Integrand size = 34, antiderivative size = 160 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=-\frac {64 c^3 (a+a \sec (e+f x))^m \tan (e+f x)}{f (5+2 m) \left (3+8 m+4 m^2\right ) \sqrt {c-c \sec (e+f x)}}-\frac {16 c^2 (a+a \sec (e+f x))^m \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{f \left (15+16 m+4 m^2\right )}-\frac {2 c (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (5+2 m)} \]
-2*c*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(5+2*m)-64*c^3 *(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(5+2*m)/(4*m^2+8*m+3)/(c-c*sec(f*x+e))^(1 /2)-16*c^2*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f/(4*m^2+1 6*m+15)
Time = 0.82 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.68 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=-\frac {2 c^3 (a (1+\sec (e+f x)))^m \left (43+24 m+4 m^2-2 \left (7+16 m+4 m^2\right ) \sec (e+f x)+\left (3+8 m+4 m^2\right ) \sec ^2(e+f x)\right ) \tan (e+f x)}{f (1+2 m) (3+2 m) (5+2 m) \sqrt {c-c \sec (e+f x)}} \]
(-2*c^3*(a*(1 + Sec[e + f*x]))^m*(43 + 24*m + 4*m^2 - 2*(7 + 16*m + 4*m^2) *Sec[e + f*x] + (3 + 8*m + 4*m^2)*Sec[e + f*x]^2)*Tan[e + f*x])/(f*(1 + 2* m)*(3 + 2*m)*(5 + 2*m)*Sqrt[c - c*Sec[e + f*x]])
Time = 0.80 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {3042, 4443, 3042, 4443, 3042, 4441}
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.
| \(\displaystyle \int \sec (e+f x) (c-c \sec (e+f x))^{5/2} (a \sec (e+f x)+a)^m \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2} \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^mdx\) |
| \(\Big \downarrow \) 4443 |
| \(\displaystyle \frac {8 c \int \sec (e+f x) (\sec (e+f x) a+a)^m (c-c \sec (e+f x))^{3/2}dx}{2 m+5}-\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8 c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^m \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}dx}{2 m+5}-\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)}\) |
| \(\Big \downarrow \) 4443 |
| \(\displaystyle \frac {8 c \left (\frac {4 c \int \sec (e+f x) (\sec (e+f x) a+a)^m \sqrt {c-c \sec (e+f x)}dx}{2 m+3}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)}\right )}{2 m+5}-\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {8 c \left (\frac {4 c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^m \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{2 m+3}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)}\right )}{2 m+5}-\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)}\) |
| \(\Big \downarrow \) 4441 |
| \(\displaystyle \frac {8 c \left (-\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^m}{f (2 m+1) (2 m+3) \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)}\right )}{2 m+5}-\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)}\) |
(-2*c*(a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*( 5 + 2*m)) + (8*c*((-8*c^2*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(1 + 2*m )*(3 + 2*m)*Sqrt[c - c*Sec[e + f*x]]) - (2*c*(a + a*Sec[e + f*x])^m*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(f*(3 + 2*m))))/(5 + 2*m)
3.2.56.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*Sq rt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)], x_Symbol] :> Simp[2*a*c*Cot[e + f *x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1)*Sqrt[c + d*Csc[e + f*x]])), x] / ; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[(-d)*Cot[e + f *x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[c*((2*n - 1)/(m + n)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b *c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && !LtQ[m, -2^(-1)] && !(IGtQ[m - 1/2, 0] && LtQ[m, n])
\[\int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
Time = 0.29 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.19 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=\frac {2 \, {\left (4 \, c^{2} m^{2} + {\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 43 \, c^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, c^{2} m - {\left (4 \, c^{2} m^{2} + 8 \, c^{2} m - 29 \, c^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, c^{2} - {\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 11 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \left (\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )} \]
2*(4*c^2*m^2 + (4*c^2*m^2 + 24*c^2*m + 43*c^2)*cos(f*x + e)^3 + 8*c^2*m - (4*c^2*m^2 + 8*c^2*m - 29*c^2)*cos(f*x + e)^2 + 3*c^2 - (4*c^2*m^2 + 24*c^ 2*m + 11*c^2)*cos(f*x + e))*((a*cos(f*x + e) + a)/cos(f*x + e))^m*sqrt((c* cos(f*x + e) - c)/cos(f*x + e))/((8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos( f*x + e)^2*sin(f*x + e))
Timed out. \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.42 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=-\frac {2 \, {\left (\frac {\sqrt {2} {\left (2^{m + 5} m + 5 \cdot 2^{m + 4}\right )} \left (-a\right )^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {\sqrt {2} {\left (2^{m + 4} m^{2} + 2^{m + 6} m + 15 \cdot 2^{m + 2}\right )} \left (-a\right )^{m} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2^{m + \frac {11}{2}} \left (-a\right )^{m} c^{\frac {5}{2}}\right )} e^{\left (-m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {5}{2}}} \]
-2*(sqrt(2)*(2^(m + 5)*m + 5*2^(m + 4))*(-a)^m*c^(5/2)*sin(f*x + e)^2/(cos (f*x + e) + 1)^2 - sqrt(2)*(2^(m + 4)*m^2 + 2^(m + 6)*m + 15*2^(m + 2))*(- a)^m*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2^(m + 11/2)*(-a)^m*c^( 5/2))*e^(-m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)/ (cos(f*x + e) + 1) - 1))/((8*m^3 + 36*m^2 + 46*m + 15)*f*(sin(f*x + e)/(co s(f*x + e) + 1) + 1)^(5/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(5/2))
\[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=\int { {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ) \,d x } \]
Timed out. \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )} \,d x \]