Integrand size = 10, antiderivative size = 44 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {5 x}{16}-\frac {5}{16} \arctan \left (\frac {\cos (x) \sin (x)}{2-\sin ^2(x)}\right )-\frac {3 \cos (x) \sin (x)}{8 \left (4-3 \sin ^2(x)\right )} \] Output:
5/16*x-5/16*arctan(cos(x)*sin(x)/(2-sin(x)^2))-3*cos(x)*sin(x)/(32-24*sin( x)^2)
Time = 3.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {-5 \arctan (2 \cot (x)) (5+3 \cos (2 x))-6 \sin (2 x)}{80+48 \cos (2 x)} \] Input:
Integrate[(4 - 3*Sin[x]^2)^(-2),x]
Output:
(-5*ArcTan[2*Cot[x]]*(5 + 3*Cos[2*x]) - 6*Sin[2*x])/(80 + 48*Cos[2*x])
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3663, 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.
\(\displaystyle \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (4-3 \sin (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle -\frac {1}{8} \int -\frac {5}{4-3 \sin ^2(x)}dx-\frac {3 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{8} \int \frac {1}{4-3 \sin ^2(x)}dx-\frac {3 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5}{8} \int \frac {1}{4-3 \sin (x)^2}dx-\frac {3 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {5}{8} \int \frac {1}{\tan ^2(x)+4}d\tan (x)-\frac {3 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{16} \arctan \left (\frac {\tan (x)}{2}\right )-\frac {3 \sin (x) \cos (x)}{8 \left (4-3 \sin ^2(x)\right )}\) |
Input:
Int[(4 - 3*Sin[x]^2)^(-2),x]
Output:
(5*ArcTan[Tan[x]/2])/16 - (3*Cos[x]*Sin[x])/(8*(4 - 3*Sin[x]^2))
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)^(-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 = 0.62 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.48
method | result | size |
default | \(-\frac {3 \tan \left (x \right )}{8 \left (\tan \left (x \right )^{2}+4\right )}+\frac {5 \arctan \left (\frac {\tan \left (x \right )}{2}\right )}{16}\) | \(21\) |
risch | \(-\frac {i \left (5 \,{\mathrm e}^{2 i x}+3\right )}{4 \left (3 \,{\mathrm e}^{4 i x}+10 \,{\mathrm e}^{2 i x}+3\right )}-\frac {5 i \ln \left ({\mathrm e}^{2 i x}+\frac {1}{3}\right )}{32}+\frac {5 i \ln \left ({\mathrm e}^{2 i x}+3\right )}{32}\) | \(54\) |
parallelrisch | \(\frac {5 i \left (-5-3 \cos \left (2 x \right )\right ) \ln \left (-i \tan \left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )^{2}-1\right )+5 i \left (5+3 \cos \left (2 x \right )\right ) \ln \left (i \tan \left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )^{2}-1\right )-12 \sin \left (2 x \right )}{160+96 \cos \left (2 x \right )}\) | \(74\) |
Input:
int(1/(4-3*sin(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
-3/8*tan(x)/(tan(x)^2+4)+5/16*arctan(1/2*tan(x))
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=-\frac {5 \, {\left (3 \, \cos \left (x\right )^{2} + 1\right )} \arctan \left (\frac {5 \, \cos \left (x\right )^{2} - 1}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) + 12 \, \cos \left (x\right ) \sin \left (x\right )}{32 \, {\left (3 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(4-3*sin(x)^2)^2,x, algorithm="fricas")
Output:
-1/32*(5*(3*cos(x)^2 + 1)*arctan(1/4*(5*cos(x)^2 - 1)/(cos(x)*sin(x))) + 1 2*cos(x)*sin(x))/(3*cos(x)^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (39) = 78\).
Time = 1.57 (sec) , antiderivative size = 357, normalized size of antiderivative = 8.11 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} - \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{4}{\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} - \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} - \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} + \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} - \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} + \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} + \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{4}{\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} - \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} + \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} + \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {x}{2} \right )} + \sqrt {3} \right )} + \pi \left \lfloor {\frac {\frac {x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} + \frac {3 \tan ^{3}{\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} - \frac {3 \tan {\left (\frac {x}{2} \right )}}{16 \tan ^{4}{\left (\frac {x}{2} \right )} - 16 \tan ^{2}{\left (\frac {x}{2} \right )} + 16} \] Input:
integrate(1/(4-3*sin(x)**2)**2,x)
Output:
5*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(16 *tan(x/2)**4 - 16*tan(x/2)**2 + 16) - 5*(atan(2*tan(x/2) - sqrt(3)) + pi*f loor((x/2 - pi/2)/pi))*tan(x/2)**2/(16*tan(x/2)**4 - 16*tan(x/2)**2 + 16) + 5*(atan(2*tan(x/2) - sqrt(3)) + pi*floor((x/2 - pi/2)/pi))/(16*tan(x/2)* *4 - 16*tan(x/2)**2 + 16) + 5*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**4/(16*tan(x/2)**4 - 16*tan(x/2)**2 + 16) - 5*(atan( 2*tan(x/2) + sqrt(3)) + pi*floor((x/2 - pi/2)/pi))*tan(x/2)**2/(16*tan(x/2 )**4 - 16*tan(x/2)**2 + 16) + 5*(atan(2*tan(x/2) + sqrt(3)) + pi*floor((x/ 2 - pi/2)/pi))/(16*tan(x/2)**4 - 16*tan(x/2)**2 + 16) + 3*tan(x/2)**3/(16* tan(x/2)**4 - 16*tan(x/2)**2 + 16) - 3*tan(x/2)/(16*tan(x/2)**4 - 16*tan(x /2)**2 + 16)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=-\frac {3 \, \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} + 4\right )}} + \frac {5}{16} \, \arctan \left (\frac {1}{2} \, \tan \left (x\right )\right ) \] Input:
integrate(1/(4-3*sin(x)^2)^2,x, algorithm="maxima")
Output:
-3/8*tan(x)/(tan(x)^2 + 4) + 5/16*arctan(1/2*tan(x))
Time = 0.44 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {5}{16} \, x - \frac {3 \, \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} + 4\right )}} - \frac {5}{16} \, \arctan \left (\frac {\sin \left (2 \, x\right )}{\cos \left (2 \, x\right ) + 3}\right ) \] Input:
integrate(1/(4-3*sin(x)^2)^2,x, algorithm="giac")
Output:
5/16*x - 3/8*tan(x)/(tan(x)^2 + 4) - 5/16*arctan(sin(2*x)/(cos(2*x) + 3))
Time = 36.44 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {5\,x}{16}-\frac {5\,\mathrm {atan}\left (\mathrm {tan}\left (x\right )\right )}{16}+\frac {5\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (x\right )}{2}\right )}{16}-\frac {3\,\mathrm {tan}\left (x\right )}{8\,\left ({\mathrm {tan}\left (x\right )}^2+4\right )} \] Input:
int(1/(3*sin(x)^2 - 4)^2,x)
Output:
(5*x)/16 - (5*atan(tan(x)))/16 + (5*atan(tan(x)/2))/16 - (3*tan(x))/(8*(ta n(x)^2 + 4))
Time = 0.18 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.68 \[ \int \frac {1}{\left (4-3 \sin ^2(x)\right )^2} \, dx=\frac {-15 \mathit {atan} \left (\sqrt {3}-2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2}+20 \mathit {atan} \left (\sqrt {3}-2 \tan \left (\frac {x}{2}\right )\right )+15 \mathit {atan} \left (\sqrt {3}+2 \tan \left (\frac {x}{2}\right )\right ) \sin \left (x \right )^{2}-20 \mathit {atan} \left (\sqrt {3}+2 \tan \left (\frac {x}{2}\right )\right )+6 \cos \left (x \right ) \sin \left (x \right )}{48 \sin \left (x \right )^{2}-64} \] Input:
int(1/(4-3*sin(x)^2)^2,x)
Output:
( - 15*atan(sqrt(3) - 2*tan(x/2))*sin(x)**2 + 20*atan(sqrt(3) - 2*tan(x/2) ) + 15*atan(sqrt(3) + 2*tan(x/2))*sin(x)**2 - 20*atan(sqrt(3) + 2*tan(x/2) ) + 6*cos(x)*sin(x))/(16*(3*sin(x)**2 - 4))