\(\int \frac {\tan (a+i \log (x))}{x^4} \, dx\) [142]

Optimal result

Integrand size = 13, antiderivative size = 45 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=\frac {i}{3 x^3}-\frac {2 i e^{-2 i a}}{x}-2 i e^{-3 i a} \arctan \left (e^{-i a} x\right ) \]

[Out]

1/3*I/x^3-2*I/exp(2*I*a)/x-2*I*arctan(x/exp(I*a))/exp(3*I*a)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {4591, 456, 464, 331, 209} \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=-2 i e^{-3 i a} \arctan \left (e^{-i a} x\right )-\frac {2 i e^{-2 i a}}{x}+\frac {i}{3 x^3} \]

[In]

Int[Tan[a + I*Log[x]]/x^4,x]

[Out]

(I/3)/x^3 - (2*I)/(E^((2*I)*a)*x) - ((2*I)*ArcTan[x/E^(I*a)])/E^((3*I)*a)

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 456

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

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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^4} \, dx \\ & = \int \frac {-i e^{2 i a}+i x^2}{x^4 \left (e^{2 i a}+x^2\right )} \, dx \\ & = \frac {i}{3 x^3}+2 i \int \frac {1}{x^2 \left (e^{2 i a}+x^2\right )} \, dx \\ & = \frac {i}{3 x^3}-\frac {2 i e^{-2 i a}}{x}-\left (2 i e^{-2 i a}\right ) \int \frac {1}{e^{2 i a}+x^2} \, dx \\ & = \frac {i}{3 x^3}-\frac {2 i e^{-2 i a}}{x}-2 i e^{-3 i a} \arctan \left (e^{-i a} x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.56 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=\frac {i}{3 x^3}-\frac {2 i \cos (2 a)}{x}-2 i \arctan (x \cos (a)-i x \sin (a)) \cos (3 a)-\frac {2 \sin (2 a)}{x}-2 \arctan (x \cos (a)-i x \sin (a)) \sin (3 a) \]

[In]

Integrate[Tan[a + I*Log[x]]/x^4,x]

[Out]

(I/3)/x^3 - ((2*I)*Cos[2*a])/x - (2*I)*ArcTan[x*Cos[a] - I*x*Sin[a]]*Cos[3*a] - (2*Sin[2*a])/x - 2*ArcTan[x*Co
s[a] - I*x*Sin[a]]*Sin[3*a]

Maple [A] (verified)

Time = 3.67 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {i}{3 x^{3}}-2 i \arctan \left (x \,{\mathrm e}^{-i a}\right ) {\mathrm e}^{-3 i a}-\frac {2 i {\mathrm e}^{-2 i a}}{x}\)	\(35\)

[In]

int(tan(a+I*ln(x))/x^4,x,method=_RETURNVERBOSE)

[Out]

1/3*I/x^3-2*I*arctan(x*exp(-I*a))*exp(-3*I*a)-2*I*exp(-2*I*a)/x

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=\frac {{\left (3 \, x^{3} \log \left (x + i \, e^{\left (i \, a\right )}\right ) - 3 \, x^{3} \log \left (x - i \, e^{\left (i \, a\right )}\right ) - 6 i \, x^{2} e^{\left (i \, a\right )} + i \, e^{\left (3 i \, a\right )}\right )} e^{\left (-3 i \, a\right )}}{3 \, x^{3}} \]

[In]

integrate(tan(a+I*log(x))/x^4,x, algorithm="fricas")

[Out]

1/3*(3*x^3*log(x + I*e^(I*a)) - 3*x^3*log(x - I*e^(I*a)) - 6*I*x^2*e^(I*a) + I*e^(3*I*a))*e^(-3*I*a)/x^3

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.18 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=\left (- \log {\left (x - i e^{i a} \right )} + \log {\left (x + i e^{i a} \right )}\right ) e^{- 3 i a} + \frac {\left (- 6 i x^{2} + i e^{2 i a}\right ) e^{- 2 i a}}{3 x^{3}} \]

[In]

integrate(tan(a+I*ln(x))/x**4,x)

[Out]

(-log(x - I*exp(I*a)) + log(x + I*exp(I*a)))*exp(-3*I*a) + (-6*I*x**2 + I*exp(2*I*a))*exp(-2*I*a)/(3*x**3)

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. 156 vs. \(2 (28) = 56\).

Time = 0.37 (sec) , antiderivative size = 156, normalized size of antiderivative = 3.47 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=-\frac {6 \, x^{3} {\left (-i \, \cos \left (3 \, a\right ) - \sin \left (3 \, a\right )\right )} \arctan \left (\frac {2 \, x \cos \left (a\right )}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}, \frac {x^{2} - \cos \left (a\right )^{2} - \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 3 \, x^{3} {\left (\cos \left (3 \, a\right ) - i \, \sin \left (3 \, a\right )\right )} \log \left (\frac {x^{2} + \cos \left (a\right )^{2} + 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}{x^{2} + \cos \left (a\right )^{2} - 2 \, x \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 12 \, x^{2} {\left (i \, \cos \left (2 \, a\right ) + \sin \left (2 \, a\right )\right )} - 2 i}{6 \, x^{3}} \]

[In]

integrate(tan(a+I*log(x))/x^4,x, algorithm="maxima")

[Out]

-1/6*(6*x^3*(-I*cos(3*a) - sin(3*a))*arctan2(2*x*cos(a)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2), (x^2 - cos(a
)^2 - sin(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) + 3*x^3*(cos(3*a) - I*sin(3*a))*log((x^2 + cos(a)^2
+ 2*x*sin(a) + sin(a)^2)/(x^2 + cos(a)^2 - 2*x*sin(a) + sin(a)^2)) + 12*x^2*(I*cos(2*a) + sin(2*a)) - 2*I)/x^3

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.62 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=-2 i \, \arctan \left (x e^{\left (-i \, a\right )}\right ) e^{\left (-3 i \, a\right )} - \frac {2 i \, e^{\left (-2 i \, a\right )}}{x} + \frac {i}{3 \, x^{3}} \]

[In]

integrate(tan(a+I*log(x))/x^4,x, algorithm="giac")

[Out]

-2*I*arctan(x*e^(-I*a))*e^(-3*I*a) - 2*I*e^(-2*I*a)/x + 1/3*I/x^3

Mupad [B] (verification not implemented)

Time = 27.74 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.89 \[ \int \frac {\tan (a+i \log (x))}{x^4} \, dx=-\frac {\mathrm {atan}\left (\frac {x}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}}}\right )\,2{}\mathrm {i}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\right )}^{3/2}}-\frac {x^2\,{\mathrm {e}}^{-a\,2{}\mathrm {i}}\,2{}\mathrm {i}-\frac {1}{3}{}\mathrm {i}}{x^3} \]

[In]

int(tan(a + log(x)*1i)/x^4,x)

[Out]

- (atan(x/exp(a*2i)^(1/2))*2i)/exp(a*2i)^(3/2) - (x^2*exp(-a*2i)*2i - 1i/3)/x^3