\(\int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx\) [320]

Optimal result

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

[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)/d^2/f/(b*tan(f*x+e))^(1/2)+2/3*(b*tan(f*x+e))^(1/2)/b/f/(d*sec(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 = {2692, 2696, 2721, 2720} \[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {4 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

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

Rule 2692

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*m)), x] + Dist[(m + n + 1)/(a^2*m), Int[(a*Sec[e + f*x])^(m + 2)*(b*Ta
n[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1] && EqQ[n, -2^(-1)])) && Integ
ersQ[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 {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}+\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 d^2} \\ & = \frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (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 d^2 \sqrt {b \tan (e+f x)}} \\ & = \frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (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 d^2 \sqrt {b \tan (e+f x)}} \\ & = \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 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 0.74 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {2 \left (1+2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4}\right ) \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.93 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.52

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

[In]

int(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3/f/(b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^(1/2)/d*(2*I*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(I*(-I-cot(f*x+e)+
csc(f*x+e)))^(1/2)*(I*(csc(f*x+e)-cot(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*sec(f*x+e)*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(I*(-I-cot(f*x+e)+csc(f*x+e)))^(1/2)*(I*(csc(f*x+e)-cot
(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)*sin(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 = 101, normalized size of antiderivative = 1.06 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + \sqrt {-2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + \sqrt {2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, b d^{2} f} \]

[In]

integrate(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),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)*weierstrassPInvers
e(4, 0, cos(f*x + e) + I*sin(f*x + e)) + sqrt(2*I*b*d)*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)
))/(b*d^2*f)

Sympy [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int \frac {1}{\sqrt {b \tan {\left (e + f x \right )}} \left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(d*sec(f*x+e))**(3/2)/(b*tan(f*x+e))**(1/2),x)

[Out]

Integral(1/(sqrt(b*tan(e + f*x))*(d*sec(e + f*x))**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[In]

integrate(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} \sqrt {b \tan \left (f x + e\right )}} \,d x } \]

[In]

integrate(1/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}} \, dx=\int \frac {1}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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