3.5.59 \(\int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx\) [459]

Optimal. Leaf size=95 \[ -\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {2 F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}} \]

[Out]

-2/3*b/a/f/(a*sin(f*x+e))^(3/2)/(b*sec(f*x+e))^(1/2)-2/3*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f*x)*Ellipti
cF(cos(e+1/4*Pi+f*x),2^(1/2))*(b*sec(f*x+e))^(1/2)*sin(2*f*x+2*e)^(1/2)/a^2/f/(a*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.11, antiderivative size = 95, 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 = {2664, 2665, 2653, 2720} \begin {gather*} \frac {2 \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)}}{3 a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \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])^(5/2),x]

[Out]

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

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 {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx &=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {2 \int \frac {\sqrt {b \sec (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{3 a^2}\\ &=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {\left (2 \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}{3 a^2}\\ &=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {\left (2 \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}{3 a^2 \sqrt {a \sin (e+f x)}}\\ &=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {2 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)}}{3 a^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 = 10.33, size = 75, normalized size = 0.79 \begin {gather*} \frac {2 \cot (e+f x) \sqrt {b \sec (e+f x)} \left (-1+\, _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 )}{3 a^2 f \sqrt {a \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(277\) vs. \(2(106)=212\).
time = 0.24, size = 278, normalized size = 2.93

method	result	size
default	\(-\frac {\left (-2 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {\sin \left (f x +e \right )-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )+\cos \left (f x +e \right ) \sqrt {2}\right ) \sin \left (f x +e \right ) \sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \sqrt {2}}{3 f \left (a \sin \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*(-2*cos(f*x+e)*sin(f*x+e)*((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))-2*((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^(1/2))*sin(f*x+e)*(b/cos(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2)*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((b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(5/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.12, size = 133, normalized size = 1.40 \begin {gather*} -\frac {2 \, {\left (\sqrt {i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm ellipticF}\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ), -1\right ) + \sqrt {-i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm ellipticF}\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ), -1\right ) - \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )\right )}}{3 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - a^{3} f\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))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6438 deep

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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))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(f*x + e))/(a*sin(f*x + e))^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \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))^(5/2),x)

[Out]

int((b/cos(e + f*x))^(1/2)/(a*sin(e + f*x))^(5/2), x)

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