Integrand size = 13, antiderivative size = 61 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}(\sin (x)) \cos (x)}{8 a^2 \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}+\frac {3 \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}} \] Output:
3/8*arctanh(sin(x))*cos(x)/a^2/(a*cos(x)^2)^(1/2)+1/4*tan(x)/a/(a*cos(x)^2 )^(3/2)+3/8*tan(x)/a^2/(a*cos(x)^2)^(1/2)
Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.59 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=\frac {3 \text {arctanh}(\sin (x)) \cos (x)+\left (3+2 \sec ^2(x)\right ) \tan (x)}{8 a^2 \sqrt {a \cos ^2(x)}} \] Input:
Integrate[(a - a*Sin[x]^2)^(-5/2),x]
Output:
(3*ArcTanh[Sin[x]]*Cos[x] + (3 + 2*Sec[x]^2)*Tan[x])/(8*a^2*Sqrt[a*Cos[x]^ 2])
Time = 0.44 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {3042, 3655, 3042, 3683, 3042, 3683, 3042, 3686, 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.
\(\displaystyle \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a-a \sin (x)^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {1}{\left (a \cos ^2(x)\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}}dx\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {3 \int \frac {1}{\left (a \cos ^2(x)\right )^{3/2}}dx}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {1}{\left (a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{3/2}}dx}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3683 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {a \cos ^2(x)}}dx}{2 a}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}}\right )}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {a \sin \left (x+\frac {\pi }{2}\right )^2}}dx}{2 a}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}}\right )}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {3 \left (\frac {\cos (x) \int \sec (x)dx}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}}\right )}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\cos (x) \int \csc \left (x+\frac {\pi }{2}\right )dx}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}}\right )}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {3 \left (\frac {\cos (x) \text {arctanh}(\sin (x))}{2 a \sqrt {a \cos ^2(x)}}+\frac {\tan (x)}{2 a \sqrt {a \cos ^2(x)}}\right )}{4 a}+\frac {\tan (x)}{4 a \left (a \cos ^2(x)\right )^{3/2}}\) |
Input:
Int[(a - a*Sin[x]^2)^(-5/2),x]
Output:
Tan[x]/(4*a*(a*Cos[x]^2)^(3/2)) + (3*((ArcTanh[Sin[x]]*Cos[x])/(2*a*Sqrt[a *Cos[x]^2]) + Tan[x]/(2*a*Sqrt[a*Cos[x]^2])))/(4*a)
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[Cot[e + f*x]* ((b*Sin[e + f*x]^2)^(p + 1)/(b*f*(2*p + 1))), x] + Simp[2*((p + 1)/(b*(2*p + 1))) Int[(b*Sin[e + f*x]^2)^(p + 1), x], x] /; FreeQ[{b, e, f}, x] && !IntegerQ[p] && LtQ[p, -1]
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03
method | result | size |
default | \(-\frac {\left (3 \ln \left (1+\sin \left (x \right )\right )-3 \ln \left (\sin \left (x \right )-1\right )\right ) \cos \left (x \right )^{4}+6 \cos \left (x \right )^{2} \sin \left (x \right )+4 \sin \left (x \right )}{16 a^{2} \left (1+\sin \left (x \right )\right ) \left (\sin \left (x \right )-1\right ) \cos \left (x \right ) \sqrt {a \cos \left (x \right )^{2}}}\) | \(63\) |
risch | \(-\frac {i \left (3 \,{\mathrm e}^{6 i x}+11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}-3\right )}{4 a^{2} \left ({\mathrm e}^{2 i x}+1\right )^{3} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {3 \ln \left ({\mathrm e}^{i x}-i\right ) \cos \left (x \right )}{4 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {3 \ln \left ({\mathrm e}^{i x}+i\right ) \cos \left (x \right )}{4 a^{2} \sqrt {a \left ({\mathrm e}^{2 i x}+1\right )^{2} {\mathrm e}^{-2 i x}}}\) | \(126\) |
Input:
int(1/(a-a*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/16/a^2*((3*ln(1+sin(x))-3*ln(sin(x)-1))*cos(x)^4+6*cos(x)^2*sin(x)+4*si n(x))/(1+sin(x))/(sin(x)-1)/cos(x)/(a*cos(x)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=-\frac {{\left (3 \, \cos \left (x\right )^{4} \log \left (-\frac {\sin \left (x\right ) - 1}{\sin \left (x\right ) + 1}\right ) - 2 \, {\left (3 \, \cos \left (x\right )^{2} + 2\right )} \sin \left (x\right )\right )} \sqrt {a \cos \left (x\right )^{2}}}{16 \, a^{3} \cos \left (x\right )^{5}} \] Input:
integrate(1/(a-a*sin(x)^2)^(5/2),x, algorithm="fricas")
Output:
-1/16*(3*cos(x)^4*log(-(sin(x) - 1)/(sin(x) + 1)) - 2*(3*cos(x)^2 + 2)*sin (x))*sqrt(a*cos(x)^2)/(a^3*cos(x)^5)
\[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (- a \sin ^{2}{\left (x \right )} + a\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a-a*sin(x)**2)**(5/2),x)
Output:
Integral((-a*sin(x)**2 + a)**(-5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 933 vs. \(2 (49) = 98\).
Time = 0.30 (sec) , antiderivative size = 933, normalized size of antiderivative = 15.30 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a-a*sin(x)^2)^(5/2),x, algorithm="maxima")
Output:
1/16*(4*(3*sin(7*x) + 11*sin(5*x) - 11*sin(3*x) - 3*sin(x))*cos(8*x) - 24* (2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*cos(7*x) + 16*(11*sin(5*x) - 11*sin (3*x) - 3*sin(x))*cos(6*x) - 88*(3*sin(4*x) + 2*sin(2*x))*cos(5*x) - 24*(1 1*sin(3*x) + 3*sin(x))*cos(4*x) + 3*(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2* x) + 1)*cos(8*x) + cos(8*x)^2 + 8*(6*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^2 + 16*cos(2*x )^2 + 4*(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16* (3*sin(4*x) + 2*sin(2*x))*sin(6*x) + 16*sin(6*x)^2 + 36*sin(4*x)^2 + 48*si n(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x) + 1)*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - 3*(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*cos(8*x) + cos(8*x)^2 + 8*(6*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x) + 36*cos(4*x)^2 + 16*cos(2*x)^2 + 4*(2*sin(6 *x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16*(3*sin(4*x) + 2* sin(2*x))*sin(6*x) + 16*sin(6*x)^2 + 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x) + 1)*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) - 4*(3*cos(7*x) + 11*cos(5*x) - 11*cos(3*x) - 3*cos(x))*sin(8*x) + 12*(4*c os(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*sin(7*x) - 16*(11*cos(5*x) - 11*cos (3*x) - 3*cos(x))*sin(6*x) + 44*(6*cos(4*x) + 4*cos(2*x) + 1)*sin(5*x) + 2 4*(11*cos(3*x) + 3*cos(x))*sin(4*x) - 44*(4*cos(2*x) + 1)*sin(3*x) + 176*c os(3*x)*sin(2*x) + 48*cos(x)*sin(2*x) - 48*cos(2*x)*sin(x) - 12*sin(x))...
Time = 0.47 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=-\frac {5 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{3} - \frac {12}{\tan \left (\frac {1}{2} \, x\right )} - 12 \, \tan \left (\frac {1}{2} \, x\right )}{4 \, {\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, x\right )} + \tan \left (\frac {1}{2} \, x\right )\right )}^{2} - 4\right )}^{2} a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{4} - 1\right )} \] Input:
integrate(1/(a-a*sin(x)^2)^(5/2),x, algorithm="giac")
Output:
-1/4*(5*(1/tan(1/2*x) + tan(1/2*x))^3 - 12/tan(1/2*x) - 12*tan(1/2*x))/((( 1/tan(1/2*x) + tan(1/2*x))^2 - 4)^2*a^(5/2)*sgn(tan(1/2*x)^4 - 1))
Timed out. \[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a-a\,{\sin \left (x\right )}^2\right )}^{5/2}} \,d x \] Input:
int(1/(a - a*sin(x)^2)^(5/2),x)
Output:
int(1/(a - a*sin(x)^2)^(5/2), x)
\[ \int \frac {1}{\left (a-a \sin ^2(x)\right )^{5/2}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\sin \left (x \right )^{2}+1}}{\sin \left (x \right )^{6}-3 \sin \left (x \right )^{4}+3 \sin \left (x \right )^{2}-1}d x \right )}{a^{3}} \] Input:
int(1/(a-a*sin(x)^2)^(5/2),x)
Output:
( - sqrt(a)*int(sqrt( - sin(x)**2 + 1)/(sin(x)**6 - 3*sin(x)**4 + 3*sin(x) **2 - 1),x))/a**3