Integrand size = 38, antiderivative size = 97 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{10 a c f (c-c \sin (e+f x))^{11/2}}+\frac {\cos (e+f x) (a+a \sin (e+f x))^{7/2}}{80 a c^2 f (c-c \sin (e+f x))^{9/2}} \]
1/10*cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c/f/(c-c*sin(f*x+e))^(11/2)+1/80* cos(f*x+e)*(a+a*sin(f*x+e))^(7/2)/a/c^2/f/(c-c*sin(f*x+e))^(9/2)
Time = 8.89 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.34 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \sqrt {a (1+\sin (e+f x))} (14-10 \cos (2 (e+f x))+35 \sin (e+f x)-5 \sin (3 (e+f x)))}{40 c^6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^6 \sqrt {c-c \sin (e+f x)}} \]
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*Sqrt[a*(1 + Sin[e + f*x])]*(1 4 - 10*Cos[2*(e + f*x)] + 35*Sin[e + f*x] - 5*Sin[3*(e + f*x)]))/(40*c^6*f *(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^6*Sqrt[c - c*Si n[e + f*x]])
Time = 0.70 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3320, 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.
| \(\displaystyle \int \frac {\cos ^2(e+f x) (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (e+f x)^2 (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{13/2}}dx\) |
| \(\Big \downarrow \) 3320 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{11/2}}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{11/2}}dx}{a c}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{9/2}}dx}{10 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{(c-c \sin (e+f x))^{9/2}}dx}{10 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\) |
| \(\Big \downarrow \) 3221 |
| \(\displaystyle \frac {\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{80 c f (c-c \sin (e+f x))^{9/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{7/2}}{10 f (c-c \sin (e+f x))^{11/2}}}{a c}\) |
((Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(10*f*(c - c*Sin[e + f*x])^(11/ 2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(80*c*f*(c - c*Sin[e + f*x ])^(9/2)))/(a*c)
3.1.29.3.1 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])
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]
Time = 0.34 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.52
| method | result | size |
| default | \(\frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )-5 \left (\cos ^{3}\left (f x +e \right )\right )-17 \cos \left (f x +e \right ) \sin \left (f x +e \right )+15 \cos \left (f x +e \right )+26 \tan \left (f x +e \right )-10 \sec \left (f x +e \right )\right )}{10 f \left (\cos ^{4}\left (f x +e \right )+4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{2}\left (f x +e \right )\right )-8 \sin \left (f x +e \right )+8\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{6}}\) | \(147\) |
1/10/f*(a*(1+sin(f*x+e)))^(1/2)*a^2/(cos(f*x+e)^4+4*cos(f*x+e)^2*sin(f*x+e )-8*cos(f*x+e)^2-8*sin(f*x+e)+8)/(-c*(sin(f*x+e)-1))^(1/2)/c^6*(cos(f*x+e) ^3*sin(f*x+e)-5*cos(f*x+e)^3-17*cos(f*x+e)*sin(f*x+e)+15*cos(f*x+e)+26*tan (f*x+e)-10*sec(f*x+e))
Time = 0.32 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.68 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=-\frac {{\left (5 \, a^{2} \cos \left (f x + e\right )^{2} - 6 \, a^{2} + 5 \, {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{10 \, {\left (5 \, c^{7} f \cos \left (f x + e\right )^{5} - 20 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right ) - {\left (c^{7} f \cos \left (f x + e\right )^{5} - 12 \, c^{7} f \cos \left (f x + e\right )^{3} + 16 \, c^{7} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \]
-1/10*(5*a^2*cos(f*x + e)^2 - 6*a^2 + 5*(a^2*cos(f*x + e)^2 - 2*a^2)*sin(f *x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(5*c^7*f*cos(f *x + e)^5 - 20*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e) - (c^7*f*cos(f *x + e)^5 - 12*c^7*f*cos(f*x + e)^3 + 16*c^7*f*cos(f*x + e))*sin(f*x + e))
Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {13}{2}}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.72 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {{\left (10 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 20 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a^{2} \sqrt {c} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a}}{80 \, c^{7} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10}} \]
1/80*(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)^6 - 20*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 - 4*a^2*sqrt(c)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(c^7*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 = 15.74 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.27 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{13/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,112{}\mathrm {i}}{5\,c^7\,f}+\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,56{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,16{}\mathrm {i}}{c^7\,f}-\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,8{}\mathrm {i}}{c^7\,f}\right )}{\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,264{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )\,220{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )\,20{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )\,330{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )\,88{}\mathrm {i}-{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\sin \left (6\,e+6\,f\,x\right )\,2{}\mathrm {i}} \]
((c - c*sin(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(a + a*sin(e + f*x))^ (1/2)*112i)/(5*c^7*f) + (a^2*exp(e*6i + f*x*6i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2)*56i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2)*16i)/(c^7*f) - (a^2*exp(e*6i + f*x*6i)*sin(3*e + 3* f*x)*(a + a*sin(e + f*x))^(1/2)*8i)/(c^7*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)