Integrand size = 24, antiderivative size = 58 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {a c \cos ^3(e+f x)}{5 f (a+a \sin (e+f x))^4}-\frac {c \cos ^3(e+f x)}{15 f (a+a \sin (e+f x))^3} \] Output:
-1/5*a*c*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^4-1/15*c*cos(f*x+e)^3/f/(a+a*sin( f*x+e))^3
Time = 3.19 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.59 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=\frac {c \left (-15 \cos \left (e+\frac {f x}{2}\right )+5 \cos \left (e+\frac {3 f x}{2}\right )+5 \sin \left (\frac {f x}{2}\right )+\sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f \left (\cos \left (\frac {e}{2}\right )+\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \] Input:
Integrate[(c - c*Sin[e + f*x])/(a + a*Sin[e + f*x])^3,x]
Output:
(c*(-15*Cos[e + (f*x)/2] + 5*Cos[e + (3*f*x)/2] + 5*Sin[(f*x)/2] + Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
Time = 0.37 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3215, 3042, 3151, 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.
\(\displaystyle \int \frac {c-c \sin (e+f x)}{(a \sin (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {c-c \sin (e+f x)}{(a \sin (e+f x)+a)^3}dx\) |
\(\Big \downarrow \) 3215 |
\(\displaystyle a c \int \frac {\cos ^2(e+f x)}{(\sin (e+f x) a+a)^4}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a c \int \frac {\cos (e+f x)^2}{(\sin (e+f x) a+a)^4}dx\) |
\(\Big \downarrow \) 3151 |
\(\displaystyle a c \left (\frac {\int \frac {\cos ^2(e+f x)}{(\sin (e+f x) a+a)^3}dx}{5 a}-\frac {\cos ^3(e+f x)}{5 f (a \sin (e+f x)+a)^4}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a c \left (\frac {\int \frac {\cos (e+f x)^2}{(\sin (e+f x) a+a)^3}dx}{5 a}-\frac {\cos ^3(e+f x)}{5 f (a \sin (e+f x)+a)^4}\right )\) |
\(\Big \downarrow \) 3150 |
\(\displaystyle a c \left (-\frac {\cos ^3(e+f x)}{15 a f (a \sin (e+f x)+a)^3}-\frac {\cos ^3(e+f x)}{5 f (a \sin (e+f x)+a)^4}\right )\) |
Input:
Int[(c - c*Sin[e + f*x])/(a + a*Sin[e + f*x])^3,x]
Output:
a*c*(-1/5*Cos[e + f*x]^3/(f*(a + a*Sin[e + f*x])^4) - Cos[e + f*x]^3/(15*a *f*(a + a*Sin[e + f*x])^3))
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[(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*Simplify[2*m + p + 1])), x] + Simp[Simplify[m + p + 1]/(a*Simpl ify[2*m + p + 1]) 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] && ILtQ[Simpli fy[m + p + 1], 0] && NeQ[2*m + p + 1, 0] && !IGtQ[m, 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]))
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.07
method | result | size |
risch | \(\frac {2 i c \left (-5 i {\mathrm e}^{2 i \left (f x +e \right )}+15 \,{\mathrm e}^{3 i \left (f x +e \right )}-i-5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(62\) |
parallelrisch | \(-\frac {2 c \left (15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+25 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4\right )}{15 f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(75\) |
derivativedivides | \(\frac {2 c \left (\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\right )}{f \,a^{3}}\) | \(86\) |
default | \(\frac {2 c \left (\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\right )}{f \,a^{3}}\) | \(86\) |
norman | \(\frac {-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a f}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {8 c}{15 a f}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {16 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 a f}-\frac {8 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {58 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 a f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(161\) |
Input:
int((c-c*sin(f*x+e))/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
Output:
2/15*I*c*(-5*I*exp(2*I*(f*x+e))+15*exp(3*I*(f*x+e))-I-5*exp(I*(f*x+e)))/f/ a^3/(exp(I*(f*x+e))+I)^5
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (54) = 108\).
Time = 0.07 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.66 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=\frac {c \cos \left (f x + e\right )^{3} - 2 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) - {\left (c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + 6 \, c\right )} \sin \left (f x + e\right ) + 6 \, c}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate((c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
Output:
1/15*(c*cos(f*x + e)^3 - 2*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) - (c*cos(f* x + e)^2 + 3*c*cos(f*x + e) + 6*c)*sin(f*x + e) + 6*c)/(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. 573 vs. \(2 (53) = 106\).
Time = 2.54 (sec) , antiderivative size = 573, normalized size of antiderivative = 9.88 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:
integrate((c-c*sin(f*x+e))/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-30*c*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a **3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*ta n(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*c*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)** 4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a **3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 50*c*tan(e/2 + f*x/2)**2/(15*a**3*f* tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 10*c*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a* *3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan (e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 8*c/(15*a**3* f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f), Ne(f, 0)), (x*(-c*sin(e) + c)/(a*sin(e) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (54) = 108\).
Time = 0.04 (sec) , antiderivative size = 387, normalized size of antiderivative = 6.67 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {c {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {3 \, c {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \] Input:
integrate((c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(c*(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) - 3*c*(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/(c os(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*si n(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 = 0.14 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.36 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 5 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, c\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate((c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x, algorithm="giac")
Output:
-2/15*(15*c*tan(1/2*f*x + 1/2*e)^4 + 15*c*tan(1/2*f*x + 1/2*e)^3 + 25*c*ta n(1/2*f*x + 1/2*e)^2 + 5*c*tan(1/2*f*x + 1/2*e) + 4*c)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 17.71 (sec) , antiderivative size = 134, normalized size of antiderivative = 2.31 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=-\frac {2\,c\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+25\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+15\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \] Input:
int((c - c*sin(e + f*x))/(a + a*sin(e + f*x))^3,x)
Output:
-(2*c*cos(e/2 + (f*x)/2)*(4*cos(e/2 + (f*x)/2)^4 + 15*sin(e/2 + (f*x)/2)^4 + 15*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^3 + 5*cos(e/2 + (f*x)/2)^3*sin (e/2 + (f*x)/2) + 25*cos(e/2 + (f*x)/2)^2*sin(e/2 + (f*x)/2)^2))/(15*a^3*f *(cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2))^5)
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.17 \[ \int \frac {c-c \sin (e+f x)}{(a+a \sin (e+f x))^3} \, dx=\frac {2 c \left (3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{15 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int((c-c*sin(f*x+e))/(a+a*sin(f*x+e))^3,x)
Output:
(2*c*(3*tan((e + f*x)/2)**5 + 15*tan((e + f*x)/2)**3 + 5*tan((e + f*x)/2)* *2 + 10*tan((e + f*x)/2) - 1))/(15*a**3*f*(tan((e + f*x)/2)**5 + 5*tan((e + f*x)/2)**4 + 10*tan((e + f*x)/2)**3 + 10*tan((e + f*x)/2)**2 + 5*tan((e + f*x)/2) + 1))