Integrand size = 14, antiderivative size = 35 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3 \text {arctanh}(\sin (x))}{8 a^2}+\frac {3 \sec (x) \tan (x)}{8 a^2}+\frac {\sec ^3(x) \tan (x)}{4 a^2} \] Output:
3/8*arctanh(sin(x))/a^2+3/8*sec(x)*tan(x)/a^2+1/4*sec(x)^3*tan(x)/a^2
Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\frac {3}{8} \text {arctanh}(\sin (x))+\frac {3}{8} \sec (x) \tan (x)+\frac {1}{4} \sec ^3(x) \tan (x)}{a^2} \] Input:
Integrate[Sec[x]/(a - a*Sin[x]^2)^2,x]
Output:
((3*ArcTanh[Sin[x]])/8 + (3*Sec[x]*Tan[x])/8 + (Sec[x]^3*Tan[x])/4)/a^2
Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3654, 3042, 4255, 3042, 4255, 3042, 4257}
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.
\(\displaystyle \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\cos (x) \left (a-a \sin (x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \frac {\int \sec ^5(x)dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (x+\frac {\pi }{2}\right )^5dx}{a^2}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {3}{4} \int \sec ^3(x)dx+\frac {1}{4} \tan (x) \sec ^3(x)}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{4} \int \csc \left (x+\frac {\pi }{2}\right )^3dx+\frac {1}{4} \tan (x) \sec ^3(x)}{a^2}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {\int \sec (x)dx}{2}+\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int \csc \left (x+\frac {\pi }{2}\right )dx+\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \text {arctanh}(\sin (x))+\frac {1}{2} \tan (x) \sec (x)\right )+\frac {1}{4} \tan (x) \sec ^3(x)}{a^2}\) |
Input:
Int[Sec[x]/(a - a*Sin[x]^2)^2,x]
Output:
((Sec[x]^3*Tan[x])/4 + (3*(ArcTanh[Sin[x]]/2 + (Sec[x]*Tan[x])/2))/4)/a^2
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ a^p Int[ActivateTrig[u*cos[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.57 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49
method | result | size |
default | \(\frac {\frac {1}{16 \left (\sin \left (x \right )-1\right )^{2}}-\frac {3}{16 \left (\sin \left (x \right )-1\right )}-\frac {3 \ln \left (\sin \left (x \right )-1\right )}{16}-\frac {1}{16 \left (1+\sin \left (x \right )\right )^{2}}-\frac {3}{16 \left (1+\sin \left (x \right )\right )}+\frac {3 \ln \left (1+\sin \left (x \right )\right )}{16}}{a^{2}}\) | \(52\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{7 i x}+11 \,{\mathrm e}^{5 i x}-11 \,{\mathrm e}^{3 i x}-3 \,{\mathrm e}^{i x}\right )}{4 \left ({\mathrm e}^{2 i x}+1\right )^{4} a^{2}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right )}{8 a^{2}}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right )}{8 a^{2}}\) | \(74\) |
parallelrisch | \(\frac {\left (-9-3 \cos \left (4 x \right )-12 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )+\left (9+3 \cos \left (4 x \right )+12 \cos \left (2 x \right )\right ) \ln \left (\tan \left (\frac {x}{2}\right )+1\right )+22 \sin \left (x \right )+6 \sin \left (3 x \right )}{8 a^{2} \left (\cos \left (4 x \right )+3+4 \cos \left (2 x \right )\right )}\) | \(75\) |
norman | \(\frac {\frac {5 \tan \left (\frac {x}{2}\right )}{4 a}+\frac {3 \tan \left (\frac {x}{2}\right )^{3}}{4 a}+\frac {3 \tan \left (\frac {x}{2}\right )^{5}}{4 a}+\frac {5 \tan \left (\frac {x}{2}\right )^{7}}{4 a}}{\left (\tan \left (\frac {x}{2}\right )^{2}-1\right )^{4} a}-\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{8 a^{2}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{8 a^{2}}\) | \(83\) |
Input:
int(sec(x)/(a-a*sin(x)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/a^2*(1/16/(sin(x)-1)^2-3/16/(sin(x)-1)-3/16*ln(sin(x)-1)-1/16/(1+sin(x)) ^2-3/16/(1+sin(x))+3/16*ln(1+sin(x)))
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.31 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3 \, \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 3 \, \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (3 \, \cos \left (x\right )^{2} + 2\right )} \sin \left (x\right )}{16 \, a^{2} \cos \left (x\right )^{4}} \] Input:
integrate(sec(x)/(a-a*sin(x)^2)^2,x, algorithm="fricas")
Output:
1/16*(3*cos(x)^4*log(sin(x) + 1) - 3*cos(x)^4*log(-sin(x) + 1) + 2*(3*cos( x)^2 + 2)*sin(x))/(a^2*cos(x)^4)
\[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {\int \frac {\sec {\left (x \right )}}{\sin ^{4}{\left (x \right )} - 2 \sin ^{2}{\left (x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(x)/(a-a*sin(x)**2)**2,x)
Output:
Integral(sec(x)/(sin(x)**4 - 2*sin(x)**2 + 1), x)/a**2
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.63 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=-\frac {3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \, {\left (a^{2} \sin \left (x\right )^{4} - 2 \, a^{2} \sin \left (x\right )^{2} + a^{2}\right )}} + \frac {3 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a^{2}} - \frac {3 \, \log \left (\sin \left (x\right ) - 1\right )}{16 \, a^{2}} \] Input:
integrate(sec(x)/(a-a*sin(x)^2)^2,x, algorithm="maxima")
Output:
-1/8*(3*sin(x)^3 - 5*sin(x))/(a^2*sin(x)^4 - 2*a^2*sin(x)^2 + a^2) + 3/16* log(sin(x) + 1)/a^2 - 3/16*log(sin(x) - 1)/a^2
Time = 0.41 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.34 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3 \, \log \left (\sin \left (x\right ) + 1\right )}{16 \, a^{2}} - \frac {3 \, \log \left (-\sin \left (x\right ) + 1\right )}{16 \, a^{2}} - \frac {3 \, \sin \left (x\right )^{3} - 5 \, \sin \left (x\right )}{8 \, {\left (\sin \left (x\right )^{2} - 1\right )}^{2} a^{2}} \] Input:
integrate(sec(x)/(a-a*sin(x)^2)^2,x, algorithm="giac")
Output:
3/16*log(sin(x) + 1)/a^2 - 3/16*log(-sin(x) + 1)/a^2 - 1/8*(3*sin(x)^3 - 5 *sin(x))/((sin(x)^2 - 1)^2*a^2)
Time = 36.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {3\,\mathrm {atanh}\left (\sin \left (x\right )\right )}{8\,a^2}+\frac {3\,\sin \left (x\right )}{8\,a^2\,{\cos \left (x\right )}^2}+\frac {\sin \left (x\right )}{4\,a^2\,{\cos \left (x\right )}^4} \] Input:
int(1/(cos(x)*(a - a*sin(x)^2)^2),x)
Output:
(3*atanh(sin(x)))/(8*a^2) + (3*sin(x))/(8*a^2*cos(x)^2) + sin(x)/(4*a^2*co s(x)^4)
Time = 0.17 (sec) , antiderivative size = 100, normalized size of antiderivative = 2.86 \[ \int \frac {\sec (x)}{\left (a-a \sin ^2(x)\right )^2} \, dx=\frac {-3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{4}+6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right ) \sin \left (x \right )^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )-1\right )+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{4}-6 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right ) \sin \left (x \right )^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {x}{2}\right )+1\right )-3 \sin \left (x \right )^{3}+5 \sin \left (x \right )}{8 a^{2} \left (\sin \left (x \right )^{4}-2 \sin \left (x \right )^{2}+1\right )} \] Input:
int(sec(x)/(a-a*sin(x)^2)^2,x)
Output:
( - 3*log(tan(x/2) - 1)*sin(x)**4 + 6*log(tan(x/2) - 1)*sin(x)**2 - 3*log( tan(x/2) - 1) + 3*log(tan(x/2) + 1)*sin(x)**4 - 6*log(tan(x/2) + 1)*sin(x) **2 + 3*log(tan(x/2) + 1) - 3*sin(x)**3 + 5*sin(x))/(8*a**2*(sin(x)**4 - 2 *sin(x)**2 + 1))