Integrand size = 30, antiderivative size = 140 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{a f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
-1/2*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(1/2)-c*tan(f* x+e)/a/f/(a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^(1/2)+c*ln(1+cos(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)
Time = 0.42 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {c \left (-\log (\cos (e+f x))-\log (1+\sec (e+f x))+\frac {1}{2 (1+\sec (e+f x))^2}+\frac {1}{1+\sec (e+f x)}\right ) \tan (e+f x)}{a^2 f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
-((c*(-Log[Cos[e + f*x]] - Log[1 + Sec[e + f*x]] + 1/(2*(1 + Sec[e + f*x]) ^2) + (1 + Sec[e + f*x])^(-1))*Tan[e + f*x])/(a^2*f*Sqrt[a*(1 + Sec[e + f* x])]*Sqrt[c - c*Sec[e + f*x]]))
Time = 0.66 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 4395, 3042, 4395, 3042, 4399, 16}
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.
| \(\displaystyle \int \frac {\sqrt {c-c \sec (e+f x)}}{(a \sec (e+f x)+a)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4395 |
| \(\displaystyle \frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{(\sec (e+f x) a+a)^{3/2}}dx}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 4395 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {\sec (e+f x) a+a}}dx}{a}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}{\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 4399 |
| \(\displaystyle \frac {\frac {c \tan (e+f x) \int \frac {1}{\cos (e+f x) a+a}d\cos (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {\frac {c \tan (e+f x) \log (\cos (e+f x)+1)}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {c \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \sqrt {c-c \sec (e+f x)}}}{a}-\frac {c \tan (e+f x)}{2 f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}}\) |
-1/2*(c*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x ]]) + (-((c*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(3/2)*Sqrt[c - c*Sec[e + f*x]])) + (c*Log[1 + Cos[e + f*x]]*Tan[e + f*x])/(a*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]))/a
3.2.27.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_ .) + (c_))^(n_), x_Symbol] :> Simp[-2*a*Cot[e + f*x]*((c + d*Csc[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[1/c Int[Sqrt[a + b*C sc[e + f*x]]*(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] && LtQ[n, -2^(-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[(b + a*x)^(m - 1/2)*((d + c*x)^(n - 1/2)/x^(m + n)), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e , f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2] && EqQ[m + n, 0]
Time = 2.08 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(\frac {\left (8 \cos \left (f x +e \right )^{2} \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+16 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+7 \cos \left (f x +e \right )^{2}+8 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \cos \left (f x +e \right )-5\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, \cot \left (f x +e \right )}{8 f \,a^{3} \left (\cos \left (f x +e \right )+1\right )^{2}}\) | \(129\) |
| risch | \(-\frac {\sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{4 i \left (f x +e \right )}+{\mathrm e}^{4 i \left (f x +e \right )} f x +8 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+2 \,{\mathrm e}^{4 i \left (f x +e \right )} e +4 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +4 i {\mathrm e}^{i \left (f x +e \right )}+8 \,{\mathrm e}^{3 i \left (f x +e \right )} e +6 \,{\mathrm e}^{2 i \left (f x +e \right )} f x +6 i {\mathrm e}^{2 i \left (f x +e \right )}+12 \,{\mathrm e}^{2 i \left (f x +e \right )} e +4 \,{\mathrm e}^{i \left (f x +e \right )} f x +12 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}+4 i {\mathrm e}^{3 i \left (f x +e \right )}+2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+8 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\) | \(348\) |
1/8/f/a^3*(8*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))+16*cos(f*x+e)*ln(2/(cos(f*x +e)+1))+7*cos(f*x+e)^2+8*ln(2/(cos(f*x+e)+1))-2*cos(f*x+e)-5)*(a*(sec(f*x+ e)+1))^(1/2)*(-c*(sec(f*x+e)-1))^(1/2)/(cos(f*x+e)+1)^2*cot(f*x+e)
\[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int { \frac {\sqrt {-c \sec \left (f x + e\right ) + c}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integral(sqrt(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)
\[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1165 vs. \(2 (126) = 252\).
Time = 0.49 (sec) , antiderivative size = 1165, normalized size of antiderivative = 8.32 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
-((f*x + e)*cos(4*f*x + 4*e)^2 + 16*(f*x + e)*cos(3*f*x + 3*e)^2 + 36*(f*x + e)*cos(2*f*x + 2*e)^2 + 16*(f*x + e)*cos(f*x + e)^2 + (f*x + e)*sin(4*f *x + 4*e)^2 + 16*(f*x + e)*sin(3*f*x + 3*e)^2 + 36*(f*x + e)*sin(2*f*x + 2 *e)^2 + 16*(f*x + e)*sin(f*x + e)^2 + f*x - 2*(2*(4*cos(3*f*x + 3*e) + 6*c os(2*f*x + 2*e) + 4*cos(f*x + e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^ 2 + 8*(6*cos(2*f*x + 2*e) + 4*cos(f*x + e) + 1)*cos(3*f*x + 3*e) + 16*cos( 3*f*x + 3*e)^2 + 12*(4*cos(f*x + e) + 1)*cos(2*f*x + 2*e) + 36*cos(2*f*x + 2*e)^2 + 16*cos(f*x + e)^2 + 4*(2*sin(3*f*x + 3*e) + 3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(4*f*x + 4*e) + sin(4*f*x + 4*e)^2 + 16*(3*sin(2*f*x + 2*e) + 2*sin(f*x + e))*sin(3*f*x + 3*e) + 16*sin(3*f*x + 3*e)^2 + 36*sin( 2*f*x + 2*e)^2 + 48*sin(2*f*x + 2*e)*sin(f*x + e) + 16*sin(f*x + e)^2 + 8* cos(f*x + e) + 1)*arctan2(sin(f*x + e), cos(f*x + e) + 1) + 2*(f*x + 4*(f* x + e)*cos(3*f*x + 3*e) + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x + e)*cos(f *x + e) + e - 2*sin(3*f*x + 3*e) - 3*sin(2*f*x + 2*e) - 2*sin(f*x + e))*co s(4*f*x + 4*e) + 8*(f*x + 6*(f*x + e)*cos(2*f*x + 2*e) + 4*(f*x + e)*cos(f *x + e) + e)*cos(3*f*x + 3*e) + 12*(f*x + 4*(f*x + e)*cos(f*x + e) + e)*co s(2*f*x + 2*e) + 8*(f*x + e)*cos(f*x + e) + 2*(4*(f*x + e)*sin(3*f*x + 3*e ) + 6*(f*x + e)*sin(2*f*x + 2*e) + 4*(f*x + e)*sin(f*x + e) + 2*cos(3*f*x + 3*e) + 3*cos(2*f*x + 2*e) + 2*cos(f*x + e))*sin(4*f*x + 4*e) + 4*(12*(f* x + e)*sin(2*f*x + 2*e) + 8*(f*x + e)*sin(f*x + e) - 1)*sin(3*f*x + 3*e...
Time = 1.46 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=-\frac {\sqrt {2} {\left (\frac {8 \, \sqrt {2} \sqrt {-a c} c \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{a^{3} {\left | c \right |}} + \frac {\sqrt {2} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} a^{3} c {\left | c \right |} - 4 \, \sqrt {2} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} \sqrt {-a c} a^{3} c^{2} {\left | c \right |}}{a^{6} c^{4}}\right )}}{16 \, f} \]
-1/16*sqrt(2)*(8*sqrt(2)*sqrt(-a*c)*c*log(abs(c*tan(1/2*f*x + 1/2*e)^2 + c ))/(a^3*abs(c)) + (sqrt(2)*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*sqrt(-a*c)*a^3 *c*abs(c) - 4*sqrt(2)*(c*tan(1/2*f*x + 1/2*e)^2 - c)*sqrt(-a*c)*a^3*c^2*ab s(c))/(a^6*c^4))/f
Timed out. \[ \int \frac {\sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^{5/2}} \, dx=\int \frac {\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]