Integrand size = 28, antiderivative size = 228 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {63 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{128 \sqrt {2} a^3 c^{5/2} f}+\frac {63 \cos (e+f x)}{128 a^3 c f (c-c \sin (e+f x))^{3/2}}+\frac {21 \sec (e+f x)}{80 a^3 c f (c-c \sin (e+f x))^{3/2}}-\frac {21 \sec (e+f x)}{32 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {3 \sec ^3(e+f x)}{10 a^3 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 a^3 c^3 f} \] Output:
63/256*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^(5/2)/f+63/128*cos(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)+21/80*s ec(f*x+e)/a^3/c/f/(c-c*sin(f*x+e))^(3/2)-21/32*sec(f*x+e)/a^3/c^2/f/(c-c*s in(f*x+e))^(1/2)-3/10*sec(f*x+e)^3/a^3/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/5*se c(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 = 8.00 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.94 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-240 \cos ^4(e+f x)-32 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-80 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+20 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+75 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-(315+315 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+40 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+150 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right )}{640 a^3 f (1+\sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 ])*(-240*Cos[e + f*x]^4 - 32*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 80* (Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/ 2])^2 + 20*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[( e + f*x)/2])^5 + 75*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x) /2] + Sin[(e + f*x)/2])^5 - (315 + 315*I)*(-1)^(1/4)*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] + Sin[(e + f*x)/2])^5 + 40*Sin[(e + f*x)/2]*(Cos[(e + f*x )/2] + Sin[(e + f*x)/2])^5 + 150*(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))/(640*a^3*f*(1 + Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2))
Time = 1.07 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.99, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 3215, 3042, 3154, 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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle \frac {\int \sec ^6(e+f x) \sqrt {c-c \sin (e+f x)}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {c-c \sin (e+f x)}}{\cos (e+f x)^6}dx}{a^3 c^3}\) |
| \(\Big \downarrow \) 3154 |
| \(\displaystyle \frac {\frac {9}{10} c \int \frac {\sec ^4(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{10} c \int \frac {1}{\cos (e+f x)^4 \sqrt {c-c \sin (e+f x)}}dx-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \int \frac {\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \int \frac {1}{\cos (e+f x)^2 (c-c \sin (e+f x))^{3/2}}dx-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3160 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \int \frac {\sec ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \int \frac {1}{\cos (e+f x)^2 \sqrt {c-c \sin (e+f x)}}dx}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3166 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \int \frac {1}{(c-c \sin (e+f x))^{3/2}}dx-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3129 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{4 c}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c f}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {9}{10} c \left (\frac {7}{6} c \left (\frac {5 \left (\frac {3}{2} c \left (\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{3/2} f}+\frac {\cos (e+f x)}{2 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\right )}{8 c}+\frac {\sec (e+f x)}{4 f (c-c \sin (e+f x))^{3/2}}\right )-\frac {\sec ^3(e+f x)}{3 f \sqrt {c-c \sin (e+f x)}}\right )-\frac {\sec ^5(e+f x) \sqrt {c-c \sin (e+f x)}}{5 f}}{a^3 c^3}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
(-1/5*(Sec[e + f*x]^5*Sqrt[c - c*Sin[e + f*x]])/f + (9*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*(Ar cTanh[(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)
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_)*((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[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && !(IntegerQ[n] && ((Lt Q[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
Time = 0.52 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.08
| method | result | size |
| default | \(-\frac {315 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \sin \left (f x +e \right )^{2} c^{2}+1176 \sin \left (f x +e \right )^{2} c^{\frac {9}{2}}-420 \sin \left (f x +e \right )^{3} c^{\frac {9}{2}}-630 \sin \left (f x +e \right )^{4} c^{\frac {9}{2}}-630 \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \sqrt {2}\, \sin \left (f x +e \right ) c^{2}+708 \sin \left (f x +e \right ) c^{\frac {9}{2}}+315 \sqrt {2}\, \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-514 c^{\frac {9}{2}}}{1280 c^{\frac {13}{2}} a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(246\) |
Input:
int(1/(a+sin(f*x+e)*a)^3/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/1280/c^(13/2)/a^3*(315*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^( 1/2))*(c*(1+sin(f*x+e)))^(5/2)*2^(1/2)*sin(f*x+e)^2*c^2+1176*sin(f*x+e)^2* c^(9/2)-420*sin(f*x+e)^3*c^(9/2)-630*sin(f*x+e)^4*c^(9/2)-630*arctanh(1/2* (c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*(c*(1+sin(f*x+e)))^(5/2)*2^(1/2) *sin(f*x+e)*c^2+708*sin(f*x+e)*c^(9/2)+315*2^(1/2)*(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))*c^2-514*c^(9/2))/( 1+sin(f*x+e))^2/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.91 \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {315 \, \sqrt {2} \sqrt {c} \cos \left (f x + e\right )^{5} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (315 \, \cos \left (f x + e\right )^{4} - 42 \, \cos \left (f x + e\right )^{2} - 6 \, {\left (35 \, \cos \left (f x + e\right )^{2} + 24\right )} \sin \left (f x + e\right ) - 16\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{2560 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas ")
Output:
1/2560*(315*sqrt(2)*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)/(cos(f*x + e )^2 + (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(315*cos(f* x + e)^4 - 42*cos(f*x + e)^2 - 6*(35*cos(f*x + e)^2 + 24)*sin(f*x + e) - 1 6)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^3*f*cos(f*x + e)^5)
\[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{5}{\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a^{3}} \] Input:
integrate(1/(a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**(5/2),x)
Output:
Integral(1/(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**5 + c**2*sqrt(-c* sin(e + f*x) + c)*sin(e + f*x)**4 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**3 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2 + c**2*sqrt( -c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sqrt(-c*sin(e + f*x) + c)), x)/a* *3
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima ")
Output:
Timed out
Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x, algorithm="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. \[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2)),x)
Output:
int(1/((a + a*sin(e + f*x))^3*(c - c*sin(e + f*x))^(5/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x))^3 (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1}d x \right )}{a^{3} c^{3}} \] Input:
int(1/(a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(5/2),x)
Output:
( - sqrt(c)*int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**6 - 3*sin(e + f*x )**4 + 3*sin(e + f*x)**2 - 1),x))/(a**3*c**3)