Integrand size = 30, antiderivative size = 169 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=-\frac {81 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{35 f \sqrt {3+3 \sin (e+f x)}}-\frac {27 \cos (e+f x) \sqrt {3+3 \sin (e+f x)} (c-c \sin (e+f x))^{9/2}}{14 f}-\frac {27 \cos (e+f x) (3+3 \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{9/2}}{28 f}-\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f} \]
-3/28*a^2*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(9/2)/f-1/8*a *cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(9/2)/f-1/35*a^4*cos(f *x+e)*(c-c*sin(f*x+e))^(9/2)/f/(a+a*sin(f*x+e))^(1/2)-1/14*a^3*cos(f*x+e)* (c-c*sin(f*x+e))^(9/2)*(a+a*sin(f*x+e))^(1/2)/f
Time = 8.95 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.06 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=\frac {27 \sqrt {3} c^4 (-1+\sin (e+f x))^4 (1+\sin (e+f x))^{7/2} \sqrt {c-c \sin (e+f x)} (1960 \cos (2 (e+f x))+980 \cos (4 (e+f x))+280 \cos (6 (e+f x))+35 \cos (8 (e+f x))+19600 \sin (e+f x)+3920 \sin (3 (e+f x))+784 \sin (5 (e+f x))+80 \sin (7 (e+f x)))}{35840 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^7} \]
(27*Sqrt[3]*c^4*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^(7/2)*Sqrt[c - c* Sin[e + f*x]]*(1960*Cos[2*(e + f*x)] + 980*Cos[4*(e + f*x)] + 280*Cos[6*(e + f*x)] + 35*Cos[8*(e + f*x)] + 19600*Sin[e + f*x] + 3920*Sin[3*(e + f*x) ] + 784*Sin[5*(e + f*x)] + 80*Sin[7*(e + f*x)]))/(35840*f*(Cos[(e + f*x)/2 ] - Sin[(e + f*x)/2])^9*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^7)
Time = 0.81 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3219, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{9/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{9/2}dx\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {3}{4} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{4} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {3}{4} a \left (\frac {4}{7} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{4} a \left (\frac {4}{7} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {3}{4} a \left (\frac {4}{7} a \left (\frac {1}{3} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {3}{4} a \left (\frac {4}{7} a \left (\frac {1}{3} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{9/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {3}{4} a \left (\frac {4}{7} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{9/2}}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{9/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{9/2}}{7 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{9/2}}{8 f}\) |
-1/8*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(9/2) )/f + (3*a*(-1/7*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(9/2))/f + (4*a*(-1/15*(a^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2) )/(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])^(9/2))/(6*f)))/7))/4
3.4.69.3.1 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])
Time = 4.65 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.65
| method | result | size |
| default | \(\frac {\sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{3} c^{4} \left (35 \left (\cos ^{7}\left (f x +e \right )\right )+40 \left (\cos ^{5}\left (f x +e \right )\right ) \sin \left (f x +e \right )+48 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+64 \sin \left (f x +e \right ) \cos \left (f x +e \right )+128 \tan \left (f x +e \right )-35 \sec \left (f x +e \right )\right )}{280 f}\) | \(110\) |
1/280/f*(a*(sin(f*x+e)+1))^(1/2)*(-c*(sin(f*x+e)-1))^(1/2)*a^3*c^4*(35*cos (f*x+e)^7+40*cos(f*x+e)^5*sin(f*x+e)+48*cos(f*x+e)^3*sin(f*x+e)+64*sin(f*x +e)*cos(f*x+e)+128*tan(f*x+e)-35*sec(f*x+e))
Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.76 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=\frac {{\left (35 \, a^{3} c^{4} \cos \left (f x + e\right )^{8} - 35 \, a^{3} c^{4} + 8 \, {\left (5 \, a^{3} c^{4} \cos \left (f x + e\right )^{6} + 6 \, a^{3} c^{4} \cos \left (f x + e\right )^{4} + 8 \, a^{3} c^{4} \cos \left (f x + e\right )^{2} + 16 \, a^{3} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \]
1/280*(35*a^3*c^4*cos(f*x + e)^8 - 35*a^3*c^4 + 8*(5*a^3*c^4*cos(f*x + e)^ 6 + 6*a^3*c^4*cos(f*x + e)^4 + 8*a^3*c^4*cos(f*x + e)^2 + 16*a^3*c^4)*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. \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=\text {Timed out} \]
\[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.21 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=-\frac {32 \, {\left (35 \, a^{3} c^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{16} - 120 \, a^{3} c^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} + 140 \, a^{3} c^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 56 \, a^{3} c^{4} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \]
-32/35*(35*a^3*c^4*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)^16 - 120*a^3*c^4*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)^14 + 140*a^3*c^4*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 - 56* a^3*c^4*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)*sqrt(a)*sqrt(c)/f
Time = 11.49 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.22 \[ \int (3+3 \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{9/2} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {7\,a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {a^3\,c^4\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]
(exp(- e*8i - f*x*8i)*(c - c*sin(e + f*x))^(1/2)*((35*a^3*c^4*exp(e*8i + f *x*8i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) + (7*a^3*c^4*exp(e* 8i + f*x*8i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) + (7*a^3* c^4*exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/(128*f ) + (a^3*c^4*exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*(a + a*sin(e + f*x))^(1/2 ))/(64*f) + (a^3*c^4*exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*(a + a*sin(e + f* x))^(1/2))/(512*f) + (7*a^3*c^4*exp(e*8i + f*x*8i)*sin(3*e + 3*f*x)*(a + a *sin(e + f*x))^(1/2))/(32*f) + (7*a^3*c^4*exp(e*8i + f*x*8i)*sin(5*e + 5*f *x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (a^3*c^4*exp(e*8i + f*x*8i)*sin( 7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos(e + f*x))