Integrand size = 30, antiderivative size = 168 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\tan (e+f x)}{2 c f (1-\cos (e+f x)) \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {3 \log (1-\cos (e+f x)) \tan (e+f x)}{4 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\log (1+\cos (e+f x)) \tan (e+f x)}{4 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
1/2*tan(f*x+e)/c/f/(1-cos(f*x+e))/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^ (1/2)+3/4*ln(1-cos(f*x+e))*tan(f*x+e)/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec( f*x+e))^(1/2)+1/4*ln(1+cos(f*x+e))*tan(f*x+e)/c/f/(a+a*sec(f*x+e))^(1/2)/( c-c*sec(f*x+e))^(1/2)
Time = 0.61 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\frac {\left (4 \log (\cos (e+f x))+3 \log (1-\sec (e+f x))+\log (1+\sec (e+f x))+\frac {2}{-1+\sec (e+f x)}\right ) \tan (e+f x)}{4 c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
((4*Log[Cos[e + f*x]] + 3*Log[1 - Sec[e + f*x]] + Log[1 + Sec[e + f*x]] + 2/(-1 + Sec[e + f*x]))*Tan[e + f*x])/(4*c*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqr t[c - c*Sec[e + f*x]])
Time = 0.36 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.57, 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.
| \(\displaystyle \int \frac {1}{\sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a} \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4400 |
| \(\displaystyle -\frac {a c \tan (e+f x) \int \frac {\cos (e+f x)}{a c^2 (1-\sec (e+f x))^2 (\sec (e+f x)+1)}d\sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\tan (e+f x) \int \frac {\cos (e+f x)}{(1-\sec (e+f x))^2 (\sec (e+f x)+1)}d\sec (e+f x)}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 93 |
| \(\displaystyle -\frac {\tan (e+f x) \int \left (\cos (e+f x)-\frac {3}{4 (\sec (e+f x)-1)}-\frac {1}{4 (\sec (e+f x)+1)}+\frac {1}{2 (\sec (e+f x)-1)^2}\right )d\sec (e+f x)}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\tan (e+f x) \left (\frac {1}{2 (1-\sec (e+f x))}-\frac {3}{4} \log (1-\sec (e+f x))+\log (\sec (e+f x))-\frac {1}{4} \log (\sec (e+f x)+1)\right )}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}\) |
-((((-3*Log[1 - Sec[e + f*x]])/4 + Log[Sec[e + f*x]] - Log[1 + Sec[e + f*x ]]/4 + 1/(2*(1 - Sec[e + f*x])))*Tan[e + f*x])/(c*f*Sqrt[a + a*Sec[e + f*x ]]*Sqrt[c - c*Sec[e + f*x]]))
3.2.15.3.1 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 = 1.99 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90
| method | result | size |
| default | \(-\frac {\left (6 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\cos \left (f x +e \right )-6 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right )}{4 f a \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \left (\sec \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )}\) | \(151\) |
| risch | \(\frac {2 i {\mathrm e}^{i \left (f x +e \right )}-3 i {\mathrm e}^{3 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-2 \,{\mathrm e}^{3 i \left (f x +e \right )} f x +3 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+3 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-4 \,{\mathrm e}^{3 i \left (f x +e \right )} e +2 \,{\mathrm e}^{2 i \left (f x +e \right )} f x -i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) {\mathrm e}^{3 i \left (f x +e \right )}+2 i {\mathrm e}^{2 i \left (f x +e \right )}+4 \,{\mathrm e}^{2 i \left (f x +e \right )} e +2 \,{\mathrm e}^{i \left (f x +e \right )} f x +i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )+i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )-i \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )-3 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+4 \,{\mathrm e}^{i \left (f x +e \right )} e -2 f x -4 e}{2 c \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\) | \(376\) |
-1/4/f/a*(6*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e))-4*cos(f*x+e)*ln(2/(cos(f *x+e)+1))-cos(f*x+e)-6*ln(-cot(f*x+e)+csc(f*x+e))+4*ln(2/(cos(f*x+e)+1))-1 )*(a*(sec(f*x+e)+1))^(1/2)/(-c*(sec(f*x+e)-1))^(1/2)/c/(sec(f*x+e)-1)/(cos (f*x+e)+1)*tan(f*x+e)
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {a \sec \left (f x + e\right ) + a} {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(a*c^2*sec(f*x + e)^3 - a*c^2*sec(f*x + e)^2 - a*c^2*sec(f*x + e) + a*c^2), x)
\[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )} \left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 818 vs. \(2 (148) = 296\).
Time = 0.40 (sec) , antiderivative size = 818, normalized size of antiderivative = 4.87 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\text {Too large to display} \]
-1/2*(2*(f*x + e)*cos(2*f*x + 2*e)^2 + 8*(f*x + e)*cos(1/2*arctan2(sin(2*f *x + 2*e), cos(2*f*x + 2*e)))^2 + 2*(f*x + e)*sin(2*f*x + 2*e)^2 + 8*(f*x + e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 2*f*x - (cos (2*f*x + 2*e)^2 - 4*(cos(2*f*x + 2*e) + 1)*cos(1/2*arctan2(sin(2*f*x + 2*e ), cos(2*f*x + 2*e))) + 4*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2* e)))^2 + sin(2*f*x + 2*e)^2 - 4*sin(2*f*x + 2*e)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f* x + 2*e)))^2 + 2*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 *e))) + 1) - 3*(cos(2*f*x + 2*e)^2 - 4*(cos(2*f*x + 2*e) + 1)*cos(1/2*arct an2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*cos(1/2*arctan2(sin(2*f*x + 2 *e), cos(2*f*x + 2*e)))^2 + sin(2*f*x + 2*e)^2 - 4*sin(2*f*x + 2*e)*sin(1/ 2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 4*sin(1/2*arctan2(sin(2*f *x + 2*e), cos(2*f*x + 2*e)))^2 + 2*cos(2*f*x + 2*e) + 1)*arctan2(sin(1/2* arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 1) + 4*(f*x + e)*cos(2*f*x + 2*e) - 2*(4*f*x + 4*(f*x + e)*cos(2*f*x + 2*e) + 4*e + sin(2*f*x + 2*e))*cos(1/2*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e))) - 2*(4*(f*x + e)*sin(2*f*x + 2*e) - cos( 2*f*x + 2*e) - 1)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 2 *e)/((c*cos(2*f*x + 2*e)^2 + 4*c*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(...
Time = 1.88 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=-\frac {\frac {3 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{\sqrt {-a c} {\left | c \right |}} - \frac {4 \, \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{\sqrt {-a c} {\left | c \right |}} - \frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}{\sqrt {-a c} c {\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}}{4 \, f} \]
-1/4*(3*log(abs(c)*tan(1/2*f*x + 1/2*e)^2)/(sqrt(-a*c)*abs(c)) - 4*log(abs (c*tan(1/2*f*x + 1/2*e)^2 + c))/(sqrt(-a*c)*abs(c)) - (3*c*tan(1/2*f*x + 1 /2*e)^2 - c)/(sqrt(-a*c)*c*abs(c)*tan(1/2*f*x + 1/2*e)^2))/f
Timed out. \[ \int \frac {1}{\sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]