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

Optimal. Leaf size=72 \[ \frac{8 \sqrt{b \tan (e+f x)}}{5 b d^2 f \sqrt{d \sec (e+f x)}}+\frac{2 \sqrt{b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}} \]

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(5*b*f*(d*Sec[e + f*x])^(5/2)) + (8*Sqrt[b*Tan[e + f*x]])/(5*b*d^2*f*Sqrt[d*Sec[e + f
*x]])

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Rubi [A]  time = 0.0973784, antiderivative size = 72, 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 = {2612, 2605} \[ \frac{8 \sqrt{b \tan (e+f x)}}{5 b d^2 f \sqrt{d \sec (e+f x)}}+\frac{2 \sqrt{b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*Tan[e + f*x]])/(5*b*f*(d*Sec[e + f*x])^(5/2)) + (8*Sqrt[b*Tan[e + f*x]])/(5*b*d^2*f*Sqrt[d*Sec[e + f
*x]])

Rule 2612

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] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[
e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integer
sQ[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}{(d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx &=\frac{2 \sqrt{b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}+\frac{4 \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{5 d^2}\\ &=\frac{2 \sqrt{b \tan (e+f x)}}{5 b f (d \sec (e+f x))^{5/2}}+\frac{8 \sqrt{b \tan (e+f x)}}{5 b d^2 f \sqrt{d \sec (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.15039, size = 112, normalized size = 1.56 \[ \frac{\sqrt{\frac{1}{\cos (e+f x)+1}} \cos (2 (e+f x)) \tan (e+f x)+9 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)} \sqrt{\sec (e+f x)+1}}{5 d^2 f \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.17, size = 60, normalized size = 0.8 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{2}+4 \right ) }{5\,f \left ( \cos \left ( fx+e \right ) \right ) ^{3}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{{\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.696, size = 140, normalized size = 1.94 \begin{align*} \frac{2 \,{\left (\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 )}}}{5 \, b d^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(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*d^3*f)

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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))**(5/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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