Integrand size = 25, antiderivative size = 96 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=\frac {6 b^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}} \]
-6/5*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)*Ell ipticE(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2))*(b*tan(f*x+e))^(1/2)/d^2/f/(d*se c(f*x+e))^(1/2)/sin(f*x+e)^(1/2)-2/5*b*(b*tan(f*x+e))^(3/2)/f/(d*sec(f*x+e ))^(5/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.84 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {b \left (1+\cos (2 (e+f x))-2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{4},\frac {7}{4},-\tan ^2(e+f x)\right ) \sqrt [4]{\sec ^2(e+f x)}\right ) (b \tan (e+f x))^{3/2}}{5 d^2 f \sqrt {d \sec (e+f x)}} \]
-1/5*(b*(1 + Cos[2*(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))/(d^2*f*Sqrt[d*Se c[e + f*x]])
Time = 0.52 (sec) , antiderivative size = 96, 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, 3090, 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.
\(\displaystyle \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3090 |
\(\displaystyle \frac {3 b^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 b^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3096 |
\(\displaystyle \frac {3 b^2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 b^2 \sqrt {b \tan (e+f x)} \int \sqrt {b \sin (e+f x)}dx}{5 d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {3 b^2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 b^2 \sqrt {b \tan (e+f x)} \int \sqrt {\sin (e+f x)}dx}{5 d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {6 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\) |
(6*b^2*EllipticE[(e - Pi/2 + f*x)/2, 2]*Sqrt[b*Tan[e + f*x]])/(5*d^2*f*Sqr t[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]]) - (2*b*(b*Tan[e + f*x])^(3/2))/(5*f* (d*Sec[e + f*x])^(5/2))
3.4.12.3.1 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^(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 = 1.85 (sec) , antiderivative size = 458, normalized size of antiderivative = 4.77
method | result | size |
default | \(\frac {\csc \left (f x +e \right ) b^{2} \sqrt {b \tan \left (f x +e \right )}\, \left (-6 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+3 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right ) \cos \left (f x +e \right )+\sqrt {2}\, \left (\cos ^{3}\left (f x +e \right )\right )-6 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, E\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )+3 \sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (-i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}\, \sqrt {i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}\, F\left (\sqrt {-i \left (i-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}, \frac {\sqrt {2}}{2}\right )-4 \sqrt {2}\, \cos \left (f x +e \right )+3 \sqrt {2}\right ) \sqrt {2}}{5 f \sqrt {d \sec \left (f x +e \right )}\, d^{2}}\) | \(458\) |
1/5/f*csc(f*x+e)*b^2*(b*tan(f*x+e))^(1/2)*(-6*(-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))*cos(f*x+ e)+3*(-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))*cos(f*x+e)+2^(1/2)*cos(f*x+e)^3-6*(-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))+3*(-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))-4*2^(1/2)*cos(f*x+e)+3*2^(1/2))/(d*sec(f*x+e))^ (1/2)/d^2*2^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.27 \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 \, b^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - 3 i \, \sqrt {-2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 i \, \sqrt {2 i \, b d} b^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}{5 \, d^{3} f} \]
-1/5*(2*b^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2*sin(f*x + e) - 3*I*sqrt(-2*I*b*d)*b^2*weierstrassZeta(4, 0, weiers trassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*I*sqrt(2*I*b*d)*b^ 2*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e))))/(d^3*f)
Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]