Integrand size = 26, antiderivative size = 103 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {64 a c^4 \cos ^3(e+f x)}{105 f (c-c \sin (e+f x))^{3/2}}+\frac {16 a c^3 \cos ^3(e+f x)}{35 f \sqrt {c-c \sin (e+f x)}}+\frac {2 a c^2 \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f} \]
64/105*a*c^4*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)+16/35*a*c^3*cos(f*x+e)^ 3/f/(c-c*sin(f*x+e))^(1/2)+2/7*a*c^2*cos(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/f
Time = 0.91 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.91 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 (-157+15 \cos (2 (e+f x))+108 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{105 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
-1/105*(a*c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(-157 + 15*Cos[2*(e + f*x)] + 108*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
Time = 0.56 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3215, 3042, 3153, 3042, 3153, 3042, 3152}
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.
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a c \int \cos ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}dx\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a c \left (\frac {8}{7} c \int \cos ^2(e+f x) \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {8}{7} c \int \cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}dx+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\) |
| \(\Big \downarrow \) 3153 |
| \(\displaystyle a c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\cos (e+f x)^2}{\sqrt {c-c \sin (e+f x)}}dx+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\) |
| \(\Big \downarrow \) 3152 |
| \(\displaystyle a c \left (\frac {8}{7} c \left (\frac {8 c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}+\frac {2 c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}\right )+\frac {2 c \cos ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{7 f}\right )\) |
a*c*((2*c*Cos[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/(7*f) + (8*c*((8*c^2*Co s[e + f*x]^3)/(15*f*(c - c*Sin[e + f*x])^(3/2)) + (2*c*Cos[e + f*x]^3)/(5* f*Sqrt[c - c*Sin[e + f*x]])))/7)
3.3.91.3.1 Defintions of rubi rules used
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x ])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) Int[(g* Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 2.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.67
| method | result | size |
| default | \(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right )^{2} a \left (15 \left (\sin ^{2}\left (f x +e \right )\right )-54 \sin \left (f x +e \right )+71\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(69\) |
| parts | \(-\frac {2 a \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )-14 \sin \left (f x +e \right )+43\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a \left (\sin \left (f x +e \right )-1\right ) c^{3} \left (\sin \left (f x +e \right )+1\right ) \left (3 \left (\sin ^{3}\left (f x +e \right )\right )-12 \left (\sin ^{2}\left (f x +e \right )\right )+23 \sin \left (f x +e \right )-46\right )}{21 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(144\) |
-2/105*(sin(f*x+e)-1)*c^3*(sin(f*x+e)+1)^2*a*(15*sin(f*x+e)^2-54*sin(f*x+e )+71)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.25 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.48 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\frac {2 \, {\left (15 \, a c^{2} \cos \left (f x + e\right )^{4} + 39 \, a c^{2} \cos \left (f x + e\right )^{3} - 8 \, a c^{2} \cos \left (f x + e\right )^{2} + 32 \, a c^{2} \cos \left (f x + e\right ) + 64 \, a c^{2} - {\left (15 \, a c^{2} \cos \left (f x + e\right )^{3} - 24 \, a c^{2} \cos \left (f x + e\right )^{2} - 32 \, a c^{2} \cos \left (f x + e\right ) - 64 \, a c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{105 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]
2/105*(15*a*c^2*cos(f*x + e)^4 + 39*a*c^2*cos(f*x + e)^3 - 8*a*c^2*cos(f*x + e)^2 + 32*a*c^2*cos(f*x + e) + 64*a*c^2 - (15*a*c^2*cos(f*x + e)^3 - 24 *a*c^2*cos(f*x + e)^2 - 32*a*c^2*cos(f*x + e) - 64*a*c^2)*sin(f*x + e))*sq rt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
\[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=a \left (\int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int \left (- c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \]
a*(Integral(c**2*sqrt(-c*sin(e + f*x) + c), x) + Integral(-c**2*sqrt(-c*si n(e + f*x) + c)*sin(e + f*x), x) + Integral(-c**2*sqrt(-c*sin(e + f*x) + c )*sin(e + f*x)**2, x) + Integral(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f* x)**3, x))
\[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Time = 0.38 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.32 \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (525 \, a c^{2} \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 35 \, a c^{2} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 63 \, a c^{2} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 15 \, a c^{2} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {c}}{420 \, f} \]
-1/420*sqrt(2)*(525*a*c^2*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 35*a*c^2*cos(-3/4*pi + 3/2*f*x + 3/2*e)*sgn(sin(-1/4* pi + 1/2*f*x + 1/2*e)) - 63*a*c^2*cos(-5/4*pi + 5/2*f*x + 5/2*e)*sgn(sin(- 1/4*pi + 1/2*f*x + 1/2*e)) + 15*a*c^2*cos(-7/4*pi + 7/2*f*x + 7/2*e)*sgn(s in(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(c)/f
Timed out. \[ \int (a+a \sin (e+f x)) (c-c \sin (e+f x))^{5/2} \, dx=\int \left (a+a\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]