Integrand size = 38, antiderivative size = 120 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {8 a^2 (9 A+B) c^4 \cos ^5(e+f x)}{315 f (c-c \sin (e+f x))^{5/2}}+\frac {2 a^2 (9 A+B) c^3 \cos ^5(e+f x)}{63 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}} \]
8/315*a^2*(9*A+B)*c^4*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)+2/63*a^2*(9*A+ B)*c^3*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(3/2)-2/9*a^2*B*c^2*cos(f*x+e)^5/f/ (c-c*sin(f*x+e))^(1/2)
Time = 9.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.88 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\frac {a^2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 (162 A-87 B+35 B \cos (2 (e+f x))+(-90 A+130 B) \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{315 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
(a^2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(162*A - 87*B + 35*B*Cos[2* (e + f*x)] + (-90*A + 130*B)*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(315* f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))
Time = 0.75 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3446, 3042, 3335, 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.
\(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^{3/2} (A+B \sin (e+f x))dx\) |
\(\Big \downarrow \) 3446 |
\(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}}dx\) |
\(\Big \downarrow \) 3335 |
\(\displaystyle a^2 c^2 \left (\frac {1}{9} (9 A+B) \int \frac {\cos ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 B \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \left (\frac {1}{9} (9 A+B) \int \frac {\cos (e+f x)^4}{\sqrt {c-c \sin (e+f x)}}dx-\frac {2 B \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle a^2 c^2 \left (\frac {1}{9} (9 A+B) \left (\frac {4}{7} c \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \left (\frac {1}{9} (9 A+B) \left (\frac {4}{7} c \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{3/2}}dx+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle a^2 c^2 \left (\frac {1}{9} (9 A+B) \left (\frac {8 c^2 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}+\frac {2 c \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{9 f \sqrt {c-c \sin (e+f x)}}\right )\) |
a^2*c^2*((-2*B*Cos[e + f*x]^5)/(9*f*Sqrt[c - c*Sin[e + f*x]]) + ((9*A + B) *((8*c^2*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) + (2*c*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^(3/2))))/9)
3.1.91.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)* (g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(f*g*(m + p + 1))), x] + S imp[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)) Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin [e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 2.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.69
method | result | size |
default | \(\frac {2 \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right )^{3} a^{2} \left (-35 B \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \left (45 A -65 B \right )-81 A +61 B \right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(83\) |
parts | \(\frac {2 A \,a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-5\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 B \,a^{2} \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (35 \left (\sin ^{4}\left (f x +e \right )\right )-85 \left (\sin ^{3}\left (f x +e \right )\right )+102 \left (\sin ^{2}\left (f x +e \right )\right )-136 \sin \left (f x +e \right )+272\right )}{315 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (A +2 B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (15 \left (\sin ^{3}\left (f x +e \right )\right )-39 \left (\sin ^{2}\left (f x +e \right )\right )+52 \sin \left (f x +e \right )-104\right )}{105 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {2 a^{2} \left (2 A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c^{2} \left (1+\sin \left (f x +e \right )\right ) \left (\sin ^{2}\left (f x +e \right )-3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(302\) |
2/315*(sin(f*x+e)-1)*c^2*(1+sin(f*x+e))^3*a^2*(-35*B*cos(f*x+e)^2+sin(f*x+ e)*(45*A-65*B)-81*A+61*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (108) = 216\).
Time = 0.26 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.90 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {2 \, {\left (35 \, B a^{2} c \cos \left (f x + e\right )^{5} + 5 \, {\left (9 \, A + 8 \, B\right )} a^{2} c \cos \left (f x + e\right )^{4} - {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{3} + 2 \, {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{2} - 8 \, {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right ) - 16 \, {\left (9 \, A + B\right )} a^{2} c + {\left (35 \, B a^{2} c \cos \left (f x + e\right )^{4} - 5 \, {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{3} - 6 \, {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right )^{2} - 8 \, {\left (9 \, A + B\right )} a^{2} c \cos \left (f x + e\right ) - 16 \, {\left (9 \, A + B\right )} a^{2} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{315 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]
-2/315*(35*B*a^2*c*cos(f*x + e)^5 + 5*(9*A + 8*B)*a^2*c*cos(f*x + e)^4 - ( 9*A + B)*a^2*c*cos(f*x + e)^3 + 2*(9*A + B)*a^2*c*cos(f*x + e)^2 - 8*(9*A + B)*a^2*c*cos(f*x + e) - 16*(9*A + B)*a^2*c + (35*B*a^2*c*cos(f*x + e)^4 - 5*(9*A + B)*a^2*c*cos(f*x + e)^3 - 6*(9*A + B)*a^2*c*cos(f*x + e)^2 - 8* (9*A + B)*a^2*c*cos(f*x + e) - 16*(9*A + B)*a^2*c)*sin(f*x + e))*sqrt(-c*s in(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*x + e) + f)
\[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=a^{2} \left (\int A c \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int A c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int \left (- A c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- A c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx + \int \left (- B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- B c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )}\right )\, dx\right ) \]
a**2*(Integral(A*c*sqrt(-c*sin(e + f*x) + c), x) + Integral(A*c*sqrt(-c*si n(e + f*x) + c)*sin(e + f*x), x) + Integral(-A*c*sqrt(-c*sin(e + f*x) + c) *sin(e + f*x)**2, x) + Integral(-A*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x )**3, x) + Integral(B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Integ ral(B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(-B*c*sqrt (-c*sin(e + f*x) + c)*sin(e + f*x)**3, x) + Integral(-B*c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**4, x))
\[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (108) = 216\).
Time = 0.65 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.98 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=-\frac {\sqrt {2} {\left (1890 \, A a^{2} c \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 35 \, B a^{2} c \cos \left (-\frac {9}{4} \, \pi + \frac {9}{2} \, f x + \frac {9}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 210 \, {\left (3 \, A a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right ) - 126 \, {\left (A a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - B a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) - 45 \, {\left (2 \, A a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a^{2} c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {7}{4} \, \pi + \frac {7}{2} \, f x + \frac {7}{2} \, e\right )\right )} \sqrt {c}}{2520 \, f} \]
-1/2520*sqrt(2)*(1890*A*a^2*c*cos(-1/4*pi + 1/2*f*x + 1/2*e)*sgn(sin(-1/4* pi + 1/2*f*x + 1/2*e)) - 35*B*a^2*c*cos(-9/4*pi + 9/2*f*x + 9/2*e)*sgn(sin (-1/4*pi + 1/2*f*x + 1/2*e)) + 210*(3*A*a^2*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a^2*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-3/4*pi + 3/2*f *x + 3/2*e) - 126*(A*a^2*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - B*a^2*c*s gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)))*cos(-5/4*pi + 5/2*f*x + 5/2*e) - 45*(2 *A*a^2*c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + B*a^2*c*sgn(sin(-1/4*pi + 1 /2*f*x + 1/2*e)))*cos(-7/4*pi + 7/2*f*x + 7/2*e))*sqrt(c)/f
Timed out. \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]