Integrand size = 26, antiderivative size = 33 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5} \]
Leaf count is larger than twice the leaf count of optimal. \(81\) vs. \(2(33)=66\).
Time = 1.62 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.45 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (10 \sin \left (\frac {1}{2} (e+f x)\right )+5 \sin \left (\frac {3}{2} (e+f x)\right )-\sin \left (\frac {5}{2} (e+f x)\right )\right )}{10 a^3 f (1+\sin (e+f x))^3} \]
(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 |
\(\displaystyle -\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}\) |
3.3.82.3.1 Defintions of rubi rules used
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 = 0.77 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61
method | result | size |
parallelrisch | \(-\frac {2 c^{2} \left (5 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+10 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+1\right )}{5 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(53\) |
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 | \(\frac {2 c^{2} \left (\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(88\) |
default | \(\frac {2 c^{2} \left (\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{f \,a^{3}}\) | \(88\) |
norman | \(\frac {-\frac {2 c^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {2 c^{2}}{5 f a}-\frac {24 c^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}-\frac {52 c^{2} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 f a}-\frac {8 c^{2} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(133\) |
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (31) = 62\).
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 5.09 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^{2} \cos \left (f x + e\right )^{3} + 3 \, c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2} - {\left (c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
-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 = 4.47 (sec) , antiderivative size = 354, normalized size of antiderivative = 10.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\begin {cases} - \frac {10 c^{2} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {20 c^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {2 c^{2}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\left (e \right )} + c\right )^{2}}{\left (a \sin {\left (e \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]
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 = 0.21 (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=-\frac {2 \, {\left (\frac {c^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, c^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {6 \, c^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]
-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 = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (5 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 10 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2}\right )}}{5 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]
-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 = 6.53 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.73 \[ \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {2\,c^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{5\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]