\(\int \frac {1}{(3+3 \sin (e+f x))^3} \, dx\) [476]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 76 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x)}{5 f (3+3 \sin (e+f x))^3}-\frac {2 \cos (e+f x)}{45 f (3+3 \sin (e+f x))^2}-\frac {2 \cos (e+f x)}{15 f (27+27 \sin (e+f x))} \]

[Out]

-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))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2729, 2727} \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac {2 \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

[In]

Int[(a + a*Sin[e + f*x])^(-3),x]

[Out]

-1/5*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^3) - (2*Cos[e + f*x])/(15*a*f*(a + a*Sin[e + f*x])^2) - (2*Cos[e + f
*x])/(15*f*(a^3 + a^3*Sin[e + f*x]))

Rule 2727

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]

Rule 2729

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] + Dist[(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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac {2 \int \frac {1}{(a+a \sin (e+f x))^2} \, dx}{5 a} \\ & = -\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 \int \frac {1}{a+a \sin (e+f x)} \, dx}{15 a^2} \\ & = -\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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=\frac {10-15 \cos (e+f x)-6 \cos (2 (e+f x))+\cos (3 (e+f x))+15 \sin (e+f x)-6 \sin (2 (e+f x))-\sin (3 (e+f x))}{810 f (1+\sin (e+f x))^3} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^(-3),x]

[Out]

(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)])/(810*f*(1 + Sin[e + f*x])^3)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.63

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 \(\frac {-\frac {14}{15}-2 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-4 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {16 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3}}{f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) \(74\)
derivativedivides \(\frac {-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) \(85\)
default \(\frac {-\frac {16}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {8}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}+\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{3} f}\) \(85\)
norman \(\frac {-\frac {2 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {14}{15 a f}-\frac {4 \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f a}-\frac {8 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 f a}-\frac {16 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 f a}}{a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) \(101\)

[In]

int(1/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.93 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )^{2} - {\left (2 \, \cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) - 3\right )} \sin \left (f x + e\right ) - 9 \, \cos \left (f x + e\right ) - 3}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

[In]

integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (73) = 146\).

Time = 1.28 (sec) , antiderivative size = 558, normalized size of antiderivative = 7.34 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=\begin {cases} - \frac {30 \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {60 \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {80 \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {40 \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} - \frac {14}{15 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 150 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 75 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 15 a^{3} f} & \text {for}\: f \neq 0 \\\frac {x}{\left (a \sin {\left (e \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+a*sin(f*x+e))**3,x)

[Out]

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 + 1
50*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*ta
n(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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 203 vs. \(2 (77) = 154\).

Time = 0.20 (sec) , antiderivative size = 203, normalized size of antiderivative = 2.67 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{15 \, {\left (a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f} \]

[In]

integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 20 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 7\right )}}{15 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]

[In]

integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 6.90 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.75 \[ \int \frac {1}{(3+3 \sin (e+f x))^3} \, dx=-\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (7\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+20\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )+40\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+30\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+15\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{15\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]

[In]

int(1/(a + a*sin(e + f*x))^3,x)

[Out]

-(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)