3.1.62 \(\int \frac {a+b \text {ArcTan}(c x)}{(d+i c d x)^3} \, dx\) [62]

Optimal. Leaf size=92 \[ -\frac {b}{8 c d^3 (i-c x)^2}+\frac {i b}{8 c d^3 (i-c x)}-\frac {i b \text {ArcTan}(c x)}{8 c d^3}+\frac {i (a+b \text {ArcTan}(c x))}{2 c d^3 (1+i c x)^2} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 92, 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 = {4972, 641, 46, 209} \begin {gather*} \frac {i (a+b \text {ArcTan}(c x))}{2 c d^3 (1+i c x)^2}-\frac {i b \text {ArcTan}(c x)}{8 c d^3}+\frac {i b}{8 c d^3 (-c x+i)}-\frac {b}{8 c d^3 (-c x+i)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*b/(c*d^3*(I - c*x)^2) + ((I/8)*b)/(c*d^3*(I - c*x)) - ((I/8)*b*ArcTan[c*x])/(c*d^3) + ((I/2)*(a + b*ArcTa
n[c*x]))/(c*d^3*(1 + I*c*x)^2)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 4972

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

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Mathematica [A]
time = 0.04, size = 55, normalized size = 0.60 \begin {gather*} -\frac {i \left (4 a+b (-2 i+c x)+b \left (3-2 i c x+c^2 x^2\right ) \text {ArcTan}(c x)\right )}{8 c d^3 (-i+c x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 82, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\frac {i a}{2 d^{3} \left (i c x +1\right )^{2}}+\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (i c x +1\right )^{2}}-\frac {i b \arctan \left (c x \right )}{8 d^{3}}-\frac {b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {i b}{8 d^{3} \left (c x -i\right )}}{c}\)	\(82\)
default	\(\frac {\frac {i a}{2 d^{3} \left (i c x +1\right )^{2}}+\frac {i b \arctan \left (c x \right )}{2 d^{3} \left (i c x +1\right )^{2}}-\frac {i b \arctan \left (c x \right )}{8 d^{3}}-\frac {b}{8 d^{3} \left (c x -i\right )^{2}}-\frac {i b}{8 d^{3} \left (c x -i\right )}}{c}\)	\(82\)
risch	\(-\frac {b \ln \left (i c x +1\right )}{4 c \,d^{3} \left (c x -i\right )^{2}}+\frac {4 b \ln \left (-i c x +1\right )-\ln \left (-c x +i\right ) b \,c^{2} x^{2}+\ln \left (c x +i\right ) b \,c^{2} x^{2}+2 i \ln \left (-c x +i\right ) b c x -2 i \ln \left (c x +i\right ) b c x +b \ln \left (-c x +i\right )-b \ln \left (c x +i\right )-2 i b c x -8 i a -4 b}{16 d^{3} \left (c x -i\right )^{2} c}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 65, normalized size = 0.71 \begin {gather*} -\frac {i \, b c x + {\left (i \, b c^{2} x^{2} + 2 \, b c x + 3 i \, b\right )} \arctan \left (c x\right ) + 4 i \, a + 2 \, b}{8 \, {\left (c^{3} d^{3} x^{2} - 2 i \, c^{2} d^{3} x - c d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.25, size = 75, normalized size = 0.82 \begin {gather*} \frac {-2 i \, b c x + {\left (b c^{2} x^{2} - 2 i \, b c x + 3 \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 8 i \, a - 4 \, b}{16 \, {\left (c^{3} d^{3} x^{2} - 2 i \, c^{2} d^{3} x - c d^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (70) = 140\).
time = 1.82, size = 158, normalized size = 1.72 \begin {gather*} \frac {b \log {\left (- i c x + 1 \right )}}{4 c^{3} d^{3} x^{2} - 8 i c^{2} d^{3} x - 4 c d^{3}} - \frac {b \log {\left (i c x + 1 \right )}}{4 c^{3} d^{3} x^{2} - 8 i c^{2} d^{3} x - 4 c d^{3}} + \frac {b \left (- \frac {\log {\left (b x - \frac {i b}{c} \right )}}{16} + \frac {\log {\left (b x + \frac {i b}{c} \right )}}{16}\right )}{c d^{3}} + \frac {- 4 i a - i b c x - 2 b}{8 c^{3} d^{3} x^{2} - 16 i c^{2} d^{3} x - 8 c d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/(d+I*c*d*x)**3,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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