3.5.54 \(\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{7/2}} \, dx\) [454]

Optimal. Leaf size=71 \[ -\frac {2 b}{5 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2}}-\frac {8 b}{5 a^3 f \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)}} \]

[Out]

-2/5*b/a/f/(a*sin(f*x+e))^(5/2)/(b*sec(f*x+e))^(1/2)-8/5*b/a^3/f/(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2664, 2658} \begin {gather*} -\frac {8 b}{5 a^3 f \sqrt {a \sin (e+f x)} \sqrt {b \sec (e+f x)}}-\frac {2 b}{5 a f (a \sin (e+f x))^{5/2} \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b)/(5*a*f*Sqrt[b*Sec[e + f*x]]*(a*Sin[e + f*x])^(5/2)) - (8*b)/(5*a^3*f*Sqrt[b*Sec[e + f*x]]*Sqrt[a*Sin[e
+ f*x]])

Rule 2658

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]

Rule 2664

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

Rubi steps

\begin {align*} \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{7/2}} \, dx &=-\frac {2 b}{5 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2}}+\frac {4 \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx}{5 a^2}\\ &=-\frac {2 b}{5 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{5/2}}-\frac {8 b}{5 a^3 f \sqrt {b \sec (e+f x)} \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.14, size = 52, normalized size = 0.73 \begin {gather*} \frac {2 (-3+2 \cos (2 (e+f x))) \cot (e+f x) \sqrt {b \sec (e+f x)}}{5 a^2 f (a \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.24, size = 52, normalized size = 0.73

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (f x +e \right )\right )-5\right ) \cos \left (f x +e \right ) \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right )}{5 f \left (a \sin \left (f x +e \right )\right )^{\frac {7}{2}}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5/f*(4*cos(f*x+e)^2-5)*cos(f*x+e)*(b/cos(f*x+e))^(1/2)*sin(f*x+e)/(a*sin(f*x+e))^(7/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.51, size = 79, normalized size = 1.11 \begin {gather*} -\frac {2 \, {\left (4 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{5 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} - a^{4} f\right )} \sin \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.81, size = 83, normalized size = 1.17 \begin {gather*} -\frac {4\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}\,\left (3\,\cos \left (e+f\,x\right )-4\,\cos \left (3\,e+3\,f\,x\right )+\cos \left (5\,e+5\,f\,x\right )\right )}{5\,a^3\,f\,\sqrt {a\,\sin \left (e+f\,x\right )}\,\left (\cos \left (4\,e+4\,f\,x\right )-4\,\cos \left (2\,e+2\,f\,x\right )+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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