3.313 \(\int \frac{(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=34 \[ \frac{2 (b \tan (e+f x))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}} \]

[Out]

(2*(b*Tan[e + f*x])^(7/2))/(7*b*f*(d*Sec[e + f*x])^(7/2))

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Rubi [A]  time = 0.0549945, 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} \[ \frac{2 (b \tan (e+f x))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*(b*Tan[e + f*x])^(7/2))/(7*b*f*(d*Sec[e + f*x])^(7/2))

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))^{5/2}}{(d \sec (e+f x))^{7/2}} \, dx &=\frac{2 (b \tan (e+f x))^{7/2}}{7 b f (d \sec (e+f x))^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.157927, size = 45, normalized size = 1.32 \[ \frac{2 b^2 \sin ^3(e+f x) \sqrt{b \tan (e+f x)}}{7 d^3 f \sqrt{d \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[e + f*x])^(5/2)/(d*Sec[e + f*x])^(7/2),x]

[Out]

(2*b^2*Sin[e + f*x]^3*Sqrt[b*Tan[e + f*x]])/(7*d^3*f*Sqrt[d*Sec[e + f*x]])

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Maple [A]  time = 0.149, size = 50, normalized size = 1.5 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) }{7\,f\cos \left ( fx+e \right ) } \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{{\frac{5}{2}}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x)

[Out]

2/7/f*sin(f*x+e)*(b*sin(f*x+e)/cos(f*x+e))^(5/2)/cos(f*x+e)/(d/cos(f*x+e))^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(7/2), x)

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Fricas [B]  time = 1.71274, size = 165, normalized size = 4.85 \begin{align*} -\frac{2 \,{\left (b^{2} \cos \left (f x + e\right )^{3} - b^{2} \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 )}} \sin \left (f x + e\right )}{7 \, d^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

-2/7*(b^2*cos(f*x + e)^3 - b^2*cos(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*sin(f*x +
e)/(d^4*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((b*tan(f*x+e))**(5/2)/(d*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (b \tan \left (f x + e\right )\right )^{\frac{5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*tan(f*x+e))^(5/2)/(d*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e))^(5/2)/(d*sec(f*x + e))^(7/2), x)