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

Optimal. Leaf size=122 \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c d^2 (-c x+i)}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{c d^2 (1+i c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2}+\frac {b^2}{2 c d^2 (-c x+i)}-\frac {b^2 \tan ^{-1}(c x)}{2 c d^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {i b \left (a+b \tan ^{-1}(c x)\right )}{c d^2 (-c x+i)}+\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{c d^2 (1+i c x)}-\frac {i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c d^2}+\frac {b^2}{2 c d^2 (-c x+i)}-\frac {b^2 \tan ^{-1}(c x)}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.20, size = 72, normalized size = 0.59 \[ -\frac {-2 a^2+b (b+2 i a) (c x+i) \tan ^{-1}(c x)+2 i a b+b^2 (-1+i c x) \tan ^{-1}(c x)^2+b^2}{2 c d^2 (c x-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 104, normalized size = 0.85 \[ \frac {{\left (i \, b^{2} c x - b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} + 8 \, a^{2} - 8 i \, a b - 4 \, b^{2} + {\left ({\left (4 \, a b - 2 i \, b^{2}\right )} c x + 4 i \, a b + 2 \, b^{2}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{8 \, c^{2} d^{2} x - 8 i \, c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.08, size = 344, normalized size = 2.82 \[ \frac {i a^{2}}{c \,d^{2} \left (i c x +1\right )}+\frac {i b^{2} \arctan \left (c x \right )^{2}}{c \,d^{2} \left (i c x +1\right )}+\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x +i\right )}{2 c \,d^{2}}-\frac {b^{2} \arctan \left (c x \right ) \ln \left (c x -i\right )}{2 c \,d^{2}}-\frac {i b^{2} \arctan \left (c x \right )}{c \,d^{2} \left (c x -i\right )}+\frac {i b^{2} \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{4 c \,d^{2}}-\frac {i b^{2} \ln \left (c x -i\right )^{2}}{8 c \,d^{2}}-\frac {b^{2} \arctan \left (c x \right )}{2 c \,d^{2}}-\frac {b^{2}}{2 c \,d^{2} \left (c x -i\right )}-\frac {i b^{2} \ln \left (c x +i\right )^{2}}{8 c \,d^{2}}+\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (c x +i\right )}{4 c \,d^{2}}-\frac {i b^{2} \ln \left (-\frac {i \left (-c x +i\right )}{2}\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{4 c \,d^{2}}+\frac {2 i a b \arctan \left (c x \right )}{c \,d^{2} \left (i c x +1\right )}-\frac {i a b \arctan \left (c x \right )}{c \,d^{2}}-\frac {i a b}{c \,d^{2} \left (c x -i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 10.37, size = 301, normalized size = 2.47 \[ \frac {b \left (2 a - i b\right ) \log {\left (- \frac {b \left (2 a - i b\right )}{c} + x \left (2 i a b + b^{2}\right ) \right )}}{4 c d^{2}} - \frac {b \left (2 a - i b\right ) \log {\left (\frac {b \left (2 a - i b\right )}{c} + x \left (2 i a b + b^{2}\right ) \right )}}{4 c d^{2}} + \frac {\left (- 2 i a b - b^{2}\right ) \log {\left (i c x + 1 \right )}}{2 c^{2} d^{2} x - 2 i c d^{2}} + \frac {\left (- b^{2} c x - i b^{2}\right ) \log {\left (- i c x + 1 \right )}^{2}}{8 i c^{2} d^{2} x + 8 c d^{2}} + \frac {\left (- 4 a b + b^{2} c x \log {\left (i c x + 1 \right )} + i b^{2} \log {\left (i c x + 1 \right )} + 2 i b^{2}\right ) \log {\left (- i c x + 1 \right )}}{4 i c^{2} d^{2} x + 4 c d^{2}} + \frac {\left (i b^{2} c x - b^{2}\right ) \log {\left (i c x + 1 \right )}^{2}}{8 c^{2} d^{2} x - 8 i c d^{2}} - \frac {- 2 a^{2} + 2 i a b + b^{2}}{2 c^{2} d^{2} x - 2 i c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))**2/(d+I*c*d*x)**2,x)

[Out]

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

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