Integrand size = 38, antiderivative size = 140 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{15 c f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}{5 c f}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}{5 c f} \] Output:
-2/15*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/c/f/(a+a*sin(f*x+e))^(1/2)-1/5 *a*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(5/2)/c/f-1/5*cos(f* x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(5/2)/c/f
Time = 1.81 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.58 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=-\frac {c \sec ^2(e+f x) (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} \left (15-10 \sin ^2(e+f x)+3 \sin ^4(e+f x)\right ) \tan (e+f x)}{15 f} \] Input:
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^( 3/2),x]
Output:
-1/15*(c*Sec[e + f*x]^2*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(3/2)*S qrt[c - c*Sin[e + f*x]]*(15 - 10*Sin[e + f*x]^2 + 3*Sin[e + f*x]^4)*Tan[e + f*x])/f
Time = 0.93 (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.
| \(\displaystyle \int \cos ^2(e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (e+f x)^2 (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3320 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {\frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4}{5} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {\frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {4}{5} a \left (\frac {1}{2} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}}{a c}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {\frac {4}{5} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}{4 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}{5 f}}{a c}\) |
Input:
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(3/2),x ]
Output:
(-1/5*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2 ))/f + (4*a*(-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)))/5)/(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]
\[\int \cos \left (f x +e \right )^{2} \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}d x\]
Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x)
Output:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x)
Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {{\left (3 \, a c \cos \left (f x + e\right )^{4} + 4 \, a c \cos \left (f x + e\right )^{2} + 8 \, a c\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )} \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x, al gorithm="fricas")
Output:
1/15*(3*a*c*cos(f*x + e)^4 + 4*a*c*cos(f*x + e)^2 + 8*a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e)/(f*cos(f*x + e))
Timed out. \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(3/2),x)
Output:
Timed out
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{2} \,d x } \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x, al gorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(3/2)*cos(f*x + e)^2, x)
Time = 0.37 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.03 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {16 \, {\left (6 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 15 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 10 \, a c \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \] Input:
integrate(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x, al gorithm="giac")
Output:
16/15*(6*a*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)^10 - 15*a*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 + 10*a*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 = 0.99 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.56 \[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\frac {a\,c\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (175\,\sin \left (2\,e+2\,f\,x\right )+28\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )\right )}{240\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(3/2),x )
Output:
(a*c*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(175*sin(2 *e + 2*f*x) + 28*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x)))/(240*f*(cos(2*e + 2*f*x) + 1))
\[ \int \cos ^2(e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2} \, dx=\sqrt {c}\, \sqrt {a}\, a c \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )^{2}d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right )^{2}d x \right ) \] Input:
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(3/2),x)
Output:
sqrt(c)*sqrt(a)*a*c*( - 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,x))