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

Optimal. Leaf size=34 \[ \frac{2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0570212, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2605} \[ \frac{2 (b \tan (e+f x))^{5/2}}{5 b f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[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 2605

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*} \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}}\\ \end{align*}

Mathematica [B]  time = 1.33815, size = 141, normalized size = 4.15 \[ -\frac{b \sec ^{\frac{3}{2}}(e+f x) \sqrt{b \tan (e+f x)} \left (-\sqrt{\sec (e+f x)+1} \sec ^2\left (\frac{1}{2} (e+f x)\right )+\sqrt{\frac{1}{\cos (e+f x)+1}} \cos (3 (e+f x)) \sec ^{\frac{3}{2}}(e+f x)+\sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{\sec (e+f x)}\right )}{10 f \sqrt{\frac{1}{\cos (e+f x)+1}} (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Sec[e + f*x]^(3/2)*(Sqrt[(1 + Cos[e + f*x])^(-1)]*Sqrt[Sec[e + f*x]] + Sqrt[(1 + Cos[e + f*x])^(-1)]*Cos[3
*(e + f*x)]*Sec[e + f*x]^(3/2) - Sec[(e + f*x)/2]^2*Sqrt[1 + Sec[e + f*x]])*Sqrt[b*Tan[e + f*x]])/(10*f*Sqrt[(
1 + Cos[e + f*x])^(-1)]*(d*Sec[e + f*x])^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.151, size = 50, normalized size = 1.5 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) }{5\,f\cos \left ( fx+e \right ) } \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{3}{2}}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Fricas [B]  time = 2.01878, size = 142, normalized size = 4.18 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)