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

Optimal. Leaf size=69 \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )}{c d^2 (1+i c x)}+\frac {i b}{2 c d^2 (-c x+i)}-\frac {i b \tan ^{-1}(c x)}{2 c d^2} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4862, 627, 44, 203} \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )}{c d^2 (1+i c x)}+\frac {i b}{2 c d^2 (-c x+i)}-\frac {i b \tan ^{-1}(c x)}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 42, normalized size = 0.61 \[ \frac {2 a+(b-i b c x) \tan ^{-1}(c x)-i b}{2 c d^2 (c x-i)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 50, normalized size = 0.72 \[ \frac {{\left (b c x + i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) + 4 \, a - 2 i \, b}{4 \, c^{2} d^{2} x - 4 i \, c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.05, size = 76, normalized size = 1.10 \[ \frac {i a}{c \,d^{2} \left (i c x +1\right )}+\frac {i b \arctan \left (c x \right )}{c \,d^{2} \left (i c x +1\right )}-\frac {i b \arctan \left (c x \right )}{2 c \,d^{2}}-\frac {i b}{2 c \,d^{2} \left (c x -i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I/c*a/d^2/(1+I*c*x)+I/c*b/d^2/(1+I*c*x)*arctan(c*x)-1/2*I*b*arctan(c*x)/c/d^2-1/2*I/c*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))/(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 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.94, size = 116, normalized size = 1.68 \[ \frac {i b \log {\left (- i c x + 1 \right )}}{2 c^{2} d^{2} x - 2 i c d^{2}} - \frac {i b \log {\left (i c x + 1 \right )}}{2 c^{2} d^{2} x - 2 i c d^{2}} - \frac {b \left (\frac {\log {\left (b x - \frac {i b}{c} \right )}}{4} - \frac {\log {\left (b x + \frac {i b}{c} \right )}}{4}\right )}{c d^{2}} - \frac {- 2 a + i b}{2 c^{2} d^{2} x - 2 i c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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