Integrand size = 28, antiderivative size = 104 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^3 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^3 f}+\frac {8 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 c^3 f} \]
2*a^(5/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^3/f+8/5*cot( f*x+e)^5*(a+a*sec(f*x+e))^(5/2)/c^3/f+2*a^2*cot(f*x+e)*(a+a*sec(f*x+e))^(1 /2)/c^3/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 1.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\frac {2 a^3 \left (4+3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},1-\sec (e+f x)\right )+5 \sec (e+f x)\right ) \tan (e+f x)}{15 c^3 f (-1+\sec (e+f x))^3 \sqrt {a (1+\sec (e+f x))}} \]
(2*a^3*(4 + 3*Hypergeometric2F1[-5/2, 1, -3/2, 1 - Sec[e + f*x]] + 5*Sec[e + f*x])*Tan[e + f*x])/(15*c^3*f*(-1 + Sec[e + f*x])^3*Sqrt[a*(1 + Sec[e + f*x])])
Time = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88, 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.
| \(\displaystyle \int \frac {(a \sec (e+f x)+a)^{5/2}}{(c-c \sec (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle -\frac {\int \cot ^6(e+f x) (\sec (e+f x) a+a)^{11/2}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{11/2}}{\cot \left (e+f x+\frac {\pi }{2}\right )^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 4375 |
| \(\displaystyle \frac {2 \int \frac {\cot ^6(e+f x) (\sec (e+f x) a+a)^3 \left (\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+2\right )^2}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c^3 f}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {2 \int \left (4 (\sec (e+f x) a+a)^3 \cot ^6(e+f x)+a^2 (\sec (e+f x) a+a) \cot ^2(e+f x)-\frac {a^3}{\frac {a \tan ^2(e+f x)}{\sec (e+f x) a+a}+1}\right )d\left (-\frac {\tan (e+f x)}{\sqrt {\sec (e+f x) a+a}}\right )}{c^3 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 \left (a^{5/2} \arctan \left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )+a^2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}+\frac {4}{5} \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}\right )}{c^3 f}\) |
(2*(a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]] + a^2* Cot[e + f*x]*Sqrt[a + a*Sec[e + f*x]] + (4*Cot[e + f*x]^5*(a + a*Sec[e + f *x])^(5/2))/5))/(c^3*f)
3.1.62.3.1 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])
Leaf count of result is larger than twice the leaf count of optimal. \(266\) vs. \(2(92)=184\).
Time = 70.08 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.57
| method | result | size |
| default | \(\frac {2 a^{2} \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (5 \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \cos \left (f x +e \right )^{2}-10 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \cos \left (f x +e \right )+9 \cos \left (f x +e \right )^{2} \cot \left (f x +e \right )+5 \,\operatorname {arctanh}\left (\frac {\sin \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-10 \cos \left (f x +e \right ) \cot \left (f x +e \right )+5 \cot \left (f x +e \right )\right )}{5 c^{3} f \left (\cos \left (f x +e \right )-1\right )^{2}}\) | \(267\) |
2/5/c^3/f*a^2*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)-1)^2*(5*(-cos(f*x+e)/(c os(f*x+e)+1))^(1/2)*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e)/(cos(f* x+e)+1))^(1/2))*cos(f*x+e)^2-10*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f* x+e)/(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*cos(f*x+e)+ 9*cos(f*x+e)^2*cot(f*x+e)+5*arctanh(sin(f*x+e)/(cos(f*x+e)+1)/(-cos(f*x+e) /(cos(f*x+e)+1))^(1/2))*(-cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-10*cos(f*x+e)*c ot(f*x+e)+5*cot(f*x+e))
Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (92) = 184\).
Time = 0.31 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.24 \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\left [\frac {5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (9 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{10 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}, \frac {5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) + a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (9 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 5 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{5 \, {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) + c^{3} f\right )} \sin \left (f x + e\right )}\right ] \]
[1/10*(5*(a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*sqrt(-a)*log(-(8* a*cos(f*x + e)^3 - 4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*co s(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*(9*a^2*cos(f*x + e)^3 - 10*a^2*cos(f*x + e)^ 2 + 5*a^2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^3*f*c os(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)), 1/5*(5*(a^2*c os(f*x + e)^2 - 2*a^2*cos(f*x + e) + a^2)*sqrt(a)*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*(9*a^2*cos(f*x + e)^3 - 10*a^ 2*cos(f*x + e)^2 + 5*a^2*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) + c^3*f)*sin(f*x + e)) ]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=- \frac {\int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx + \int \frac {a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} - 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} - 1}\, dx}{c^{3}} \]
-(Integral(a**2*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**3 - 3*sec(e + f*x) **2 + 3*sec(e + f*x) - 1), x) + Integral(2*a**2*sqrt(a*sec(e + f*x) + a)*s ec(e + f*x)/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x) + Integral(a**2*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)**2/(sec(e + f*x)**3 - 3*sec(e + f*x)**2 + 3*sec(e + f*x) - 1), x))/c**3
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\int { -\frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (c \sec \left (f x + e\right ) - c\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^3} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^3} \,d x \]