Integrand size = 30, antiderivative size = 133 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{40 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (3+3 \sin (e+f x))^{5/2}}{240 c^2 f (c-c \sin (e+f x))^{7/2}} \]
1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f/(c-c*sin(f*x+e))^(11/2)+1/40*cos( f*x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c*sin(f*x+e))^(9/2)+1/240*cos(f*x+e)* (a+a*sin(f*x+e))^(5/2)/c^2/f/(c-c*sin(f*x+e))^(7/2)
Time = 5.67 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.89 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {3 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2} (9-5 \cos (2 (e+f x))+10 \sin (e+f x))}{10 c^5 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-1+\sin (e+f x))^5 \sqrt {c-c \sin (e+f x)}} \]
(-3*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(5/2) *(9 - 5*Cos[2*(e + f*x)] + 10*Sin[e + f*x]))/(10*c^5*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(-1 + Sin[e + f*x])^5*Sqrt[c - c*Sin[e + f*x]])
Time = 0.65 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3222, 3042, 3222, 3042, 3221}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{11/2}}dx\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{9/2}}dx}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{5/2}}{(c-c \sin (e+f x))^{7/2}}dx}{8 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}\) |
| \(\Big \downarrow \) 3221 |
| \(\displaystyle \frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{10 f (c-c \sin (e+f x))^{11/2}}+\frac {\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{48 c f (c-c \sin (e+f x))^{7/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{8 f (c-c \sin (e+f x))^{9/2}}}{5 c}\) |
(Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(10*f*(c - c*Sin[e + f*x])^(11/2 )) + ((Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(8*f*(c - c*Sin[e + f*x])^ (9/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(48*c*f*(c - c*Sin[e + f*x])^(7/2)))/(5*c)
3.4.67.3.1 Defintions of rubi rules used
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*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne Q[m, -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*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
Time = 3.04 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11
| method | result | size |
| default | \(\frac {\sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2} \left (2 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-10 \left (\cos ^{3}\left (f x +e \right )\right )-24 \sin \left (f x +e \right ) \cos \left (f x +e \right )+35 \cos \left (f x +e \right )+37 \tan \left (f x +e \right )-25 \sec \left (f x +e \right )\right )}{15 f \left (\cos ^{4}\left (f x +e \right )+4 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{5}}\) | \(148\) |
1/15/f*(a*(sin(f*x+e)+1))^(1/2)*a^2/(cos(f*x+e)^4+4*sin(f*x+e)*cos(f*x+e)^ 2-8*cos(f*x+e)^2-8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(1/2)/c^5*(2*cos(f*x+ e)^3*sin(f*x+e)-10*cos(f*x+e)^3-24*sin(f*x+e)*cos(f*x+e)+35*cos(f*x+e)+37* tan(f*x+e)-25*sec(f*x+e))
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.11 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 5 \, a^{2} \sin \left (f x + e\right ) - 7 \, a^{2}\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (5 \, c^{6} f \cos \left (f x + e\right )^{5} - 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right ) - {\left (c^{6} f \cos \left (f x + e\right )^{5} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} + 16 \, c^{6} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]
-1/15*(5*a^2*cos(f*x + e)^2 - 5*a^2*sin(f*x + e) - 7*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^6*f*cos(f*x + e)^5 - 20*c^6*f*cos( f*x + e)^3 + 16*c^6*f*cos(f*x + e) - (c^6*f*cos(f*x + e)^5 - 12*c^6*f*cos( f*x + e)^3 + 16*c^6*f*cos(f*x + e))*sin(f*x + e))
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.99 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=-\frac {{\left (10 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{240 \, c^{6} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}} \]
-1/240*(10*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1 /2*f*x + 1/2*e)^4 - 15*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin (-1/4*pi + 1/2*f*x + 1/2*e)^2 + 6*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^6*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10)
Time = 12.45 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.05 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{11/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,96{}\mathrm {i}}{5\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,64{}\mathrm {i}}{3\,c^6\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,32{}\mathrm {i}}{3\,c^6\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]
((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^ (1/2)*96i)/(5*c^6*f) + (a^2*exp(e*6i + f*x*6i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*64i)/(3*c^6*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2)*32i)/(3*c^6*f)))/(cos(e + f*x)*exp(e*6i + f*x*6i)* 264i - exp(e*6i + f*x*6i)*cos(3*e + 3*f*x)*220i + exp(e*6i + f*x*6i)*cos(5 *e + 5*f*x)*20i - exp(e*6i + f*x*6i)*sin(2*e + 2*f*x)*330i + exp(e*6i + f* x*6i)*sin(4*e + 4*f*x)*88i - exp(e*6i + f*x*6i)*sin(6*e + 6*f*x)*2i)