Integrand size = 13, antiderivative size = 86 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=\frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac {2 \cos (c+d x)}{15 a d (a-a \sin (c+d x))^2}+\frac {2 \cos (c+d x)}{15 d \left (a^3-a^3 \sin (c+d x)\right )} \] Output:
1/5*cos(d*x+c)/d/(a-a*sin(d*x+c))^3+2/15*cos(d*x+c)/a/d/(a-a*sin(d*x+c))^2 +2/15*cos(d*x+c)/d/(a^3-a^3*sin(d*x+c))
Time = 0.46 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=\frac {10-15 \cos (c+d x)-6 \cos (2 (c+d x))+\cos (3 (c+d x))-15 \sin (c+d x)+6 \sin (2 (c+d x))+\sin (3 (c+d x))}{30 a^3 d (-1+\sin (c+d x))^3} \] Input:
Integrate[(a - a*Sin[c + d*x])^(-3),x]
Output:
(10 - 15*Cos[c + d*x] - 6*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] - 15*Sin[c + d*x] + 6*Sin[2*(c + d*x)] + Sin[3*(c + d*x)])/(30*a^3*d*(-1 + Sin[c + d*x ])^3)
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, 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.
\(\displaystyle \int \frac {1}{(a-a \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a-a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \int \frac {1}{(a-a \sin (c+d x))^2}dx}{5 a}+\frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{(a-a \sin (c+d x))^2}dx}{5 a}+\frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{a-a \sin (c+d x)}dx}{3 a}+\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )}{5 a}+\frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{a-a \sin (c+d x)}dx}{3 a}+\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )}{5 a}+\frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\cos (c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac {2 \left (\frac {\cos (c+d x)}{3 a d (a-a \sin (c+d x))}+\frac {\cos (c+d x)}{3 d (a-a \sin (c+d x))^2}\right )}{5 a}\) |
Input:
Int[(a - a*Sin[c + d*x])^(-3),x]
Output:
Cos[c + d*x]/(5*d*(a - a*Sin[c + d*x])^3) + (2*(Cos[c + d*x]/(3*d*(a - a*S in[c + d*x])^2) + Cos[c + d*x]/(3*a*d*(a - a*Sin[c + d*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 = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56
method | result | size |
risch | \(-\frac {4 \left (-1+10 \,{\mathrm e}^{2 i \left (d x +c \right )}-5 i {\mathrm e}^{i \left (d x +c \right )}\right )}{15 \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d \,a^{3}}\) | \(48\) |
parallelrisch | \(\frac {-\frac {14}{15}-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{d \,a^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\) | \(74\) |
derivativedivides | \(\frac {-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{d \,a^{3}}\) | \(85\) |
default | \(\frac {-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {4}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}}{d \,a^{3}}\) | \(85\) |
norman | \(\frac {-\frac {14}{15 a d}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d a}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d a}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a d}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 a d}}{a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}\) | \(101\) |
Input:
int(1/(a-a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-4/15*(-1+10*exp(2*I*(d*x+c))-5*I*exp(I*(d*x+c)))/(exp(I*(d*x+c))-I)^5/d/a ^3
Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.71 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=\frac {2 \, \cos \left (d x + c\right )^{3} - 4 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) - 3\right )} \sin \left (d x + c\right ) - 9 \, \cos \left (d x + c\right ) - 3}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d - {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(1/(a-a*sin(d*x+c))^3,x, algorithm="fricas")
Output:
1/15*(2*cos(d*x + c)^3 - 4*cos(d*x + c)^2 + (2*cos(d*x + c)^2 + 6*cos(d*x + c) - 3)*sin(d*x + c) - 9*cos(d*x + c) - 3)/(a^3*d*cos(d*x + c)^3 + 3*a^3 *d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d - (a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))
Leaf count of result is larger than twice the leaf count of optimal. 556 vs. \(2 (71) = 142\).
Time = 1.32 (sec) , antiderivative size = 556, normalized size of antiderivative = 6.47 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate(1/(a-a*sin(d*x+c))**3,x)
Output:
Piecewise((-30*tan(c/2 + d*x/2)**4/(15*a**3*d*tan(c/2 + d*x/2)**5 - 75*a** 3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3 - 150*a**3*d*tan( c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) - 15*a**3*d) + 60*tan(c/2 + d *x/2)**3/(15*a**3*d*tan(c/2 + d*x/2)**5 - 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3 - 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3* d*tan(c/2 + d*x/2) - 15*a**3*d) - 80*tan(c/2 + d*x/2)**2/(15*a**3*d*tan(c/ 2 + d*x/2)**5 - 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2 )**3 - 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) - 15*a* *3*d) + 40*tan(c/2 + d*x/2)/(15*a**3*d*tan(c/2 + d*x/2)**5 - 75*a**3*d*tan (c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2)**3 - 150*a**3*d*tan(c/2 + d *x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) - 15*a**3*d) - 14/(15*a**3*d*tan(c/2 + d*x/2)**5 - 75*a**3*d*tan(c/2 + d*x/2)**4 + 150*a**3*d*tan(c/2 + d*x/2) **3 - 150*a**3*d*tan(c/2 + d*x/2)**2 + 75*a**3*d*tan(c/2 + d*x/2) - 15*a** 3*d), Ne(d, 0)), (x/(-a*sin(c) + a)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (83) = 166\).
Time = 0.04 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.37 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (\frac {20 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 7\right )}}{15 \, {\left (a^{3} - \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )} d} \] Input:
integrate(1/(a-a*sin(d*x+c))^3,x, algorithm="maxima")
Output:
-2/15*(20*sin(d*x + c)/(cos(d*x + c) + 1) - 40*sin(d*x + c)^2/(cos(d*x + c ) + 1)^2 + 30*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 15*sin(d*x + c)^4/(cos (d*x + c) + 1)^4 - 7)/((a^3 - 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1) + 10*a ^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a^3*sin(d*x + c)^5 /(cos(d*x + c) + 1)^5)*d)
Time = 0.13 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=-\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}} \] Input:
integrate(1/(a-a*sin(d*x+c))^3,x, algorithm="giac")
Output:
-2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/2*d*x + 1/2*c)^3 + 40*tan(1/2* d*x + 1/2*c)^2 - 20*tan(1/2*d*x + 1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2*c) - 1)^5)
Time = 25.91 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.57 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=\frac {2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-20\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-30\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+15\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}{15\,a^3\,d\,{\left (\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}^5} \] Input:
int(1/(a - a*sin(c + d*x))^3,x)
Output:
(2*cos(c/2 + (d*x)/2)*(7*cos(c/2 + (d*x)/2)^4 + 15*sin(c/2 + (d*x)/2)^4 - 30*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2)^3 - 20*cos(c/2 + (d*x)/2)^3*sin(c /2 + (d*x)/2) + 40*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^2))/(15*a^3*d*( cos(c/2 + (d*x)/2) - sin(c/2 + (d*x)/2))^5)
Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(a-a \sin (c+d x))^3} \, dx=\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}-\frac {8}{15}}{a^{3} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )} \] Input:
int(1/(a-a*sin(d*x+c))^3,x)
Output:
(2*( - 3*tan((c + d*x)/2)**5 - 10*tan((c + d*x)/2)**2 + 5*tan((c + d*x)/2) - 4))/(15*a**3*d*(tan((c + d*x)/2)**5 - 5*tan((c + d*x)/2)**4 + 10*tan((c + d*x)/2)**3 - 10*tan((c + d*x)/2)**2 + 5*tan((c + d*x)/2) - 1))