Integrand size = 38, antiderivative size = 193 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {8 a^3 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {4 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{c f \sqrt {c-c \sin (e+f x)}}-\frac {\cos (e+f x) (a+a \sin (e+f x))^{5/2}}{3 c f \sqrt {c-c \sin (e+f x)}} \]
-a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(1/2)-1/3*cos(f* x+e)*(a+a*sin(f*x+e))^(5/2)/c/f/(c-c*sin(f*x+e))^(1/2)-8*a^3*cos(f*x+e)*ln (1-sin(f*x+e))/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-4*a^2*cos (f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(1/2)
Time = 8.80 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (a (1+\sin (e+f x)))^{5/2} \left (-12 \cos (2 (e+f x))+192 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+87 \sin (e+f x)-\sin (3 (e+f x))\right )}{12 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \sqrt {c-c \sin (e+f x)}} \]
-1/12*((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(a*(1 + Sin[e + f*x]))^(5/2)* (-12*Cos[2*(e + f*x)] + 192*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + 87* Sin[e + f*x] - Sin[3*(e + f*x)]))/(c*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])^5*Sqrt[c - c*Sin[e + f*x]])
Time = 1.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.03, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3320, 3042, 3219, 3042, 3219, 3042, 3219, 3042, 3216, 3042, 3146, 16}
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.
| \(\displaystyle \int \frac {\cos ^2(e+f x) (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (e+f x)^2 (a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3320 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{\sqrt {c-c \sin (e+f x)}}dx}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^{7/2}}{\sqrt {c-c \sin (e+f x)}}dx}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {2 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \int \frac {(\sin (e+f x) a+a)^{5/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {2 a \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {2 a \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {2 a \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 a \left (2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {2 a \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {2 a \left (2 a \left (-\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f \sqrt {c-c \sin (e+f x)}}}{a c}\) |
(-1/3*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(f*Sqrt[c - c*Sin[e + f* x]]) + 2*a*(-1/2*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f*Sqrt[c - c *Sin[e + f*x]]) + 2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(f*Sq rt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(f*Sqrt[c - c*Sin[e + f*x]]))))/(a*c)
3.1.24.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ]
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])
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 = 0.21 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.40
| method | result | size |
| default | \(\frac {\left (\cos ^{4}\left (f x +e \right )+\left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+6 \left (\cos ^{3}\left (f x +e \right )\right )-5 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+24 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \cos \left (f x +e \right )-24 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \sin \left (f x +e \right )-48 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \cos \left (f x +e \right )+48 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-17 \left (\cos ^{2}\left (f x +e \right )\right )-22 \cos \left (f x +e \right ) \sin \left (f x +e \right )+24 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-48 \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )-1\right )-6 \cos \left (f x +e \right )-16 \sin \left (f x +e \right )+16\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, a^{2}}{3 f \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c}\) | \(270\) |
1/3/f*(cos(f*x+e)^4+cos(f*x+e)^3*sin(f*x+e)+6*cos(f*x+e)^3-5*cos(f*x+e)^2* sin(f*x+e)+24*ln(2/(1+cos(f*x+e)))*cos(f*x+e)-24*ln(2/(1+cos(f*x+e)))*sin( f*x+e)-48*ln(csc(f*x+e)-cot(f*x+e)-1)*cos(f*x+e)+48*ln(csc(f*x+e)-cot(f*x+ e)-1)*sin(f*x+e)-17*cos(f*x+e)^2-22*cos(f*x+e)*sin(f*x+e)+24*ln(2/(1+cos(f *x+e)))-48*ln(csc(f*x+e)-cot(f*x+e)-1)-6*cos(f*x+e)-16*sin(f*x+e)+16)*(a*( 1+sin(f*x+e)))^(1/2)*a^2/(cos(f*x+e)+sin(f*x+e)+1)/(-c*(sin(f*x+e)-1))^(1/ 2)/c
\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integral((a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2*sin(f*x + e) - 2*a^2*c os(f*x + e)^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^2*cos (f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2), x)
Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {4 \, a^{\frac {5}{2}} \sqrt {c} {\left (\frac {6 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, c^{4} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} + 3 \, c^{4} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 6 \, c^{4} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c^{6} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{3 \, f} \]
4/3*a^(5/2)*sqrt(c)*(6*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*sgn (sin(-1/4*pi + 1/2*f*x + 1/2*e))) + (2*c^4*cos(-1/4*pi + 1/2*f*x + 1/2*e)^ 6 + 3*c^4*cos(-1/4*pi + 1/2*f*x + 1/2*e)^4 + 6*c^4*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2)/(c^6*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))*sgn(cos(-1/4*pi + 1/ 2*f*x + 1/2*e))/f
Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]