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

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=-\frac {2 b}{3 a f \sqrt {b \sec (e+f x)} (a \sin (e+f x))^{3/2}}+\frac {2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {b \sec (e+f x)} \sqrt {\sin (2 e+2 f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}} \]

[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)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2664, 2665, 2653, 2720} \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right ) \sqrt {b \sec (e+f x)}}{3 a^2 f \sqrt {a \sin (e+f x)}}-\frac {2 b}{3 a f (a \sin (e+f x))^{3/2} \sqrt {b \sec (e+f x)}} \]

[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*} \text {integral}& = -\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 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,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*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.70 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {2 \cot (e+f x) \sqrt {b \sec (e+f x)} \left (-1+\operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{3 a^2 f \sqrt {a \sin (e+f x)}} \]

[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]])

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.18

method	result	size
default	\(-\frac {\sqrt {2}\, \sqrt {b \sec \left (f x +e \right )}\, \left (-2 \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )-2 \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )+1}\, \sqrt {\cot \left (f x +e \right )-\csc \left (f x +e \right )}\, F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \cot \left (f x +e \right )\right )}{3 f \sqrt {a \sin \left (f x +e \right )}\, a^{2}}\)	\(207\)

[In]

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

[Out]

-1/3/f*2^(1/2)*(b*sec(f*x+e))^(1/2)/(a*sin(f*x+e))^(1/2)/a^2*(-2*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-
csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*EllipticF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*cos(
f*x+e)-2*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e)+1)^(1/2)*(cot(f*x+e)-csc(f*x+e))^(1/2)*Ellipt
icF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))+2^(1/2)*cot(f*x+e))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (\sqrt {i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + \sqrt {-i \, a b} {\left (\cos \left (f x + e\right )^{2} - 1\right )} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )\right )}}{3 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - a^{3} f\right )}} \]

[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)*elliptic_f(arcsin(cos(f*x + e) + I*sin(f*x + e)), -1) + sqrt(-I*a*b)*(c
os(f*x + e)^2 - 1)*elliptic_f(arcsin(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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate((b*sec(f*x+e))**(1/2)/(a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b \sec (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{{\left (a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

[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)