Integrand size = 24, antiderivative size = 60 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=\frac {a c \cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}+\frac {a \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^3} \]
Time = 2.70 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.60 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=\frac {a \left (15 \cos \left (e+\frac {f x}{2}\right )-5 \cos \left (e+\frac {3 f x}{2}\right )+5 \sin \left (\frac {f x}{2}\right )+\sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac {e}{2}\right )-\sin \left (\frac {e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
(a*(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*c^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 = 63, 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.
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a \sin (e+f x)+a}{(c-c \sin (e+f x))^3}dx\) |
| \(\Big \downarrow \) 3215 |
| \(\displaystyle a c \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^4}dx\) |
| \(\Big \downarrow \) 3151 |
| \(\displaystyle a c \left (\frac {\int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^3}dx}{5 c}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {\int \frac {\cos (e+f x)^2}{(c-c \sin (e+f x))^3}dx}{5 c}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )\) |
| \(\Big \downarrow \) 3150 |
| \(\displaystyle a c \left (\frac {\cos ^3(e+f x)}{15 c f (c-c \sin (e+f x))^3}+\frac {\cos ^3(e+f x)}{5 f (c-c \sin (e+f x))^4}\right )\) |
a*c*(Cos[e + f*x]^3/(5*f*(c - c*Sin[e + f*x])^4) + Cos[e + f*x]^3/(15*c*f* (c - c*Sin[e + f*x])^3))
3.3.33.3.1 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[(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 = 0.79 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.03
| method | result | size |
| risch | \(\frac {2 i a \left (5 i {\mathrm e}^{2 i \left (f x +e \right )}+15 \,{\mathrm e}^{3 i \left (f x +e \right )}+i-5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{15 f \,c^{3} \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5}}\) | \(62\) |
| parallelrisch | \(-\frac {2 a \left (15 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-15 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+25 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4\right )}{15 f \,c^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(75\) |
| derivativedivides | \(\frac {2 a \left (-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{3}}\) | \(86\) |
| default | \(\frac {2 a \left (-\frac {3}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {14}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}\right )}{f \,c^{3}}\) | \(86\) |
| norman | \(\frac {\frac {2 a \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {8 a}{15 c f}-\frac {2 a \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {8 a \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {16 a \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {58 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\) | \(161\) |
2/15*I*a*(5*I*exp(2*I*(f*x+e))+15*exp(3*I*(f*x+e))+I-5*exp(I*(f*x+e)))/f/c ^3/(exp(I*(f*x+e))-I)^5
Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (58) = 116\).
Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.57 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=-\frac {a \cos \left (f x + e\right )^{3} - 2 \, a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) + 6 \, a\right )} \sin \left (f x + e\right ) + 6 \, a}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \]
-1/15*(a*cos(f*x + e)^3 - 2*a*cos(f*x + e)^2 + 3*a*cos(f*x + e) + (a*cos(f *x + e)^2 + 3*a*cos(f*x + e) + 6*a)*sin(f*x + e) + 6*a)/(c^3*f*cos(f*x + e )^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos (f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))
Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (51) = 102\).
Time = 2.38 (sec) , antiderivative size = 571, normalized size of antiderivative = 9.52 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=\begin {cases} - \frac {30 a \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} + \frac {30 a \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} - \frac {50 a \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} + \frac {10 a \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} - \frac {8 a}{15 c^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 75 c^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 c^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} - 150 c^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 c^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - 15 c^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )}{\left (- c \sin {\left (e \right )} + c\right )^{3}} & \text {otherwise} \end {cases} \]
Piecewise((-30*a*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c **3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*ta n(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 30*a*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)** 4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c **3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 50*a*tan(e/2 + f*x/2)**2/(15*c**3*f* tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 10*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c* *3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan (e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 8*a/(15*c**3* f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f), Ne(f, 0)), (x*(a*sin(e) + a)/(-c*sin(e) + c)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 389 vs. \(2 (58) = 116\).
Time = 0.21 (sec) , antiderivative size = 389, normalized size of antiderivative = 6.48 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {a {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 7\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {3 \, a {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]
-2/15*(a*(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)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10 *c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e) ^5/(cos(f*x + e) + 1)^5) - 3*a*(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)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(c os(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*si n(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^ 5))/f
Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 25 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a\right )}}{15 \, c^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}} \]
-2/15*(15*a*tan(1/2*f*x + 1/2*e)^4 - 15*a*tan(1/2*f*x + 1/2*e)^3 + 25*a*ta n(1/2*f*x + 1/2*e)^2 - 5*a*tan(1/2*f*x + 1/2*e) + 4*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)
Time = 6.86 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.27 \[ \int \frac {a+a \sin (e+f x)}{(c-c \sin (e+f x))^3} \, dx=\frac {2\,a\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-5\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+25\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-15\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,c^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )-\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]