Integrand size = 30, antiderivative size = 274 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\frac {\log (\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {7 \log (1-\sec (e+f x)) \tan (e+f x)}{8 c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\log (1+\sec (e+f x)) \tan (e+f x)}{8 c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{4 c^2 f (1-\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {3 \tan (e+f x)}{4 c^2 f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
ln(cos(f*x+e))*tan(f*x+e)/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1 /2)+7/8*ln(1-sec(f*x+e))*tan(f*x+e)/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec( f*x+e))^(1/2)+1/8*ln(1+sec(f*x+e))*tan(f*x+e)/c^2/f/(a+a*sec(f*x+e))^(1/2) /(c-c*sec(f*x+e))^(1/2)-1/4*tan(f*x+e)/c^2/f/(1-sec(f*x+e))^2/(a+a*sec(f*x +e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-3/4*tan(f*x+e)/c^2/f/(1-sec(f*x+e))/(a+a *sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
Time = 1.60 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.37 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\frac {\left (8 \log (\cos (e+f x))+7 \log (1-\sec (e+f x))+\log (1+\sec (e+f x))-\frac {2}{(-1+\sec (e+f x))^2}+\frac {6}{-1+\sec (e+f x)}\right ) \tan (e+f x)}{8 c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
((8*Log[Cos[e + f*x]] + 7*Log[1 - Sec[e + f*x]] + Log[1 + Sec[e + f*x]] - 2/(-1 + Sec[e + f*x])^2 + 6/(-1 + Sec[e + f*x]))*Tan[e + f*x])/(8*c^2*f*Sq rt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.41, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 4400, 27, 93, 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.
| \(\displaystyle \int \frac {1}{\sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4400 |
| \(\displaystyle -\frac {a c \tan (e+f x) \int \frac {\cos (e+f x)}{a c^3 (1-\sec (e+f x))^3 (\sec (e+f x)+1)}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(1-\sec (e+f x))^3 (\sec (e+f x)+1)}d\sec (e+f x)}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 93 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\cos (e+f x)-\frac {7}{8 (\sec (e+f x)-1)}-\frac {1}{8 (\sec (e+f x)+1)}+\frac {3}{4 (\sec (e+f x)-1)^2}-\frac {1}{2 (\sec (e+f x)-1)^3}\right )d\sec (e+f x)}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {3}{4 (1-\sec (e+f x))}+\frac {1}{4 (1-\sec (e+f x))^2}-\frac {7}{8} \log (1-\sec (e+f x))+\log (\sec (e+f x))-\frac {1}{8} \log (\sec (e+f x)+1)\right )}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((((-7*Log[1 - Sec[e + f*x]])/8 + Log[Sec[e + f*x]] - Log[1 + Sec[e + f*x ]]/8 + 1/(4*(1 - Sec[e + f*x])^2) + 3/(4*(1 - Sec[e + f*x])))*Tan[e + f*x] )/(c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))
3.2.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Int[ExpandIntegrand[(e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; Fre eQ[{a, b, c, d, e, f}, x] && IntegerQ[p]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[a*c*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]])) Subst[Int[(a + b*x)^(m - 1/2)*((c + d *x)^(n - 1/2)/x), x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Time = 2.34 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.88
| method | result | size |
| default | \(-\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (16 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-28 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-8 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \csc \left (f x +e \right )}{32 f a \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\) | \(240\) |
| risch | \(\frac {\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{c^{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 )}}}\, \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 )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}-\frac {2 \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{c^{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 )}}}\, \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 )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) f}+\frac {i \left (5 \,{\mathrm e}^{2 i \left (f x +e \right )}-8 \,{\mathrm e}^{i \left (f x +e \right )}+5\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{2 c^{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 )^{3} \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 )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) f}-\frac {7 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{4 c^{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 )}}}\, \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 )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) f}-\frac {i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{4 c^{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 )}}}\, \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 )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) f}\) | \(580\) |
-1/32/f*2^(1/2)/a*(-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)^2/(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))*(16*ln((1-cos(f*x+e))^2*csc(f*x+ e)^2+1)*(1-cos(f*x+e))^4*csc(f*x+e)^4-28*ln(-cot(f*x+e)+csc(f*x+e))*(1-cos (f*x+e))^4*csc(f*x+e)^4-8*(1-cos(f*x+e))^2*csc(f*x+e)^2+1)*csc(f*x+e)
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a*c^3*sec(f* x + e)^4 - 2*a*c^3*sec(f*x + e)^3 + 2*a*c^3*sec(f*x + e) - a*c^3), x)
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 2206 vs. \(2 (242) = 484\).
Time = 0.52 (sec) , antiderivative size = 2206, normalized size of antiderivative = 8.05 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
-1/4*(4*(f*x + e)*cos(4*f*x + 4*e)^2 + 144*(f*x + e)*cos(2*f*x + 2*e)^2 + 64*(f*x + e)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 64*( f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*(f*x + e)*sin(4*f*x + 4*e)^2 + 144*(f*x + e)*sin(2*f*x + 2*e)^2 + 64*(f*x + e)*s in(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 64*(f*x + e)*sin(1 /2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 4*f*x - (2*(6*cos(2*f* x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 36*cos(2*f*x + 2*e)^ 2 - 8*(cos(4*f*x + 4*e) + 6*cos(2*f*x + 2*e) - 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f *x + 2*e))) + 16*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(cos(4*f*x + 4*e) + 6*cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(4*f*x + 4*e)^2 + 12*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 3 6*sin(2*f*x + 2*e)^2 - 8*(sin(4*f*x + 4*e) + 6*sin(2*f*x + 2*e) - 4*sin(1/ 2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f* x + 2*e)))^2 - 8*(sin(4*f*x + 4*e) + 6*sin(2*f*x + 2*e))*sin(1/2*arctan2(s in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 16*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 12*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2*arctan2(s in(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), ...
Time = 1.70 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {14 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} c {\left | c \right |}} - \frac {16 \, \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{\sqrt {-a c} c {\left | c \right |}} - \frac {21 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} + 34 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + 14 \, c^{2}}{\sqrt {-a c} c^{3} {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}}{16 \, f} \]
-1/16*(14*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*c*abs(c)) - 16*lo g(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/(sqrt(-a*c)*c*abs(c)) - (21*(c*tan(1/ 2*f*x + 1/2*e)^2 - c)^2 + 34*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + 14*c^2)/(s qrt(-a*c)*c^3*abs(c)*tan(1/2*f*x + 1/2*e)^4))/f
Timed out. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]