Integrand size = 25, antiderivative size = 131 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {4 b^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{15 d^4 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}+\frac {2 b (b \tan (e+f x))^{3/2}}{15 d^2 f (d \sec (e+f x))^{5/2}} \]
-4/15*b^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)*El lipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/d^4/f/(d*s ec(f*x+e))^(1/2)/sin(f*x+e)^(1/2)-2/9*b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+ e))^(9/2)+2/15*b*(b*tan(f*x+e))^(3/2)/d^2/f/(d*sec(f*x+e))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 1.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.69 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\frac {b^3 \left (1-5 \cos (2 (e+f x))+4 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/4}\right ) \sin ^2(e+f x)}{45 d^4 f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}} \]
(b^3*(1 - 5*Cos[2*(e + f*x)] + 4*Hypergeometric2F1[3/4, 5/4, 7/4, -Tan[e + f*x]^2]*(Sec[e + f*x]^2)^(5/4))*Sin[e + f*x]^2)/(45*d^4*f*Sqrt[d*Sec[e + f*x]]*Sqrt[b*Tan[e + f*x]])
Time = 0.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3090, 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.
| \(\displaystyle \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3090 |
| \(\displaystyle \frac {b^2 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \int \frac {\sqrt {b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}}dx}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3092 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3096 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b^2 \left (\frac {2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {b^2 \left (\frac {4 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 (b \tan (e+f x))^{3/2}}{5 b f (d \sec (e+f x))^{5/2}}\right )}{3 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{9 f (d \sec (e+f x))^{9/2}}\) |
(-2*b*(b*Tan[e + f*x])^(3/2))/(9*f*(d*Sec[e + f*x])^(9/2)) + (b^2*((4*Elli pticE[(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*Sec[e + f*x])^(5/2))))/(3*d^2)
3.4.14.3.1 Defintions of rubi rules used
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[b*(a*Sec[e + f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*m)) , x] - Simp[b^2*((n - 1)/(a^2*m)) Int[(a*Sec[e + f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[n, 1] && (LtQ[m, -1 ] || (EqQ[m, -1] && EqQ[n, 3/2])) && IntegersQ[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 = 2.19 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.60
| method | result | size |
| default | \(\frac {\csc \left (f x +e \right ) \left (5 \left (\cos ^{5}\left (f x +e \right )\right ) \sqrt {2}-12 \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+6 \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )-8 \sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right )-12 \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, E\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+6 \sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )+i\right )}\, \sqrt {-i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}\, F\left (\sqrt {i \left (-i+\cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {2}\, \cos \left (f x +e \right )+6 \sqrt {2}\right ) \sqrt {b \tan \left (f x +e \right )}\, b^{2} \sqrt {2}}{45 f \sqrt {d \sec \left (f x +e \right )}\, d^{4}}\) | \(472\) |
1/45/f*csc(f*x+e)*(5*cos(f*x+e)^5*2^(1/2)-12*(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))*cos(f*x+ e)+6*(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)-8*2^(1/2)*cos(f*x+e)^3-12*(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))+6*(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))-3*2^(1/2)*cos(f*x+e)+6*2^(1/2))*(b*tan(f* x+e))^(1/2)*b^2/(d*sec(f*x+e))^(1/2)/d^4*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.05 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=-\frac {2 \, {\left (-3 i \, \sqrt {-2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + {\left (5 \, b^{2} \cos \left (f x + e\right )^{4} - 3 \, b^{2} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )\right )}}{45 \, d^{5} f} \]
-2/45*(-3*I*sqrt(-2*I*b*d)*b^2*weierstrassZeta(4, 0, weierstrassPInverse(4 , 0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(2*I*b*d)*b^2*weierstrassZe ta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))) + (5*b^ 2*cos(f*x + e)^4 - 3*b^2*cos(f*x + e)^2)*sqrt(b*sin(f*x + e)/cos(f*x + e)) *sqrt(d/cos(f*x + e))*sin(f*x + e))/(d^5*f)
Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \]
\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {9}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{9/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{9/2}} \,d x \]