Integrand size = 10, antiderivative size = 30 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=-\frac {3}{16} \text {arctanh}\left (\frac {\tan (x)}{2}\right )+\frac {5 \cos (x) \sin (x)}{8 \left (4-5 \sin ^2(x)\right )} \] Output:
-3/16*arctanh(1/2*tan(x))+5*cos(x)*sin(x)/(32-40*sin(x)^2)
Time = 3.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=\frac {1}{32} \left (3 \log (2 \cos (x)-\sin (x))-3 \log (2 \cos (x)+\sin (x))+\frac {20 \sin (2 x)}{3+5 \cos (2 x)}\right ) \] Input:
Integrate[(4 - 5*Sin[x]^2)^(-2),x]
Output:
(3*Log[2*Cos[x] - Sin[x]] - 3*Log[2*Cos[x] + Sin[x]] + (20*Sin[2*x])/(3 + 5*Cos[2*x]))/32
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, 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, 219}
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.
\(\displaystyle \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (4-5 \sin (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3663 |
\(\displaystyle \frac {1}{8} \int -\frac {3}{4-5 \sin ^2(x)}dx+\frac {5 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}-\frac {3}{8} \int \frac {1}{4-5 \sin ^2(x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {5 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}-\frac {3}{8} \int \frac {1}{4-5 \sin (x)^2}dx\) |
\(\Big \downarrow \) 3660 |
\(\displaystyle \frac {5 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}-\frac {3}{8} \int \frac {1}{4-\tan ^2(x)}d\tan (x)\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}-\frac {3}{16} \text {arctanh}\left (\frac {\tan (x)}{2}\right )\) |
Input:
Int[(4 - 5*Sin[x]^2)^(-2),x]
Output:
(-3*ArcTanh[Tan[x]/2])/16 + (5*Cos[x]*Sin[x])/(8*(4 - 5*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]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[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 = 0.39 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07
method | result | size |
default | \(-\frac {5}{16 \left (\tan \left (x \right )-2\right )}+\frac {3 \ln \left (\tan \left (x \right )-2\right )}{32}-\frac {5}{16 \left (\tan \left (x \right )+2\right )}-\frac {3 \ln \left (\tan \left (x \right )+2\right )}{32}\) | \(32\) |
risch | \(\frac {i \left (3 \,{\mathrm e}^{2 i x}+5\right )}{20 \,{\mathrm e}^{4 i x}+24 \,{\mathrm e}^{2 i x}+20}-\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}+\frac {4 i}{5}\right )}{32}+\frac {3 \ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}-\frac {4 i}{5}\right )}{32}\) | \(56\) |
parallelrisch | \(\frac {\left (-9-15 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )-1\right )+\left (9+15 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2}+\tan \left (\frac {x}{2}\right )-1\right )+20 \sin \left (2 x \right )}{96+160 \cos \left (2 x \right )}\) | \(65\) |
norman | \(\frac {-\frac {5 \tan \left (\frac {x}{2}\right )^{3}}{16}+\frac {5 \tan \left (\frac {x}{2}\right )}{16}}{\tan \left (\frac {x}{2}\right )^{4}-3 \tan \left (\frac {x}{2}\right )^{2}+1}-\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )-1\right )}{32}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )^{2}+\tan \left (\frac {x}{2}\right )-1\right )}{32}\) | \(68\) |
Input:
int(1/(4-5*sin(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
-5/16/(tan(x)-2)+3/32*ln(tan(x)-2)-5/16/(tan(x)+2)-3/32*ln(tan(x)+2)
Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.27 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=-\frac {3 \, {\left (5 \, \cos \left (x\right )^{2} - 1\right )} \log \left (\frac {3}{4} \, \cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4}\right ) - 3 \, {\left (5 \, \cos \left (x\right )^{2} - 1\right )} \log \left (\frac {3}{4} \, \cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4}\right ) - 40 \, \cos \left (x\right ) \sin \left (x\right )}{64 \, {\left (5 \, \cos \left (x\right )^{2} - 1\right )}} \] Input:
integrate(1/(4-5*sin(x)^2)^2,x, algorithm="fricas")
Output:
-1/64*(3*(5*cos(x)^2 - 1)*log(3/4*cos(x)^2 + cos(x)*sin(x) + 1/4) - 3*(5*c os(x)^2 - 1)*log(3/4*cos(x)^2 - cos(x)*sin(x) + 1/4) - 40*cos(x)*sin(x))/( 5*cos(x)^2 - 1)
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (26) = 52\).
Time = 0.79 (sec) , antiderivative size = 292, normalized size of antiderivative = 9.73 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=- \frac {3 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} - \tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan ^{4}{\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} + \frac {9 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} - \tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} - \frac {3 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} - \tan {\left (\frac {x}{2} \right )} - 1 \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} + \frac {3 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + \tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan ^{4}{\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} - \frac {9 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + \tan {\left (\frac {x}{2} \right )} - 1 \right )} \tan ^{2}{\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} + \frac {3 \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + \tan {\left (\frac {x}{2} \right )} - 1 \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} - \frac {10 \tan ^{3}{\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} + \frac {10 \tan {\left (\frac {x}{2} \right )}}{32 \tan ^{4}{\left (\frac {x}{2} \right )} - 96 \tan ^{2}{\left (\frac {x}{2} \right )} + 32} \] Input:
integrate(1/(4-5*sin(x)**2)**2,x)
Output:
-3*log(tan(x/2)**2 - tan(x/2) - 1)*tan(x/2)**4/(32*tan(x/2)**4 - 96*tan(x/ 2)**2 + 32) + 9*log(tan(x/2)**2 - tan(x/2) - 1)*tan(x/2)**2/(32*tan(x/2)** 4 - 96*tan(x/2)**2 + 32) - 3*log(tan(x/2)**2 - tan(x/2) - 1)/(32*tan(x/2)* *4 - 96*tan(x/2)**2 + 32) + 3*log(tan(x/2)**2 + tan(x/2) - 1)*tan(x/2)**4/ (32*tan(x/2)**4 - 96*tan(x/2)**2 + 32) - 9*log(tan(x/2)**2 + tan(x/2) - 1) *tan(x/2)**2/(32*tan(x/2)**4 - 96*tan(x/2)**2 + 32) + 3*log(tan(x/2)**2 + tan(x/2) - 1)/(32*tan(x/2)**4 - 96*tan(x/2)**2 + 32) - 10*tan(x/2)**3/(32* tan(x/2)**4 - 96*tan(x/2)**2 + 32) + 10*tan(x/2)/(32*tan(x/2)**4 - 96*tan( x/2)**2 + 32)
Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=-\frac {5 \, \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} - 4\right )}} - \frac {3}{32} \, \log \left (\tan \left (x\right ) + 2\right ) + \frac {3}{32} \, \log \left (\tan \left (x\right ) - 2\right ) \] Input:
integrate(1/(4-5*sin(x)^2)^2,x, algorithm="maxima")
Output:
-5/8*tan(x)/(tan(x)^2 - 4) - 3/32*log(tan(x) + 2) + 3/32*log(tan(x) - 2)
Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=-\frac {5 \, \tan \left (x\right )}{8 \, {\left (\tan \left (x\right )^{2} - 4\right )}} - \frac {3}{32} \, \log \left ({\left | \tan \left (x\right ) + 2 \right |}\right ) + \frac {3}{32} \, \log \left ({\left | \tan \left (x\right ) - 2 \right |}\right ) \] Input:
integrate(1/(4-5*sin(x)^2)^2,x, algorithm="giac")
Output:
-5/8*tan(x)/(tan(x)^2 - 4) - 3/32*log(abs(tan(x) + 2)) + 3/32*log(abs(tan( x) - 2))
Time = 36.98 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=-\frac {3\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (x\right )}{2}\right )}{16}-\frac {5\,\mathrm {tan}\left (x\right )}{8\,\left ({\mathrm {tan}\left (x\right )}^2-4\right )} \] Input:
int(1/(5*sin(x)^2 - 4)^2,x)
Output:
- (3*atanh(tan(x)/2))/16 - (5*tan(x))/(8*(tan(x)^2 - 4))
Time = 0.18 (sec) , antiderivative size = 146, normalized size of antiderivative = 4.87 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^2} \, dx=\frac {-20 \cos \left (x \right ) \sin \left (x \right )-15 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}+12 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )+15 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-12 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right )-15 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}+12 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )+15 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-12 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right )}{160 \sin \left (x \right )^{2}-128} \] Input:
int(1/(4-5*sin(x)^2)^2,x)
Output:
( - 20*cos(x)*sin(x) - 15*log( - sqrt(5) + 2*tan(x/2) - 1)*sin(x)**2 + 12* log( - sqrt(5) + 2*tan(x/2) - 1) + 15*log( - sqrt(5) + 2*tan(x/2) + 1)*sin (x)**2 - 12*log( - sqrt(5) + 2*tan(x/2) + 1) - 15*log(sqrt(5) + 2*tan(x/2) - 1)*sin(x)**2 + 12*log(sqrt(5) + 2*tan(x/2) - 1) + 15*log(sqrt(5) + 2*ta n(x/2) + 1)*sin(x)**2 - 12*log(sqrt(5) + 2*tan(x/2) + 1))/(32*(5*sin(x)**2 - 4))