\(\int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx\) [331]

Optimal result

Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \]

[Out]

4/3*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))
*(d*sec(f*x+e))^(1/2)*sin(f*x+e)^(1/2)/b^2/f/(b*tan(f*x+e))^(1/2)-2/3*(d*sec(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^
(3/2)

Rubi [A] (verified)

Time = 0.13 (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 = {2689, 2696, 2721, 2720} \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {4 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}}-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

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

Rule 2689

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

Rule 2696

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

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 b^2} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}} \\ & = -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.82 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 d^2 \left (\csc ^2(e+f x)+2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

(-2*d^2*(Csc[e + f*x]^2 + 2*Hypergeometric2F1[1/4, 3/4, 5/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(3/4))*Sqrt[b*T
an[e + f*x]])/(3*b^3*f*(d*Sec[e + f*x])^(3/2))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.07 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.52

method	result	size
default	\(-\frac {\sqrt {d \sec \left (f x +e \right )}\, \left (2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+2 i \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \cot \left (f x +e \right )\right ) \sqrt {2}}{3 f \sqrt {b \tan \left (f x +e \right )}\, b^{2}}\)	\(239\)

[In]

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

[Out]

-1/3/f*(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)/b^2*(2*I*cos(f*x+e)*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*
(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(
1/2),1/2*2^(1/2))+2*I*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(-I*(cot(f*x+e
)-csc(f*x+e)))^(1/2)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))+2^(1/2)*cot(f*x+e))*2^(1/2)

Fricas [C] (verification not implemented)

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

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.43 \[ \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - \sqrt {-2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - \sqrt {2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \]

[In]

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

[Out]

2/3*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 - sqrt(-2*I*b*d)*(cos(f*x + e)^2 -
1)*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) - sqrt(2*I*b*d)*(cos(f*x + e)^2 - 1)*weierstrassPI
nverse(4, 0, cos(f*x + e) - I*sin(f*x + e)))/(b^3*f*cos(f*x + e)^2 - b^3*f)

Sympy [F]

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

[In]

integrate((d*sec(f*x+e))**(1/2)/(b*tan(f*x+e))**(5/2),x)

[Out]

Integral(sqrt(d*sec(e + f*x))/(b*tan(e + f*x))**(5/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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