Integrand size = 30, antiderivative size = 345 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\log (\cos (e+f x)) \tan (e+f x)}{a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {5 \log (1-\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {11 \log (1+\sec (e+f x)) \tan (e+f x)}{16 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1-\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{8 a^2 c f (1+\sec (e+f x))^2 \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {\tan (e+f x)}{2 a^2 c f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
ln(cos(f*x+e))*tan(f*x+e)/a^2/c/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^2/c/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^2/c/f/(a+a*sec(f*x+ e))^(1/2)/(c-c*sec(f*x+e))^(1/2)-1/8*tan(f*x+e)/a^2/c/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^2/c/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^2/ c/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)
Time = 0.89 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.34 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\left (16 \log (\cos (e+f x))+5 \log (1-\sec (e+f x))+11 \log (1+\sec (e+f x))+\frac {2}{-1+\sec (e+f x)}-\frac {2}{(1+\sec (e+f x))^2}-\frac {8}{1+\sec (e+f x)}\right ) \tan (e+f x)}{16 a^2 c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
((16*Log[Cos[e + f*x]] + 5*Log[1 - Sec[e + f*x]] + 11*Log[1 + Sec[e + f*x] ] + 2/(-1 + Sec[e + f*x]) - 2/(1 + Sec[e + f*x])^2 - 8/(1 + Sec[e + f*x])) *Tan[e + f*x])/(16*a^2*c*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f *x]])
Time = 0.36 (sec) , antiderivative size = 126, 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.
| \(\displaystyle \int \frac {1}{(a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4400 |
| \(\displaystyle -\frac {a c \tan (e+f x) \int \frac {\cos (e+f x)}{a^3 c^2 (1-\sec (e+f x))^2 (\sec (e+f x)+1)^3}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(1-\sec (e+f x))^2 (\sec (e+f x)+1)^3}d\sec (e+f x)}{a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\cos (e+f x)-\frac {5}{16 (\sec (e+f x)-1)}-\frac {11}{16 (\sec (e+f x)+1)}+\frac {1}{8 (\sec (e+f x)-1)^2}-\frac {1}{2 (\sec (e+f x)+1)^2}-\frac {1}{4 (\sec (e+f x)+1)^3}\right )d\sec (e+f x)}{a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {1}{8 (1-\sec (e+f x))}+\frac {1}{2 (\sec (e+f x)+1)}+\frac {1}{8 (\sec (e+f x)+1)^2}-\frac {5}{16} \log (1-\sec (e+f x))+\log (\sec (e+f x))-\frac {11}{16} \log (\sec (e+f x)+1)\right )}{a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((((-5*Log[1 - Sec[e + f*x]])/16 + Log[Sec[e + f*x]] - (11*Log[1 + Sec[e + f*x]])/16 + 1/(8*(1 - Sec[e + f*x])) + 1/(8*(1 + Sec[e + f*x])^2) + 1/(2 *(1 + Sec[e + f*x])))*Tan[e + f*x])/(a^2*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt [c - c*Sec[e + f*x]]))
3.2.29.3.1 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 = 2.26 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.76
| method | result | size |
| default | \(\frac {\sqrt {2}\, \sqrt {-\frac {2 a}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}}\, \left (1-\cos \left (f x +e \right )\right ) \left (-\left (1-\cos \left (f x +e \right )\right )^{6} \csc \left (f x +e \right )^{6}+10 \left (1-\cos \left (f x +e \right )\right )^{4} \csc \left (f x +e \right )^{4}+20 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-32 \ln \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+1\right ) \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}+2\right ) \csc \left (f x +e \right )}{64 f \,a^{3} \left (\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1\right ) \left (\frac {c \left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}}{\left (1-\cos \left (f x +e \right )\right )^{2} \csc \left (f x +e \right )^{2}-1}\right )^{\frac {3}{2}}}\) | \(262\) |
| 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^{2} c \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, \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^{2} c \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, \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^{2} c \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 {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 ) \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^{2} c \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, \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^{2} c \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\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 )}}}\, \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^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)/(c*(1-cos(f*x+e))^2/((1-cos(f*x+e))^2*csc(f*x+e) ^2-1)*csc(f*x+e)^2)^(3/2)*(1-cos(f*x+e))*(-(1-cos(f*x+e))^6*csc(f*x+e)^6+1 0*(1-cos(f*x+e))^4*csc(f*x+e)^4+20*ln(-cot(f*x+e)+csc(f*x+e))*(1-cos(f*x+e ))^2*csc(f*x+e)^2-32*ln((1-cos(f*x+e))^2*csc(f*x+e)^2+1)*(1-cos(f*x+e))^2* csc(f*x+e)^2+2)*csc(f*x+e)
\[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a^3*c^2*sec(f *x + e)^5 + a^3*c^2*sec(f*x + e)^4 - 2*a^3*c^2*sec(f*x + e)^3 - 2*a^3*c^2* sec(f*x + e)^2 + a^3*c^2*sec(f*x + e) + a^3*c^2), x)
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 4272 vs. \(2 (309) = 618\).
Time = 1.90 (sec) , antiderivative size = 4272, normalized size of antiderivative = 12.38 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
-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 + 11*(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(si n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2*cos(1/2*arctan2(sin(2*f*x + 2*e), c os(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*a rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 4*cos(1/2*arctan2(sin(2*f...
Time = 2.00 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=-\frac {\frac {10 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} a^{2} {\left | c \right |}} - \frac {32 \, \log \left ({\left | -c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c \right |}\right )}{\sqrt {-a c} a^{2} {\left | c \right |}} - \frac {2 \, {\left (5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}}{\sqrt {-a c} a^{2} c {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{2} c^{2} {\left | c \right |} - 8 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{2} c^{3} {\left | c \right |}}{a^{5} c^{7}}}{32 \, f} \]
-1/32*(10*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*a^2*abs(c)) - 32* log(abs(-c*tan(1/2*f*x + 1/2*e)^2 - c))/(sqrt(-a*c)*a^2*abs(c)) - 2*(5*c*t an(1/2*f*x + 1/2*e)^2 - c)/(sqrt(-a*c)*a^2*c*abs(c)*tan(1/2*f*x + 1/2*e)^2 ) + ((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^2*c^2*abs(c) - 8*(c*tan (1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^2*c^3*abs(c))/(a^5*c^7))/f
Timed out. \[ \int \frac {1}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]