Integrand size = 10, antiderivative size = 62 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\frac {59 x}{256}-\frac {59}{256} \arctan \left (\frac {\cos (x) \sin (x)}{2-\sin ^2(x)}\right )-\frac {3 \cos (x) \sin (x)}{16 \left (4-3 \sin ^2(x)\right )^2}-\frac {45 \cos (x) \sin (x)}{128 \left (4-3 \sin ^2(x)\right )} \] Output:
59/256*x-59/256*arctan(cos(x)*sin(x)/(2-sin(x)^2))-3/16*cos(x)*sin(x)/(4-3 *sin(x)^2)^2-45*cos(x)*sin(x)/(512-384*sin(x)^2)
Time = 5.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.60 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\frac {1}{256} \left (-59 \arctan (2 \cot (x))-\frac {3 (182 \sin (2 x)+45 \sin (4 x))}{(5+3 \cos (2 x))^2}\right ) \] Input:
Integrate[(4 - 3*Sin[x]^2)^(-3),x]
Output:
(-59*ArcTan[2*Cot[x]] - (3*(182*Sin[2*x] + 45*Sin[4*x]))/(5 + 3*Cos[2*x])^ 2)/256
Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3663, 25, 3042, 3652, 27, 3042, 3660, 216}
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.
\(\displaystyle \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (4-3 \sin (x)^2\right )^3}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle -\frac {1}{16} \int -\frac {6 \sin ^2(x)+7}{\left (4-3 \sin ^2(x)\right )^2}dx-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{16} \int \frac {6 \sin ^2(x)+7}{\left (4-3 \sin ^2(x)\right )^2}dx-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{16} \int \frac {6 \sin (x)^2+7}{\left (4-3 \sin (x)^2\right )^2}dx-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 3652 |
\(\displaystyle \frac {1}{16} \left (\frac {1}{8} \int \frac {59}{4-3 \sin ^2(x)}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\right )-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \left (\frac {59}{8} \int \frac {1}{4-3 \sin ^2(x)}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\right )-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{16} \left (\frac {59}{8} \int \frac {1}{4-3 \sin (x)^2}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\right )-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {1}{16} \left (\frac {59}{8} \int \frac {1}{\tan ^2(x)+4}d\tan (x)-\frac {45 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\right )-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{16} \left (\frac {59}{16} \arctan \left (\frac {\tan (x)}{2}\right )-\frac {45 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\right )-\frac {3 \sin (x) \cos (x)}{16 \left (4-3 \sin ^2(x)\right )^2}\) |
Input:
Int[(4 - 3*Sin[x]^2)^(-3),x]
Output:
(-3*Cos[x]*Sin[x])/(16*(4 - 3*Sin[x]^2)^2) + ((59*ArcTan[Tan[x]/2])/16 - ( 45*Cos[x]*Sin[x])/(8*(4 - 3*Sin[x]^2)))/16
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*C os[e + f*x]*Sin[e + f*x]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(p + 1)*(a + b))), x] + Simp[1/(2*a*(p + 1)*(a + b)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[2*a*(p + 1) + b*(2*p + 3) - 2*b*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && LtQ[p, -1]
Time = 1.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {-\frac {69 \tan \left (x \right )^{3}}{128}-\frac {51 \tan \left (x \right )}{32}}{\left (\tan \left (x \right )^{2}+4\right )^{2}}+\frac {59 \arctan \left (\frac {\tan \left (x \right )}{2}\right )}{256}\) | \(29\) |
risch | \(-\frac {3 i \left (59 \,{\mathrm e}^{6 i x}+295 \,{\mathrm e}^{4 i x}+241 \,{\mathrm e}^{2 i x}+45\right )}{64 \left (3 \,{\mathrm e}^{4 i x}+10 \,{\mathrm e}^{2 i x}+3\right )^{2}}+\frac {59 i \ln \left ({\mathrm e}^{2 i x}+3\right )}{512}-\frac {59 i \ln \left ({\mathrm e}^{2 i x}+\frac {1}{3}\right )}{512}\) | \(68\) |
parallelrisch | \(\frac {59 i \left (-59-9 \cos \left (4 x \right )-60 \cos \left (2 x \right )\right ) \ln \left (-i \tan \left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )^{2}-1\right )+59 i \left (9 \cos \left (4 x \right )+59+60 \cos \left (2 x \right )\right ) \ln \left (i \tan \left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )^{2}-1\right )-2184 \sin \left (2 x \right )-540 \sin \left (4 x \right )}{4608 \cos \left (4 x \right )+30208+30720 \cos \left (2 x \right )}\) | \(98\) |
Input:
int(1/(4-3*sin(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
(-69/128*tan(x)^3-51/32*tan(x))/(tan(x)^2+4)^2+59/256*arctan(1/2*tan(x))
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=-\frac {59 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )} \arctan \left (\frac {5 \, \cos \left (x\right )^{2} - 1}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 12 \, {\left (45 \, \cos \left (x\right )^{3} + 23 \, \cos \left (x\right )\right )} \sin \left (x\right )}{512 \, {\left (9 \, \cos \left (x\right )^{4} + 6 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(4-3*sin(x)^2)^3,x, algorithm="fricas")
Output:
-1/512*(59*(9*cos(x)^4 + 6*cos(x)^2 + 1)*arctan(1/4*(5*cos(x)^2 - 1)/(cos( x)*sin(x))) + 12*(45*cos(x)^3 + 23*cos(x))*sin(x))/(9*cos(x)^4 + 6*cos(x)^ 2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 860 vs. \(2 (60) = 120\).
Time = 5.34 (sec) , antiderivative size = 860, normalized size of antiderivative = 13.87 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(4-3*sin(x)**2)**3,x)
Output:
59*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**8/(2 56*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/2)**2 + 256 ) - 118*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)* *6/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/2)**2 + 256) + 177*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan( x/2)**4/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/2 )**2 + 256) - 118*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi)) *tan(x/2)**2/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*ta n(x/2)**2 + 256) + 59*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/ pi))/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/2)** 2 + 256) + 59*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan (x/2)**8/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/ 2)**2 + 256) - 118*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2)/pi) )*tan(x/2)**6/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*t an(x/2)**2 + 256) + 177*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2 )/pi))*tan(x/2)**4/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - 512*tan(x/2)**2 + 256) - 118*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**2/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)* *4 - 512*tan(x/2)**2 + 256) + 59*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x /2 - pi/2)/pi))/(256*tan(x/2)**8 - 512*tan(x/2)**6 + 768*tan(x/2)**4 - ...
Time = 0.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.56 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=-\frac {3 \, {\left (23 \, \tan \left (x\right )^{3} + 68 \, \tan \left (x\right )\right )}}{128 \, {\left (\tan \left (x\right )^{4} + 8 \, \tan \left (x\right )^{2} + 16\right )}} + \frac {59}{256} \, \arctan \left (\frac {1}{2} \, \tan \left (x\right )\right ) \] Input:
integrate(1/(4-3*sin(x)^2)^3,x, algorithm="maxima")
Output:
-3/128*(23*tan(x)^3 + 68*tan(x))/(tan(x)^4 + 8*tan(x)^2 + 16) + 59/256*arc tan(1/2*tan(x))
Time = 0.45 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\frac {59}{256} \, x - \frac {3 \, {\left (23 \, \tan \left (x\right )^{3} + 68 \, \tan \left (x\right )\right )}}{128 \, {\left (\tan \left (x\right )^{2} + 4\right )}^{2}} - \frac {59}{256} \, \arctan \left (\frac {\sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 3}\right ) \] Input:
integrate(1/(4-3*sin(x)^2)^3,x, algorithm="giac")
Output:
59/256*x - 3/128*(23*tan(x)^3 + 68*tan(x))/(tan(x)^2 + 4)^2 - 59/256*arcta n(sin(2*x)/(cos(2*x) + 3))
Time = 36.02 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.69 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\frac {59\,x}{256}-\frac {59\,\mathrm {atan}\left (\mathrm {tan}\left (x\right )\right )}{256}+\frac {59\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )}{2}\right )}{256}-\frac {\frac {69\,{\mathrm {tan}\left (x\right )}^3}{128}+\frac {51\,\mathrm {tan}\left (x\right )}{32}}{{\mathrm {tan}\left (x\right )}^4+8\,{\mathrm {tan}\left (x\right )}^2+16} \] Input:
int(-1/(3*sin(x)^2 - 4)^3,x)
Output:
(59*x)/256 - (59*atan(tan(x)))/256 + (59*atan(tan(x)/2))/256 - ((51*tan(x) )/32 + (69*tan(x)^3)/128)/(8*tan(x)^2 + tan(x)^4 + 16)
Time = 0.17 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.94 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^3} \, dx=\frac {-531 \mathit {atan} \left (\sqrt {3}-2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{4}+1416 \mathit {atan} \left (\sqrt {3}-2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2}-944 \mathit {atan} \left (\sqrt {3}-2 \tan \left (\frac {x}{2}\right )\right )+531 \mathit {atan} \left (\sqrt {3}+2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{4}-1416 \mathit {atan} \left (\sqrt {3}+2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2}+944 \mathit {atan} \left (\sqrt {3}+2 \tan \left (\frac {x}{2}\right )\right )+270 \cos \left (x \right ) \sin \left (x \right )^{3}-408 \cos \left (x \right ) \sin \left (x \right )}{2304 \sin \left (x \right )^{4}-6144 \sin \left (x \right )^{2}+4096} \] Input:
int(1/(4-3*sin(x)^2)^3,x)
Output:
( - 531*atan(sqrt(3) - 2*tan(x/2))*sin(x)**4 + 1416*atan(sqrt(3) - 2*tan(x /2))*sin(x)**2 - 944*atan(sqrt(3) - 2*tan(x/2)) + 531*atan(sqrt(3) + 2*tan (x/2))*sin(x)**4 - 1416*atan(sqrt(3) + 2*tan(x/2))*sin(x)**2 + 944*atan(sq rt(3) + 2*tan(x/2)) + 270*cos(x)*sin(x)**3 - 408*cos(x)*sin(x))/(256*(9*si n(x)**4 - 24*sin(x)**2 + 16))