Integrand size = 30, antiderivative size = 141 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {4 c^3 \cos (e+f x) \log (1+\sin (e+f x))}{3 f \sqrt {3+3 \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {2 c^2 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{3 f \sqrt {3+3 \sin (e+f x)}}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (3+3 \sin (e+f x))^{3/2}} \]
-c*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^(3/2)-4*c^3*cos(f* x+e)*ln(1+sin(f*x+e))/a/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)-2* c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f/(a+a*sin(f*x+e))^(1/2)
Time = 6.66 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.11 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} \left (7+\cos (2 (e+f x))+16 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 \left (-1+8 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{6 \sqrt {3} f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^{3/2}} \]
-1/6*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*( 7 + Cos[2*(e + f*x)] + 16*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + 2*(-1 + 8*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]])*Sin[e + f*x]))/(Sqrt[3]*f*( Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^(3/2))
Time = 0.72 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2}}{(a \sin (e+f x)+a)^{3/2}}dx\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle -\frac {2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a}}dx}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle -\frac {2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 c \left (2 c \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {\sin (e+f x) a+a}}dx+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle -\frac {2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 c \left (\frac {2 a c^2 \cos (e+f x) \int \frac {\cos (e+f x)}{\sin (e+f x) a+a}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle -\frac {2 c \left (\frac {2 c^2 \cos (e+f x) \int \frac {1}{\sin (e+f x) a+a}d(a \sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle -\frac {2 c \left (\frac {2 c^2 \cos (e+f x) \log (a \sin (e+f x)+a)}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {c \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{f \sqrt {a \sin (e+f x)+a}}\right )}{a}-\frac {c \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{f (a \sin (e+f x)+a)^{3/2}}\) |
-((c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(f*(a + a*Sin[e + f*x])^(3/2 ))) - (2*c*((2*c^2*Cos[e + f*x]*Log[a + a*Sin[e + f*x]])/(f*Sqrt[a + a*Sin [e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) + (c*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]])))/a
3.4.90.3.1 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 = 3.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(-\frac {\sec \left (f x +e \right ) \left (8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right ) \sin \left (f x +e \right )-4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right ) \sin \left (f x +e \right )+\cos ^{2}\left (f x +e \right )+8 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1\right )-4 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-5 \sin \left (f x +e \right )-1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, c^{2}}{f \sqrt {a \left (\sin \left (f x +e \right )+1\right )}\, a}\) | \(141\) |
-1/f*sec(f*x+e)*(8*ln(-cot(f*x+e)+csc(f*x+e)+1)*sin(f*x+e)-4*ln(2/(cos(f*x +e)+1))*sin(f*x+e)+cos(f*x+e)^2+8*ln(-cot(f*x+e)+csc(f*x+e)+1)-4*ln(2/(cos (f*x+e)+1))-5*sin(f*x+e)-1)*(-c*(sin(f*x+e)-1))^(1/2)*c^2/(a*(sin(f*x+e)+1 ))^(1/2)/a
\[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integral((c^2*cos(f*x + e)^2 + 2*c^2*sin(f*x + e) - 2*c^2)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x + e) - 2*a^2), x)
Timed out. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int { \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.60 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.93 \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=-\frac {2 \, {\left (\sqrt {a} c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 4 \, \sqrt {a} c^{2} \log \left ({\left | \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {\sqrt {a} c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}}\right )} \sqrt {c}}{a^{2} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]
-2*(sqrt(a)*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 4*sqrt(a)*c^2*log(abs(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sgn(si n(-1/4*pi + 1/2*f*x + 1/2*e)) - sqrt(a)*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/ 2*e))/cos(-1/4*pi + 1/2*f*x + 1/2*e)^2)*sqrt(c)/(a^2*f*sgn(cos(-1/4*pi + 1 /2*f*x + 1/2*e)))
Timed out. \[ \int \frac {(c-c \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^{3/2}} \, dx=\int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \]