*b**2))/(32*c*d**3) - b*(4*a - 3*I*b)*log(b*(4*a - 3*I*
b)/c + x*(4*I*a*b + 3*b**2))/(32*c*d**3) + (4*I*a*b + I*b**2*c*x + 2*b**2)*log(I*c*x + 1)/(-8*I*c**3*d**3*x**2
- 16*c**2*d**3*x + 8*I*c*d**3) + (-b**2*c**2*x**2 + 2*I*b**2*c*x - 3*b**2)*log(I*c*x + 1)**2/(32*I*c**3*d**3*
x**2 + 64*c**2*d**3*x - 32*I*c*d**3) + (8*I*a**2 + 8*a*b - 4*I*b**2 + x*(4*I*a*b*c + 3*b**2*c))/(-16*c**3*d**3
*x**2 + 32*I*c**2*d**3*x + 16*c*d**3) + (-8*a*b + I*b**2*c**2*x**2*log(I*c*x + 1) + 2*b**2*c*x*log(I*c*x + 1)
- 2*b**2*c*x + 3*I*b**2*log(I*c*x + 1) + 4*I*b**2)*log(-I*c*x + 1)/(-16*c**3*d**3*x**2 + 32*I*c**2*d**3*x + 16
*c*d**3) + (-I*b**2*c**2*x**2 - 2*b**2*c*x - 3*I*b**2)*log(-I*c*x + 1)**2/(-32*c**3*d**3*x**2 + 64*I*c**2*d**3
*x + 32*c*d**3)

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