Integrand size = 25, antiderivative size = 101 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {8 a^2 (5 A+3 B) \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a (5 A+3 B) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f} \]
-2/5*B*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-8/15*a^2*(5*A+3*B)*cos(f*x+e)/f /(a+a*sin(f*x+e))^(1/2)-2/15*a*(5*A+3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2) /f
Time = 1.09 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (50 A+39 B-3 B \cos (2 (e+f x))+2 (5 A+9 B) \sin (e+f x))}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
-1/15*(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]* (50*A + 39*B - 3*B*Cos[2*(e + f*x)] + 2*(5*A + 9*B)*Sin[e + f*x]))/(f*(Cos [(e + f*x)/2] + Sin[(e + f*x)/2]))
Time = 0.39 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3230, 3042, 3126, 3042, 3125}
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.
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {1}{5} (5 A+3 B) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} (5 A+3 B) \int (\sin (e+f x) a+a)^{3/2}dx-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{5} (5 A+3 B) \left (\frac {4}{3} a \int \sqrt {\sin (e+f x) a+a}dx-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {1}{5} (5 A+3 B) \left (-\frac {8 a^2 \cos (e+f x)}{3 f \sqrt {a \sin (e+f x)+a}}-\frac {2 a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}\right )-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}\) |
(-2*B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f) + ((5*A + 3*B)*((-8*a ^2*Cos[e + f*x])/(3*f*Sqrt[a + a*Sin[e + f*x]]) - (2*a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f)))/5
3.3.96.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Time = 1.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.76
method | result | size |
default | \(\frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (-3 B \left (\cos ^{2}\left (f x +e \right )\right )+\sin \left (f x +e \right ) \left (5 A +9 B \right )+25 A +21 B \right )}{15 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(77\) |
parts | \(\frac {2 A \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin \left (f x +e \right )+5\right )}{3 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}+\frac {2 B \left (1+\sin \left (f x +e \right )\right ) a^{2} \left (\sin \left (f x +e \right )-1\right ) \left (\sin ^{2}\left (f x +e \right )+3 \sin \left (f x +e \right )+6\right )}{5 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\) | \(118\) |
2/15*(1+sin(f*x+e))*a^2*(sin(f*x+e)-1)*(-3*B*cos(f*x+e)^2+sin(f*x+e)*(5*A+ 9*B)+25*A+21*B)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f
Time = 0.25 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.36 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {2 \, {\left (3 \, B a \cos \left (f x + e\right )^{3} - {\left (5 \, A + 6 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (25 \, A + 21 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a - {\left (3 \, B a \cos \left (f x + e\right )^{2} + {\left (5 \, A + 9 \, B\right )} a \cos \left (f x + e\right ) - 4 \, {\left (5 \, A + 3 \, B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]
2/15*(3*B*a*cos(f*x + e)^3 - (5*A + 6*B)*a*cos(f*x + e)^2 - (25*A + 21*B)* a*cos(f*x + e) - 4*(5*A + 3*B)*a - (3*B*a*cos(f*x + e)^2 + (5*A + 9*B)*a*c os(f*x + e) - 4*(5*A + 3*B)*a)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/(f*c os(f*x + e) + f*sin(f*x + e) + f)
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sin {\left (e + f x \right )}\right )\, dx \]
\[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.38 \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\frac {\sqrt {2} {\left (3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) + 30 \, {\left (3 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 2 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 5 \, {\left (2 \, A a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sin \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {a}}{30 \, f} \]
1/30*sqrt(2)*(3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-5/4*pi + 5/2* f*x + 5/2*e) + 30*(3*A*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 2*B*a*sgn(c os(-1/4*pi + 1/2*f*x + 1/2*e)))*sin(-1/4*pi + 1/2*f*x + 1/2*e) + 5*(2*A*a* sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*B*a*sgn(cos(-1/4*pi + 1/2*f*x + 1/ 2*e)))*sin(-3/4*pi + 3/2*f*x + 3/2*e))*sqrt(a)/f
Timed out. \[ \int (a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x)) \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2} \,d x \]