Integrand size = 25, antiderivative size = 55 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \]
-2*(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)*EllipticE (cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/f/(d*sec(f*x+e))^ (1/2)/sin(f*x+e)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.52 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} (b \tan (e+f x))^{3/2}}{3 b f \sqrt {d \sec (e+f x)}} \]
(2*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(1/4 )*(b*Tan[e + f*x])^(3/2))/(3*b*f*Sqrt[d*Sec[e + f*x]])
Time = 0.36 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3096 |
\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{\sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{\sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{\sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}\) |
(2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])
3.3.94.3.1 Defintions of rubi rules used
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 = 2.74 (sec) , antiderivative size = 342, normalized size of antiderivative = 6.22
method | result | size |
default | \(-\frac {\sqrt {-\frac {b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}}\, \left (2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}\right ) \sin \left (f x +e \right ) \sqrt {2}}{f \sqrt {-\frac {d \left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right )}\) | \(342\) |
risch | \(-\frac {i \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}+\frac {i \left (\frac {2 i \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}{b d \sqrt {{\mathrm e}^{i \left (f x +e \right )} \left (-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d b \right )}}-\frac {\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}\, \sqrt {-2 \,{\mathrm e}^{i \left (f x +e \right )}+2}\, \sqrt {-{\mathrm e}^{i \left (f x +e \right )}}\, \left (-2 E\left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )+F\left (\sqrt {{\mathrm e}^{i \left (f x +e \right )}+1}, \frac {\sqrt {2}}{2}\right )\right )}{\sqrt {-i b d \,{\mathrm e}^{3 i \left (f x +e \right )}+i d b \,{\mathrm e}^{i \left (f x +e \right )}}}\right ) \sqrt {2}\, \sqrt {-\frac {i b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {-i d \,{\mathrm e}^{i \left (f x +e \right )} b \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}}{f \sqrt {\frac {d \,{\mathrm e}^{i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}\) | \(371\) |
-1/f*(-b/(csc(f*x+e)^2*(1-cos(f*x+e))^2-1)*(csc(f*x+e)-cot(f*x+e)))^(1/2)* (2*(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*2^(1/2)*(-I*(cot(f*x+e)-csc(f*x+e) +I))^(1/2)*(I*(csc(f*x+e)-cot(f*x+e)))^(1/2)*EllipticE((-I*(I-cot(f*x+e)+c sc(f*x+e)))^(1/2),1/2*2^(1/2))-(-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2)*2^(1/2 )*(-I*(cot(f*x+e)-csc(f*x+e)+I))^(1/2)*(I*(csc(f*x+e)-cot(f*x+e)))^(1/2)*E llipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))-2*csc(f*x+e)^2* (1-cos(f*x+e))^2)/(-d*(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)/(csc(f*x+e)^2*(1-c os(f*x+e))^2-1))^(1/2)/(1-cos(f*x+e))*sin(f*x+e)*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\frac {i \, \sqrt {-2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{d f} \]
(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, weierstras sPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d*f)
\[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {\sqrt {b \tan {\left (e + f x \right )}}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
\[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\int { \frac {\sqrt {b \tan \left (f x + e\right )}}{\sqrt {d \sec \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx=\int \frac {\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]