Integrand size = 25, antiderivative size = 95 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {4 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}} \]
-4/5*(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)*Ellipti cE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/d^2/f/(d*sec(f* x+e))^(1/2)/sin(f*x+e)^(1/2)+2/5*(b*tan(f*x+e))^(3/2)/b/f/(d*sec(f*x+e))^( 5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.92 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.74 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \left (3+2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/4}\right ) (b \tan (e+f x))^{3/2}}{15 b f (d \sec (e+f x))^{5/2}} \]
(2*(3 + 2*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^ 2)^(5/4))*(b*Tan[e + f*x])^(3/2))/(15*b*f*(d*Sec[e + f*x])^(5/2))
Time = 0.50 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3092, 3042, 3096, 3042, 3121, 3042, 3119}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3092 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\) |
(4*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqrt[d* Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) + (2*(b*Tan[e + f*x])^(3/2))/(5*b*f*(d*S ec[e + f*x])^(5/2))
3.3.96.3.1 Defintions of rubi rules used
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] + Simp[(m + n + 1)/(a^2*m) Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (LtQ[m, -1] || (EqQ[m, -1 ] && EqQ[n, -2^(-1)])) && IntegersQ[2*m, 2*n]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[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]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Result contains complex when optimal does not.
Time = 2.90 (sec) , antiderivative size = 454, normalized size of antiderivative = 4.78
| method | result | size |
| default | \(-\frac {\csc \left (f x +e \right ) \left (-2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+4 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+\sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right )-2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+4 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \cos \left (f x +e \right )-2 \sqrt {2}\right ) \sqrt {b \tan \left (f x +e \right )}\, \sqrt {2}}{5 f \sqrt {d \sec \left (f x +e \right )}\, d^{2}}\) | \(454\) |
-1/5/f*csc(f*x+e)*(-2*(-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))*cos(f*x+e)+4*(-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)-cs c(f*x+e)))^(1/2)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2 ))*cos(f*x+e)+2^(1/2)*cos(f*x+e)^3-2*(-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)*El lipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))+4*(-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)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^ (1/2))+2^(1/2)*cos(f*x+e)-2*2^(1/2))*(b*tan(f*x+e))^(1/2)/(d*sec(f*x+e))^( 1/2)/d^2*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + i \, \sqrt {-2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{5 \, d^{3} f} \]
2/5*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 *sin(f*x + e) + I*sqrt(-2*I*b*d)*weierstrassZeta(4, 0, weierstrassPInverse (4, 0, cos(f*x + e) + I*sin(f*x + e))) - I*sqrt(2*I*b*d)*weierstrassZeta(4 , 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d^3*f)
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {b \tan {\left (e + f x \right )}}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]