Integrand size = 28, antiderivative size = 68 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {8 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 c f} \] Output:
8/3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^2/f-2*sec(f*x+e)^3*(c-c*sin(f*x+ e))^(5/2)/a^2/c/f
Time = 2.75 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.35 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+3 \sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \] Input:
Integrate[(c - c*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^2,x]
Output:
(2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(1 + 3*Sin[e + f*x])*Sqrt[c - c *Sin[e + f*x]])/(3*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)
Time = 0.48 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3042, 3215, 3042, 3153, 3042, 3152}
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.
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{(a \sin (e+f x)+a)^2}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle \frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^4}dx}{a^2 c^2}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {-4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}}{a^2 c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-4 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^4}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}}{a^2 c^2}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {\frac {8 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}}{a^2 c^2}\) |
Input:
Int[(c - c*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x])^2,x]
Output:
((8*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*f) - (2*c*Sec[e + f* x]^3*(c - c*Sin[e + f*x])^(5/2))/f)/(a^2*c^2)
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 - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
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 - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p)) 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] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
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 = 0.47 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {2 c^{2} \left (-1+\sin \left (f x +e \right )\right ) \left (3 \sin \left (f x +e \right )+1\right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(61\) |
Input:
int((c-c*sin(f*x+e))^(3/2)/(a+sin(f*x+e)*a)^2,x,method=_RETURNVERBOSE)
Output:
-2/3*c^2/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(3*sin(f*x+e)+1)/cos(f*x+e)/(c -c*sin(f*x+e))^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.84 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {2 \, {\left (3 \, c \sin \left (f x + e\right ) + c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
Output:
2/3*(3*c*sin(f*x + e) + c)*sqrt(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*s in(f*x + e) + a^2*f*cos(f*x + e))
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {\int \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\right )\, dx}{a^{2}} \] Input:
integrate((c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e))**2,x)
Output:
(Integral(c*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x) + Integral(-c*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x)* *2 + 2*sin(e + f*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (62) = 124\).
Time = 0.13 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.51 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (c^{\frac {3}{2}} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {12 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {c^{\frac {3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
Output:
-2/3*(c^(3/2) + 6*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^(3/2)*sin( f*x + e)^2/(cos(f*x + e) + 1)^2 + 12*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 6*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 )/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(co s(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))
Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.63 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {4 \, \sqrt {2} {\left (c \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {3 \, c {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )} \sqrt {c}}{3 \, a^{2} f {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{3}} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")
Output:
-4/3*sqrt(2)*(c*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + 3*c*(cos(-1/4*pi + 1 /2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/ 2*f*x + 1/2*e) + 1))*sqrt(c)/(a^2*f*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/ (cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^3)
Time = 24.43 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.76 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {4\,c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\sqrt {c-c\,\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}\,\left (2\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,3{}\mathrm {i}+3{}\mathrm {i}\right )}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1{}\mathrm {i}\right )}^3\,\left (1+{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}\right )} \] Input:
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x))^2,x)
Output:
-(4*c*exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*(2*exp(e*1i + f*x*1i) - exp(e*2i + f*x*2i)*3i + 3i) )/(3*a^2*f*(exp(e*1i + f*x*1i) + 1i)^3*(exp(e*1i + f*x*1i)*1i + 1))
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {\sqrt {c}\, c \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{2}+2 \sin \left (f x +e \right )+1}d x \right )\right )}{a^{2}} \] Input:
int((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x)
Output:
(sqrt(c)*c*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**2 + 2*sin(e + f*x ) + 1),x) - int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)))/a**2