Integrand size = 25, antiderivative size = 91 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=-\frac {2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {4 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}} \]
-2/b/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)+4*(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)/b^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 = 0.64 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \left (3+2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/4} \sin ^2(e+f x)\right )}{3 b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]
(-2*(3 + 2*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x] ^2)^(5/4)*Sin[e + f*x]^2))/(3*b*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x] ])
Time = 0.50 (sec) , antiderivative size = 91, 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, 3089, 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.
\(\displaystyle \int \frac {1}{(b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(b \tan (e+f x))^{3/2} \sqrt {d \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3089 |
\(\displaystyle -\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3096 |
\(\displaystyle -\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{b^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{b^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle -\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{b^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{b^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle -\frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}}\) |
-2/(b*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]]) - (4*EllipticE[(e - Pi/ 2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(b^2*f*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[ e + f*x]])
3.4.25.3.1 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*(n + 1))), x] - Simp[(m + n + 1)/(b^2*(n + 1)) Int[(a*Sec[e + f*x])^m*(b*Ta n[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && I ntegersQ[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 = 2.24 (sec) , antiderivative size = 365, normalized size of antiderivative = 4.01
method | result | size |
default | \(\frac {\csc \left (f x +e \right ) \left (1-\cos \left (f x +e \right )\right ) \left (4 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-2 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {2}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1\right ) \sqrt {2}}{f {\left (-\frac {b \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}\right )}^{\frac {3}{2}} {\left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1\right )}^{2} \sqrt {-\frac {d \left (\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}+1\right )}{\left (\csc ^{2}\left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2}-1}}}\) | \(365\) |
1/f*csc(f*x+e)*(1-cos(f*x+e))*(4*(-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)+csc(f*x+e)))^(1/2),1/2*2^(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)*EllipticF((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1 /2),1/2*2^(1/2))-3*csc(f*x+e)^2*(1-cos(f*x+e))^2-1)/(-b/(csc(f*x+e)^2*(1-c os(f*x+e))^2-1)*(csc(f*x+e)-cot(f*x+e)))^(3/2)/(csc(f*x+e)^2*(1-cos(f*x+e) )^2-1)^2/(-d*(csc(f*x+e)^2*(1-cos(f*x+e))^2+1)/(csc(f*x+e)^2*(1-cos(f*x+e) )^2-1))^(1/2)*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.42 \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + i \, \sqrt {-2 i \, b d} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - i \, \sqrt {2 i \, b d} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )\right )}}{b^{2} d f \sin \left (f x + e\right )} \]
-2*(sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 + I*sqrt(-2*I*b*d)*sin(f*x + e)*weierstrassZeta(4, 0, weierstrassPInverse( 4, 0, cos(f*x + e) + I*sin(f*x + e))) - I*sqrt(2*I*b*d)*sin(f*x + e)*weier strassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))) )/(b^2*d*f*sin(f*x + e))
\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}} \sqrt {d \sec {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {d \sec \left (f x + e\right )} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\sqrt {d \sec (e+f x)} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]