Integrand size = 23, antiderivative size = 92 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {2 a d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}+\frac {2 a d (d \sec (e+f x))^{3/2} \sin (e+f x)}{3 f} \]
2/5*b*(d*sec(f*x+e))^(5/2)/f+2/3*a*d*(d*sec(f*x+e))^(3/2)*sin(f*x+e)/f+2/3 *a*d^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f *x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(d*sec(f*x+e))^(1/2)/f
Time = 0.90 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.63 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {(d \sec (e+f x))^{5/2} \left (6 b+10 a \cos ^{\frac {5}{2}}(e+f x) \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )+5 a \sin (2 (e+f x))\right )}{15 f} \]
((d*Sec[e + f*x])^(5/2)*(6*b + 10*a*Cos[e + f*x]^(5/2)*EllipticF[(e + f*x) /2, 2] + 5*a*Sin[2*(e + f*x)]))/(15*f)
Time = 0.46 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3967, 3042, 4255, 3042, 4258, 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.
\(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x))dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int (d \sec (e+f x))^{5/2}dx+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}dx+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle a \left (\frac {1}{3} d^2 \int \sqrt {d \sec (e+f x)}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{3} d^2 \int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle a \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {1}{3} d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle a \left (\frac {2 d^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 f}+\frac {2 d \sin (e+f x) (d \sec (e+f x))^{3/2}}{3 f}\right )+\frac {2 b (d \sec (e+f x))^{5/2}}{5 f}\) |
(2*b*(d*Sec[e + f*x])^(5/2))/(5*f) + a*((2*d^2*Sqrt[Cos[e + f*x]]*Elliptic F[(e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*f) + (2*d*(d*Sec[e + f*x])^(3/2 )*Sin[e + f*x])/(3*f))
3.6.79.3.1 Defintions of rubi rules used
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[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Result contains complex when optimal does not.
Time = 24.30 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.76
method | result | size |
default | \(-\frac {2 a \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\) | \(162\) |
parts | \(-\frac {2 a \sqrt {d \sec \left (f x +e \right )}\, d^{2} \left (i \cos \left (f x +e \right ) F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\tan \left (f x +e \right )\right )}{3 f}+\frac {2 b \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\) | \(162\) |
-2/3*a/f*(d*sec(f*x+e))^(1/2)*d^2*(I*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I )*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+I* (cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)*( 1/(cos(f*x+e)+1))^(1/2)-tan(f*x+e))+2/5*b*(d*sec(f*x+e))^(5/2)/f
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.34 \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\frac {-5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + 5 i \, \sqrt {2} a d^{\frac {5}{2}} \cos \left (f x + e\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) + 2 \, {\left (5 \, a d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, b d^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{15 \, f \cos \left (f x + e\right )^{2}} \]
1/15*(-5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierstrassPInverse(-4, 0, cos (f*x + e) + I*sin(f*x + e)) + 5*I*sqrt(2)*a*d^(5/2)*cos(f*x + e)^2*weierst rassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) + 2*(5*a*d^2*cos(f*x + e)*sin(f*x + e) + 3*b*d^2)*sqrt(d/cos(f*x + e)))/(f*cos(f*x + e)^2)
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\int \left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (a + b \tan {\left (e + f x \right )}\right )\, dx \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )} \,d x } \]
\[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}} {\left (b \tan \left (f x + e\right ) + a\right )} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{5/2} (a+b \tan (e+f x)) \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right ) \,d x \]