Integrand size = 38, antiderivative size = 224 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {(7 A+3 B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}+\frac {(7 A+3 B) \cos (e+f x)}{16 a^3 f (c-c \sin (e+f x))^{3/2}}-\frac {(7 A+3 B) \sec (e+f x)}{12 a^3 c f \sqrt {c-c \sin (e+f x)}}-\frac {(7 A+3 B) \sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{30 a^3 c^2 f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 a^3 c^3 f} \] Output:
1/32*(7*A+3*B)*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/ 2))*2^(1/2)/a^3/c^(3/2)/f+1/16*(7*A+3*B)*cos(f*x+e)/a^3/f/(c-c*sin(f*x+e)) ^(3/2)-1/12*(7*A+3*B)*sec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(1/2)-1/30*(7*A+ 3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(1/2)/a^3/c^2/f-1/5*(A-B)*sec(f*x+e)^5* (c-c*sin(f*x+e))^(3/2)/a^3/c^3/f
Result contains complex when optimal does not.
Time = 6.02 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.59 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-40 A \cos ^2(e+f x)+24 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-30 (3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+15 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(15+15 i) \sqrt [4]{-1} (7 A+3 B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+30 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{240 a^3 f (1+\sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \] Input:
Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x] )^(3/2)),x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(-40*A*Cos[e + f*x]^2 + 24*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2 ])^2 - 30*(3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*(Cos[(e + f*x) /2] + Sin[(e + f*x)/2])^4 + 15*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2 ])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - (15 + 15*I)*(-1)^(1/4)*(7*A + 3*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])^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 30*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5))/(240*a^3*f *(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(3/2))
Time = 1.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.90, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {3042, 3446, 3042, 3334, 3042, 3154, 3042, 3166, 3042, 3129, 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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3334 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \int \sec ^4(e+f x) \sqrt {c-c \sin (e+f x)}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \int \frac {\sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{10} c (7 A+3 B) \left (\frac {5}{6} c \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^3(e+f x) \sqrt {c-c \sin (e+f x)}}{3 f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{3/2}}{5 f}}{a^3 c^3}\) |
Input:
Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(3/2 )),x]
Output:
(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(3/2))/f + ((7*A + 3*B) *c*(-1/3*(Sec[e + f*x]^3*Sqrt[c - c*Sin[e + f*x]])/f + (5*c*(-(Sec[e + f*x ]/(f*Sqrt[c - c*Sin[e + f*x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sq rt[2]*Sqrt[c - c*Sin[e + f*x]])]/(2*Sqrt[2]*c^(3/2)*f) + Cos[e + f*x]/(2*f *(c - c*Sin[e + f*x])^(3/2))))/2))/6))/10)/(a^3*c^3)
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[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
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/(a*f*g*(p + 1))), x] + Simp[a*((m + p + 1)/(g^2*(p + 1))) 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] && GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ [m + 1/2, 2*p]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]], x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*S qrt[a + b*Sin[e + f*x]])), x] + Simp[a*((2*p + 1)/(2*g^2*(p + 1))) Int[(g *Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e , f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[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[(-(b* c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) , x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))) Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
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 = 0.38 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.38
| method | result | size |
| default | \(-\frac {278 A \,c^{\frac {7}{2}}-18 B \,c^{\frac {7}{2}}+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c +45 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, \sin \left (f x +e \right ) c -105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c -45 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c +42 A \sin \left (f x +e \right ) c^{\frac {7}{2}}+18 B \sin \left (f x +e \right ) c^{\frac {7}{2}}-210 A \sin \left (f x +e \right )^{3} c^{\frac {7}{2}}-90 B \sin \left (f x +e \right )^{3} c^{\frac {7}{2}}-350 A \sin \left (f x +e \right )^{2} c^{\frac {7}{2}}-150 B \sin \left (f x +e \right )^{2} c^{\frac {7}{2}}}{480 c^{\frac {9}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(308\) |
Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x,method=_R ETURNVERBOSE)
Output:
-1/480/c^(9/2)/a^3*(278*A*c^(7/2)-18*B*c^(7/2)+105*A*(c*(1+sin(f*x+e)))^(5 /2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*sin(f*x+ e)*c+45*B*(c*(1+sin(f*x+e)))^(5/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^ (1/2)/c^(1/2))*2^(1/2)*sin(f*x+e)*c-105*A*(c*(1+sin(f*x+e)))^(5/2)*arctanh (1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c-45*B*(c*(1+sin(f* x+e)))^(5/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2) *c+42*A*sin(f*x+e)*c^(7/2)+18*B*sin(f*x+e)*c^(7/2)-210*A*sin(f*x+e)^3*c^(7 /2)-90*B*sin(f*x+e)^3*c^(7/2)-350*A*sin(f*x+e)^2*c^(7/2)-150*B*sin(f*x+e)^ 2*c^(7/2))/(1+sin(f*x+e))^2/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.10 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.24 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {15 \, \sqrt {2} {\left ({\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{3}\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (25 \, {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, {\left (7 \, A + 3 \, B\right )} \cos \left (f x + e\right )^{2} - 28 \, A - 12 \, B\right )} \sin \left (f x + e\right ) - 36 \, A - 84 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{960 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, al gorithm="fricas")
Output:
1/960*(15*sqrt(2)*((7*A + 3*B)*cos(f*x + e)^3*sin(f*x + e) + (7*A + 3*B)*c os(f*x + e)^3)*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + ( c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(25*(7*A + 3*B)*cos(f*x + e)^2 + 3*(5*(7*A + 3*B)*cos(f*x + e)^2 - 28*A - 12*B)*sin(f*x + e) - 36*A - 84* B)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3 *c^2*f*cos(f*x + e)^3)
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(3/2),x)
Output:
Timed out
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x, al gorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),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. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(3/2 )),x)
Output:
int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(3/2 )), x)
\[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \left (\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}+\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )^{2}+\sin \left (f x +e \right )+1}d x \right ) a +\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{5}+\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )^{2}+\sin \left (f x +e \right )+1}d x \right ) b \right )}{a^{3} c^{2}} \] Input:
int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(3/2),x)
Output:
(sqrt(c)*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**5 + sin(e + f*x)**4 - 2*sin(e + f*x)**3 - 2*sin(e + f*x)**2 + sin(e + f*x) + 1),x)*a + int((s qrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**5 + sin(e + f*x)**4 - 2*sin(e + f*x)**3 - 2*sin(e + f*x)**2 + sin(e + f*x) + 1),x)*b))/(a**3*c **2)