Integrand size = 25, antiderivative size = 132 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2}{3 b f (d \sec (e+f x))^{3/2} (b \tan (e+f x))^{3/2}}-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 d^2 f \sqrt {b \tan (e+f x)}}-\frac {4 \sqrt {b \tan (e+f x)}}{3 b^3 f (d \sec (e+f x))^{3/2}} \]
8/3*(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)*Elliptic F(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/d^2/f/(b*tan(f*x+e))^(1/2)-4/3*(b*tan(f*x+e))^(1/2)/b^3/f/(d*sec(f*x+ e))^(3/2)-2/3/b/f/(d*sec(f*x+e))^(3/2)/(b*tan(f*x+e))^(3/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=\frac {(-3+\cos (2 (e+f x))) \csc (e+f x)-8 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{3/4} \sin (e+f x)}{3 b^2 d f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]
((-3 + Cos[2*(e + f*x)])*Csc[e + f*x] - 8*Hypergeometric2F1[1/4, 3/4, 5/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(3/4)*Sin[e + f*x])/(3*b^2*d*f*Sqrt[d*S ec[e + f*x]]*Sqrt[b*Tan[e + f*x]])
Time = 0.72 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3089, 3042, 3092, 3042, 3096, 3042, 3121, 3042, 3120}
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.
| \(\displaystyle \int \frac {1}{(b \tan (e+f x))^{5/2} (d \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(b \tan (e+f x))^{5/2} (d \sec (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3089 |
| \(\displaystyle -\frac {2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}dx}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \int \frac {1}{(d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}dx}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3092 |
| \(\displaystyle -\frac {2 \left (\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{3 d^2}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{3 d^2}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle -\frac {2 \left (\frac {2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {b \sin (e+f x)}}dx}{3 d^2 \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (\frac {2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {b \sin (e+f x)}}dx}{3 d^2 \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {2 \left (\frac {2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin (e+f x)}}dx}{3 d^2 \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \left (\frac {2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin (e+f x)}}dx}{3 d^2 \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {2 \left (\frac {4 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f \sqrt {b \tan (e+f x)}}+\frac {2 \sqrt {b \tan (e+f x)}}{3 b f (d \sec (e+f x))^{3/2}}\right )}{b^2}-\frac {2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{3/2}}\) |
-2/(3*b*f*(d*Sec[e + f*x])^(3/2)*(b*Tan[e + f*x])^(3/2)) - (2*((4*Elliptic F[(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))))/b^2
3.4.33.3.1 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*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[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]
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 = 2.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.96
| method | result | size |
| default | \(\frac {\left (-4 i \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-4 i \sec \left (f x +e \right ) \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+\cos \left (f x +e \right ) \cot \left (f x +e \right ) \sqrt {2}-2 \csc \left (f x +e \right ) \sqrt {2}\right ) \sqrt {2}}{3 f \sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b^{2} d}\) | \(259\) |
1/3/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)/b^2/d*(-4*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))-4*I*sec(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))+cos(f*x+e)*cot(f*x+e)*2^(1/2)-2 *csc(f*x+e)*2^(1/2))*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.16 \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (2 \, \sqrt {-2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 2 \, \sqrt {2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + {\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}\right )}}{3 \, {\left (b^{3} d^{2} f \cos \left (f x + e\right )^{2} - b^{3} d^{2} f\right )}} \]
-2/3*(2*sqrt(-2*I*b*d)*(cos(f*x + e)^2 - 1)*weierstrassPInverse(4, 0, cos( f*x + e) + I*sin(f*x + e)) + 2*sqrt(2*I*b*d)*(cos(f*x + e)^2 - 1)*weierstr assPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)) + (cos(f*x + e)^4 - 2*cos (f*x + e)^2)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(b^3* d^2*f*cos(f*x + e)^2 - b^3*d^2*f)
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{3/2} (b \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]