3.332 \(\int \frac{1}{\sqrt{d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac{8 \sqrt{b \tan (e+f x)}}{3 b^3 f \sqrt{d \sec (e+f x)}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}} \]

[Out]

-2/(3*b*f*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2)) - (8*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*Sqrt[d*Sec[e + f*x]
])

________________________________________________________________________________________

Rubi [A]  time = 0.10089, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {2609, 2605} \[ -\frac{8 \sqrt{b \tan (e+f x)}}{3 b^3 f \sqrt{d \sec (e+f x)}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)),x]

[Out]

-2/(3*b*f*Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(3/2)) - (8*Sqrt[b*Tan[e + f*x]])/(3*b^3*f*Sqrt[d*Sec[e + f*x]
])

Rule 2609

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*(n + 1)), x] - Dist[(m + n + 1)/(b^2*(n + 1)), Int[(a*Sec[e + f*x])^m*(
b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && IntegersQ[2*m, 2*n]

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{1}{\sqrt{d \sec (e+f x)} (b \tan (e+f x))^{5/2}} \, dx &=-\frac{2}{3 b f \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac{4 \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac{2}{3 b f \sqrt{d \sec (e+f x)} (b \tan (e+f x))^{3/2}}-\frac{8 \sqrt{b \tan (e+f x)}}{3 b^3 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.851484, size = 110, normalized size = 1.59 \[ -\frac{2 \left (3 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)+1} \sqrt{\sec (e+f x)}+\sqrt{\frac{1}{\cos (e+f x)+1}} \csc (e+f x) \sec (e+f x)\right )}{3 b^2 f \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[d*Sec[e + f*x]]*(b*Tan[e + f*x])^(5/2)),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.177, size = 62, normalized size = 0.9 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-4 \right ) }{3\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{\frac{1}{\sqrt{{\frac{d}{\cos \left ( fx+e \right ) }}}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/f*sin(f*x+e)*(3*cos(f*x+e)^2-4)/cos(f*x+e)^3/(d/cos(f*x+e))^(1/2)/(b*sin(f*x+e)/cos(f*x+e))^(5/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \sec \left (f x + e\right )} \left (b \tan \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(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.74557, size = 178, normalized size = 2.58 \begin{align*} -\frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{3} - 4 \, \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 )}}}{3 \,{\left (b^{3} d f \cos \left (f x + e\right )^{2} - b^{3} d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-2/3*(3*cos(f*x + e)^3 - 4*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))/(b^3*d*f*cos(f
*x + e)^2 - b^3*d*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(1/(d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d \sec \left (f x + e\right )} \left (b \tan \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(1/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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