Integrand size = 42, antiderivative size = 237 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {6 (g \cos (e+f x))^{5/2}}{5 a f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {6 (g \cos (e+f x))^{5/2}}{5 a c f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {6 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:
-2*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2)+ 6/5*(g*cos(f*x+e))^(5/2)/a/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/ 2)+6/5*(g*cos(f*x+e))^(5/2)/a/c/f/g/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e) )^(3/2)-6/5*g*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)*EllipticE(sin(1/2*f*x+ 1/2*e),2^(1/2))/a/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 4.90 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.57 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {\cos (e+f x)} (g \cos (e+f x))^{3/2} \left (\sqrt {\cos (e+f x)} (1-3 \cos (2 (e+f x))-6 \sin (e+f x))+E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) (6 \cos (e+f x)-3 \sin (2 (e+f x)))\right )}{5 c^2 f (-1+\sin (e+f x))^2 (a (1+\sin (e+f x)))^{3/2} \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[(g*Cos[e + f*x])^(3/2)/((a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
-1/5*(Sqrt[Cos[e + f*x]]*(g*Cos[e + f*x])^(3/2)*(Sqrt[Cos[e + f*x]]*(1 - 3 *Cos[2*(e + f*x)] - 6*Sin[e + f*x]) + EllipticE[(e + f*x)/2, 2]*(6*Cos[e + f*x] - 3*Sin[2*(e + f*x)])))/(c^2*f*(-1 + Sin[e + f*x])^2*(a*(1 + Sin[e + f*x]))^(3/2)*Sqrt[c - c*Sin[e + f*x]])
Time = 1.91 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3331, 3042, 3331, 3042, 3331, 3042, 3321, 3042, 3121, 3042, 3119}
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.
\(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 3331 |
\(\displaystyle \frac {3 \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3331 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3331 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3321 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3121 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {3 \left (\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}+\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}\) |
Input:
Int[(g*Cos[e + f*x])^(3/2)/((a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x] )^(5/2)),x]
Output:
(-2*(g*Cos[e + f*x])^(5/2))/(f*g*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(5/2)) + (3*((2*(g*Cos[e + f*x])^(5/2))/(5*f*g*Sqrt[a + a*Sin[e + f *x]]*(c - c*Sin[e + f*x])^(5/2)) + ((2*(g*Cos[e + f*x])^(5/2))/(f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) - (2*g*Sqrt[Cos[e + f*x]]*S qrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(c*f*Sqrt[a + a*Sin[e + f*x ]]*Sqrt[c - c*Sin[e + f*x]]))/(5*c)))/a
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_ .)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[g* (Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])) Int[(g *Cos[e + f*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && EqQ [b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b* (g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a* f*g*(2*m + p + 1))), x] + Simp[(m + n + p + 1)/(a*(2*m + p + 1)) Int[(g*C os[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && !LtQ[m, n, -1] && Integers Q[2*m, 2*n, 2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(784\) vs. \(2(207)=414\).
Time = 23.99 (sec) , antiderivative size = 785, normalized size of antiderivative = 3.31
Input:
int((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
Output:
1/5/f*g/(2*2^(1/2)-3)/(1+2^(1/2))/a/c^2*(-2+3*(-1+2*cos(1/2*f*x+1/2*e)*(co s(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e)+1)*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+ 1/2*e)^2-2*cos(1/2*f*x+1/2*e))*2^(1/2)*((2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2 )+2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))^(1/2)*(-2*(2^(1/2)*cos(1 /2*f*x+1/2*e)-2^(1/2)-2*cos(1/2*f*x+1/2*e)+1)/(cos(1/2*f*x+1/2*e)+1))^(1/2 )*EllipticE((1+2^(1/2))*(csc(1/2*f*x+1/2*e)-cot(1/2*f*x+1/2*e)),-2*2^(1/2) +3)+6*(2+(-1+2*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2* e)+1)*sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e)^2-2*cos(1/2*f*x+1/2*e))*2^(1/2 )-4*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)^2+2*cos(1/2*f*x+1/2*e)+1)*sin(1 /2*f*x+1/2*e)+2*cos(1/2*f*x+1/2*e)^2+4*cos(1/2*f*x+1/2*e))*((2^(1/2)*cos(1 /2*f*x+1/2*e)-2^(1/2)+2*cos(1/2*f*x+1/2*e)-1)/(cos(1/2*f*x+1/2*e)+1))^(1/2 )*(-2*(2^(1/2)*cos(1/2*f*x+1/2*e)-2^(1/2)-2*cos(1/2*f*x+1/2*e)+1)/(cos(1/2 *f*x+1/2*e)+1))^(1/2)*EllipticF((1+2^(1/2))*(csc(1/2*f*x+1/2*e)-cot(1/2*f* x+1/2*e)),-2*2^(1/2)+3)+2*(1+3*(2*cos(1/2*f*x+1/2*e)+1)*sin(1/2*f*x+1/2*e) +12*cos(1/2*f*x+1/2*e)^4+6*cos(1/2*f*x+1/2*e)^3-12*cos(1/2*f*x+1/2*e)^2-5* cos(1/2*f*x+1/2*e))*2^(1/2)+6*(-2*cos(1/2*f*x+1/2*e)-1)*sin(1/2*f*x+1/2*e) -24*cos(1/2*f*x+1/2*e)^4-12*cos(1/2*f*x+1/2*e)^3+24*cos(1/2*f*x+1/2*e)^2+1 0*cos(1/2*f*x+1/2*e))*(g*(-1+2*cos(1/2*f*x+1/2*e)^2))^(1/2)/((2*cos(1/2*f* x+1/2*e)*sin(1/2*f*x+1/2*e)+1)*a)^(1/2)/(-(2*cos(1/2*f*x+1/2*e)*sin(1/2*f* x+1/2*e)-1)*c)^(1/2)/(2*cos(1/2*f*x+1/2*e)*(cos(1/2*f*x+1/2*e)+1)*sin(1...
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.96 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 \, {\left (3 \, \sqrt {\frac {1}{2}} {\left (-i \, g \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) + i \, g \cos \left (f x + e\right )^{2}\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, g \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - i \, g \cos \left (f x + e\right )^{2}\right )} \sqrt {a c g} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + {\left (3 \, g \cos \left (f x + e\right )^{2} + 3 \, g \sin \left (f x + e\right ) - 2 \, g\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}\right )}}{5 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{2} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{2}\right )}} \] Input:
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="fricas")
Output:
-2/5*(3*sqrt(1/2)*(-I*g*cos(f*x + e)^2*sin(f*x + e) + I*g*cos(f*x + e)^2)* sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) + 3*sqrt(1/2)*(I*g*cos(f*x + e)^2*sin(f*x + e) - I*g*c os(f*x + e)^2)*sqrt(a*c*g)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) + (3*g*cos(f*x + e)^2 + 3*g*sin(f*x + e ) - 2*g)*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e ) + c))/(a^2*c^3*f*cos(f*x + e)^2*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^2)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((g*cos(f*x+e))**(3/2)/(a+a*sin(f*x+e))**(3/2)/(c-c*sin(f*x+e))** (5/2),x)
Output:
Timed out
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="maxima")
Output:
integrate((g*cos(f*x + e))^(3/2)/((a*sin(f*x + e) + a)^(3/2)*(-c*sin(f*x + e) + c)^(5/2)), x)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int((g*cos(e + f*x))^(3/2)/((a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x) )^(5/2)),x)
Output:
int((g*cos(e + f*x))^(3/2)/((a + a*sin(e + f*x))^(3/2)*(c - c*sin(e + f*x) )^(5/2)), x)
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:
int((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2),x)
Output:
(sqrt(g)*sqrt(c)*sqrt(a)*g*( - 4*sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f* x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2 + 4*sqrt(sin(e + f*x) + 1)*sqrt ( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x) + 2*sqrt(sin(e + f*x ) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x)) - 2*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**3)/(c os(e + f*x)*sin(e + f*x)**3 - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)* sin(e + f*x) + cos(e + f*x)),x)*sin(e + f*x)**3*f + 2*int((sqrt(sin(e + f* x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**3)/(cos (e + f*x)*sin(e + f*x)**3 - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)*si n(e + f*x) + cos(e + f*x)),x)*sin(e + f*x)**2*f + 2*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**3)/(cos(e + f*x)*sin(e + f*x)**3 - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)*sin( e + f*x) + cos(e + f*x)),x)*sin(e + f*x)*f - 2*int((sqrt(sin(e + f*x) + 1) *sqrt( - sin(e + f*x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**3)/(cos(e + f* x)*sin(e + f*x)**3 - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)*sin(e + f *x) + cos(e + f*x)),x)*f + 2*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f *x) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2)/(cos(e + f*x)*sin(e + f*x)**3 - cos(e + f*x)*sin(e + f*x)**2 - cos(e + f*x)*sin(e + f*x) + cos(e + f*x) ),x)*sin(e + f*x)**3*f - 2*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x ) + 1)*sqrt(cos(e + f*x))*sin(e + f*x)**2)/(cos(e + f*x)*sin(e + f*x)**...