Integrand size = 23, antiderivative size = 94 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}+\frac {2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 a \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}} \]
-2/3*b/f/(d*sec(f*x+e))^(3/2)+2/3*a*sin(f*x+e)/d/f/(d*sec(f*x+e))^(1/2)+2/ 3*a*(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)/d^2/f
Time = 0.70 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=-\frac {\sqrt {d \sec (e+f x)} \left (b+b \cos (2 (e+f x))-2 a \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )-a \sin (2 (e+f x))\right )}{3 d^2 f} \]
-1/3*(Sqrt[d*Sec[e + f*x]]*(b + b*Cos[2*(e + f*x)] - 2*a*Sqrt[Cos[e + f*x] ]*EllipticF[(e + f*x)/2, 2] - a*Sin[2*(e + f*x)]))/(d^2*f)
Time = 0.44 (sec) , antiderivative size = 95, 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, 4256, 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.
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 3967 |
\(\displaystyle a \int \frac {1}{(d \sec (e+f x))^{3/2}}dx-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4256 |
\(\displaystyle a \left (\frac {\int \sqrt {d \sec (e+f x)}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\int \sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle a \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\cos (e+f x)}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (\frac {\sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)} \int \frac {1}{\sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}}dx}{3 d^2}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle a \left (\frac {2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right ) \sqrt {d \sec (e+f x)}}{3 d^2 f}+\frac {2 \sin (e+f x)}{3 d f \sqrt {d \sec (e+f x)}}\right )-\frac {2 b}{3 f (d \sec (e+f x))^{3/2}}\) |
(-2*b)/(3*f*(d*Sec[e + f*x])^(3/2)) + a*((2*Sqrt[Cos[e + f*x]]*EllipticF[( e + f*x)/2, 2]*Sqrt[d*Sec[e + f*x]])/(3*d^2*f) + (2*Sin[e + f*x])/(3*d*f*S qrt[d*Sec[e + f*x]]))
3.6.83.3.1 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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 = 10.85 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.65
method | result | size |
default | \(\frac {\frac {2 i F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, a}{3}+\frac {2 i \sec \left (f x +e \right ) F\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), i\right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}\, a}{3}+\frac {2 \sin \left (f x +e \right ) a}{3}-\frac {2 b \cos \left (f x +e \right )}{3}}{d f \sqrt {d \sec \left (f x +e \right )}}\) | \(155\) |
parts | \(-\frac {2 a \left (i \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}+i \sec \left (f x +e \right ) \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}-\sin \left (f x +e \right )\right )}{3 f \sqrt {d \sec \left (f x +e \right )}\, d}-\frac {2 b}{3 f \left (d \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\) | \(162\) |
2/3/d/f/(d*sec(f*x+e))^(1/2)*(I*EllipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(co s(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/(cos(f*x+e)+1))^(1/2)*a+I*sec(f*x+e)*Ell ipticF(I*(cot(f*x+e)-csc(f*x+e)),I)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*(1/( cos(f*x+e)+1))^(1/2)*a+sin(f*x+e)*a-b*cos(f*x+e))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.10 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\frac {-i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) + i \, \sqrt {2} a \sqrt {d} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{2} - a \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d^{2} f} \]
1/3*(-I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin( f*x + e)) + I*sqrt(2)*a*sqrt(d)*weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e)) - 2*(b*cos(f*x + e)^2 - a*cos(f*x + e)*sin(f*x + e))*sqrt( d/cos(f*x + e)))/(d^2*f)
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{3/2}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]