Integrand size = 30, antiderivative size = 89 \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a+a \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{5 f} \] Output:
-1/10*a^2*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/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))^(7/2)/f
Time = 7.62 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.64 \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 (-1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)} (20 \cos (2 (e+f x))+5 \cos (4 (e+f x))+70 \sin (e+f x)+5 \sin (3 (e+f x))-\sin (5 (e+f x)))}{80 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:
Integrate[(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2),x]
Output:
-1/80*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*S in[e + f*x]]*(20*Cos[2*(e + f*x)] + 5*Cos[4*(e + f*x)] + 70*Sin[e + f*x] + 5*Sin[3*(e + f*x)] - Sin[5*(e + f*x)]))/(f*(Cos[(e + f*x)/2] - Sin[(e + f *x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
Time = 0.45 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {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.
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx\) |
\(\Big \downarrow \) 3219 |
\(\displaystyle \frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\) |
\(\Big \downarrow \) 3217 |
\(\displaystyle -\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2),x]
Output:
-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)
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])
Time = 1.82 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.40
method | result | size |
default | \(\frac {4 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (4 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a \,c^{3} \sqrt {4}}{5 f}\) | \(125\) |
Input:
int((a+sin(f*x+e)*a)^(3/2)*(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
4/5/f*sin(1/4*Pi+1/2*f*x+1/2*e)^2*tan(1/4*Pi+1/2*f*x+1/2*e)*(a*sin(1/4*Pi+ 1/2*f*x+1/2*e)^2)^(1/2)*(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(4*cos(1/4*P i+1/2*f*x+1/2*e)^6+3*cos(1/4*Pi+1/2*f*x+1/2*e)^4+2*cos(1/4*Pi+1/2*f*x+1/2* e)^2+1)*a*c^3*4^(1/2)
Time = 0.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (5 \, a c^{3} \cos \left (f x + e\right )^{4} - 5 \, a c^{3} - 2 \, {\left (a c^{3} \cos \left (f x + e\right )^{4} - 2 \, a c^{3} \cos \left (f x + e\right )^{2} - 4 \, a c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="fric as")
Output:
1/10*(5*a*c^3*cos(f*x + e)^4 - 5*a*c^3 - 2*(a*c^3*cos(f*x + e)^4 - 2*a*c^3 *cos(f*x + e)^2 - 4*a*c^3)*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. \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(3/2)*(c-c*sin(f*x+e))**(7/2),x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxi ma")
Output:
integrate((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(7/2), x)
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17 \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {8 \, {\left (4 \, a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 5 \, a c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{5 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac ")
Output:
-8/5*(4*a*c^3*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 - 5*a*c^3*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 = 18.88 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.25 \[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {a\,c^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (20\,\cos \left (e+f\,x\right )+25\,\cos \left (3\,e+3\,f\,x\right )+5\,\cos \left (5\,e+5\,f\,x\right )+75\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-\sin \left (6\,e+6\,f\,x\right )\right )}{80\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x))^(7/2),x)
Output:
(a*c^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(20*cos( e + f*x) + 25*cos(3*e + 3*f*x) + 5*cos(5*e + 5*f*x) + 75*sin(2*e + 2*f*x) + 4*sin(4*e + 4*f*x) - sin(6*e + 6*f*x)))/(80*f*(cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2} \, dx=\sqrt {c}\, \sqrt {a}\, a \,c^{3} \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x \right )+2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right )-2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) \] Input:
int((a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2),x)
Output:
sqrt(c)*sqrt(a)*a*c**3*( - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**4,x) + 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* x) + 1)*sin(e + f*x)**3,x) - 2*int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x),x) + int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x ) + 1),x))