Integrand size = 26, antiderivative size = 33 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5} \] Output:
-1/5*a^2*c^2*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^5
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(33)=66\).
Time = 3.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.45 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (10 \sin \left (\frac {1}{2} (e+f x)\right )+5 \sin \left (\frac {3}{2} (e+f x)\right )-\sin \left (\frac {5}{2} (e+f x)\right )\right )}{10 a^3 f (1+\sin (e+f x))^3} \] Input:
Integrate[(c - c*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^3,x]
Output:
(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(10*Sin[(e + f*x)/2] + 5*Sin[(3 *(e + f*x))/2] - Sin[(5*(e + f*x))/2]))/(10*a^3*f*(1 + Sin[e + f*x])^3)
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3215, 3042, 3150}
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.
\(\displaystyle \int \frac {(c-c \sin (e+f x))^2}{(a \sin (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-c \sin (e+f x))^2}{(a \sin (e+f x)+a)^3}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^5}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^5}dx\) |
\(\Big \downarrow \) 3150 |
\(\displaystyle -\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}\) |
Input:
Int[(c - c*Sin[e + f*x])^2/(a + a*Sin[e + f*x])^3,x]
Output:
-1/5*(a^2*c^2*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^5)
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*m)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && EqQ[Simplify[m + p + 1], 0] && !ILtQ[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 = 0.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(-\frac {2 c^{2} \left (5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right )}{5 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(53\) |
risch | \(-\frac {2 \left (5 c^{2} {\mathrm e}^{4 i \left (f x +e \right )}-10 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+c^{2}\right )}{5 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(55\) |
derivativedivides | \(\frac {2 c^{2} \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}\right )}{f \,a^{3}}\) | \(88\) |
default | \(\frac {2 c^{2} \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}\right )}{f \,a^{3}}\) | \(88\) |
norman | \(\frac {-\frac {2 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a f}-\frac {2 c^{2}}{5 a f}-\frac {24 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{5 a f}-\frac {52 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f}-\frac {8 c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(133\) |
Input:
int((c-c*sin(f*x+e))^2/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
Output:
-2/5*c^2*(5*tan(1/2*f*x+1/2*e)^4+10*tan(1/2*f*x+1/2*e)^2+1)/f/a^3/(tan(1/2 *f*x+1/2*e)+1)^5
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (31) = 62\).
Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 5.09 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^{2} \cos \left (f x + e\right )^{3} + 3 \, c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2} - {\left (c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/5*(c^2*cos(f*x + e)^3 + 3*c^2*cos(f*x + e)^2 - 2*c^2*cos(f*x + e) - 4*c ^2 - (c^2*cos(f*x + e)^2 - 2*c^2*cos(f*x + e) - 4*c^2)*sin(f*x + e))/(a^3* f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (31) = 62\).
Time = 4.53 (sec) , antiderivative size = 354, normalized size of antiderivative = 10.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\begin {cases} - \frac {10 c^{2} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {20 c^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {2 c^{2}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\left (e \right )} + c\right )^{2}}{\left (a \sin {\left (e \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate((c-c*sin(f*x+e))**2/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-10*c**2*tan(e/2 + f*x/2)**4/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25 *a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f*ta n(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f) - 20*c**2*tan(e /2 + f*x/2)**2/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)* *4 + 50*a**3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a* *3*f*tan(e/2 + f*x/2) + 5*a**3*f) - 2*c**2/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f *tan(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f), Ne(f, 0)), (x*(-c*sin(e) + c)**2/(a*sin(e) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (31) = 62\).
Time = 0.05 (sec) , antiderivative size = 554, normalized size of antiderivative = 16.79 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f* x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4 /(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f* x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10 *sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f *x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos (f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^ 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e ) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/ (cos(f*x + e) + 1)^5))/f
Time = 0.15 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (5 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 10 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2}\right )}}{5 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algorithm="giac")
Output:
-2/5*(5*c^2*tan(1/2*f*x + 1/2*e)^4 + 10*c^2*tan(1/2*f*x + 1/2*e)^2 + c^2)/ (a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 17.66 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {2\,c^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{5\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \] Input:
int((c - c*sin(e + f*x))^2/(a + a*sin(e + f*x))^3,x)
Output:
-(2*c^2*cos(e/2 + (f*x)/2)*(cos(e/2 + (f*x)/2)^4 + 5*sin(e/2 + (f*x)/2)^4 + 10*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^2))/(5*a^3*f*(cos(e/2 + (f*x) /2) + sin(e/2 + (f*x)/2))^5)
Time = 0.19 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.36 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {2 c^{2} \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+\cos \left (f x +e \right )-3 \sin \left (f x +e \right )^{2}-1\right )}{5 a^{3} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )-1\right )} \] Input:
int((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x)
Output:
(2*c**2*(cos(e + f*x)*sin(e + f*x)**2 + cos(e + f*x) - 3*sin(e + f*x)**2 - 1))/(5*a**3*f*(cos(e + f*x)*sin(e + f*x)**2 + 2*cos(e + f*x)*sin(e + f*x) + cos(e + f*x) - sin(e + f*x)**3 - 3*sin(e + f*x)**2 - 3*sin(e + f*x) - 1 ))