Integrand size = 36, antiderivative size = 95 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {2 a^2 \log (1-\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {a \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{f \sqrt {c-c \sec (e+f x)}} \]
2*a^2*ln(1-sec(f*x+e))*tan(f*x+e)/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e) )^(1/2)+a*(a+a*sec(f*x+e))^(1/2)*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)
Time = 0.41 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\frac {a^2 (2 \log (1-\sec (e+f x))+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a (1+\sec (e+f x))} \sqrt {c-c \sec (e+f x)}} \]
(a^2*(2*Log[1 - Sec[e + f*x]] + Sec[e + f*x])*Tan[e + f*x])/(f*Sqrt[a*(1 + Sec[e + f*x])]*Sqrt[c - c*Sec[e + f*x]])
Time = 0.56 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 4443, 3042, 4440}
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.
\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^{3/2}}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx\) |
\(\Big \downarrow \) 4443 |
\(\displaystyle 2 a \int \frac {\sec (e+f x) \sqrt {\sec (e+f x) a+a}}{\sqrt {c-c \sec (e+f x)}}dx+\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 a \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \sqrt {\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}}{\sqrt {c-c \csc \left (e+f x+\frac {\pi }{2}\right )}}dx+\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}}\) |
\(\Big \downarrow \) 4440 |
\(\displaystyle \frac {2 a^2 \tan (e+f x) \log (1-\sec (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{f \sqrt {c-c \sec (e+f x)}}\) |
(2*a^2*Log[1 - Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqr t[c - c*Sec[e + f*x]]) + (a*Sqrt[a + a*Sec[e + f*x]]*Tan[e + f*x])/(f*Sqrt [c - c*Sec[e + f*x]])
3.2.17.3.1 Defintions of rubi rules used
Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)])/Sq rt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*c*Log[1 + (b/ a)*Csc[e + f*x]]*(Cot[e + f*x]/(b*f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc [e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && Eq Q[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(c sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[(-d)*Cot[e + f *x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(f*(m + n))), x] + Simp[c*((2*n - 1)/(m + n)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[b *c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n - 1/2, 0] && !LtQ[m, -2^(-1)] && !(IGtQ[m - 1/2, 0] && LtQ[m, n])
Time = 3.83 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.35
method | result | size |
default | \(\frac {a \sqrt {a \left (\sec \left (f x +e \right )+1\right )}\, \left (4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \sin \left (f x +e \right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )+\sin \left (f x +e \right )+\tan \left (f x +e \right )\right )}{f \left (\cos \left (f x +e \right )+1\right ) \sqrt {-c \left (\sec \left (f x +e \right )-1\right )}}\) | \(128\) |
risch | \(-\frac {2 i a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, \left (2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) {\mathrm e}^{3 i \left (f x +e \right )}-\ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) {\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )-{\mathrm e}^{i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )-2 \,{\mathrm e}^{2 i \left (f x +e \right )} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+{\mathrm e}^{2 i \left (f x +e \right )} \ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )-{\mathrm e}^{i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}-2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )+\ln \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )\right )}{\left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{1+{\mathrm e}^{2 i \left (f x +e \right )}}}\, f \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}\) | \(278\) |
1/f*a*(a*(sec(f*x+e)+1))^(1/2)/(cos(f*x+e)+1)/(-c*(sec(f*x+e)-1))^(1/2)*(4 *ln(-cot(f*x+e)+csc(f*x+e))*sin(f*x+e)-2*ln(-cot(f*x+e)+csc(f*x+e)-1)*sin( f*x+e)-2*ln(-cot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)+sin(f*x+e)+tan(f*x+e))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
integral(-(a*sec(f*x + e)^2 + a*sec(f*x + e))*sqrt(a*sec(f*x + e) + a)*sqr t(-c*sec(f*x + e) + c)/(c*sec(f*x + e) - c), x)
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sec {\left (e + f x \right )}}{\sqrt {- c \left (\sec {\left (e + f x \right )} - 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (87) = 174\).
Time = 0.39 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.89 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=-\frac {2 \, {\left (a \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) \sin \left (2 \, f x + 2 \, e\right ) + {\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right ) - 2 \, {\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 1\right ) - {\left (a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{{\left (c \cos \left (2 \, f x + 2 \, e\right )^{2} + c \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, c \cos \left (2 \, f x + 2 \, e\right ) + c\right )} f} \]
-2*(a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))*sin(2*f*x + 2*e ) + (a*cos(2*f*x + 2*e)^2 + a*sin(2*f*x + 2*e)^2 + 2*a*cos(2*f*x + 2*e) + a)*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1) - 2*(a*cos(2*f*x + 2*e) ^2 + a*sin(2*f*x + 2*e)^2 + 2*a*cos(2*f*x + 2*e) + a)*arctan2(sin(1/2*arct an2(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) - (a*cos(2*f*x + 2*e) + a)*sin(1/2*arctan2(sin(2 *f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((c*cos(2*f*x + 2*e)^2 + c*sin(2*f*x + 2*e)^2 + 2*c*cos(2*f*x + 2*e) + c)*f)
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\int { \frac {{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )}{\sqrt {-c \sec \left (f x + e\right ) + c}} \,d x } \]
Timed out. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{\sqrt {c-c \sec (e+f x)}} \, dx=\int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\cos \left (e+f\,x\right )\,\sqrt {c-\frac {c}{\cos \left (e+f\,x\right )}}} \,d x \]