\(\int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx\) [304]

Optimal result

Integrand size = 25, antiderivative size = 34 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

[Out]

2/5*(b*tan(f*x+e))^(5/2)/b/f/(d*sec(f*x+e))^(5/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2685} \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

[In]

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

[Out]

(2*(b*Tan[e + f*x])^(5/2))/(5*b*f*(d*Sec[e + f*x])^(5/2))

Rule 2685

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-(a*Sec[e
+ f*x])^m)*((b*Tan[e + f*x])^(n + 1)/(b*f*m)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.26 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 b \sin ^2(e+f x) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)}} \]

[In]

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

[Out]

(2*b*Sin[e + f*x]^2*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d*Sec[e + f*x]])

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12

method	result	size
default	\(\frac {2 \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {b \tan \left (f x +e \right )}\, b}{5 f \,d^{2} \sqrt {d \sec \left (f x +e \right )}}\)	\(38\)

[In]

int((b*tan(f*x+e))^(3/2)/(d*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/5/f*sin(f*x+e)^2*(b*tan(f*x+e))^(1/2)*b/d^2/(d*sec(f*x+e))^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (28) = 56\).

Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.71 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (b \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \]

[In]

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

[Out]

-2/5*(b*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(d^3*f)

Sympy [A] (verification not implemented)

Time = 45.79 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\begin {cases} \frac {2 \left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \tan {\left (e + f x \right )}}{5 f \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}} & \text {for}\: f \neq 0 \\\frac {x \left (b \tan {\left (e \right )}\right )^{\frac {3}{2}}}{\left (d \sec {\left (e \right )}\right )^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b*tan(e + f*x))**(3/2)*tan(e + f*x)/(5*f*(d*sec(e + f*x))**(5/2)), Ne(f, 0)), (x*(b*tan(e))**(3/
2)/(d*sec(e))**(5/2), True))

Maxima [F]

\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.91 \[ \int \frac {(b \tan (e+f x))^{3/2}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {b\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}\,\left (\cos \left (e+f\,x\right )-\cos \left (3\,e+3\,f\,x\right )\right )\,\sqrt {\frac {b\,\sin \left (2\,e+2\,f\,x\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}}{10\,d^3\,f} \]

[In]

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

[Out]

(b*(d/cos(e + f*x))^(1/2)*(cos(e + f*x) - cos(3*e + 3*f*x))*((b*sin(2*e + 2*f*x))/(cos(2*e + 2*f*x) + 1))^(1/2
))/(10*d^3*f)