Integrand size = 30, antiderivative size = 140 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {9 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {27 \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
1/2*a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(5/2)-a^2*cos(f *x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(3/2)-a^3*cos(f*x+e)*ln( 1-sin(f*x+e))/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 6.63 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.36 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {9 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2} \left (-2-3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 \left (1+\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \]
(9*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(5/2)* (-2 - 3*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Cos[2*(e + f*x)]*Log[Co s[(e + f*x)/2] - Sin[(e + f*x)/2]] + 4*(1 + Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(c^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5* (-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])
Time = 0.73 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3218, 3042, 3218, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}\right )}{c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{2 f (c-c \sin (e+f x))^{5/2}}-\frac {a \left (\frac {a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}\right )}{c}\) |
(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(2*f*(c - c*Sin[e + f*x])^(5/2 )) - (a*((a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(f*(c - c*Sin[e + f*x]) ^(3/2)) + (a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])))/c
3.4.64.3.1 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[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 2*n + 1))) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b ^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] && !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Time = 3.03 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.39
| method | result | size |
| default | \(-\frac {\sec \left (f x +e \right ) \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )-2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )-2 \left (\cos ^{2}\left (f x +e \right )\right )-4 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2}}{f \left (\sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}\) | \(195\) |
-1/f*sec(f*x+e)*(2*cos(f*x+e)^2*ln(-cot(f*x+e)+csc(f*x+e)-1)-cos(f*x+e)^2* ln(2/(cos(f*x+e)+1))+4*ln(-cot(f*x+e)+csc(f*x+e)-1)*sin(f*x+e)-2*ln(2/(cos (f*x+e)+1))*sin(f*x+e)-2*cos(f*x+e)^2-4*ln(-cot(f*x+e)+csc(f*x+e)-1)+2*ln( 2/(cos(f*x+e)+1))+2)*(a*(sin(f*x+e)+1))^(1/2)*a^2/(sin(f*x+e)-1)/(-c*(sin( f*x+e)-1))^(1/2)/c^2
\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integral((a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*c os(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.31 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\frac {8 \, a^{\frac {5}{2}} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (c^{3} - \frac {4 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {4 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} + \frac {a^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}}}{f} \]
-(8*a^(5/2)*sqrt(c)*sin(f*x + e)^2/((c^3 - 4*c^3*sin(f*x + e)/(cos(f*x + e ) + 1) + 6*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^3*sin(f*x + e)^3/ (cos(f*x + e) + 1)^3 + c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)*(cos(f*x + e) + 1)^2) - 2*a^(5/2)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/c^(5/2) + a^(5/2)*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/c^(5/2))/f
Time = 0.35 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {{\left (4 \, a^{2} \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {4 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )} \sqrt {a}}{2 \, c^{\frac {5}{2}} f \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
1/2*(4*a^2*log(abs(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*sgn(cos(-1/4*pi + 1/2* f*x + 1/2*e)) + (4*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1 /2*f*x + 1/2*e)^2 - a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))/sin(-1/4*pi + 1/2*f*x + 1/2*e)^4)*sqrt(a)/(c^(5/2)*f*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e) ))
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]