Integrand size = 30, antiderivative size = 96 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {a \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \log (1-\cos (e+f x)) \tan (e+f x)}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]
-a*tan(f*x+e)/f/(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(1/2)+a*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 = 0.68 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=\frac {a \left (\log (\cos (e+f x))+\log (1-\sec (e+f x))+\frac {1}{-1+\sec (e+f x)}\right ) \tan (e+f x)}{c f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
(a*(Log[Cos[e + f*x]] + Log[1 - Sec[e + f*x]] + (-1 + Sec[e + f*x])^(-1))* Tan[e + f*x])/(c*f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {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.
| \(\displaystyle \int \frac {\sqrt {a \sec (e+f x)+a}}{(c-c \sec (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \csc \left (e+f x+\frac {\pi }{2}\right )+a}}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 4395 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (e+f x) a+a}}{\sqrt {c-c \sec (e+f x)}}dx}{c}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx}{c}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 4399 |
| \(\displaystyle \frac {a \tan (e+f x) \int \frac {1}{c \cos (e+f x)-c}d\cos (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}\) |
-((a*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) ) + (a*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(c*f*Sqrt[a + a*Sec[e + f*x]]*S qrt[c - c*Sec[e + f*x]])
3.1.91.3.1 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 = 2.41 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.54
| method | result | size |
| default | \(-\frac {\left (4 \cos \left (f x +e \right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )-2 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\cos \left (f x +e \right )-1\right ) \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \tan \left (f x +e \right )}{2 f \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}\, c \left (\sec \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right )}\) | \(148\) |
| risch | \(-\frac {\sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (2 i {\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+{\mathrm e}^{2 i \left (f x +e \right )} f x -4 i {\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+2 \,{\mathrm e}^{2 i \left (f x +e \right )} e -2 \,{\mathrm e}^{i \left (f x +e \right )} f x -2 i {\mathrm e}^{i \left (f x +e \right )}+2 i \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-4 \,{\mathrm e}^{i \left (f x +e \right )} e +f x +2 e \right )}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f}\) | \(226\) |
-1/2/f*(4*cos(f*x+e)*ln(-cot(f*x+e)+csc(f*x+e))-2*cos(f*x+e)*ln(2/(cos(f*x +e)+1))-4*ln(-cot(f*x+e)+csc(f*x+e))+2*ln(2/(cos(f*x+e)+1))-cos(f*x+e)-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)
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a}}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2), x)
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a \left (\sec {\left (e + f x \right )} + 1\right )}}{\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (88) = 176\).
Time = 0.37 (sec) , antiderivative size = 399, normalized size of antiderivative = 4.16 \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=-\frac {{\left ({\left (f x + e\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \cos \left (f x + e\right )^{2} + {\left (f x + e\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, {\left (f x + e\right )} \sin \left (f x + e\right )^{2} + f x + 2 \, {\left (2 \, {\left (2 \, \cos \left (f x + e\right ) - 1\right )} \cos \left (2 \, f x + 2 \, e\right ) - \cos \left (2 \, f x + 2 \, e\right )^{2} - 4 \, \cos \left (f x + e\right )^{2} - \sin \left (2 \, f x + 2 \, e\right )^{2} + 4 \, \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 4 \, \sin \left (f x + e\right )^{2} + 4 \, \cos \left (f x + e\right ) - 1\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) + 2 \, {\left (f x - 2 \, {\left (f x + e\right )} \cos \left (f x + e\right ) + e + \sin \left (f x + e\right )\right )} \cos \left (2 \, f x + 2 \, e\right ) - 4 \, {\left (f x + e\right )} \cos \left (f x + e\right ) - 2 \, {\left (2 \, {\left (f x + e\right )} \sin \left (f x + e\right ) + \cos \left (f x + e\right )\right )} \sin \left (2 \, f x + 2 \, e\right ) + e + 2 \, \sin \left (f x + e\right )\right )} \sqrt {a} \sqrt {c}}{{\left (c^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, c^{2} \cos \left (f x + e\right )^{2} + c^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 4 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) + 4 \, c^{2} \sin \left (f x + e\right )^{2} - 4 \, c^{2} \cos \left (f x + e\right ) + c^{2} - 2 \, {\left (2 \, c^{2} \cos \left (f x + e\right ) - c^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right )} f} \]
-((f*x + e)*cos(2*f*x + 2*e)^2 + 4*(f*x + e)*cos(f*x + e)^2 + (f*x + e)*si n(2*f*x + 2*e)^2 + 4*(f*x + e)*sin(f*x + e)^2 + f*x + 2*(2*(2*cos(f*x + e) - 1)*cos(2*f*x + 2*e) - cos(2*f*x + 2*e)^2 - 4*cos(f*x + e)^2 - sin(2*f*x + 2*e)^2 + 4*sin(2*f*x + 2*e)*sin(f*x + e) - 4*sin(f*x + e)^2 + 4*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) + 2*(f*x - 2*(f*x + e)* cos(f*x + e) + e + sin(f*x + e))*cos(2*f*x + 2*e) - 4*(f*x + e)*cos(f*x + e) - 2*(2*(f*x + e)*sin(f*x + e) + cos(f*x + e))*sin(2*f*x + 2*e) + e + 2* sin(f*x + e))*sqrt(a)*sqrt(c)/((c^2*cos(2*f*x + 2*e)^2 + 4*c^2*cos(f*x + e )^2 + c^2*sin(2*f*x + 2*e)^2 - 4*c^2*sin(2*f*x + 2*e)*sin(f*x + e) + 4*c^2 *sin(f*x + e)^2 - 4*c^2*cos(f*x + e) + c^2 - 2*(2*c^2*cos(f*x + e) - c^2)* cos(2*f*x + 2*e))*f)
\[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=\int { \frac {\sqrt {a \sec \left (f x + e\right ) + a}}{{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx=\int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]