Integrand size = 28, antiderivative size = 103 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c f}+\frac {8 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c f}-\frac {2 a^3 \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)}} \]
2*a^(5/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c/f+8*a^2*cot( f*x+e)*(a+a*sec(f*x+e))^(1/2)/c/f-2*a^3*tan(f*x+e)/c/f/(a+a*sec(f*x+e))^(1 /2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.40 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.64 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\frac {2 a^2 \csc (e+f x) \left (-1+4 \cos (e+f x)+\cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\sec (e+f x)\right )\right ) \sqrt {a (1+\sec (e+f x))}}{c f} \]
(2*a^2*Csc[e + f*x]*(-1 + 4*Cos[e + f*x] + Cos[e + f*x]*Hypergeometric2F1[ -1/2, 1, 1/2, 1 - Sec[e + f*x]])*Sqrt[a*(1 + Sec[e + f*x])])/(c*f)
Time = 0.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 4392, 3042, 4375, 364, 2009}
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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{5/2}}{c-c \sec (e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle -\frac {\int \cot ^2(e+f x) (\sec (e+f x) a+a)^{7/2}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{7/2}}{\cot \left (e+f x+\frac {\pi }{2}\right )^2}dx}{a c}\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle \frac {2 a^2 \int \frac {\cot ^2(e+f x) (\sec (e+f x) a+a) \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^2}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c f}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {2 a^2 \int \left (4 (\sec (e+f x) a+a) \cot ^2(e+f x)+a-\frac {a}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}\right )d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 a^2 \left (\sqrt {a} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )+4 \cot (e+f x) \sqrt {a \sec (e+f x)+a}-\frac {a \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c f}\) |
(2*a^2*(Sqrt[a]*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]] + 4*Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]] - (a*Tan[e + f*x])/Sqrt[a + a*Sec[ e + f*x]]))/(c*f)
3.1.60.3.1 Defintions of rubi rules used
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d) Subst[Int[x^m*((2 + a*x^2 )^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] ]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I ntegerQ[n - 1/2]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 5.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.97
| method | result | size |
| default | \(\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+5 \cot \left (f x +e \right )-\csc \left (f x +e \right )\right )}{c f}\) | \(100\) |
2/c/f*a^2*(a*(sec(f*x+e)+1))^(1/2)*(arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-co s(f*x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)+5*cot(f *x+e)-csc(f*x+e))
Time = 0.31 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.83 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\left [\frac {\sqrt {-a} a^{2} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (5 \, a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{2 \, c f \sin \left (f x + e\right )}, \frac {a^{\frac {5}{2}} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (5 \, a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{c f \sin \left (f x + e\right )}\right ] \]
[1/2*(sqrt(-a)*a^2*log(-(8*a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 - cos(f* x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7* a*cos(f*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) + 4*(5*a^2*cos(f*x + e) - a^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/(c*f*sin(f*x + e)), (a^ (5/2)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e )*sin(f*x + e)/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 2 *(5*a^2*cos(f*x + e) - a^2)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/(c*f* sin(f*x + e))]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=- \frac {\int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx + \int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )}}{\sec {\left (e + f x \right )} - 1}\, dx}{c} \]
-(Integral(a**2*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x) - 1), x) + Integral (2*a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)/(sec(e + f*x) - 1), x) + Int egral(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2/(sec(e + f*x) - 1), x) )/c
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{c \sec \left (f x + e\right ) - c} \,d x } \]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{c \sec \left (f x + e\right ) - c} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{c-c \sec (e+f x)} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{c-\frac {c}{\cos \left (e+f\,x\right )}} \,d x \]