3.3.5 \(\int \frac {e^{-2 i \text {ArcTan}(a+b x)}}{x^3} \, dx\) [205]

Optimal. Leaf size=83 \[ \frac {-i-a}{2 (i-a) x^2}-\frac {2 i b}{(i-a)^2 x}-\frac {2 b^2 \log (x)}{(1+i a)^3}+\frac {2 b^2 \log (i-a-b x)}{(1+i a)^3} \]

[Out]

1/2*(-I-a)/(I-a)/x^2-2*I*b/(I-a)^2/x-2*b^2*ln(x)/(1+I*a)^3+2*b^2*ln(I-a-b*x)/(1+I*a)^3

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 81, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5203, 78} \begin {gather*} -\frac {2 b^2 \log (x)}{(1+i a)^3}+\frac {2 b^2 \log (-a-b x+i)}{(1+i a)^3}-\frac {2 i b}{(-a+i)^2 x}-\frac {a+i}{2 (-a+i) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]

[Out]

-1/2*(I + a)/((I - a)*x^2) - ((2*I)*b)/((I - a)^2*x) - (2*b^2*Log[x])/(1 + I*a)^3 + (2*b^2*Log[I - a - b*x])/(
1 + I*a)^3

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 5203

Int[E^(ArcTan[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((1 -
I*a*c - I*b*c*x)^(I*(n/2))/(1 + I*a*c + I*b*c*x)^(I*(n/2))), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int \frac {e^{-2 i \tan ^{-1}(a+b x)}}{x^3} \, dx &=\int \frac {1-i a-i b x}{x^3 (1+i a+i b x)} \, dx\\ &=\int \left (\frac {-i-a}{(-i+a) x^3}+\frac {2 i b}{(-i+a)^2 x^2}-\frac {2 i b^2}{(-i+a)^3 x}+\frac {2 i b^3}{(-i+a)^3 (-i+a+b x)}\right ) \, dx\\ &=-\frac {i+a}{2 (i-a) x^2}-\frac {2 i b}{(i-a)^2 x}-\frac {2 b^2 \log (x)}{(1+i a)^3}+\frac {2 b^2 \log (i-a-b x)}{(1+i a)^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 66, normalized size = 0.80 \begin {gather*} \frac {(-i+a) \left (1+a^2-4 i b x\right )-4 i b^2 x^2 \log (x)+4 i b^2 x^2 \log (i-a-b x)}{2 (-i+a)^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((2*I)*ArcTan[a + b*x])*x^3),x]

[Out]

((-I + a)*(1 + a^2 - (4*I)*b*x) - (4*I)*b^2*x^2*Log[x] + (4*I)*b^2*x^2*Log[I - a - b*x])/(2*(-I + a)^3*x^2)

________________________________________________________________________________________

Maple [A]
time = 0.13, size = 109, normalized size = 1.31

method	result	size
default	\(-\frac {-a^{4}+2 i a^{3}+2 i a +1}{2 \left (i-a \right )^{4} x^{2}}-\frac {2 b^{2} \left (i a +1\right ) \ln \left (x \right )}{\left (i-a \right )^{4}}-\frac {2 b \left (i a^{2}+2 a -i\right )}{\left (i-a \right )^{4} x}+\frac {2 b^{2} \left (i a +1\right ) \ln \left (-b x -a +i\right )}{\left (i-a \right )^{4}}\)	\(109\)
risch	\(\frac {-\frac {2 i b x}{a^{2}-2 i a -1}+\frac {i+a}{2 a -2 i}}{x^{2}}+\frac {2 b^{2} \ln \left (\left (2 a^{4} b +4 a^{2} b +2 b \right ) x \right )}{i a^{3}+3 a^{2}-3 i a -1}-\frac {b^{2} \ln \left (4 a^{8} b^{2} x^{2}+8 a^{9} b x +4 a^{10}+16 a^{6} b^{2} x^{2}+32 a^{7} b x +20 a^{8}+24 a^{4} b^{2} x^{2}+48 a^{5} b x +40 a^{6}+16 a^{2} b^{2} x^{2}+32 a^{3} b x +40 a^{4}+4 b^{2} x^{2}+8 a b x +20 a^{2}+4\right )}{i a^{3}+3 a^{2}-3 i a -1}+\frac {2 i b^{2} \arctan \left (\frac {\left (-2 a^{4} b -4 a^{2} b -2 b \right ) x -2 a^{5}-4 a^{3}-2 a}{2 a^{4}+4 a^{2}+2}\right )}{i a^{3}+3 a^{2}-3 i a -1}\)	\(288\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2/(I-a)^4*(2*I*a^3-a^4+2*I*a+1)/x^2-2*b^2*(1+I*a)/(I-a)^4*ln(x)-2*b*(I*a^2-I+2*a)/(I-a)^4/x+2*b^2*(1+I*a)/(
I-a)^4*ln(I-a-b*x)

________________________________________________________________________________________

Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (61) = 122\).
time = 0.28, size = 160, normalized size = 1.93 \begin {gather*} -\frac {2 \, {\left (-i \, a - 1\right )} b^{2} \log \left (i \, b x + i \, a + 1\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} - \frac {2 \, {\left (i \, a + 1\right )} b^{2} \log \left (x\right )}{a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1} + \frac {4 \, {\left (-i \, a - 1\right )} b^{2} x^{2} + a^{4} - 2 i \, a^{3} + {\left (a^{3} - 5 i \, a^{2} - 7 \, a + 3 i\right )} b x - 2 i \, a - 1}{2 \, {\left ({\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b x^{3} + {\left (a^{4} - 4 i \, a^{3} - 6 \, a^{2} + 4 i \, a + 1\right )} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="maxima")

[Out]

-2*(-I*a - 1)*b^2*log(I*b*x + I*a + 1)/(a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1) - 2*(I*a + 1)*b^2*log(x)/(a^4 - 4*I
*a^3 - 6*a^2 + 4*I*a + 1) + 1/2*(4*(-I*a - 1)*b^2*x^2 + a^4 - 2*I*a^3 + (a^3 - 5*I*a^2 - 7*a + 3*I)*b*x - 2*I*
a - 1)/((a^3 - 3*I*a^2 - 3*a + I)*b*x^3 + (a^4 - 4*I*a^3 - 6*a^2 + 4*I*a + 1)*x^2)

________________________________________________________________________________________

Fricas [A]
time = 2.30, size = 69, normalized size = 0.83 \begin {gather*} \frac {-4 i \, b^{2} x^{2} \log \left (x\right ) + 4 i \, b^{2} x^{2} \log \left (\frac {b x + a - i}{b}\right ) + a^{3} - 4 \, {\left (i \, a + 1\right )} b x - i \, a^{2} + a - i}{2 \, {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (61) = 122\).
time = 0.51, size = 226, normalized size = 2.72 \begin {gather*} - \frac {2 i b^{2} \log {\left (- \frac {2 a^{4} b^{2}}{\left (a - i\right )^{3}} + \frac {8 i a^{3} b^{2}}{\left (a - i\right )^{3}} + \frac {12 a^{2} b^{2}}{\left (a - i\right )^{3}} + 2 a b^{2} - \frac {8 i a b^{2}}{\left (a - i\right )^{3}} + 4 b^{3} x - 2 i b^{2} - \frac {2 b^{2}}{\left (a - i\right )^{3}} \right )}}{\left (a - i\right )^{3}} + \frac {2 i b^{2} \log {\left (\frac {2 a^{4} b^{2}}{\left (a - i\right )^{3}} - \frac {8 i a^{3} b^{2}}{\left (a - i\right )^{3}} - \frac {12 a^{2} b^{2}}{\left (a - i\right )^{3}} + 2 a b^{2} + \frac {8 i a b^{2}}{\left (a - i\right )^{3}} + 4 b^{3} x - 2 i b^{2} + \frac {2 b^{2}}{\left (a - i\right )^{3}} \right )}}{\left (a - i\right )^{3}} - \frac {- a^{2} + 4 i b x - 1}{x^{2} \cdot \left (2 a^{2} - 4 i a - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))**2*(1+(b*x+a)**2)/x**3,x)

[Out]

-2*I*b**2*log(-2*a**4*b**2/(a - I)**3 + 8*I*a**3*b**2/(a - I)**3 + 12*a**2*b**2/(a - I)**3 + 2*a*b**2 - 8*I*a*
b**2/(a - I)**3 + 4*b**3*x - 2*I*b**2 - 2*b**2/(a - I)**3)/(a - I)**3 + 2*I*b**2*log(2*a**4*b**2/(a - I)**3 -
8*I*a**3*b**2/(a - I)**3 - 12*a**2*b**2/(a - I)**3 + 2*a*b**2 + 8*I*a*b**2/(a - I)**3 + 4*b**3*x - 2*I*b**2 +
2*b**2/(a - I)**3)/(a - I)**3 - (-a**2 + 4*I*b*x - 1)/(x**2*(2*a**2 - 4*I*a - 2))

________________________________________________________________________________________

Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (61) = 122\).
time = 0.44, size = 142, normalized size = 1.71 \begin {gather*} \frac {2 \, b^{3} \log \left (-\frac {i \, a}{i \, b x + i \, a + 1} - \frac {1}{i \, b x + i \, a + 1} + 1\right )}{i \, a^{3} b + 3 \, a^{2} b - 3 i \, a b - b} + \frac {\frac {i \, a b^{2} - 5 \, b^{2}}{-i \, a - 1} + \frac {2 i \, {\left (a b^{3} + 3 i \, b^{3}\right )}}{{\left (i \, b x + i \, a + 1\right )} b}}{2 \, {\left (a - i\right )}^{2} {\left (\frac {i \, a}{i \, b x + i \, a + 1} + \frac {1}{i \, b x + i \, a + 1} - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+I*(b*x+a))^2*(1+(b*x+a)^2)/x^3,x, algorithm="giac")

[Out]

2*b^3*log(-I*a/(I*b*x + I*a + 1) - 1/(I*b*x + I*a + 1) + 1)/(I*a^3*b + 3*a^2*b - 3*I*a*b - b) + 1/2*((I*a*b^2
- 5*b^2)/(-I*a - 1) + 2*I*(a*b^3 + 3*I*b^3)/((I*b*x + I*a + 1)*b))/((a - I)^2*(I*a/(I*b*x + I*a + 1) + 1/(I*b*
x + I*a + 1) - 1)^2)

________________________________________________________________________________________

Mupad [B]
time = 0.70, size = 156, normalized size = 1.88 \begin {gather*} \frac {\frac {a+1{}\mathrm {i}}{2\,\left (a-\mathrm {i}\right )}-\frac {b\,x\,2{}\mathrm {i}}{{\left (a-\mathrm {i}\right )}^2}}{x^2}-\frac {b^2\,\mathrm {atanh}\left (\frac {-a^3+a^2\,3{}\mathrm {i}+3\,a-\mathrm {i}}{{\left (a-\mathrm {i}\right )}^3}-\frac {x\,\left (2\,a^8\,b^2+8\,a^6\,b^2+12\,a^4\,b^2+8\,a^2\,b^2+2\,b^2\right )}{{\left (a-\mathrm {i}\right )}^3\,\left (b\,a^6+2{}\mathrm {i}\,b\,a^5+b\,a^4+4{}\mathrm {i}\,b\,a^3-b\,a^2+2{}\mathrm {i}\,b\,a-b\right )}\right )\,4{}\mathrm {i}}{{\left (a-\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + 1i)/(2*(a - 1i)) - (b*x*2i)/(a - 1i)^2)/x^2 - (b^2*atanh((3*a + a^2*3i - a^3 - 1i)/(a - 1i)^3 - (x*(2*b^
2 + 8*a^2*b^2 + 12*a^4*b^2 + 8*a^6*b^2 + 2*a^8*b^2))/((a - 1i)^3*(a*b*2i - b - a^2*b + a^3*b*4i + a^4*b + a^5*
b*2i + a^6*b)))*4i)/(a - 1i)^3

________________________________________________________________________________________