Integrand size = 38, antiderivative size = 140 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 c f \sqrt {a+a \sin (e+f x)}}-\frac {2 a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 c f} \] Output:
-1/15*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/c/f/(a+a*sin(f*x+e))^(1/2)-2/1 5*a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(7/2)/c/f-1/6*cos(f *x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/c/f
Time = 7.70 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {c^2 (-1+\sin (e+f x))^2 (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (75 \cos (2 (e+f x))+30 \cos (4 (e+f x))+5 \cos (6 (e+f x))+600 \sin (e+f x)+100 \sin (3 (e+f x))+12 \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^( 5/2),x]
Output:
(c^2*(-1 + Sin[e + f*x])^2*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]]*(75*Cos[2*(e + f*x)] + 30*Cos[4*(e + f*x)] + 5*Cos[6*(e + f*x)] + 6 00*Sin[e + f*x] + 100*Sin[3*(e + f*x)] + 12*Sin[5*(e + f*x)]))/(960*f*(Cos [(e + f*x)/2] - Sin[(e + f*x)/2])^5*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^ 3)
Time = 0.94 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3320, 3042, 3219, 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.
\(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 3320 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx}{a c}\) |
\(\Big \downarrow \) 3219 |
\(\displaystyle \frac {\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}}{a c}\) |
\(\Big \downarrow \) 3219 |
\(\displaystyle \frac {\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}}{a c}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle \frac {\frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}}{a c}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2),x ]
Output:
(-1/6*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2 ))/f + (2*a*(-1/10*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(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])^(7/2))/(5*f)))/3)/(a*c)
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]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x)
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.73 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {{\left (5 \, a c^{2} \cos \left (f x + e\right )^{6} - 5 \, a c^{2} + 2 \, {\left (3 \, a c^{2} \cos \left (f x + e\right )^{4} + 4 \, a c^{2} \cos \left (f x + e\right )^{2} + 8 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, al gorithm="fricas")
Output:
1/30*(5*a*c^2*cos(f*x + e)^6 - 5*a*c^2 + 2*(3*a*c^2*cos(f*x + e)^4 + 4*a*c ^2*cos(f*x + e)^2 + 8*a*c^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(- c*sin(f*x + e) + c)/(f*cos(f*x + e))
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, al gorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2)*cos(f*x + e)^2, x)
Time = 0.38 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.07 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {16 \, {\left (10 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 24 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 15 \, a c^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x, al gorithm="giac")
Output:
16/15*(10*a*c^2*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)^12 - 24*a*c^2*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)^10 + 15*a*c^2*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)*sqrt(a)*sqrt(c )/f
Time = 19.63 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\frac {a\,c^2\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,\cos \left (e+f\,x\right )+105\,\cos \left (3\,e+3\,f\,x\right )+35\,\cos \left (5\,e+5\,f\,x\right )+5\,\cos \left (7\,e+7\,f\,x\right )+700\,\sin \left (2\,e+2\,f\,x\right )+112\,\sin \left (4\,e+4\,f\,x\right )+12\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(5/2),x )
Output:
(a*c^2*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(75*cos( e + f*x) + 105*cos(3*e + 3*f*x) + 35*cos(5*e + 5*f*x) + 5*cos(7*e + 7*f*x) + 700*sin(2*e + 2*f*x) + 112*sin(4*e + 4*f*x) + 12*sin(6*e + 6*f*x)))/(96 0*f*(cos(2*e + 2*f*x) + 1))
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, a \,c^{2} \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{3}d x -\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right )-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}d x \right ) \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2),x)
Output:
sqrt(c)*sqrt(a)*a*c**2*(int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x)**3,x) - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x)**2,x) - int(sqrt(sin(e + f *x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2*sin(e + f*x),x) + int(s qrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)**2,x))