3.66 \(\int \frac {a+b \tan ^{-1}(c x)}{(1+i c x)^4} \, dx\)

Optimal. Leaf size=100 \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}+\frac {i b}{24 c (-c x+i)}-\frac {b}{24 c (-c x+i)^2}-\frac {i b}{18 c (-c x+i)^3}-\frac {i b \tan ^{-1}(c x)}{24 c} \]

[Out]

-1/18*I*b/c/(I-c*x)^3-1/24*b/c/(I-c*x)^2+1/24*I*b/c/(I-c*x)-1/24*I*b*arctan(c*x)/c+1/3*I*(a+b*arctan(c*x))/c/(
1+I*c*x)^3

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Rubi [A]  time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4862, 627, 44, 203} \[ \frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}+\frac {i b}{24 c (-c x+i)}-\frac {b}{24 c (-c x+i)^2}-\frac {i b}{18 c (-c x+i)^3}-\frac {i b \tan ^{-1}(c x)}{24 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(1 + I*c*x)^4,x]

[Out]

((-I/18)*b)/(c*(I - c*x)^3) - b/(24*c*(I - c*x)^2) + ((I/24)*b)/(c*(I - c*x)) - ((I/24)*b*ArcTan[c*x])/c + ((I
/3)*(a + b*ArcTan[c*x]))/(c*(1 + I*c*x)^3)

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)}{(1+i c x)^4} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}-\frac {1}{3} (i b) \int \frac {1}{(1+i c x)^3 \left (1+c^2 x^2\right )} \, dx\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}-\frac {1}{3} (i b) \int \frac {1}{(1-i c x) (1+i c x)^4} \, dx\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}-\frac {1}{3} (i b) \int \left (\frac {1}{2 (-i+c x)^4}+\frac {i}{4 (-i+c x)^3}-\frac {1}{8 (-i+c x)^2}+\frac {1}{8 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {i b}{18 c (i-c x)^3}-\frac {b}{24 c (i-c x)^2}+\frac {i b}{24 c (i-c x)}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}-\frac {1}{24} (i b) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {i b}{18 c (i-c x)^3}-\frac {b}{24 c (i-c x)^2}+\frac {i b}{24 c (i-c x)}-\frac {i b \tan ^{-1}(c x)}{24 c}+\frac {i \left (a+b \tan ^{-1}(c x)\right )}{3 c (1+i c x)^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 73, normalized size = 0.73 \[ \frac {-24 a+b \left (-3 i c^2 x^2-9 c x+10 i\right )+3 b \left (-i c^3 x^3-3 c^2 x^2+3 i c x-7\right ) \tan ^{-1}(c x)}{72 c (c x-i)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/(1 + I*c*x)^4,x]

[Out]

(-24*a + b*(10*I - 9*c*x - (3*I)*c^2*x^2) + 3*b*(-7 + (3*I)*c*x - 3*c^2*x^2 - I*c^3*x^3)*ArcTan[c*x])/(72*c*(-
I + c*x)^3)

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fricas [A]  time = 0.43, size = 93, normalized size = 0.93 \[ \frac {-6 i \, b c^{2} x^{2} - 18 \, b c x + {\left (3 \, b c^{3} x^{3} - 9 i \, b c^{2} x^{2} - 9 \, b c x - 21 i \, b\right )} \log \left (-\frac {c x + i}{c x - i}\right ) - 48 \, a + 20 i \, b}{144 \, c^{4} x^{3} - 432 i \, c^{3} x^{2} - 432 \, c^{2} x + 144 i \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/(1+I*c*x)^4,x, algorithm="fricas")

[Out]

(-6*I*b*c^2*x^2 - 18*b*c*x + (3*b*c^3*x^3 - 9*I*b*c^2*x^2 - 9*b*c*x - 21*I*b)*log(-(c*x + I)/(c*x - I)) - 48*a
+ 20*I*b)/(144*c^4*x^3 - 432*I*c^3*x^2 - 432*c^2*x + 144*I*c)

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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))/(1+I*c*x)^4,x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 93, normalized size = 0.93 \[ \frac {i a}{3 c \left (i c x +1\right )^{3}}+\frac {i b \arctan \left (c x \right )}{3 c \left (i c x +1\right )^{3}}-\frac {i b \arctan \left (c x \right )}{24 c}-\frac {b}{24 c \left (c x -i\right )^{2}}+\frac {i b}{18 c \left (c x -i\right )^{3}}-\frac {i b}{24 c \left (c x -i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/(1+I*c*x)^4,x)

[Out]

1/3*I/c*a/(1+I*c*x)^3+1/3*I/c*b/(1+I*c*x)^3*arctan(c*x)-1/24*I*b*arctan(c*x)/c-1/24/c*b/(c*x-I)^2+1/18*I/c*b/(
c*x-I)^3-1/24*I/c*b/(c*x-I)

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maxima [A]  time = 0.33, size = 83, normalized size = 0.83 \[ -\frac {3 i \, b c^{2} x^{2} + 9 \, b c x + {\left (3 i \, b c^{3} x^{3} + 9 \, b c^{2} x^{2} - 9 i \, b c x + 21 \, b\right )} \arctan \left (c x\right ) + 24 \, a - 10 i \, b}{72 \, c^{4} x^{3} - 216 i \, c^{3} x^{2} - 216 \, c^{2} x + 72 i \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/(1+I*c*x)^4,x, algorithm="maxima")

[Out]

-(3*I*b*c^2*x^2 + 9*b*c*x + (3*I*b*c^3*x^3 + 9*b*c^2*x^2 - 9*I*b*c*x + 21*b)*arctan(c*x) + 24*a - 10*I*b)/(72*
c^4*x^3 - 216*I*c^3*x^2 - 216*c^2*x + 72*I*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(c*x*1i + 1)^4,x)

[Out]

int((a + b*atan(c*x))/(c*x*1i + 1)^4, x)

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sympy [B]  time = 4.27, size = 168, normalized size = 1.68 \[ - \frac {i b \log {\left (- i c x + 1 \right )}}{6 c^{4} x^{3} - 18 i c^{3} x^{2} - 18 c^{2} x + 6 i c} + \frac {i b \log {\left (i c x + 1 \right )}}{6 c^{4} x^{3} - 18 i c^{3} x^{2} - 18 c^{2} x + 6 i c} + \frac {b \left (- \frac {\log {\left (b x - \frac {i b}{c} \right )}}{48} + \frac {\log {\left (b x + \frac {i b}{c} \right )}}{48}\right )}{c} + \frac {24 a + 3 i b c^{2} x^{2} + 9 b c x - 10 i b}{- 72 c^{4} x^{3} + 216 i c^{3} x^{2} + 216 c^{2} x - 72 i c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/(1+I*c*x)**4,x)

[Out]

-I*b*log(-I*c*x + 1)/(6*c**4*x**3 - 18*I*c**3*x**2 - 18*c**2*x + 6*I*c) + I*b*log(I*c*x + 1)/(6*c**4*x**3 - 18
*I*c**3*x**2 - 18*c**2*x + 6*I*c) + b*(-log(b*x - I*b/c)/48 + log(b*x + I*b/c)/48)/c + (24*a + 3*I*b*c**2*x**2
+ 9*b*c*x - 10*I*b)/(-72*c**4*x**3 + 216*I*c**3*x**2 + 216*c**2*x - 72*I*c)

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