Integrand size = 10, antiderivative size = 48 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {43}{256} \text {arctanh}\left (\frac {\tan (x)}{2}\right )+\frac {5 \cos (x) \sin (x)}{16 \left (4-5 \sin ^2(x)\right )^2}-\frac {45 \cos (x) \sin (x)}{128 \left (4-5 \sin ^2(x)\right )} \] Output:
43/256*arctanh(1/2*tan(x))+5/16*cos(x)*sin(x)/(4-5*sin(x)^2)^2-45*cos(x)*s in(x)/(512-640*sin(x)^2)
Time = 5.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.35 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {1}{512} \left (-43 \log (2 \cos (x)-\sin (x))+43 \log (2 \cos (x)+\sin (x))+\frac {20}{(-2 \cos (x)+\sin (x))^2}-\frac {20}{(2 \cos (x)+\sin (x))^2}-\frac {180 \sin (2 x)}{3+5 \cos (2 x)}\right ) \] Input:
Integrate[(4 - 5*Sin[x]^2)^(-3),x]
Output:
(-43*Log[2*Cos[x] - Sin[x]] + 43*Log[2*Cos[x] + Sin[x]] + 20/(-2*Cos[x] + Sin[x])^2 - 20/(2*Cos[x] + Sin[x])^2 - (180*Sin[2*x])/(3 + 5*Cos[2*x]))/51 2
Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10, 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, 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.
| \(\displaystyle \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (4-5 \sin (x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3663 |
| \(\displaystyle \frac {1}{16} \int -\frac {10 \sin ^2(x)+1}{\left (4-5 \sin ^2(x)\right )^2}dx+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}-\frac {1}{16} \int \frac {10 \sin ^2(x)+1}{\left (4-5 \sin ^2(x)\right )^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}-\frac {1}{16} \int \frac {10 \sin (x)^2+1}{\left (4-5 \sin (x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {1}{16} \left (\frac {1}{8} \int \frac {43}{4-5 \sin ^2(x)}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\right )+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{16} \left (\frac {43}{8} \int \frac {1}{4-5 \sin ^2(x)}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\right )+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{16} \left (\frac {43}{8} \int \frac {1}{4-5 \sin (x)^2}dx-\frac {45 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\right )+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 3660 |
| \(\displaystyle \frac {1}{16} \left (\frac {43}{8} \int \frac {1}{4-\tan ^2(x)}d\tan (x)-\frac {45 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\right )+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{16} \left (\frac {43}{16} \text {arctanh}\left (\frac {\tan (x)}{2}\right )-\frac {45 \sin (x) \cos (x)}{8 \left (4-5 \sin ^2(x)\right )}\right )+\frac {5 \sin (x) \cos (x)}{16 \left (4-5 \sin ^2(x)\right )^2}\) |
Input:
Int[(4 - 5*Sin[x]^2)^(-3),x]
Output:
(5*Cos[x]*Sin[x])/(16*(4 - 5*Sin[x]^2)^2) + ((43*ArcTanh[Tan[x]/2])/16 - ( 45*Cos[x]*Sin[x])/(8*(4 - 5*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]))* 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)^(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 = 0.53 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00
| method | result | size |
| default | \(-\frac {25}{128 \left (\tan \left (x \right )+2\right )^{2}}+\frac {85}{256 \left (\tan \left (x \right )+2\right )}+\frac {43 \ln \left (\tan \left (x \right )+2\right )}{512}+\frac {25}{128 \left (\tan \left (x \right )-2\right )^{2}}+\frac {85}{256 \left (\tan \left (x \right )-2\right )}-\frac {43 \ln \left (\tan \left (x \right )-2\right )}{512}\) | \(48\) |
| risch | \(-\frac {i \left (215 \,{\mathrm e}^{6 i x}+387 \,{\mathrm e}^{4 i x}+325 \,{\mathrm e}^{2 i x}+225\right )}{64 \left (5 \,{\mathrm e}^{4 i x}+6 \,{\mathrm e}^{2 i x}+5\right )^{2}}+\frac {43 \ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}+\frac {4 i}{5}\right )}{512}-\frac {43 \ln \left ({\mathrm e}^{2 i x}+\frac {3}{5}-\frac {4 i}{5}\right )}{512}\) | \(70\) |
| norman | \(\frac {\frac {95 \tan \left (\frac {x}{2}\right )^{3}}{128}-\frac {95 \tan \left (\frac {x}{2}\right )^{5}}{128}+\frac {35 \tan \left (\frac {x}{2}\right )^{7}}{256}-\frac {35 \tan \left (\frac {x}{2}\right )}{256}}{\left (\tan \left (\frac {x}{2}\right )^{4}-3 \tan \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+\frac {43 \ln \left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )-1\right )}{512}-\frac {43 \ln \left (\tan \left (\frac {x}{2}\right )^{2}+\tan \left (\frac {x}{2}\right )-1\right )}{512}\) | \(84\) |
| parallelrisch | \(\frac {\left (1849+1075 \cos \left (4 x \right )+2580 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2}-\tan \left (\frac {x}{2}\right )-1\right )+\left (-1849-1075 \cos \left (4 x \right )-2580 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )^{2}+\tan \left (\frac {x}{2}\right )-1\right )-440 \sin \left (2 x \right )-900 \sin \left (4 x \right )}{12800 \cos \left (4 x \right )+22016+30720 \cos \left (2 x \right )}\) | \(89\) |
Input:
int(1/(4-5*sin(x)^2)^3,x,method=_RETURNVERBOSE)
Output:
-25/128/(tan(x)+2)^2+85/256/(tan(x)+2)+43/512*ln(tan(x)+2)+25/128/(tan(x)- 2)^2+85/256/(tan(x)-2)-43/512*ln(tan(x)-2)
Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (40) = 80\).
Time = 0.10 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.98 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {43 \, {\left (25 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 1\right )} \log \left (\frac {3}{4} \, \cos \left (x\right )^{2} + \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4}\right ) - 43 \, {\left (25 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 1\right )} \log \left (\frac {3}{4} \, \cos \left (x\right )^{2} - \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4}\right ) - 40 \, {\left (45 \, \cos \left (x\right )^{3} - 17 \, \cos \left (x\right )\right )} \sin \left (x\right )}{1024 \, {\left (25 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:
integrate(1/(4-5*sin(x)^2)^3,x, algorithm="fricas")
Output:
1/1024*(43*(25*cos(x)^4 - 10*cos(x)^2 + 1)*log(3/4*cos(x)^2 + cos(x)*sin(x ) + 1/4) - 43*(25*cos(x)^4 - 10*cos(x)^2 + 1)*log(3/4*cos(x)^2 - cos(x)*si n(x) + 1/4) - 40*(45*cos(x)^3 - 17*cos(x))*sin(x))/(25*cos(x)^4 - 10*cos(x )^2 + 1)
Leaf count of result is larger than twice the leaf count of optimal. 755 vs. \(2 (46) = 92\).
Time = 2.53 (sec) , antiderivative size = 755, normalized size of antiderivative = 15.73 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\text {Too large to display} \] Input:
integrate(1/(4-5*sin(x)**2)**3,x)
Output:
43*log(tan(x/2)**2 - tan(x/2) - 1)*tan(x/2)**8/(512*tan(x/2)**8 - 3072*tan (x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) - 258*log(tan(x/2)** 2 - tan(x/2) - 1)*tan(x/2)**6/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*t an(x/2)**4 - 3072*tan(x/2)**2 + 512) + 473*log(tan(x/2)**2 - tan(x/2) - 1) *tan(x/2)**4/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072 *tan(x/2)**2 + 512) - 258*log(tan(x/2)**2 - tan(x/2) - 1)*tan(x/2)**2/(512 *tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 51 2) + 43*log(tan(x/2)**2 - tan(x/2) - 1)/(512*tan(x/2)**8 - 3072*tan(x/2)** 6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) - 43*log(tan(x/2)**2 + tan( x/2) - 1)*tan(x/2)**8/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)* *4 - 3072*tan(x/2)**2 + 512) + 258*log(tan(x/2)**2 + tan(x/2) - 1)*tan(x/2 )**6/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2 )**2 + 512) - 473*log(tan(x/2)**2 + tan(x/2) - 1)*tan(x/2)**4/(512*tan(x/2 )**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) + 258 *log(tan(x/2)**2 + tan(x/2) - 1)*tan(x/2)**2/(512*tan(x/2)**8 - 3072*tan(x /2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) - 43*log(tan(x/2)**2 + tan(x/2) - 1)/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 30 72*tan(x/2)**2 + 512) + 70*tan(x/2)**7/(512*tan(x/2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) - 380*tan(x/2)**5/(512*tan(x /2)**8 - 3072*tan(x/2)**6 + 5632*tan(x/2)**4 - 3072*tan(x/2)**2 + 512) ...
Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {5 \, {\left (17 \, \tan \left (x\right )^{3} - 28 \, \tan \left (x\right )\right )}}{128 \, {\left (\tan \left (x\right )^{4} - 8 \, \tan \left (x\right )^{2} + 16\right )}} + \frac {43}{512} \, \log \left (\tan \left (x\right ) + 2\right ) - \frac {43}{512} \, \log \left (\tan \left (x\right ) - 2\right ) \] Input:
integrate(1/(4-5*sin(x)^2)^3,x, algorithm="maxima")
Output:
5/128*(17*tan(x)^3 - 28*tan(x))/(tan(x)^4 - 8*tan(x)^2 + 16) + 43/512*log( tan(x) + 2) - 43/512*log(tan(x) - 2)
Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {5 \, {\left (17 \, \tan \left (x\right )^{3} - 28 \, \tan \left (x\right )\right )}}{128 \, {\left (\tan \left (x\right )^{2} - 4\right )}^{2}} + \frac {43}{512} \, \log \left ({\left | \tan \left (x\right ) + 2 \right |}\right ) - \frac {43}{512} \, \log \left ({\left | \tan \left (x\right ) - 2 \right |}\right ) \] Input:
integrate(1/(4-5*sin(x)^2)^3,x, algorithm="giac")
Output:
5/128*(17*tan(x)^3 - 28*tan(x))/(tan(x)^2 - 4)^2 + 43/512*log(abs(tan(x) + 2)) - 43/512*log(abs(tan(x) - 2))
Time = 36.94 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {43\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (x\right )}{2}\right )}{256}-\frac {\frac {35\,\mathrm {tan}\left (x\right )}{32}-\frac {85\,{\mathrm {tan}\left (x\right )}^3}{128}}{{\mathrm {tan}\left (x\right )}^4-8\,{\mathrm {tan}\left (x\right )}^2+16} \] Input:
int(-1/(5*sin(x)^2 - 4)^3,x)
Output:
(43*atanh(tan(x)/2))/256 - ((35*tan(x))/32 - (85*tan(x)^3)/128)/(tan(x)^4 - 8*tan(x)^2 + 16)
Time = 0.19 (sec) , antiderivative size = 232, normalized size of antiderivative = 4.83 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^3} \, dx=\frac {900 \cos \left (x \right ) \sin \left (x \right )^{3}-560 \cos \left (x \right ) \sin \left (x \right )+1075 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{4}-1720 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}+688 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )-1075 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{4}+1720 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-688 \,\mathrm {log}\left (-\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right )+1075 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{4}-1720 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}+688 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )-1\right )-1075 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{4}+1720 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}-688 \,\mathrm {log}\left (\sqrt {5}+2 \tan \left (\frac {x}{2}\right )+1\right )}{12800 \sin \left (x \right )^{4}-20480 \sin \left (x \right )^{2}+8192} \] Input:
int(1/(4-5*sin(x)^2)^3,x)
Output:
(900*cos(x)*sin(x)**3 - 560*cos(x)*sin(x) + 1075*log( - sqrt(5) + 2*tan(x/ 2) - 1)*sin(x)**4 - 1720*log( - sqrt(5) + 2*tan(x/2) - 1)*sin(x)**2 + 688* log( - sqrt(5) + 2*tan(x/2) - 1) - 1075*log( - sqrt(5) + 2*tan(x/2) + 1)*s in(x)**4 + 1720*log( - sqrt(5) + 2*tan(x/2) + 1)*sin(x)**2 - 688*log( - sq rt(5) + 2*tan(x/2) + 1) + 1075*log(sqrt(5) + 2*tan(x/2) - 1)*sin(x)**4 - 1 720*log(sqrt(5) + 2*tan(x/2) - 1)*sin(x)**2 + 688*log(sqrt(5) + 2*tan(x/2) - 1) - 1075*log(sqrt(5) + 2*tan(x/2) + 1)*sin(x)**4 + 1720*log(sqrt(5) + 2*tan(x/2) + 1)*sin(x)**2 - 688*log(sqrt(5) + 2*tan(x/2) + 1))/(512*(25*si n(x)**4 - 40*sin(x)**2 + 16))