Integrand size = 25, antiderivative size = 101 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}+\frac {2 d^2 \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \] Output:
-2/3*d^2*(d*sec(f*x+e))^(1/2)/b/f/(b*tan(f*x+e))^(3/2)+2/3*d^2*InverseJaco biAM(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/f/(b*tan(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.82 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {2 d^2 \sqrt {d \sec (e+f x)} \left (-\cot ^2(e+f x)+\frac {\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\tan ^2(e+f x)\right )}{\sqrt [4]{\sec ^2(e+f x)}}\right ) \sqrt {b \tan (e+f x)}}{3 b^3 f} \] Input:
Integrate[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(5/2),x]
Output:
(2*d^2*Sqrt[d*Sec[e + f*x]]*(-Cot[e + f*x]^2 + Hypergeometric2F1[1/4, 3/4, 5/4, -Tan[e + f*x]^2]/(Sec[e + f*x]^2)^(1/4))*Sqrt[b*Tan[e + f*x]])/(3*b^ 3*f)
Time = 0.53 (sec) , antiderivative size = 101, 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, 3088, 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.
\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3088 |
\(\displaystyle \frac {d^2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{3 b^2}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d^2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}}dx}{3 b^2}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3096 |
\(\displaystyle \frac {d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {b \sin (e+f x)}}dx}{3 b^2 \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d^2 \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {b \sin (e+f x)}}dx}{3 b^2 \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin (e+f x)}}dx}{3 b^2 \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {d^2 \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin (e+f x)}}dx}{3 b^2 \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {2 d^2 \sqrt {\sin (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),2\right ) \sqrt {d \sec (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}}-\frac {2 d^2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}\) |
Input:
Int[(d*Sec[e + f*x])^(5/2)/(b*Tan[e + f*x])^(5/2),x]
Output:
(-2*d^2*Sqrt[d*Sec[e + f*x]])/(3*b*f*(b*Tan[e + f*x])^(3/2)) + (2*d^2*Elli pticF[(e - Pi/2 + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]]*Sqrt[Sin[e + f*x]])/(3*b ^2*f*Sqrt[b*Tan[e + f*x]])
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_), x_Symbol] :> Simp[a^2*(a*Sec[e + f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(n + 1))), x] - Simp[a^2*((m - 2)/(b^2*(n + 1))) Int[(a*Sec[e + f *x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[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[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 = 2.02 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.49
method | result | size |
default | \(\frac {d^{2} \sqrt {d \sec \left (f x +e \right )}\, \left (i \left (1+\cos \left (f x +e \right )\right ) \sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}\, \sqrt {1-i \cot \left (f x +e \right )+i \csc \left (f x +e \right )}\, \sqrt {-i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1+i \cot \left (f x +e \right )-i \csc \left (f x +e \right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {2}\, \cot \left (f x +e \right )\right ) \sqrt {2}}{3 f \,b^{2} \sqrt {b \tan \left (f x +e \right )}}\) | \(150\) |
Input:
int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3/f*d^2*(d*sec(f*x+e))^(1/2)/b^2/(b*tan(f*x+e))^(1/2)*(I*(1+cos(f*x+e))* (1+I*cot(f*x+e)-I*csc(f*x+e))^(1/2)*(1-I*cot(f*x+e)+I*csc(f*x+e))^(1/2)*(- I*(-csc(f*x+e)+cot(f*x+e)))^(1/2)*EllipticF((1+I*cot(f*x+e)-I*csc(f*x+e))^ (1/2),1/2*2^(1/2))-2^(1/2)*cot(f*x+e))*2^(1/2)
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.52 \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {2 \, d^{2} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2}\right )} \sqrt {-2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2}\right )} \sqrt {2 i \, b d} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \] Input:
integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/3*(2*d^2*sqrt(b*sin(f*x + e)/cos(f*x + e))*sqrt(d/cos(f*x + e))*cos(f*x + e)^2 + (d^2*cos(f*x + e)^2 - d^2)*sqrt(-2*I*b*d)*weierstrassPInverse(4, 0, cos(f*x + e) + I*sin(f*x + e)) + (d^2*cos(f*x + e)^2 - d^2)*sqrt(2*I*b* d)*weierstrassPInverse(4, 0, cos(f*x + e) - I*sin(f*x + e)))/(b^3*f*cos(f* x + e)^2 - b^3*f)
Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))**(5/2)/(b*tan(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((d*sec(f*x + e))^(5/2)/(b*tan(f*x + e))^(5/2), x)
Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(5/2),x)
Output:
int((d/cos(e + f*x))^(5/2)/(b*tan(e + f*x))^(5/2), x)
\[ \int \frac {(d \sec (e+f x))^{5/2}}{(b \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right )}\, \sec \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3}}d x \right ) d^{2}}{b^{3}} \] Input:
int((d*sec(f*x+e))^(5/2)/(b*tan(f*x+e))^(5/2),x)
Output:
(sqrt(d)*sqrt(b)*int((sqrt(tan(e + f*x))*sqrt(sec(e + f*x))*sec(e + f*x)** 2)/tan(e + f*x)**3,x)*d**2)/b**3