Integrand size = 30, antiderivative size = 81 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=\frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{4 f} \] Output:
9/2*I*a^2*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)+3/4*I*a*(d*sec(f *x+e))^(2/3)*(a+I*a*tan(f*x+e))^(2/3)/f
Time = 0.99 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.86 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=-\frac {3 a d (\cos (e)-i \sin (e)) (\cos (f x)-i \sin (f x)) (-7 i+\tan (e+f x)) (a+i a \tan (e+f x))^{2/3}}{4 f \sqrt [3]{d \sec (e+f x)}} \] Input:
Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]
Output:
(-3*a*d*(Cos[e] - I*Sin[e])*(Cos[f*x] - I*Sin[f*x])*(-7*I + Tan[e + f*x])* (a + I*a*Tan[e + f*x])^(2/3))/(4*f*(d*Sec[e + f*x])^(1/3))
Time = 0.42 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 3975, 3042, 3974}
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.
\(\displaystyle \int (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}dx\) |
\(\Big \downarrow \) 3975 |
\(\displaystyle \frac {3}{2} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3}{2} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\) |
\(\Big \downarrow \) 3974 |
\(\displaystyle \frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\) |
Input:
Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3),x]
Output:
(((9*I)/2)*a^2*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((3*I)/4)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(2/3))/f
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^ (n - 1)/(f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && EqQ[Simplify[m/2 + n - 1], 0]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] && IGtQ[Simplify[m/2 + n - 1], 0] && !Inte gerQ[n]
\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {5}{3}}d x\]
Input:
int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)
Output:
int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.72 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=-\frac {3 \cdot 2^{\frac {1}{3}} {\left (-4 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a\right )} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}}}{2 \, f} \] Input:
integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="fric as")
Output:
-3/2*2^(1/3)*(-4*I*a*e^(2*I*f*x + 2*I*e) - 3*I*a)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)/f
Timed out. \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))**(2/3)*(a+I*a*tan(f*x+e))**(5/3),x)
Output:
Timed out
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (61) = 122\).
Time = 0.21 (sec) , antiderivative size = 316, normalized size of antiderivative = 3.90 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=\frac {3 \, {\left ({\left (-i \cdot 2^{\frac {1}{3}} a \cos \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - 2^{\frac {1}{3}} a \sin \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} \sqrt {\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} a^{\frac {2}{3}} d^{\frac {2}{3}} + 4 \, {\left ({\left (i \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} + i \cdot 2^{\frac {1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 i \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) + i \cdot 2^{\frac {1}{3}} a\right )} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + {\left (2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right )^{2} + 2^{\frac {1}{3}} a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \cdot 2^{\frac {1}{3}} a \cos \left (2 \, f x + 2 \, e\right ) + 2^{\frac {1}{3}} a\right )} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} a^{\frac {2}{3}} d^{\frac {2}{3}}\right )}}{2 \, {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {7}{6}} f} \] Input:
integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="maxi ma")
Output:
3/2*((-I*2^(1/3)*a*cos(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) ) - 2^(1/3)*a*sin(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sq rt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)*a^(2/ 3)*d^(2/3) + 4*((I*2^(1/3)*a*cos(2*f*x + 2*e)^2 + I*2^(1/3)*a*sin(2*f*x + 2*e)^2 + 2*I*2^(1/3)*a*cos(2*f*x + 2*e) + I*2^(1/3)*a)*cos(1/3*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (2^(1/3)*a*cos(2*f*x + 2*e)^2 + 2^ (1/3)*a*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a*cos(2*f*x + 2*e) + 2^(1/3)*a)*sin (1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*a^(2/3)*d^(2/3))/(( cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(7/6)*f)
\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{3}} \,d x } \] Input:
integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x, algorithm="giac ")
Output:
integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(5/3), x)
Time = 1.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=\frac {3\,a\,{\left (\frac {d}{2\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left ({\cos \left (e+f\,x\right )}^2\,6{}\mathrm {i}+3\,\sin \left (2\,e+2\,f\,x\right )+1{}\mathrm {i}\right )\,{\left (\frac {a\,\left (2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}\right )}^{2/3}}{4\,f} \] Input:
int((d/cos(e + f*x))^(2/3)*(a + a*tan(e + f*x)*1i)^(5/3),x)
Output:
(3*a*(d/(2*cos(e/2 + (f*x)/2)^2 - 1))^(2/3)*(3*sin(2*e + 2*f*x) + cos(e + f*x)^2*6i + 1i)*((a*(sin(2*e + 2*f*x)*1i + 2*cos(e + f*x)^2))/(2*cos(e + f *x)^2))^(2/3))/(4*f)
\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx=d^{\frac {2}{3}} a^{\frac {5}{3}} \left (\left (\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}} \tan \left (f x +e \right )d x \right ) i +\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}}d x \right ) \] Input:
int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3),x)
Output:
d**(2/3)*a**(2/3)*a*(int(sec(e + f*x)**(2/3)*(tan(e + f*x)*i + 1)**(2/3)*t an(e + f*x),x)*i + int(sec(e + f*x)**(2/3)*(tan(e + f*x)*i + 1)**(2/3),x))