\(\int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx\) [119]

Optimal result

Integrand size = 25, antiderivative size = 86 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2679, 2681, 2720} \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}}-\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \]

[In]

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

[Out]

-(b/(a^2*f*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2]*Sqrt[b*
Tan[e + f*x]])/(a^2*f*Sqrt[a*Sin[e + f*x]])

Rule 2679

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

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, 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 {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {\int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {a \sin (e+f x)}} \, dx}{2 a^2} \\ & = -\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {\left (\sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{2 a^2 \sqrt {a \sin (e+f x)}} \\ & = -\frac {b}{a^2 f \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}}+\frac {\sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {b \tan (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {b \left (-\sqrt [4]{\cos ^2(e+f x)}+\operatorname {EllipticF}\left (\frac {1}{2} \arcsin (\sin (e+f x)),2\right ) \sin (e+f x)\right )}{a^2 f \sqrt [4]{\cos ^2(e+f x)} \sqrt {a \sin (e+f x)} \sqrt {b \tan (e+f x)}} \]

[In]

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

[Out]

(b*(-(Cos[e + f*x]^2)^(1/4) + EllipticF[ArcSin[Sin[e + f*x]]/2, 2]*Sin[e + f*x]))/(a^2*f*(Cos[e + f*x]^2)^(1/4
)*Sqrt[a*Sin[e + f*x]]*Sqrt[b*Tan[e + f*x]])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.78

method	result	size
default	\(\frac {\sqrt {b \tan \left (f x +e \right )}\, \left (i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )+i \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right )-\cot \left (f x +e \right )\right )}{f \,a^{2} \sqrt {\sin \left (f x +e \right ) a}}\)	\(153\)

[In]

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

[Out]

1/f*(b*tan(f*x+e))^(1/2)/a^2/(sin(f*x+e)*a)^(1/2)*(I*(1/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)*(cos(f*x+e)/(cos(f*x+
e)+1))^(1/2)*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)+I*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/
2)*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)-cot(f*x+e))

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.09 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.71 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{5/2}} \, dx=\frac {{\left (\sqrt {2} \cos \left (f x + e\right )^{2} - \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + {\left (\sqrt {2} \cos \left (f x + e\right )^{2} - \sqrt {2}\right )} \sqrt {-a b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} - a^{3} f\right )}} \]

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/2*((sqrt(2)*cos(f*x + e)^2 - sqrt(2))*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e)) +
(sqrt(2)*cos(f*x + e)^2 - sqrt(2))*sqrt(-a*b)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*s
qrt(a*sin(f*x + e))*sqrt(b*sin(f*x + e)/cos(f*x + e))*cos(f*x + e))/(a^3*f*cos(f*x + e)^2 - a^3*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate((b*tan(f*x+e))^(1/2)/(a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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