\(\int \frac {1}{(a-a \cos ^2(x))^{5/2}} \, dx\) [46]

Optimal result

Integrand size = 13, antiderivative size = 61 \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}-\frac {3 \cot (x)}{8 a^2 \sqrt {a \sin ^2(x)}}-\frac {3 \text {arctanh}(\cos (x)) \sin (x)}{8 a^2 \sqrt {a \sin ^2(x)}} \] Output:

-1/4*cot(x)/a/(a*sin(x)^2)^(3/2)-3/8*cot(x)/a^2/(a*sin(x)^2)^(1/2)-3/8*arc 
tanh(cos(x))*sin(x)/a^2/(a*sin(x)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=-\frac {\csc (x) \left (6 \csc ^2\left (\frac {x}{2}\right )+\csc ^4\left (\frac {x}{2}\right )+24 \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )-6 \sec ^2\left (\frac {x}{2}\right )-\sec ^4\left (\frac {x}{2}\right )\right ) \sqrt {a \sin ^2(x)}}{64 a^3} \] Input:

Integrate[(a - a*Cos[x]^2)^(-5/2),x]
 

Output:

-1/64*(Csc[x]*(6*Csc[x/2]^2 + Csc[x/2]^4 + 24*(Log[Cos[x/2]] - Log[Sin[x/2 
]]) - 6*Sec[x/2]^2 - Sec[x/2]^4)*Sqrt[a*Sin[x]^2])/a^3
 

Rubi [A] (verified)

Time = 0.43 (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.

\(\displaystyle \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a-a \sin \left (x+\frac {\pi }{2}\right )^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3655

\(\displaystyle \int \frac {1}{\left (a \sin ^2(x)\right )^{5/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a \sin (x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3683

\(\displaystyle \frac {3 \int \frac {1}{\left (a \sin ^2(x)\right )^{3/2}}dx}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{\left (a \sin (x)^2\right )^{3/2}}dx}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3683

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {a \sin ^2(x)}}dx}{2 a}-\frac {\cot (x)}{2 a \sqrt {a \sin ^2(x)}}\right )}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {a \sin (x)^2}}dx}{2 a}-\frac {\cot (x)}{2 a \sqrt {a \sin ^2(x)}}\right )}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3686

\(\displaystyle \frac {3 \left (\frac {\sin (x) \int \csc (x)dx}{2 a \sqrt {a \sin ^2(x)}}-\frac {\cot (x)}{2 a \sqrt {a \sin ^2(x)}}\right )}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\sin (x) \int \csc (x)dx}{2 a \sqrt {a \sin ^2(x)}}-\frac {\cot (x)}{2 a \sqrt {a \sin ^2(x)}}\right )}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {3 \left (-\frac {\sin (x) \text {arctanh}(\cos (x))}{2 a \sqrt {a \sin ^2(x)}}-\frac {\cot (x)}{2 a \sqrt {a \sin ^2(x)}}\right )}{4 a}-\frac {\cot (x)}{4 a \left (a \sin ^2(x)\right )^{3/2}}\)

Input:

Int[(a - a*Cos[x]^2)^(-5/2),x]
 

Output:

-1/4*Cot[x]/(a*(a*Sin[x]^2)^(3/2)) + (3*(-1/2*Cot[x]/(a*Sqrt[a*Sin[x]^2]) 
- (ArcTanh[Cos[x]]*Sin[x])/(2*a*Sqrt[a*Sin[x]^2])))/(4*a)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3655

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]
 

rule 3683

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]
 

rule 3686

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]])
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {6 \sin \left (x \right )^{2} \cos \left (x \right )+4 \cos \left (x \right )+\left (3 \ln \left (1+\cos \left (x \right )\right )-3 \ln \left (-1+\cos \left (x \right )\right )\right ) \sin \left (x \right )^{4}}{16 a^{2} \left (1+\cos \left (x \right )\right ) \left (-1+\cos \left (x \right )\right ) \sin \left (x \right ) \sqrt {a \sin \left (x \right )^{2}}}\)	\(63\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{6 i x}-11 \,{\mathrm e}^{4 i x}-11 \,{\mathrm e}^{2 i x}+3\right )}{4 a^{2} \left ({\mathrm e}^{2 i x}-1\right )^{3} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}-\frac {3 \ln \left ({\mathrm e}^{i x}+1\right ) \sin \left (x \right )}{4 a^{2} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}+\frac {3 \ln \left ({\mathrm e}^{i x}-1\right ) \sin \left (x \right )}{4 a^{2} \sqrt {-a \left ({\mathrm e}^{2 i x}-1\right )^{2} {\mathrm e}^{-2 i x}}}\)	\(127\)

Input:

int(1/(a-a*cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/16/a^2*(6*sin(x)^2*cos(x)+4*cos(x)+(3*ln(1+cos(x))-3*ln(-1+cos(x)))*sin( 
x)^4)/(1+cos(x))/(-1+cos(x))/sin(x)/(a*sin(x)^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=\frac {\sqrt {-a \cos \left (x\right )^{2} + a} {\left (6 \, \cos \left (x\right )^{3} + 3 \, {\left (\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1\right )} \log \left (-\frac {\cos \left (x\right ) - 1}{\cos \left (x\right ) + 1}\right ) - 10 \, \cos \left (x\right )\right )}}{16 \, {\left (a^{3} \cos \left (x\right )^{4} - 2 \, a^{3} \cos \left (x\right )^{2} + a^{3}\right )} \sin \left (x\right )} \] Input:

integrate(1/(a-a*cos(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

1/16*sqrt(-a*cos(x)^2 + a)*(6*cos(x)^3 + 3*(cos(x)^4 - 2*cos(x)^2 + 1)*log 
(-(cos(x) - 1)/(cos(x) + 1)) - 10*cos(x))/((a^3*cos(x)^4 - 2*a^3*cos(x)^2 
+ a^3)*sin(x))
 

Sympy [F]

\[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (- a \cos ^{2}{\left (x \right )} + a\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a-a*cos(x)**2)**(5/2),x)
 

Output:

Integral((-a*cos(x)**2 + a)**(-5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 931 vs. \(2 (49) = 98\).

Time = 0.31 (sec) , antiderivative size = 931, normalized size of antiderivative = 15.26 \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a-a*cos(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/8*(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))*si 
n(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)*arctan2(sin(x), cos(x) + 1) - 3*(2*(4*cos(6*x) - 6*c 
os(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)*arctan 
2(sin(x), cos(x) - 1) + 2*(3*sin(7*x) - 11*sin(5*x) - 11*sin(3*x) + 3*sin( 
x))*cos(8*x) + 12*(2*sin(6*x) - 3*sin(4*x) + 2*sin(2*x))*cos(7*x) + 8*(11* 
sin(5*x) + 11*sin(3*x) - 3*sin(x))*cos(6*x) + 44*(3*sin(4*x) - 2*sin(2*x)) 
*cos(5*x) - 12*(11*sin(3*x) - 3*sin(x))*cos(4*x) - 2*(3*cos(7*x) - 11*cos( 
5*x) - 11*cos(3*x) + 3*cos(x))*sin(8*x) - 6*(4*cos(6*x) - 6*cos(4*x) + 4*c 
os(2*x) - 1)*sin(7*x) - 8*(11*cos(5*x) + 11*cos(3*x) - 3*cos(x))*sin(6*x) 
- 22*(6*cos(4*x) - 4*cos(2*x) + 1)*sin(5*x) + 12*(11*cos(3*x) - 3*cos(x))* 
sin(4*x) + 22*(4*cos(2*x) - 1)*sin(3*x) - 88*cos(3*x)*sin(2*x) + 24*cos(x) 
*sin(2*x) - 24*cos(2*x)*sin(x) + 6*sin(x))*sqrt(-a)/(a^3*cos(8*x)^2 + 1...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (49) = 98\).

Time = 0.16 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.98 \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=\frac {3 \, \log \left (\tan \left (\frac {1}{2} \, x\right )^{2}\right )}{16 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right )} + \frac {a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4} + 8 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{2}}{64 \, a^{5}} - \frac {18 \, \tan \left (\frac {1}{2} \, x\right )^{4} + 8 \, \tan \left (\frac {1}{2} \, x\right )^{2} + 1}{64 \, a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, x\right )^{3} + \tan \left (\frac {1}{2} \, x\right )\right ) \tan \left (\frac {1}{2} \, x\right )^{4}} \] Input:

integrate(1/(a-a*cos(x)^2)^(5/2),x, algorithm="giac")
 

Output:

3/16*log(tan(1/2*x)^2)/(a^(5/2)*sgn(tan(1/2*x)^3 + tan(1/2*x))) + 1/64*(a^ 
(5/2)*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^4 + 8*a^(5/2)*sgn(tan(1/2* 
x)^3 + tan(1/2*x))*tan(1/2*x)^2)/a^5 - 1/64*(18*tan(1/2*x)^4 + 8*tan(1/2*x 
)^2 + 1)/(a^(5/2)*sgn(tan(1/2*x)^3 + tan(1/2*x))*tan(1/2*x)^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (a-a\,{\cos \left (x\right )}^2\right )}^{5/2}} \,d x \] Input:

int(1/(a - a*cos(x)^2)^(5/2),x)
 

Output:

int(1/(a - a*cos(x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a-a \cos ^2(x)\right )^{5/2}} \, dx=-\frac {\sqrt {a}\, \left (\int \frac {\sqrt {-\cos \left (x \right )^{2}+1}}{\cos \left (x \right )^{6}-3 \cos \left (x \right )^{4}+3 \cos \left (x \right )^{2}-1}d x \right )}{a^{3}} \] Input:

int(1/(a-a*cos(x)^2)^(5/2),x)
 

Output:

( - sqrt(a)*int(sqrt( - cos(x)**2 + 1)/(cos(x)**6 - 3*cos(x)**4 + 3*cos(x) 
**2 - 1),x))/a**3