Integrand size = 30, antiderivative size = 138 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\frac {3 \cos (e+f x) (3+3 \sin (e+f x))^{3/2}}{f (c-c \sin (e+f x))^{3/2}}+\frac {108 \cos (e+f x) \log (1-\sin (e+f x))}{c f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}+\frac {18 \cos (e+f x) \sqrt {3+3 \sin (e+f x)}}{c f \sqrt {c-c \sin (e+f x)}} \]
a*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(3/2)+4*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)+2*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 = 6.85 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.22 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {9 \sqrt {3} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{5/2} \left (7+\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+\left (2-16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{2 c f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (-1+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \]
(-9*Sqrt[3]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(5/2) *(7 + Cos[2*(e + f*x)] + 16*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + (2 - 16*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(2*c*f*(Cos[ (e + f*x)/2] + Sin[(e + f*x)/2])^5*(-1 + Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])
Time = 0.73 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.04, 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, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{5/2}}{(c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \int \frac {(\sin (e+f x) a+a)^{3/2}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (2 a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (\frac {2 a^2 c \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (-\frac {2 a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{f (c-c \sin (e+f x))^{3/2}}-\frac {2 a \left (-\frac {2 a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(f*(c - c*Sin[e + f*x])^(3/2)) - (2*a*((-2*a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(f*Sqrt[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]])))/c
3.4.63.3.1 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])
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 = 3.02 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(\frac {\sec \left (f x +e \right ) \left (4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )-8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )-4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )-1\right )+5 \sin \left (f x +e \right )-1\right ) \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a^{2}}{f \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c}\) | \(140\) |
1/f*sec(f*x+e)*(4*ln(2/(cos(f*x+e)+1))*sin(f*x+e)-8*ln(-cot(f*x+e)+csc(f*x +e)-1)*sin(f*x+e)+cos(f*x+e)^2-4*ln(2/(cos(f*x+e)+1))+8*ln(-cot(f*x+e)+csc (f*x+e)-1)+5*sin(f*x+e)-1)*(a*(sin(f*x+e)+1))^(1/2)*a^2/(-c*(sin(f*x+e)-1) )^(1/2)/c
\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
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)/(c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2), x)
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98 \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, a^{\frac {5}{2}} \sqrt {c} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {2 \, \log \left (-\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}{c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {1}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}\right )} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{f} \]
-2*a^(5/2)*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e)^2/(c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 2*log(-cos(-1/4*pi + 1/2*f*x + 1/2*e)^2 + 1)/(c^2*s gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 1/((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))))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))/f
Timed out. \[ \int \frac {(3+3 \sin (e+f x))^{5/2}}{(c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]