Integrand size = 17, antiderivative size = 184 \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=-\frac {i (1-m) m x (e x)^m}{2 (1+m)}+\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^2 x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}+\frac {i e^{-2 i a} \left (e^{2 i a} (3+m)+\frac {e^{4 i a} (1-m)}{x^2}\right ) x (e x)^m}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )}-\frac {i \left (3+2 m+m^2\right ) x (e x)^m \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-1-m),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{1+m} \]
-1/2*I*(1-m)*m*x*(e*x)^m/(1+m)+1/2*I*(1-exp(2*I*a)/x^2)^2*x*(e*x)^m/(1+exp (2*I*a)/x^2)^2+1/2*I*(exp(2*I*a)*(3+m)+exp(4*I*a)*(1-m)/x^2)*x*(e*x)^m/exp (2*I*a)/(1+exp(2*I*a)/x^2)-I*(m^2+2*m+3)*x*(e*x)^m*hypergeom([1, -1/2-1/2* m],[-1/2*m+1/2],-exp(2*I*a)/x^2)/(1+m)
Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.68 \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\frac {i x (e x)^m \left (-1+6 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )-12 \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )+8 \operatorname {Hypergeometric2F1}\left (3,\frac {1+m}{2},\frac {3+m}{2},-x^2 (\cos (2 a)-i \sin (2 a))\right )\right )}{1+m} \]
(I*x*(e*x)^m*(-1 + 6*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -(x^2*(Cos [2*a] - I*Sin[2*a]))] - 12*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -(x^ 2*(Cos[2*a] - I*Sin[2*a]))] + 8*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -(x^2*(Cos[2*a] - I*Sin[2*a]))]))/(1 + m)
Time = 0.43 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.22, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {5006, 999, 26, 370, 27, 439, 27, 363, 278}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int (e x)^m \tan ^3(a+i \log (x)) \, dx\) |
| \(\Big \downarrow \) 5006 |
| \(\displaystyle \int \frac {\left (i-\frac {i e^{2 i a}}{x^2}\right )^3 (e x)^m}{\left (1+\frac {e^{2 i a}}{x^2}\right )^3}dx\) |
| \(\Big \downarrow \) 999 |
| \(\displaystyle -\left (\frac {1}{x}\right )^m (e x)^m \int -\frac {i \left (1-\frac {e^{2 i a}}{x^2}\right )^3 \left (\frac {1}{x}\right )^{-m-2}}{\left (1+\frac {e^{2 i a}}{x^2}\right )^3}d\frac {1}{x}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \int \frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^3 \left (\frac {1}{x}\right )^{-m-2}}{\left (1+\frac {e^{2 i a}}{x^2}\right )^3}d\frac {1}{x}\) |
| \(\Big \downarrow \) 370 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}-\frac {1}{4} e^{-2 i a} \int -\frac {2 \left (1-\frac {e^{2 i a}}{x^2}\right ) \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) \left (\frac {1}{x}\right )^{-m-2}}{\left (1+\frac {e^{2 i a}}{x^2}\right )^2}d\frac {1}{x}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} e^{-2 i a} \int \frac {\left (1-\frac {e^{2 i a}}{x^2}\right ) \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right ) \left (\frac {1}{x}\right )^{-m-2}}{\left (1+\frac {e^{2 i a}}{x^2}\right )^2}d\frac {1}{x}+\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}\right )\) |
| \(\Big \downarrow \) 439 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} e^{-2 i a} \left (\frac {\left (\frac {1}{x}\right )^{-m-1} \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right )}{1+\frac {e^{2 i a}}{x^2}}-\frac {1}{2} e^{-2 i a} \int -\frac {2 \left (\frac {e^{6 i a} (1-m) m}{x^2}+e^{4 i a} (m+2) (m+3)\right ) \left (\frac {1}{x}\right )^{-m-2}}{1+\frac {e^{2 i a}}{x^2}}d\frac {1}{x}\right )+\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} e^{-2 i a} \left (e^{-2 i a} \int \frac {\left (\frac {e^{6 i a} (1-m) m}{x^2}+e^{4 i a} (m+2) (m+3)\right ) \left (\frac {1}{x}\right )^{-m-2}}{1+\frac {e^{2 i a}}{x^2}}d\frac {1}{x}+\frac {\left (\frac {1}{x}\right )^{-m-1} \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right )}{1+\frac {e^{2 i a}}{x^2}}\right )+\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}\right )\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} e^{-2 i a} \left (e^{-2 i a} \left (2 e^{4 i a} \left (m^2+2 m+3\right ) \int \frac {\left (\frac {1}{x}\right )^{-m-2}}{1+\frac {e^{2 i a}}{x^2}}d\frac {1}{x}-\frac {e^{4 i a} (1-m) m \left (\frac {1}{x}\right )^{-m-1}}{m+1}\right )+\frac {\left (\frac {1}{x}\right )^{-m-1} \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right )}{1+\frac {e^{2 i a}}{x^2}}\right )+\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}\right )\) |
| \(\Big \downarrow \) 278 |
| \(\displaystyle i \left (\frac {1}{x}\right )^m (e x)^m \left (\frac {1}{2} e^{-2 i a} \left (e^{-2 i a} \left (-\frac {2 e^{4 i a} \left (m^2+2 m+3\right ) \left (\frac {1}{x}\right )^{-m-1} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (-m-1),\frac {1-m}{2},-\frac {e^{2 i a}}{x^2}\right )}{m+1}-\frac {e^{4 i a} (1-m) m \left (\frac {1}{x}\right )^{-m-1}}{m+1}\right )+\frac {\left (\frac {1}{x}\right )^{-m-1} \left (\frac {e^{4 i a} (1-m)}{x^2}+e^{2 i a} (m+3)\right )}{1+\frac {e^{2 i a}}{x^2}}\right )+\frac {\left (1-\frac {e^{2 i a}}{x^2}\right )^2 \left (\frac {1}{x}\right )^{-m-1}}{2 \left (1+\frac {e^{2 i a}}{x^2}\right )^2}\right )\) |
I*(x^(-1))^m*(e*x)^m*(((1 - E^((2*I)*a)/x^2)^2*(x^(-1))^(-1 - m))/(2*(1 + E^((2*I)*a)/x^2)^2) + (((E^((2*I)*a)*(3 + m) + (E^((4*I)*a)*(1 - m))/x^2)* (x^(-1))^(-1 - m))/(1 + E^((2*I)*a)/x^2) + (-((E^((4*I)*a)*(1 - m)*m*(x^(- 1))^(-1 - m))/(1 + m)) - (2*E^((4*I)*a)*(3 + 2*m + m^2)*(x^(-1))^(-1 - m)* Hypergeometric2F1[1, (-1 - m)/2, (1 - m)/2, -(E^((2*I)*a)/x^2)])/(1 + m))/ E^((2*I)*a))/(2*E^((2*I)*a)))
3.2.52.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( -b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] && !IGtQ[p, 0] && (ILtQ[p, 0 ] || GtQ[a, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(a*b*e*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1)) Int[(e*x) ^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*Simp[c*(b*c*2*(p + 1) + (b*c - a *d)*(m + 1)) + d*(b*c*2*(p + 1) + (b*c - a*d)*(m + 2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ .)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2*a*b*g*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(2*b*e*( p + 1) + (b*e - a*f)*(m + 1)) + d*(2*b*e*(p + 1) + (b*e - a*f)*(m + 2*q + 1 ))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && G tQ[q, 0] && !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> Simp[(-(e*x)^m)*(x^(-1))^m Subst[Int[(a + b/x^n)^p*( (c + d/x^n)^q/x^(m + 2)), x], x, 1/x], x] /; FreeQ[{a, b, c, d, e, m, p, q} , x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0] && !RationalQ[m]
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]
\[\int \left (e x \right )^{m} \tan \left (a +i \ln \left (x \right )\right )^{3}d x\]
\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
integral((-I*x^6 + 3*I*x^4*e^(2*I*a) - 3*I*x^2*e^(4*I*a) + I*e^(6*I*a))*e^ (m*log(e) + m*log(x))/(x^6 + 3*x^4*e^(2*I*a) + 3*x^2*e^(4*I*a) + e^(6*I*a) ), x)
\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int \left (e x\right )^{m} \tan ^{3}{\left (a + i \log {\left (x \right )} \right )}\, dx \]
\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
\[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int { \left (e x\right )^{m} \tan \left (a + i \, \log \left (x\right )\right )^{3} \,d x } \]
Timed out. \[ \int (e x)^m \tan ^3(a+i \log (x)) \, dx=\int {\mathrm {tan}\left (a+\ln \left (x\right )\,1{}\mathrm {i}\right )}^3\,{\left (e\,x\right )}^m \,d x \]