3.476 \(\int \frac{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=35 \[ -\frac{2 b}{5 a f (a \sin (e+f x))^{5/2} (b \sec (e+f x))^{5/2}} \]

[Out]

(-2*b)/(5*a*f*(b*Sec[e + f*x])^(5/2)*(a*Sin[e + f*x])^(5/2))

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Rubi [A]  time = 0.0575665, antiderivative size = 35, 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 = {2578} \[ -\frac{2 b}{5 a f (a \sin (e+f x))^{5/2} (b \sec (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(7/2)),x]

[Out]

(-2*b)/(5*a*f*(b*Sec[e + f*x])^(5/2)*(a*Sin[e + f*x])^(5/2))

Rule 2578

Int[((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 1)*(b*Sec[e + f*x])^(n - 1))/(a*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m - n + 2,
0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{(b \sec (e+f x))^{3/2} (a \sin (e+f x))^{7/2}} \, dx &=-\frac{2 b}{5 a f (b \sec (e+f x))^{5/2} (a \sin (e+f x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.109555, size = 45, normalized size = 1.29 \[ -\frac{2 \cot ^3(e+f x) \sqrt{a \sin (e+f x)} \sqrt{b \sec (e+f x)}}{5 a^4 b^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((b*Sec[e + f*x])^(3/2)*(a*Sin[e + f*x])^(7/2)),x]

[Out]

(-2*Cot[e + f*x]^3*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e + f*x]])/(5*a^4*b^2*f)

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Maple [A]  time = 0.085, size = 40, normalized size = 1.1 \begin{align*} -{\frac{2\,\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) }{5\,f} \left ({\frac{b}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}} \left ( a\sin \left ( fx+e \right ) \right ) ^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(7/2),x)

[Out]

-2/5/f*sin(f*x+e)*cos(f*x+e)/(b/cos(f*x+e))^(3/2)/(a*sin(f*x+e))^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (a \sin \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(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(7/2)), x)

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Fricas [B]  time = 2.46116, size = 157, normalized size = 4.49 \begin{align*} \frac{2 \, \sqrt{a \sin \left (f x + e\right )} \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{3}}{5 \,{\left (a^{4} b^{2} f \cos \left (f x + e\right )^{2} - a^{4} b^{2} f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

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

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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/(b*sec(f*x+e))**(3/2)/(a*sin(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{1}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}} \left (a \sin \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(1/(b*sec(f*x+e))^(3/2)/(a*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(f*x + e))^(3/2)*(a*sin(f*x + e))^(7/2)), x)