\(\int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx\) [225]

Optimal result

Integrand size = 29, antiderivative size = 29 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\text {Int}\left (\frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}},x\right ) \]

[Out]

Unintegrable((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x)

Rubi [N/A]

Not integrable

Time = 0.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx \]

[In]

Int[(a + b*Sec[e + f*x])^(2/3)/(c + d*Sec[e + f*x])^(5/3),x]

[Out]

Defer[Int][(a + b*Sec[e + f*x])^(2/3)/(c + d*Sec[e + f*x])^(5/3), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx \\ \end{align*}

Mathematica [N/A]

Not integrable

Time = 89.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx \]

[In]

Integrate[(a + b*Sec[e + f*x])^(2/3)/(c + d*Sec[e + f*x])^(5/3),x]

[Out]

Integrate[(a + b*Sec[e + f*x])^(2/3)/(c + d*Sec[e + f*x])^(5/3), x]

Maple [N/A] (verified)

Not integrable

Time = 0.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86

\[\int \frac {\left (a +b \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (c +d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}d x\]

[In]

int((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x)

[Out]

int((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x)

Fricas [F(-1)]

Timed out. \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x, algorithm="fricas")

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 61.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int \frac {\left (a + b \sec {\left (e + f x \right )}\right )^{\frac {2}{3}}}{\left (c + d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}\, dx \]

[In]

integrate((a+b*sec(f*x+e))**(2/3)/(c+d*sec(f*x+e))**(5/3),x)

[Out]

Integral((a + b*sec(e + f*x))**(2/3)/(c + d*sec(e + f*x))**(5/3), x)

Maxima [N/A]

Not integrable

Time = 1.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int { \frac {{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {2}{3}}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{\frac {5}{3}}} \,d x } \]

[In]

integrate((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e) + a)^(2/3)/(d*sec(f*x + e) + c)^(5/3), x)

Giac [N/A]

Not integrable

Time = 2.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int { \frac {{\left (b \sec \left (f x + e\right ) + a\right )}^{\frac {2}{3}}}{{\left (d \sec \left (f x + e\right ) + c\right )}^{\frac {5}{3}}} \,d x } \]

[In]

integrate((a+b*sec(f*x+e))^(2/3)/(c+d*sec(f*x+e))^(5/3),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e) + a)^(2/3)/(d*sec(f*x + e) + c)^(5/3), x)

Mupad [N/A]

Not integrable

Time = 107.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {(a+b \sec (e+f x))^{2/3}}{(c+d \sec (e+f x))^{5/3}} \, dx=\int \frac {{\left (a+\frac {b}{\cos \left (e+f\,x\right )}\right )}^{2/3}}{{\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}} \,d x \]

[In]

int((a + b/cos(e + f*x))^(2/3)/(c + d/cos(e + f*x))^(5/3),x)

[Out]

int((a + b/cos(e + f*x))^(2/3)/(c + d/cos(e + f*x))^(5/3), x)