Integrand size = 36, antiderivative size = 145 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {c^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2}}+\frac {c (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}} \]
1/2*c*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)-c^3*ln(1+ sec(f*x+e))*tan(f*x+e)/a^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2) -c^2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^(3/2)
Time = 0.72 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.61 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c \left (c^2 \log (1+\sec (e+f x))-\frac {2 c^2}{(1+\sec (e+f x))^2}+\frac {4 c^2}{1+\sec (e+f x)}\right ) \tan (e+f x)}{a^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
-((c*(c^2*Log[1 + Sec[e + f*x]] - (2*c^2)/(1 + Sec[e + f*x])^2 + (4*c^2)/( 1 + Sec[e + f*x]))*Tan[e + f*x])/(a^2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]]))
Time = 0.82 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4442, 3042, 4442, 3042, 4440}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a \sec (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4442 |
| \(\displaystyle \frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}}-\frac {c \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{(\sec (e+f x) a+a)^{3/2}}dx}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}}-\frac {c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{a}\) |
| \(\Big \downarrow \) 4442 |
| \(\displaystyle \frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}}-\frac {c \left (\frac {c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)^{3/2}}-\frac {c \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{\sqrt {\sec (e+f x) a+a}}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}}-\frac {c \left (\frac {c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)^{3/2}}-\frac {c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}dx}{a}\right )}{a}\) |
| \(\Big \downarrow \) 4440 |
| \(\displaystyle \frac {c \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{2 f (a \sec (e+f x)+a)^{5/2}}-\frac {c \left (\frac {c^2 \tan (e+f x) \log (\sec (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{f (a \sec (e+f x)+a)^{3/2}}\right )}{a}\) |
(c*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(2*f*(a + a*Sec[e + f*x])^(5/2 )) - (c*((c^2*Log[1 + Sec[e + f*x]]*Tan[e + f*x])/(a*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (c*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x] )/(f*(a + a*Sec[e + f*x])^(3/2))))/a
3.2.45.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)])/Sq rt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*c*Log[1 + (b/ a)*Csc[e + f*x]]*(Cot[e + f*x]/(b*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc [e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && Eq Q[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), x] - Simp[d*((2*n - 1)/(b*(2*m + 1))) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && LtQ[ m, -2^(-1)]
Time = 3.54 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.50
| method | result | size |
| default | \(\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{3} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{5} \left (\left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+2 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )\right )}{4 f \,a^{3} \left (1-\cos \left (f x +e \right )\right )^{5}}\) | \(218\) |
| risch | \(-\frac {8 i c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}+\frac {2 i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}{a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\) | \(325\) |
1/4/f*2^(1/2)/a^3*(-2*a/((1-cos(f*x+e))^2*csc(f*x+e)^2-1))^(1/2)*((1-cos(f *x+e))^2*csc(f*x+e)^2-1)^3*(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e )^2-1)*csc(f*x+e)^2)^(5/2)/(1-cos(f*x+e))^5*sin(f*x+e)^5*((1-cos(f*x+e))^4 *csc(f*x+e)^4+2*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*ln(-cot(f*x+e)+csc(f*x+e)- 1)+2*ln(-cot(f*x+e)+csc(f*x+e)+1))
\[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \sec \left (f x + e\right )}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral((c^2*sec(f*x + e)^3 - 2*c^2*sec(f*x + e)^2 + c^2*sec(f*x + e))*sq rt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^3*sec(f*x + e)^3 + 3*a ^3*sec(f*x + e)^2 + 3*a^3*sec(f*x + e) + a^3), x)
Timed out. \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.92 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{\sqrt {-a} a^{2}} + \frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {-a} a^{2}} - \frac {\frac {2 \, \sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \]
-1/2*(2*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(sqrt(-a)*a^2) + 2*c^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(sqrt(-a)*a^2) - (2*sqr t(-a)*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + sqrt(-a)*c^(5/2)*sin(f *x + e)^4/(cos(f*x + e) + 1)^4)/a^3)/f
Time = 1.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.79 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c^{4} {\left (\frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{2} + 4 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{3}}{c^{4}} + 2 \, \log \left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )\right )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{2 \, \sqrt {-a c} a^{2} f {\left | c \right |}} \]
-1/2*c^4*(((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^2 + 4*(c*tan(1/2*f*x + 1/2*e )^2 - c)*c^3)/c^4 + 2*log(c*tan(1/2*f*x + 1/2*e)^2 - c))*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e))/(sqrt(-a*c)*a^2*f*abs(c))
Timed out. \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{\cos \left (e+f\,x\right )\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]