Integrand size = 25, antiderivative size = 130 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {2}{b f (d \sec (e+f x))^{5/2} \sqrt {b \tan (e+f x)}}-\frac {24 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 b^2 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}-\frac {12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}} \]
-2/b/f/(d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(1/2)+24/5*(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)*EllipticE(cos(1/2*e+1/4*Pi+1/2*f *x),2^(1/2))*(b*tan(f*x+e))^(1/2)/b^2/d^2/f/(d*sec(f*x+e))^(1/2)/sin(f*x+e )^(1/2)-12/5*(b*tan(f*x+e))^(3/2)/b^3/f/(d*sec(f*x+e))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.89 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\frac {-11+\cos (2 (e+f x))-8 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)} \tan ^2(e+f x)}{5 b d^2 f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]
(-11 + Cos[2*(e + f*x)] - 8*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x] ^2]*(Sec[e + f*x]^2)^(1/4)*Tan[e + f*x]^2)/(5*b*d^2*f*Sqrt[d*Sec[e + f*x]] *Sqrt[b*Tan[e + f*x]])
Time = 0.71 (sec) , antiderivative size = 133, 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, 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.
| \(\displaystyle \int \frac {1}{(b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3089 |
| \(\displaystyle -\frac {6 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {6 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3092 |
| \(\displaystyle -\frac {6 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {6 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle -\frac {6 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {6 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle -\frac {6 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {6 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {6 \left (\frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{b^2}-\frac {2}{b f \sqrt {b \tan (e+f x)} (d \sec (e+f x))^{5/2}}\) |
-2/(b*f*(d*Sec[e + f*x])^(5/2)*Sqrt[b*Tan[e + f*x]]) - (6*((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]]*S qrt[Sin[e + f*x]]) + (2*(b*Tan[e + f*x])^(3/2))/(5*b*f*(d*Sec[e + f*x])^(5 /2))))/b^2
3.4.27.3.1 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[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 = 2.45 (sec) , antiderivative size = 452, normalized size of antiderivative = 3.48
| method | result | size |
| default | \(\frac {\left (24 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-12 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+\sqrt {2}\, \left (\cos ^{2}\left (f x +e \right )\right )+24 \sec \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-12 \sec \left (f x +e \right ) \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+6 \sqrt {2}-12 \sec \left (f x +e \right ) \sqrt {2}\right ) \sqrt {2}}{5 f \sqrt {d \sec \left (f x +e \right )}\, \sqrt {b \tan \left (f x +e \right )}\, b \,d^{2}}\) | \(452\) |
1/5/f/(d*sec(f*x+e))^(1/2)/(b*tan(f*x+e))^(1/2)/b/d^2*(24*(-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)-c ot(f*x+e)))^(1/2)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/ 2))-12*(-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)*cos(f*x+e)^2+24*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)*EllipticE((-I*(I-cot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2 ^(1/2))-12*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-c ot(f*x+e)+csc(f*x+e)))^(1/2),1/2*2^(1/2))+6*2^(1/2)-12*sec(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 = 141, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (6 i \, \sqrt {-2 i \, b d} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 6 i \, \sqrt {2 i \, b d} \sin \left (f x + e\right ) {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - {\left (\cos \left (f x + e\right )^{4} - 6 \, \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}\right )}}{5 \, b^{2} d^{3} f \sin \left (f x + e\right )} \]
-2/5*(6*I*sqrt(-2*I*b*d)*sin(f*x + e)*weierstrassZeta(4, 0, weierstrassPIn verse(4, 0, cos(f*x + e) + I*sin(f*x + e))) - 6*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))) - (cos(f*x + e)^4 - 6*cos(f*x + e)^2)*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e)))/(b^2*d^3*f*sin(f*x + e))
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]