Integrand size = 38, antiderivative size = 258 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {7 (9 A+B) \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {7 (9 A+B) \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {7 (9 A+B) \sec (e+f x)}{240 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {7 (9 A+B) \sec (e+f x)}{96 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(9 A+B) \sec ^3(e+f x)}{30 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f} \]
7/128*(9*A+B)*cos(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+7/240*(9*A+B)*sec( f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+7/256*(9*A+B)*arctanh(1/2*cos(f*x+e) *c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^3/c^(5/2)/f*2^(1/2)-7/96*(9*A+B )*sec(f*x+e)/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/30*(9*A+B)*sec(f*x+e)^3/a^ 3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(1/ 2)/a^3/c^3/f
Result contains complex when optimal does not.
Time = 3.57 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.86 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-720 A \cos ^4(e+f x)+96 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+80 (-3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+60 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+15 (15 A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(105+105 i) \sqrt [4]{-1} (9 A+B) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+120 (A+B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+30 (15 A+7 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{1920 a^3 f (1+\sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \]
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(-720*A*Cos[e + f*x]^4 + 96*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/ 2])^4 + 80*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e + f* x)/2] + Sin[(e + f*x)/2])^2 + 60*(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*A + 7*B)*(Cos[(e + f *x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - (10 5 + 105*I)*(-1)^(1/4)*(9*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])^4*(Cos[(e + f*x)/2] + Si n[(e + f*x)/2])^5 + 120*(A + B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[( e + f*x)/2])^5 + 30*(15*A + 7*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*S in[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5))/(1920*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))
Time = 1.23 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.91, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 3446, 3042, 3334, 3042, 3166, 3042, 3160, 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.
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin (e+f x)}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3334 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \int \frac {\sec ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \int \frac {1}{\cos (e+f x)^4 \sqrt {c-c \sin (e+f x)}}dx-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {1}{10} c (9 A+B) \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {(A-B) \sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
(-1/5*((A - B)*Sec[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/f + ((9*A + B)*c*( -1/3*Sec[e + f*x]^3/(f*Sqrt[c - c*Sin[e + f*x]]) + (7*c*(Sec[e + f*x]/(4*f *(c - c*Sin[e + f*x])^(3/2)) + (5*(-(Sec[e + f*x]/(f*Sqrt[c - c*Sin[e + f* x]])) + (3*c*(ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[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))/(8*c)))/6))/10)/(a^3*c^3)
3.2.30.3.1 Defintions of rubi rules used
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*(2*m + p + 1))), x] + Simp[(m + p + 1)/(a*(2*m + p + 1)) 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] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] & & IntegersQ[2*m, 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 = 0.99 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.59
| method | result | size |
| default | \(-\frac {\left (-1890 c^{\frac {9}{2}} A -210 c^{\frac {9}{2}} B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (1260 c^{\frac {9}{2}} A +140 c^{\frac {9}{2}} B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-945 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +252 c^{\frac {9}{2}} A -105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +28 c^{\frac {9}{2}} B \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (-1890 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +864 c^{\frac {9}{2}} A -210 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +96 c^{\frac {9}{2}} B \right ) \sin \left (f x +e \right )+1890 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} A +96 c^{\frac {9}{2}} A +210 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c +c \sin \left (f x +e \right )\right )^{\frac {5}{2}} c^{2} B +864 c^{\frac {9}{2}} B}{3840 c^{\frac {13}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(410\) |
-1/3840/c^(13/2)/a^3*((-1890*c^(9/2)*A-210*c^(9/2)*B)*cos(f*x+e)^4+(1260*c ^(9/2)*A+140*c^(9/2)*B)*cos(f*x+e)^2*sin(f*x+e)+(-945*2^(1/2)*arctanh(1/2* (c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*A+252*c ^(9/2)*A-105*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*( c+c*sin(f*x+e))^(5/2)*c^2*B+28*c^(9/2)*B)*cos(f*x+e)^2+(-1890*2^(1/2)*arct anh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2 *A+864*c^(9/2)*A-210*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^ (1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*B+96*c^(9/2)*B)*sin(f*x+e)+1890*2^(1/2)* arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*(c+c*sin(f*x+e))^(5/2) *c^2*A+96*c^(9/2)*A+210*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2) /c^(1/2))*(c+c*sin(f*x+e))^(5/2)*c^2*B+864*c^(9/2)*B)/(1+sin(f*x+e))^2/(si n(f*x+e)-1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.29 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.92 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {105 \, \sqrt {2} {\left (9 \, A + B\right )} \sqrt {c} \cos \left (f x + e\right )^{5} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (105 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{4} - 14 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (35 \, {\left (9 \, A + B\right )} \cos \left (f x + e\right )^{2} + 216 \, A + 24 \, B\right )} \sin \left (f x + e\right ) - 48 \, A - 432 \, B\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{7680 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \]
1/7680*(105*sqrt(2)*(9*A + B)*sqrt(c)*cos(f*x + e)^5*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)/(c os(f*x + e)^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*( 105*(9*A + B)*cos(f*x + e)^4 - 14*(9*A + B)*cos(f*x + e)^2 - 2*(35*(9*A + B)*cos(f*x + e)^2 + 216*A + 24*B)*sin(f*x + e) - 48*A - 432*B)*sqrt(-c*sin (f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5)
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 901 vs. \(2 (227) = 454\).
Time = 0.56 (sec) , antiderivative size = 901, normalized size of antiderivative = 3.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]
1/30720*(420*sqrt(2)*(9*A*sqrt(c) + B*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1))/(a^3*c^3*sgn(sin(-1/4 *pi + 1/2*f*x + 1/2*e))) - 15*sqrt(2)*(A*sqrt(c) + B*sqrt(c) - 32*A*sqrt(c )*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1 ) - 16*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f *x + 1/2*e) + 1) + 378*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(c os(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 42*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/ 2*f*x + 1/2*e) + 1)^2/(a^3*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn( sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 256*sqrt(2)*(54*A*sqrt(c) - 4*B*sqrt(c) + 195*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f *x + 1/2*e) + 1) - 5*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(- 1/4*pi + 1/2*f*x + 1/2*e) + 1) + 315*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/ 2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 25*B*sqrt(c)*(cos(-1/ 4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 22 5*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 - 15*B*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3/(cos (-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 75*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4 - 15*B*sqrt(c)*(cos( -1/4*pi + 1/2*f*x + 1/2*e) - 1)^4/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^...
Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {A+B\,\sin \left (e+f\,x\right )}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]