Integrand size = 30, antiderivative size = 347 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\frac {\log (\cos (e+f x)) \tan (e+f x)}{a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {11 \log (1-\sec (e+f x)) \tan (e+f x)}{16 a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {5 \log (1+\sec (e+f x)) \tan (e+f x)}{16 a c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1-\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a c^2 f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a c^2 f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
ln(cos(f*x+e))*tan(f*x+e)/a/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^ (1/2)+11/16*ln(1-sec(f*x+e))*tan(f*x+e)/a/c^2/f/(a+a*sec(f*x+e))^(1/2)/(c- c*sec(f*x+e))^(1/2)+5/16*ln(1+sec(f*x+e))*tan(f*x+e)/a/c^2/f/(a+a*sec(f*x+ e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/a/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)-1/2*tan(f*x+e)/a/c^2/f/(1-sec (f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/a/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 = 0.99 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.34 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\frac {\left (16 \log (\cos (e+f x))+11 \log (1-\sec (e+f x))+5 \log (1+\sec (e+f x))-\frac {2}{(-1+\sec (e+f x))^2}+\frac {8}{-1+\sec (e+f x)}-\frac {2}{1+\sec (e+f x)}\right ) \tan (e+f x)}{16 a c^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
((16*Log[Cos[e + f*x]] + 11*Log[1 - Sec[e + f*x]] + 5*Log[1 + Sec[e + f*x] ] - 2/(-1 + Sec[e + f*x])^2 + 8/(-1 + Sec[e + f*x]) - 2/(1 + Sec[e + f*x]) )*Tan[e + f*x])/(16*a*c^2*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.37 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.37, 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, 99, 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.
| \(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4400 |
| \(\displaystyle -\frac {a c \tan (e+f x) \int \frac {\cos (e+f x)}{a^2 c^3 (1-\sec (e+f x))^3 (\sec (e+f x)+1)^2}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(1-\sec (e+f x))^3 (\sec (e+f x)+1)^2}d\sec (e+f x)}{a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\cos (e+f x)-\frac {11}{16 (\sec (e+f x)-1)}-\frac {5}{16 (\sec (e+f x)+1)}+\frac {1}{2 (\sec (e+f x)-1)^2}-\frac {1}{8 (\sec (e+f x)+1)^2}-\frac {1}{4 (\sec (e+f x)-1)^3}\right )d\sec (e+f x)}{a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {1}{2 (1-\sec (e+f x))}+\frac {1}{8 (\sec (e+f x)+1)}+\frac {1}{8 (1-\sec (e+f x))^2}-\frac {11}{16} \log (1-\sec (e+f x))+\log (\sec (e+f x))-\frac {5}{16} \log (\sec (e+f x)+1)\right )}{a c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((((-11*Log[1 - Sec[e + f*x]])/16 + Log[Sec[e + f*x]] - (5*Log[1 + Sec[e + f*x]])/16 + 1/(8*(1 - Sec[e + f*x])^2) + 1/(2*(1 - Sec[e + f*x])) + 1/(8 *(1 + Sec[e + f*x])))*Tan[e + f*x])/(a*c^2*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt [c - c*Sec[e + f*x]]))
3.2.23.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
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 = 2.00 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(-\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (-2 \left (1-\cos \left (f x +e \right )\right )^{6} \csc \left (f x +e \right )^{6}+32 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-44 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}-10 \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \csc \left (f x +e \right )}{64 f \,a^{2} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right )^{2} \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {5}{2}}}\) | \(262\) |
| 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}{a \,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 (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}}-\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 )}{a \,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 (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, f}+\frac {i \left (5 \,{\mathrm e}^{5 i \left (f x +e \right )}+6 \,{\mathrm e}^{4 i \left (f x +e \right )}-14 \,{\mathrm e}^{3 i \left (f x +e \right )}+6 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a \,c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \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 (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, f}-\frac {11 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 )}{8 a \,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 (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, f}-\frac {5 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 )}{8 a \,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 (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, f}\) | \(621\) |
-1/64/f*2^(1/2)/a^2*(-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))*(-2*(1-cos(f*x+e))^6*csc(f*x+e )^6+32*ln((1-cos(f*x+e))^2*csc(f*x+e)^2+1)*(1-cos(f*x+e))^4*csc(f*x+e)^4-4 4*ln(-cot(f*x+e)+csc(f*x+e))*(1-cos(f*x+e))^4*csc(f*x+e)^4-10*(1-cos(f*x+e ))^2*csc(f*x+e)^2+1)*csc(f*x+e)
\[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral(-sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^2*c^3*sec( f*x + e)^5 - a^2*c^3*sec(f*x + e)^4 - 2*a^2*c^3*sec(f*x + e)^3 + 2*a^2*c^3 *sec(f*x + e)^2 + a^2*c^3*sec(f*x + e) - a^2*c^3), x)
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 4272 vs. \(2 (309) = 618\).
Time = 1.88 (sec) , antiderivative size = 4272, normalized size of antiderivative = 12.31 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
-1/8*(8*(f*x + e)*cos(6*f*x + 6*e)^2 + 8*(f*x + e)*cos(4*f*x + 4*e)^2 + 8* (f*x + e)*cos(2*f*x + 2*e)^2 + 32*(f*x + e)*cos(5/2*arctan2(sin(2*f*x + 2* e), cos(2*f*x + 2*e)))^2 + 128*(f*x + e)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 32*(f*x + e)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos (2*f*x + 2*e)))^2 + 8*(f*x + e)*sin(6*f*x + 6*e)^2 + 8*(f*x + e)*sin(4*f*x + 4*e)^2 + 8*(f*x + e)*sin(2*f*x + 2*e)^2 + 32*(f*x + e)*sin(5/2*arctan2( sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 128*(f*x + e)*sin(3/2*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 32*(f*x + e)*sin(1/2*arctan2(sin(2*f *x + 2*e), cos(2*f*x + 2*e)))^2 + 8*f*x + 5*(2*(cos(4*f*x + 4*e) + cos(2*f *x + 2*e) - 1)*cos(6*f*x + 6*e) - cos(6*f*x + 6*e)^2 - 2*(cos(2*f*x + 2*e) - 1)*cos(4*f*x + 4*e) - cos(4*f*x + 4*e)^2 - cos(2*f*x + 2*e)^2 + 4*(cos( 6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 4*cos(3/2*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*cos(1/2*arctan2(sin(2*f*x + 2*e), co s(2*f*x + 2*e))) + 1)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 8*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) - 2*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 + 4*(cos(6*f*x + 6*e) - cos(4*f*x + 4*e) - cos(2*f*x + 2*e) + 1)*cos(1/2*ar ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*cos(1/2*arctan2(sin(2*f*...
Time = 1.99 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.53 \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=-\frac {\frac {22 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} a c {\left | c \right |}} - \frac {32 \, \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{\sqrt {-a c} a c {\left | c \right |}} + \frac {2 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}}{\sqrt {-a c} a c^{2} {\left | c \right |}} - \frac {33 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} + 56 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + 24 \, c^{2}}{\sqrt {-a c} a c^{3} {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}}{32 \, f} \]
-1/32*(22*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*a*c*abs(c)) - 32* log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c))/(sqrt(-a*c)*a*c*abs(c)) + 2*(c*tan( 1/2*f*x + 1/2*e)^2 - c)/(sqrt(-a*c)*a*c^2*abs(c)) - (33*(c*tan(1/2*f*x + 1 /2*e)^2 - c)^2 + 56*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + 24*c^2)/(sqrt(-a*c) *a*c^3*abs(c)*tan(1/2*f*x + 1/2*e)^4))/f
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]