3.5.79 \(\int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx\) [479]

Optimal. Leaf size=94 \[ \frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{2 b^2 f \sqrt {a \sin (e+f x)}} \]

[Out]

(a*sin(f*x+e))^(1/2)/a/b/f/(b*sec(f*x+e))^(1/2)-1/2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*EllipticF(co
s(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/b^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2662, 2665, 2653, 2720} \begin {gather*} \frac {\sqrt {\sin (2 e+2 f x)} F\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {b \sec (e+f x)}}{2 b^2 f \sqrt {a \sin (e+f x)}}+\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[a*Sin[e + f*x]]/(a*b*f*Sqrt[b*Sec[e + f*x]]) + (EllipticF[e - Pi/4 + f*x, 2]*Sqrt[b*Sec[e + f*x]]*Sqrt[Si
n[2*e + 2*f*x]])/(2*b^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2662

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

Rule 2665

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

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {1}{(b \sec (e+f x))^{3/2} \sqrt {a \sin (e+f x)}} \, dx &=\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {\int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{2 b^2}\\ &=\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {\left (\sqrt {b \cos (e+f x)} \sqrt {b \sec (e+f x)}\right ) \int \frac {1}{\sqrt {b \cos (e+f x)} \sqrt {a \sin (e+f x)}} \, dx}{2 b^2}\\ &=\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {\left (\sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}\right ) \int \frac {1}{\sqrt {\sin (2 e+2 f x)}} \, dx}{2 b^2 \sqrt {a \sin (e+f x)}}\\ &=\frac {\sqrt {a \sin (e+f x)}}{a b f \sqrt {b \sec (e+f x)}}+\frac {F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{2 b^2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.41, size = 84, normalized size = 0.89 \begin {gather*} -\frac {\cot (e+f x) \sqrt {b \sec (e+f x)} \left (-1+\cos (2 (e+f x))-\, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{2 b^2 f \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Cot[e + f*x]*Sqrt[b*Sec[e + f*x]]*(-1 + Cos[2*(e + f*x)] - Hypergeometric2F1[1/2, 3/4, 3/2, Sec[e + f*x]
^2]*(-Tan[e + f*x]^2)^(3/4)))/(b^2*f*Sqrt[a*Sin[e + f*x]])

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Maple [A]
time = 0.27, size = 190, normalized size = 2.02

method	result	size
default	\(-\frac {\left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )-\left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}+\cos \left (f x +e \right ) \sqrt {2}\right ) \sin \left (f x +e \right ) \sqrt {2}}{2 f \left (-1+\cos \left (f x +e \right )\right ) \cos \left (f x +e \right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {a \sin \left (f x +e \right )}}\)	\(190\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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

[Out]

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

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

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sqrt {a\,\sin \left (e+f\,x\right )}\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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