Integrand size = 30, antiderivative size = 134 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{15 f}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f} \] Output:
-1/15*a^3*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-2/15* a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)*(c-c*sin(f*x+e))^(7/2)/f-1/6*a*cos(f *x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f
Time = 8.77 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.16 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 (-1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)} (75 \cos (2 (e+f x))+30 \cos (4 (e+f x))+5 \cos (6 (e+f x))+600 \sin (e+f x)+100 \sin (3 (e+f x))+12 \sin (5 (e+f x)))}{960 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:
Integrate[(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),x]
Output:
-1/960*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*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)]))/(f*( Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])^5)
Time = 0.64 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\) |
Input:
Int[(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2),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
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 = 3.36 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.95
| method | result | size |
| default | \(\frac {8 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (10 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+6 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) a^{2} c^{3} \sqrt {4}}{15 f}\) | \(127\) |
Input:
int((a+sin(f*x+e)*a)^(5/2)*(c-c*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)
Output:
8/15/f*sin(1/4*Pi+1/2*f*x+1/2*e)^4*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)*(10*cos(1/4 *Pi+1/2*f*x+1/2*e)^6+6*cos(1/4*Pi+1/2*f*x+1/2*e)^4+3*cos(1/4*Pi+1/2*f*x+1/ 2*e)^2+1)*a^2*c^3*4^(1/2)
Time = 0.11 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (5 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} - 5 \, a^{2} c^{3} + 2 \, {\left (3 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} + 4 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 8 \, a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{30 \, f \cos \left (f x + e\right )} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="fric as")
Output:
1/30*(5*a^2*c^3*cos(f*x + e)^6 - 5*a^2*c^3 + 2*(3*a^2*c^3*cos(f*x + e)^4 + 4*a^2*c^3*cos(f*x + e)^2 + 8*a^2*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. \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \] Input:
integrate((a+a*sin(f*x+e))**(5/2)*(c-c*sin(f*x+e))**(7/2),x)
Output:
Timed out
\[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="maxi ma")
Output:
integrate((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(7/2), x)
Time = 0.15 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.16 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {16 \, {\left (10 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 24 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 15 \, a^{2} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{15 \, f} \] Input:
integrate((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x, algorithm="giac ")
Output:
16/15*(10*a^2*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)^12 - 24*a^2*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 + 15*a^2*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 = 20.50 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.93 \[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\frac {a^2\,c^3\,\sqrt {a\,\left (\sin \left (e+f\,x\right )+1\right )}\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (75\,\cos \left (e+f\,x\right )+105\,\cos \left (3\,e+3\,f\,x\right )+35\,\cos \left (5\,e+5\,f\,x\right )+5\,\cos \left (7\,e+7\,f\,x\right )+700\,\sin \left (2\,e+2\,f\,x\right )+112\,\sin \left (4\,e+4\,f\,x\right )+12\,\sin \left (6\,e+6\,f\,x\right )\right )}{960\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \] Input:
int((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(7/2),x)
Output:
(a^2*c^3*(a*(sin(e + f*x) + 1))^(1/2)*(-c*(sin(e + f*x) - 1))^(1/2)*(75*co s(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)))/( 960*f*(cos(2*e + 2*f*x) + 1))
\[ \int (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} c^{3} \left (-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{5}d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{4}d x +2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}d x \right )-2 \left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}d x \right )-\left (\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )d x \right )+\int \sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}d x \right ) \] Input:
int((a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2),x)
Output:
sqrt(c)*sqrt(a)*a**2*c**3*( - int(sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f *x) + 1)*sin(e + f*x)**5,x) + 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)**2,x) - 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))