Integrand size = 38, antiderivative size = 51 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) \log (1+\sin (e+f x))}{a f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
Time = 2.49 (sec) , antiderivative size = 102, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)}} \]
(2*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] - Sin[(e + f *x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/(f*(a*(1 + Sin[e + f*x])) ^(3/2)*Sqrt[c - c*Sin[e + f*x]])
Time = 0.58 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {3042, 3320, 3042, 3216, 3042, 3146, 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.
\(\displaystyle \int \frac {\cos ^2(e+f x)}{(a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (e+f x)^2}{(a \sin (e+f x)+a)^{3/2} \sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx}{a c}\) |
\(\Big \downarrow \) 3216 |
\(\displaystyle \frac {\cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\cos (e+f x) \log (a \sin (e+f x)+a)}{a f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\) |
(Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(a*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[ c - c*Sin[e + f*x]])
3.1.54.3.1 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[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(a^(p/ 2)*c^(p/2)) Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p/2]
Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43
method | result | size |
default | \(-\frac {\left (\ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-2 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )\right ) \cos \left (f x +e \right )}{f \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a}\) | \(73\) |
-1/f*(ln(2/(1+cos(f*x+e)))-2*ln(-cot(f*x+e)+csc(f*x+e)+1))*cos(f*x+e)/(a*( 1+sin(f*x+e)))^(1/2)/(-c*(sin(f*x+e)-1))^(1/2)/a
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\cos ^{2}{\left (e + f x \right )}}{\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]
\[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {\cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 \, \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right )}{a^{\frac {3}{2}} \sqrt {c} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
-2*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/(a^(3/2)*sqrt(c)*f*sgn(cos(-1/ 4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))
Timed out. \[ \int \frac {\cos ^2(e+f x)}{(a+a \sin (e+f x))^{3/2} \sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]