Integrand size = 12, antiderivative size = 83 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac {2 \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac {2 \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:
-1/5*cos(f*x+e)/f/(a+a*sin(f*x+e))^3-2/15*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^ 2-2/15*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e))
Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=\frac {10-15 \cos (e+f x)-6 \cos (2 (e+f x))+\cos (3 (e+f x))+15 \sin (e+f x)-6 \sin (2 (e+f x))-\sin (3 (e+f x))}{30 a^3 f (1+\sin (e+f x))^3} \] Input:
Integrate[(a + a*Sin[e + f*x])^(-3),x]
Output:
(10 - 15*Cos[e + f*x] - 6*Cos[2*(e + f*x)] + Cos[3*(e + f*x)] + 15*Sin[e + f*x] - 6*Sin[2*(e + f*x)] - Sin[3*(e + f*x)])/(30*a^3*f*(1 + Sin[e + f*x] )^3)
Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3129, 3042, 3129, 3042, 3127}
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.
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \int \frac {1}{(\sin (e+f x) a+a)^2}dx}{5 a}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{(\sin (e+f x) a+a)^2}dx}{5 a}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{\sin (e+f x) a+a}dx}{3 a}-\frac {\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}\right )}{5 a}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{\sin (e+f x) a+a}dx}{3 a}-\frac {\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}\right )}{5 a}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {2 \left (-\frac {\cos (e+f x)}{3 a f (a \sin (e+f x)+a)}-\frac {\cos (e+f x)}{3 f (a \sin (e+f x)+a)^2}\right )}{5 a}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3}\) |
Input:
Int[(a + a*Sin[e + f*x])^(-3),x]
Output:
-1/5*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^3) + (2*(-1/3*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^2) - Cos[e + f*x]/(3*a*f*(a + a*Sin[e + f*x]))))/(5*a)
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b ^2, 0]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d*(2*n + 1))), x] + Simp[(n + 1)/(a*(2*n + 1)) Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]
Result contains complex when optimal does not.
Time = 0.45 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {-\frac {4}{15}+\frac {8 \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+\frac {4 i {\mathrm e}^{i \left (f x +e \right )}}{3}}{f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\) | \(48\) |
parallelrisch | \(\frac {-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}-\frac {16 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}-\frac {14}{15}}{f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(74\) |
derivativedivides | \(\frac {\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) | \(85\) |
default | \(\frac {\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) | \(85\) |
norman | \(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f a}-\frac {14}{15 f a}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{f a}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}-\frac {16 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 a f}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(101\) |
Input:
int(1/(a+sin(f*x+e)*a)^3,x,method=_RETURNVERBOSE)
Output:
4/15*(-1+10*exp(2*I*(f*x+e))+5*I*exp(I*(f*x+e)))/f/a^3/(exp(I*(f*x+e))+I)^ 5
Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )^{2} - {\left (2 \, \cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) - 3\right )} \sin \left (f x + e\right ) - 9 \, \cos \left (f x + e\right ) - 3}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:
integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
Output:
-1/15*(2*cos(f*x + e)^3 - 4*cos(f*x + e)^2 - (2*cos(f*x + e)^2 + 6*cos(f*x + e) - 3)*sin(f*x + e) - 9*cos(f*x + e) - 3)/(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. 558 vs. \(2 (73) = 146\).
Time = 1.42 (sec) , antiderivative size = 558, normalized size of antiderivative = 6.72 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+a*sin(f*x+e))**3,x)
Output:
Piecewise((-30*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*tan( e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*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) - 80*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) - 40*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) - 14/(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/(a*sin(e) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (77) = 154\).
Time = 0.04 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.45 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f} \] Input:
integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
Output:
-2/15*(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)*f)
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \] Input:
integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="giac")
Output:
-2/15*(15*tan(1/2*f*x + 1/2*e)^4 + 30*tan(1/2*f*x + 1/2*e)^3 + 40*tan(1/2* f*x + 1/2*e)^2 + 20*tan(1/2*f*x + 1/2*e) + 7)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)
Time = 16.50 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.60 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=-\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (7\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+20\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+40\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+30\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \] Input:
int(1/(a + a*sin(e + f*x))^3,x)
Output:
-(2*cos(e/2 + (f*x)/2)*(7*cos(e/2 + (f*x)/2)^4 + 15*sin(e/2 + (f*x)/2)^4 + 30*cos(e/2 + (f*x)/2)*sin(e/2 + (f*x)/2)^3 + 20*cos(e/2 + (f*x)/2)^3*sin( e/2 + (f*x)/2) + 40*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 = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}-\frac {8}{15}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:
int(1/(a+a*sin(f*x+e))^3,x)
Output:
(2*(3*tan((e + f*x)/2)**5 - 10*tan((e + f*x)/2)**2 - 5*tan((e + f*x)/2) - 4))/(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))