Integrand size = 38, antiderivative size = 161 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {4 \sqrt {2} a^2 (A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a^2 B c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}-\frac {2 a^2 (A+B) c \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^{3/2}}-\frac {4 a^2 (A+B) \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \] Output:
4*2^(1/2)*a^2*(A+B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e) )^(1/2))/c^(1/2)/f-2/5*a^2*B*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(5/2)-2/3 *a^2*(A+B)*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(3/2)-4*a^2*(A+B)*cos(f*x+e)/ f/(c-c*sin(f*x+e))^(1/2)
Result contains complex when optimal does not.
Time = 12.05 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.09 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2 \left ((120+120 i) \sqrt [4]{-1} (A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right )+\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (70 A+79 B-3 B \cos (2 (e+f x))+2 (5 A+11 B) \sin (e+f x))\right )}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + f*x]],x]
Output:
-1/15*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2*((12 0 + 120*I)*(-1)^(1/4)*(A + B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])] + (Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(70*A + 79*B - 3*B*Cos[2 *(e + f*x)] + 2*(5*A + 11*B)*Sin[e + f*x])))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*Sqrt[c - c*Sin[e + f*x]])
Time = 0.92 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules used = {3042, 3446, 3042, 3339, 3042, 3158, 3042, 3158, 3042, 3128, 219}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle a^2 c^2 \left ((A+B) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left ((A+B) \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{5/2}}dx-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{3/2}}dx}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3158 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \left (\frac {2 \int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{c}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \left (-\frac {4 \int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^2 c^2 \left ((A+B) \left (\frac {2 \left (\frac {2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{c^{3/2} f}-\frac {2 \cos (e+f x)}{c f \sqrt {c-c \sin (e+f x)}}\right )}{c}-\frac {2 \cos ^3(e+f x)}{3 c f (c-c \sin (e+f x))^{3/2}}\right )-\frac {2 B \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{5/2}}\right )\) |
Input:
Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/Sqrt[c - c*Sin[e + f*x]] ,x]
Output:
a^2*c^2*((-2*B*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(5/2)) + (A + B)* ((-2*Cos[e + f*x]^3)/(3*c*f*(c - c*Sin[e + f*x])^(3/2)) + (2*((2*Sqrt[2]*A rcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(c^(3/2 )*f) - (2*Cos[e + f*x])/(c*f*Sqrt[c - c*Sin[e + f*x]])))/c))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In tegersQ[2*m, 2*p]
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] && 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 = 0.73 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.22
| method | result | size |
| default | \(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a^{2} \left (30 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) A +30 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) B -3 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}-5 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c -5 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c -30 A \,c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}-30 B \,c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(197\) |
| parts | \(-\frac {a^{2} A \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{\sqrt {c}\, \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} B \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (15 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}+10 c \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}-30 c^{2} \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{15 c^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}+\frac {a^{2} \left (A +2 B \right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (-3 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )+2 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}}\right )}{3 c^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {a^{2} \left (2 A +B \right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \left (\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-2 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(409\) |
Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x,method=_R ETURNVERBOSE)
Output:
-2/15*(sin(f*x+e)-1)*(c*(1+sin(f*x+e)))^(1/2)*a^2*(30*c^(5/2)*2^(1/2)*arct anh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*A+30*c^(5/2)*2^(1/2)*arc tanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*B-3*B*(c*(1+sin(f*x+e)) )^(5/2)-5*A*(c*(1+sin(f*x+e)))^(3/2)*c-5*B*(c*(1+sin(f*x+e)))^(3/2)*c-30*A *c^2*(c*(1+sin(f*x+e)))^(1/2)-30*B*c^2*(c*(1+sin(f*x+e)))^(1/2))/c^3/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. 310 vs. \(2 (142) = 284\).
Time = 0.10 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.93 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {2 \, {\left (\frac {15 \, \sqrt {2} {\left ({\left (A + B\right )} a^{2} c \cos \left (f x + e\right ) - {\left (A + B\right )} a^{2} c \sin \left (f x + e\right ) + {\left (A + B\right )} a^{2} c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{3} + {\left (5 \, A + 14 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - {\left (35 \, A + 41 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2} + {\left (3 \, B a^{2} \cos \left (f x + e\right )^{2} - {\left (5 \, A + 11 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \, {\left (10 \, A + 13 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{15 \, {\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al gorithm="fricas")
Output:
2/15*(15*sqrt(2)*((A + B)*a^2*c*cos(f*x + e) - (A + B)*a^2*c*sin(f*x + e) + (A + B)*a^2*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt (c) + 3*cos(f*x + e) + 2)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e ) - cos(f*x + e) - 2))/sqrt(c) + (3*B*a^2*cos(f*x + e)^3 + (5*A + 14*B)*a^ 2*cos(f*x + e)^2 - (35*A + 41*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)*a^2 + (3*B*a^2*cos(f*x + e)^2 - (5*A + 11*B)*a^2*cos(f*x + e) - 4*(10*A + 13*B)* a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin( f*x + e) + c*f)
\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=a^{2} \left (\int \frac {A}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 A \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {A \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin {\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {2 B \sin ^{2}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {B \sin ^{3}{\left (e + f x \right )}}{\sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(1/2),x)
Output:
a**2*(Integral(A/sqrt(-c*sin(e + f*x) + c), x) + Integral(2*A*sin(e + f*x) /sqrt(-c*sin(e + f*x) + c), x) + Integral(A*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*sin(e + f*x)/sqrt(-c*sin(e + f*x) + c), x) + Integral(2*B*sin(e + f*x)**2/sqrt(-c*sin(e + f*x) + c), x) + Integral(B*si n(e + f*x)**3/sqrt(-c*sin(e + f*x) + c), x))
\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al gorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2/sqrt(-c*sin(f*x + e) + c), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x, al gorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(1/ 2),x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2)/(c - c*sin(e + f*x))^(1/ 2), x)
\[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\sqrt {c}\, a^{2} \left (-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )-1}d x \right ) a -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{3}}{\sin \left (f x +e \right )-1}d x \right ) b -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )-1}d x \right ) a -2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )-1}d x \right ) b -2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )-1}d x \right ) a -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )-1}d x \right ) b \right )}{c} \] Input:
int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(1/2),x)
Output:
(sqrt(c)*a**2*( - int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) - 1),x)*a - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + f*x) - 1),x)*b - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) - 1),x)*a - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x) - 1),x)*b - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) - 1),x)*a - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) - 1),x)*b))/c