3.2.78 \(\int \frac {e^{2 i \text {ArcTan}(a+b x)}}{x^3} \, dx\) [178]

Optimal. Leaf size=76 \[ -\frac {i-a}{2 (i+a) x^2}+\frac {2 i b}{(i+a)^2 x}-\frac {2 b^2 \log (x)}{(1-i a)^3}+\frac {2 b^2 \log (i+a+b x)}{(1-i a)^3} \]

[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

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Rubi [A]
time = 0.04, antiderivative size = 76, normalized size of antiderivative = 1.00, 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[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*}

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Mathematica [A]
time = 0.03, size = 63, normalized size = 0.83 \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[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)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (65 ) = 130\).
time = 0.14, size = 211, normalized size = 2.78

method	result	size
default	\(-\frac {2 b^{3} \left (\frac {\left (i a^{3} b +3 a^{2} b -3 i a b -b \right ) \ln \left (b^{2} x^{2}+2 a b x +a^{2}+1\right )}{2 b^{2}}+\frac {\left (i a^{4}-6 i a^{2}+4 a^{3}+i-4 a -\frac {\left (i a^{3} b +3 a^{2} b -3 i a b -b \right ) a}{b}\right ) \arctan \left (\frac {2 b^{2} x +2 a b}{2 b}\right )}{b}\right )}{\left (a^{2}+1\right )^{3}}-\frac {-a^{2}+2 i a +1}{2 \left (a^{2}+1\right ) x^{2}}+\frac {2 b \left (i a^{2}+2 a -i\right )}{\left (a^{2}+1\right )^{2} x}+\frac {2 b^{2} \left (i a^{3}+3 a^{2}-3 i a -1\right ) \ln \left (x \right )}{\left (a^{2}+1\right )^{3}}\)	\(211\)
risch	\(\frac {\frac {2 i b x}{a^{2}+2 i a -1}+\frac {a -i}{2 i+2 a}}{x^{2}}-\frac {2 b^{2} \ln \left (\left (2 a^{4} b +4 a^{2} b +2 b \right ) x \right )}{i a^{3}-3 a^{2}-3 i a +1}+\frac {b^{2} \ln \left (4 a^{8} b^{2} x^{2}+8 a^{9} b x +4 a^{10}+16 a^{6} b^{2} x^{2}+32 a^{7} b x +20 a^{8}+24 a^{4} b^{2} x^{2}+48 a^{5} b x +40 a^{6}+16 a^{2} b^{2} x^{2}+32 a^{3} b x +40 a^{4}+4 b^{2} x^{2}+8 a b x +20 a^{2}+4\right )}{i a^{3}-3 a^{2}-3 i a +1}-\frac {2 i b^{2} \arctan \left (\frac {\left (-2 a^{4} b -4 a^{2} b -2 b \right ) x -2 a^{5}-4 a^{3}-2 a}{-2 a^{4}-4 a^{2}-2}\right )}{i a^{3}-3 a^{2}-3 i a +1}\)	\(287\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b^3/(a^2+1)^3*(1/2*(I*b*a^3-3*I*a*b+3*a^2*b-b)/b^2*ln(b^2*x^2+2*a*b*x+a^2+1)+(I*a^4-6*I*a^2+4*a^3+I-4*a-(I*
b*a^3-3*I*a*b+3*a^2*b-b)*a/b)/b*arctan(1/2*(2*b^2*x+2*a*b)/b))-1/2*(2*I*a-a^2+1)/(a^2+1)/x^2+2*b*(I*a^2-I+2*a)
/(a^2+1)^2/x+2*b^2*(I*a^3-3*I*a+3*a^2-1)/(a^2+1)^3*ln(x)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (58) = 116\).
time = 0.48, size = 188, normalized size = 2.47 \begin {gather*} -\frac {2 \, {\left (a^{3} - 3 i \, a^{2} - 3 \, a + i\right )} b^{2} \arctan \left (\frac {b^{2} x + a b}{b}\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac {{\left (i \, a^{3} + 3 \, a^{2} - 3 i \, a - 1\right )} b^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} - \frac {2 \, {\left (-i \, a^{3} - 3 \, a^{2} + 3 i \, a + 1\right )} b^{2} \log \left (x\right )}{a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1} + \frac {a^{4} - 2 i \, a^{3} - 4 \, {\left (-i \, a^{2} - 2 \, a + i\right )} b x - 2 i \, a - 1}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.10, size = 69, normalized size = 0.91 \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+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)

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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. 228 vs. \(2 (60) = 120\).
time = 0.46, size = 228, normalized size = 3.00 \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+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))

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Giac [A]
time = 0.42, size = 89, normalized size = 1.17 \begin {gather*} \frac {2 \, b^{3} \log \left (b x + a + i\right )}{i \, a^{3} b - 3 \, a^{2} b - 3 i \, a b + b} + \frac {2 \, b^{2} \log \left ({\left | x \right |}\right )}{-i \, a^{3} + 3 \, a^{2} + 3 i \, a - 1} + \frac {a^{3} + i \, a^{2} + 4 i \, {\left (a b + i \, b\right )} x + a + i}{2 \, {\left (a + i\right )}^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.69, size = 154, normalized size = 2.03 \begin {gather*} \frac {\frac {a-\mathrm {i}}{2\,\left (a+1{}\mathrm {i}\right )}+\frac {b\,x\,2{}\mathrm {i}}{{\left (a+1{}\mathrm {i}\right )}^2}}{x^2}+\frac {b^2\,\mathrm {atanh}\left (\frac {-a^3-a^2\,3{}\mathrm {i}+3\,a+1{}\mathrm {i}}{{\left (a+1{}\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+1{}\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+1{}\mathrm {i}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*1i + b*x*1i + 1)^2/(x^3*((a + b*x)^2 + 1)),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*(b + a*b*2i + a^2*b + a^3*b*4i - a^4*b + a^5*
b*2i - a^6*b)))*4i)/(a + 1i)^3

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