Integrand size = 28, antiderivative size = 60 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {8 c \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a f}+\frac {2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f} \] Output:
-8*c*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a/f+2*sec(f*x+e)*(c-c*sin(f*x+e))^( 3/2)/a/f
Time = 2.48 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {2 c \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (3+\sin (e+f x)) \sqrt {c-c \sin (e+f x)}}{a f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))} \] Input:
Integrate[(c - c*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x]),x]
Output:
(-2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3 + Sin[e + f*x])*Sqrt[c - c* Sin[e + f*x]])/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x ]))
Time = 0.51 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{a \sin (e+f x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-c \sin (e+f x))^{3/2}}{a \sin (e+f x)+a}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle \frac {\int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx}{a c}\) |
\(\Big \downarrow \) 3153 |
\(\displaystyle \frac {4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}}{a c}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}}{a c}\) |
\(\Big \downarrow \) 3152 |
\(\displaystyle \frac {\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}}{a c}\) |
Input:
Int[(c - c*Sin[e + f*x])^(3/2)/(a + a*Sin[e + f*x]),x]
Output:
((-8*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/f + (2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/f)/(a*c)
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 = 0.45 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {2 c^{2} \left (-1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )+3\right )}{a \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(49\) |
Input:
int((c-c*sin(f*x+e))^(3/2)/(a+sin(f*x+e)*a),x,method=_RETURNVERBOSE)
Output:
2*c^2/a*(-1+sin(f*x+e))*(sin(f*x+e)+3)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.68 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {2 \, {\left (c \sin \left (f x + e\right ) + 3 \, c\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\right )} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")
Output:
-2*(c*sin(f*x + e) + 3*c)*sqrt(-c*sin(f*x + e) + c)/(a*f*cos(f*x + e))
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {\int \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {c \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} + 1}\right )\, dx}{a} \] Input:
integrate((c-c*sin(f*x+e))**(3/2)/(a+a*sin(f*x+e)),x)
Output:
(Integral(c*sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x) + 1), x) + Integral(-c *sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)/(sin(e + f*x) + 1), x))/a
Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (56) = 112\).
Time = 0.12 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.43 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {2 \, {\left (3 \, c^{\frac {3}{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {6 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {2 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a + \frac {a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")
Output:
2*(3*c^(3/2) + 2*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 6*c^(3/2)*sin(f *x + e)^2/(cos(f*x + e) + 1)^2 + 2*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)/((a + a*sin(f*x + e) /(cos(f*x + e) + 1))*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))
Time = 0.16 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {8 \, \sqrt {2} c^{\frac {3}{2}} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{a f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}} \] Input:
integrate((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x, algorithm="giac")
Output:
-8*sqrt(2)*c^(3/2)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a*f*((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 - 1))
Time = 18.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=-\frac {2\,c\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (22\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+4\,\sin \left (2\,e+2\,f\,x\right )-12\right )}{a\,f\,\left (4\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )-4\right )} \] Input:
int((c - c*sin(e + f*x))^(3/2)/(a + a*sin(e + f*x)),x)
Output:
-(2*c*(-c*(sin(e + f*x) - 1))^(1/2)*(4*sin(2*e + 2*f*x) + 22*sin(e/2 + (f* x)/2)^2 + 2*sin((3*e)/2 + (3*f*x)/2)^2 - 12))/(a*f*(sin(e + f*x) + sin(3*e + 3*f*x) + 4*sin(e + f*x)^2 - 4))
\[ \int \frac {(c-c \sin (e+f x))^{3/2}}{a+a \sin (e+f x)} \, dx=\frac {\sqrt {c}\, c \left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )+1}d x -\left (\int \frac {\sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )+1}d x \right )\right )}{a} \] Input:
int((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e)),x)
Output:
(sqrt(c)*c*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x) + 1),x) - int((sqr t( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x) + 1),x)))/a