Integrand size = 38, antiderivative size = 92 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {a \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 c f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{4 c f} \]
-1/6*a*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/c/f/(a+a*sin(f*x+e))^(1/2)-1/4*co s(f*x+e)*(c-c*sin(f*x+e))^(5/2)*(a+a*sin(f*x+e))^(1/2)/c/f
Time = 1.73 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.90 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {c \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (12 \cos (2 (e+f x))+3 \cos (4 (e+f x))+8 (9 \sin (e+f x)+\sin (3 (e+f x))))}{96 f} \]
(c*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(12*Co s[2*(e + f*x)] + 3*Cos[4*(e + f*x)] + 8*(9*Sin[e + f*x] + Sin[3*(e + f*x)] )))/(96*f)
Time = 0.68 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3320, 3042, 3219, 3042, 3217}
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.
\(\displaystyle \int \cos ^2(e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}dx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
\(\Big \downarrow \) 3219 |
\(\displaystyle \frac {\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}}{a c}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle \frac {-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}}{a c}\) |
(-1/6*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/ 2))/(4*f))/(a*c)
3.1.3.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f _.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 1, 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 = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.79
method | result | size |
default | \(\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c \left (3 \left (\cos ^{3}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \sin \left (f x +e \right )+8 \tan \left (f x +e \right )-3 \sec \left (f x +e \right )\right )}{12 f}\) | \(73\) |
1/12/f*(a*(1+sin(f*x+e)))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)*c*(3*cos(f*x+e)^ 3+4*cos(f*x+e)*sin(f*x+e)+8*tan(f*x+e)-3*sec(f*x+e))
Time = 0.33 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {{\left (3 \, c \cos \left (f x + e\right )^{4} + 4 \, {\left (c \cos \left (f x + e\right )^{2} + 2 \, c\right )} \sin \left (f x + e\right ) - 3 \, c\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{12 \, f \cos \left (f x + e\right )} \]
1/12*(3*c*cos(f*x + e)^4 + 4*(c*cos(f*x + e)^2 + 2*c)*sin(f*x + e) - 3*c)* sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e))
\[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\int \sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {3}{2}} \cos ^{2}{\left (e + f x \right )}\, dx \]
\[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\int { \sqrt {a \sin \left (f x + e\right ) + a} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.07 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {4 \, {\left (3 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} - 4 \, c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{3 \, f} \]
-4/3*(3*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8 - 4*c*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e )^6)*sqrt(a)*sqrt(c)/f
Time = 1.87 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int \cos ^2(e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2} \, dx=\frac {c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (12\,\cos \left (e+f\,x\right )+15\,\cos \left (3\,e+3\,f\,x\right )+3\,\cos \left (5\,e+5\,f\,x\right )+80\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )\right )}{96\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]