Integrand size = 23, antiderivative size = 94 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}+\frac {6 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 a \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}} \] Output:
-2/5*b/f/(d*sec(f*x+e))^(5/2)+6/5*a*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))/ d^2/f/cos(f*x+e)^(1/2)/(d*sec(f*x+e))^(1/2)+2/5*a*sin(f*x+e)/d/f/(d*sec(f* x+e))^(3/2)
Time = 0.67 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.79 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\frac {2 \sqrt {d \sec (e+f x)} \left (3 a \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+\cos ^2(e+f x) (-b \cos (e+f x)+a \sin (e+f x))\right )}{5 d^3 f} \] Input:
Integrate[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]
Output:
(2*Sqrt[d*Sec[e + f*x]]*(3*a*Sqrt[Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2] + Cos[e + f*x]^2*(-(b*Cos[e + f*x]) + a*Sin[e + f*x])))/(5*d^3*f)
Time = 0.45 (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, 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.
| \(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle a \int \frac {1}{(d \sec (e+f x))^{5/2}}dx-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \frac {1}{\left (d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle a \left (\frac {3 \int \frac {1}{\sqrt {d \sec (e+f x)}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3 \int \frac {1}{\sqrt {d \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{5 d^2}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle a \left (\frac {3 \int \sqrt {\cos (e+f x)}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \left (\frac {3 \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{5 d^2 \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle a \left (\frac {6 E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt {\cos (e+f x)} \sqrt {d \sec (e+f x)}}+\frac {2 \sin (e+f x)}{5 d f (d \sec (e+f x))^{3/2}}\right )-\frac {2 b}{5 f (d \sec (e+f x))^{5/2}}\) |
Input:
Int[(a + b*Tan[e + f*x])/(d*Sec[e + f*x])^(5/2),x]
Output:
(-2*b)/(5*f*(d*Sec[e + f*x])^(5/2)) + a*((6*EllipticE[(e + f*x)/2, 2])/(5* d^2*f*Sqrt[Cos[e + f*x]]*Sqrt[d*Sec[e + f*x]]) + (2*Sin[e + f*x])/(5*d*f*( d*Sec[e + f*x])^(3/2)))
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[((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 = 19.63 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.22
| method | result | size |
| parts | \(\frac {2 a \left (\sin \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+3\right )-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) \operatorname {EllipticF}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )+3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \operatorname {EllipticE}\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}-\frac {2 b}{5 f \left (d \sec \left (f x +e \right )\right )^{\frac {5}{2}}}\) | \(209\) |
| default | \(\frac {-\frac {6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) a \operatorname {EllipticE}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}+\frac {6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \left (2+\cos \left (f x +e \right )+\sec \left (f x +e \right )\right ) a \operatorname {EllipticF}\left (i \left (-\csc \left (f x +e \right )+\cot \left (f x +e \right )\right ), i\right )}{5}+\frac {2 \sin \left (f x +e \right ) \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )+3\right ) a}{5}+\frac {2 b \left (-\cos \left (f x +e \right )^{3}-\cos \left (f x +e \right )^{2}\right )}{5}}{f \left (1+\cos \left (f x +e \right )\right ) \sqrt {d \sec \left (f x +e \right )}\, d^{2}}\) | \(217\) |
Input:
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
2/5*a/f/(1+cos(f*x+e))/(d*sec(f*x+e))^(1/2)/d^2*(sin(f*x+e)*(cos(f*x+e)^2+ cos(f*x+e)+3)-3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/ 2)*(2+cos(f*x+e)+sec(f*x+e))*EllipticF(I*(csc(f*x+e)-cot(f*x+e)),I)+3*I*(1 /(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc( f*x+e)-cot(f*x+e)),I)*(2+cos(f*x+e)+sec(f*x+e)))-2/5*b/f/(d*sec(f*x+e))^(5 /2)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.18 \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\frac {3 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {2} a \sqrt {d} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (b \cos \left (f x + e\right )^{3} - a \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{5 \, d^{3} f} \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/5*(3*I*sqrt(2)*a*sqrt(d)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 3*I*sqrt(2)*a*sqrt(d)*weierstrassZeta (-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(b* cos(f*x + e)^3 - a*cos(f*x + e)^2*sin(f*x + e))*sqrt(d/cos(f*x + e)))/(d^3 *f)
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {a + b \tan {\left (e + f x \right )}}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))**(5/2),x)
Output:
Integral((a + b*tan(e + f*x))/(d*sec(e + f*x))**(5/2), x)
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int { \frac {b \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x, algorithm="giac")
Output:
integrate((b*tan(f*x + e) + a)/(d*sec(f*x + e))^(5/2), x)
Timed out. \[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\int \frac {a+b\,\mathrm {tan}\left (e+f\,x\right )}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2),x)
Output:
int((a + b*tan(e + f*x))/(d/cos(e + f*x))^(5/2), x)
\[ \int \frac {a+b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx=\frac {\sqrt {d}\, \left (\left (\int \frac {\sqrt {\sec \left (f x +e \right )}}{\sec \left (f x +e \right )^{3}}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \tan \left (f x +e \right )}{\sec \left (f x +e \right )^{3}}d x \right ) b \right )}{d^{3}} \] Input:
int((a+b*tan(f*x+e))/(d*sec(f*x+e))^(5/2),x)
Output:
(sqrt(d)*(int(sqrt(sec(e + f*x))/sec(e + f*x)**3,x)*a + int((sqrt(sec(e + f*x))*tan(e + f*x))/sec(e + f*x)**3,x)*b))/d**3