Integrand size = 28, antiderivative size = 122 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{4 \sqrt {2} c^{5/2} f}+\frac {a^2 c \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^2 \cos (e+f x)}{4 c f (c-c \sin (e+f x))^{3/2}} \] Output:
3/8*a^2*arctanh(1/2*c^(1/2)*cos(f*x+e)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))*2^( 1/2)/c^(5/2)/f+1/2*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(7/2)-3/4*a^2*cos (f*x+e)/c/f/(c-c*sin(f*x+e))^(3/2)
Result contains complex when optimal does not.
Time = 8.64 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.34 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 \cos \left (\frac {1}{2} (e+f x)\right )-5 \cos \left (\frac {3}{2} (e+f x)\right )+3 \sin \left (\frac {1}{2} (e+f x)\right )+(3+3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) (-3+\cos (2 (e+f x))+4 \sin (e+f x))+5 \sin \left (\frac {3}{2} (e+f x)\right )\right )}{8 c^2 f (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(5/2),x]
Output:
(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(3*Cos[(e + f*x)/2] - 5*Cos[(3* (e + f*x))/2] + 3*Sin[(e + f*x)/2] + (3 + 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/ 2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(-3 + Cos[2*(e + f*x)] + 4*Sin[e + f *x]) + 5*Sin[(3*(e + f*x))/2]))/(8*c^2*f*(-1 + Sin[e + f*x])^2*Sqrt[c - c* Sin[e + f*x]])
Time = 0.63 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 3215, 3042, 3159, 3042, 3159, 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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(c-c \sin (e+f x))^{9/2}}dx\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^{5/2}}dx}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3159 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}}dx}{2 c^2}\right )}{4 c^2}\right )\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\int \frac {1}{2 c-\frac {c^2 \cos ^2(e+f x)}{c-c \sin (e+f x)}}d\left (-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{c^2 f}+\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}\right )}{4 c^2}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle a^2 c^2 \left (\frac {\cos ^3(e+f x)}{2 c f (c-c \sin (e+f x))^{7/2}}-\frac {3 \left (\frac {\cos (e+f x)}{c f (c-c \sin (e+f x))^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} c^{5/2} f}\right )}{4 c^2}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(5/2),x]
Output:
a^2*c^2*(Cos[e + f*x]^3/(2*c*f*(c - c*Sin[e + f*x])^(7/2)) - (3*(-(ArcTanh [(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])]/(Sqrt[2]*c^(5/ 2)*f)) + Cos[e + f*x]/(c*f*(c - c*Sin[e + f*x])^(3/2))))/(4*c^2))
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[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f *x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 ))) Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & & NeQ[2*m + p + 1, 0] && !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 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 = 1.06 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.57
| method | result | size |
| default | \(\frac {a^{2} \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )+12 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}-10 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{8 c^{\frac {9}{2}} \left (-1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(191\) |
| parts | \(\frac {a^{2} \left (-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}+6 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}} \sin \left (f x +e \right )+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )-14 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}-3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{32 c^{\frac {9}{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}+\frac {a^{2} \left (-19 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right )^{2} c^{2}+38 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} \sin \left (f x +e \right )+44 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {3}{2}}-26 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c}-19 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{32 c^{\frac {9}{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}+\frac {a^{2} \left (5 \sin \left (f x +e \right )^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-10 \sin \left (f x +e \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}-10 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {3}{2}}+5 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3}+12 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {5}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{16 c^{\frac {11}{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}\) | \(578\) |
Input:
int((a+sin(f*x+e)*a)^2/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/8*a^2*(-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))* sin(f*x+e)^2*c^2+6*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^ (1/2))*c^2*sin(f*x+e)+12*(c*(1+sin(f*x+e)))^(1/2)*c^(3/2)-10*(c*(1+sin(f*x +e)))^(3/2)*c^(1/2)-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2) /c^(1/2))*c^2)*(c*(1+sin(f*x+e)))^(1/2)/c^(9/2)/(-1+sin(f*x+e))/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. 362 vs. \(2 (103) = 206\).
Time = 0.11 (sec) , antiderivative size = 362, normalized size of antiderivative = 2.97 \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{3} + 3 \, a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \cos \left (f x + e\right ) - 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (5 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right ) - 4 \, a^{2} - {\left (5 \, a^{2} \cos \left (f x + e\right ) + 4 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")
Output:
1/16*(3*sqrt(2)*(a^2*cos(f*x + e)^3 + 3*a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) - 4*a^2 - (a^2*cos(f*x + e)^2 - 2*a^2*cos(f*x + e) - 4*a^2)*sin(f*x + e))*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*(5*a^2*cos(f*x + e)^2 + a^2*cos(f*x + e) - 4*a^2 - (5*a^2*cos(f*x + e) + 4*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin (f*x + e))
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=a^{2} \left (\int \frac {2 \sin {\left (e + f x \right )}}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {\sin ^{2}{\left (e + f x \right )}}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx + \int \frac {1}{c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + c^{2} \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx\right ) \] Input:
integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(5/2),x)
Output:
a**2*(Integral(2*sin(e + f*x)/(c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) **2 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sqrt(-c*sin(e + f*x) + c)), x) + Integral(sin(e + f*x)**2/(c**2*sqrt(-c*sin(e + f*x) + c) *sin(e + f*x)**2 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sq rt(-c*sin(e + f*x) + c)), x) + Integral(1/(c**2*sqrt(-c*sin(e + f*x) + c)* sin(e + f*x)**2 - 2*c**2*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x) + c**2*sqr t(-c*sin(e + f*x) + c)), x))
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")
Output:
integrate((a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(5/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^2/(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. \[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(5/2),x)
Output:
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(5/2), x)
\[ \int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, a^{2} \left (-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )-\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )-2 \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right )\right )}{c^{3}} \] Input:
int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(5/2),x)
Output:
(sqrt(c)*a**2*( - int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x) - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x) - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 - 3*sin( e + f*x)**2 + 3*sin(e + f*x) - 1),x)))/c**3