\(\int \frac {\tan (a+i \log (x))}{x^3} \, dx\) [141]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=\frac {i}{2 x^2}-i e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4591, 455, 45} \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=\frac {i}{2 x^2}-i e^{-2 i a} \log \left (1+\frac {e^{2 i a}}{x^2}\right ) \]

[In]

Int[Tan[a + I*Log[x]]/x^3,x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 4591

Int[((e_.)*(x_))^(m_.)*Tan[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Int[(e*x)^m*((I - I*E^(2*I*a*d)*
x^(2*I*b*d))/(1 + E^(2*I*a*d)*x^(2*I*b*d)))^p, x] /; FreeQ[{a, b, d, e, m, p}, x]

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(132\) vs. \(2(35)=70\).

Time = 0.04 (sec) , antiderivative size = 132, normalized size of antiderivative = 3.77 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=\frac {i}{2 x^2}-\arctan \left (\frac {\left (1+x^2\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right ) \cos (2 a)+2 i \cos (2 a) \log (x)-\frac {1}{2} i \cos (2 a) \log \left (1+x^4+2 x^2 \cos (2 a)\right )+i \arctan \left (\frac {\left (1+x^2\right ) \cos (a)}{\sin (a)-x^2 \sin (a)}\right ) \sin (2 a)+2 \log (x) \sin (2 a)-\frac {1}{2} \log \left (1+x^4+2 x^2 \cos (2 a)\right ) \sin (2 a) \]

[In]

Integrate[Tan[a + I*Log[x]]/x^3,x]

[Out]

(I/2)/x^2 - ArcTan[((1 + x^2)*Cos[a])/(Sin[a] - x^2*Sin[a])]*Cos[2*a] + (2*I)*Cos[2*a]*Log[x] - (I/2)*Cos[2*a]
*Log[1 + x^4 + 2*x^2*Cos[2*a]] + I*ArcTan[((1 + x^2)*Cos[a])/(Sin[a] - x^2*Sin[a])]*Sin[2*a] + 2*Log[x]*Sin[2*
a] - (Log[1 + x^4 + 2*x^2*Cos[2*a]]*Sin[2*a])/2

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {i}{2 x^{2}}-i {\mathrm e}^{-2 i a} \ln \left ({\mathrm e}^{2 i a}+x^{2}\right )+2 i {\mathrm e}^{-2 i a} \ln \left (x \right )\)	\(36\)

[In]

int(tan(a+I*ln(x))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.06 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=\frac {{\left (-2 i \, x^{2} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) + 4 i \, x^{2} \log \left (x\right ) + i \, e^{\left (2 i \, a\right )}\right )} e^{\left (-2 i \, a\right )}}{2 \, x^{2}} \]

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.11 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=2 i e^{- 2 i a} \log {\left (x \right )} - i e^{- 2 i a} \log {\left (x^{2} + e^{2 i a} \right )} + \frac {i}{2 x^{2}} \]

[In]

integrate(tan(a+I*ln(x))/x**3,x)

[Out]

2*I*exp(-2*I*a)*log(x) - I*exp(-2*I*a)*log(x**2 + exp(2*I*a)) + I/(2*x**2)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 94 vs. \(2 (23) = 46\).

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 2.69 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=-\frac {x^{2} {\left (i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \left (x^{4} + 2 \, x^{2} \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right ) - 2 \, {\left ({\left (\cos \left (2 \, a\right ) - i \, \sin \left (2 \, a\right )\right )} \arctan \left (\sin \left (2 \, a\right ), x^{2} + \cos \left (2 \, a\right )\right ) + 2 \, {\left (i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} \log \left (x\right )\right )} x^{2} - i}{2 \, x^{2}} \]

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="maxima")

[Out]

-1/2*(x^2*(I*cos(2*a) + sin(2*a))*log(x^4 + 2*x^2*cos(2*a) + cos(2*a)^2 + sin(2*a)^2) - 2*((cos(2*a) - I*sin(2
*a))*arctan2(sin(2*a), x^2 + cos(2*a)) + 2*(I*cos(2*a) + sin(2*a))*log(x))*x^2 - I)/x^2

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=-\frac {1}{2} \, \pi e^{\left (-2 i \, a\right )} - i \, e^{\left (-2 i \, a\right )} \log \left (x^{2} + e^{\left (2 i \, a\right )}\right ) + 2 i \, e^{\left (-2 i \, a\right )} \log \left (x\right ) + \frac {i}{2 \, x^{2}} \]

[In]

integrate(tan(a+I*log(x))/x^3,x, algorithm="giac")

[Out]

-1/2*pi*e^(-2*I*a) - I*e^(-2*I*a)*log(x^2 + e^(2*I*a)) + 2*I*e^(-2*I*a)*log(x) + 1/2*I/x^2

Mupad [B] (verification not implemented)

Time = 27.79 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (a+i \log (x))}{x^3} \, dx=-{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \left (x^2+{\mathrm {e}}^{a\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}+{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,\ln \left (x\right )\,2{}\mathrm {i}+\frac {1{}\mathrm {i}}{2\,x^2} \]

[In]

int(tan(a + log(x)*1i)/x^3,x)

[Out]

exp(-a*2i)*log(x)*2i - exp(-a*2i)*log(exp(a*2i) + x^2)*1i + 1i/(2*x^2)