\(\int \frac {1}{(4-5 \sin ^2(x))^{5/2}} \, dx\) [130]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [C] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 12, antiderivative size = 59 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=\frac {E\left (x\left |\frac {5}{4}\right .\right )}{4}+\frac {\operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{24}+\frac {5 \cos (x) \sin (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}-\frac {5 \cos (x) \sin (x)}{8 \sqrt {4-5 \sin ^2(x)}} \] Output:

1/4*EllipticE(sin(x),1/2*5^(1/2))+1/24*InverseJacobiAM(x,1/2*5^(1/2))+5/12 
*cos(x)*sin(x)/(4-5*sin(x)^2)^(3/2)-5/8*cos(x)*sin(x)/(4-5*sin(x)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=\frac {1}{96} \left (24 E\left (x\left |\frac {5}{4}\right .\right )+4 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )-\frac {25 \sqrt {2} (2 \sin (2 x)+3 \sin (4 x))}{(3+5 \cos (2 x))^{3/2}}\right ) \] Input:

Integrate[(4 - 5*Sin[x]^2)^(-5/2),x]
 

Output:

(24*EllipticE[x, 5/4] + 4*EllipticF[x, 5/4] - (25*Sqrt[2]*(2*Sin[2*x] + 3* 
Sin[4*x]))/(3 + 5*Cos[2*x])^(3/2))/96
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3663, 25, 3042, 3652, 27, 3042, 3651, 3042, 3656, 3661}

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.

\(\displaystyle \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (4-5 \sin (x)^2\right )^{5/2}}dx\)

\(\Big \downarrow \) 3663

\(\displaystyle \frac {1}{12} \int -\frac {5 \sin ^2(x)+2}{\left (4-5 \sin ^2(x)\right )^{3/2}}dx+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}-\frac {1}{12} \int \frac {5 \sin ^2(x)+2}{\left (4-5 \sin ^2(x)\right )^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}-\frac {1}{12} \int \frac {5 \sin (x)^2+2}{\left (4-5 \sin (x)^2\right )^{3/2}}dx\)

\(\Big \downarrow \) 3652

\(\displaystyle \frac {1}{12} \left (\frac {1}{4} \int \frac {2 \left (14-15 \sin ^2(x)\right )}{\sqrt {4-5 \sin ^2(x)}}dx-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{12} \left (\frac {1}{2} \int \frac {14-15 \sin ^2(x)}{\sqrt {4-5 \sin ^2(x)}}dx-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{12} \left (\frac {1}{2} \int \frac {14-15 \sin (x)^2}{\sqrt {4-5 \sin (x)^2}}dx-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3651

\(\displaystyle \frac {1}{12} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {4-5 \sin ^2(x)}}dx+3 \int \sqrt {4-5 \sin ^2(x)}dx\right )-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{12} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {4-5 \sin (x)^2}}dx+3 \int \sqrt {4-5 \sin (x)^2}dx\right )-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3656

\(\displaystyle \frac {1}{12} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {4-5 \sin (x)^2}}dx+6 E\left (x\left |\frac {5}{4}\right .\right )\right )-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )+\frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}\)

\(\Big \downarrow \) 3661

\(\displaystyle \frac {5 \sin (x) \cos (x)}{12 \left (4-5 \sin ^2(x)\right )^{3/2}}+\frac {1}{12} \left (\frac {1}{2} \left (\operatorname {EllipticF}\left (x,\frac {5}{4}\right )+6 E\left (x\left |\frac {5}{4}\right .\right )\right )-\frac {15 \sin (x) \cos (x)}{2 \sqrt {4-5 \sin ^2(x)}}\right )\)

Input:

Int[(4 - 5*Sin[x]^2)^(-5/2),x]
 

Output:

(5*Cos[x]*Sin[x])/(12*(4 - 5*Sin[x]^2)^(3/2)) + ((6*EllipticE[x, 5/4] + El 
lipticF[x, 5/4])/2 - (15*Cos[x]*Sin[x])/(2*Sqrt[4 - 5*Sin[x]^2]))/12
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3651
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + 
 (f_.)*(x_)]^2], x_Symbol] :> Simp[B/b   Int[Sqrt[a + b*Sin[e + f*x]^2], x] 
, x] + Simp[(A*b - a*B)/b   Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /; Fre 
eQ[{a, b, e, f, A, B}, x]
 

rule 3652
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]
 

rule 3656
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(Sqrt[a 
]/f)*EllipticE[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]
 

rule 3661
Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(S 
qrt[a]*f))*EllipticF[e + f*x, -b/a], x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 
 0]
 

rule 3663
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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(192\) vs. \(2(52)=104\).

Time = 0.75 (sec) , antiderivative size = 193, normalized size of antiderivative = 3.27

method result size
default \(\frac {\sqrt {-\left (-4+5 \sin \left (x \right )^{2}\right ) \cos \left (x \right )^{2}}\, \left (5 \sqrt {5 \cos \left (x \right )^{2}-1}\, \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right ) \cos \left (x \right )^{2}+30 \sqrt {5 \cos \left (x \right )^{2}-1}\, \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right ) \cos \left (x \right )^{2}-75 \cos \left (x \right )^{4} \sin \left (x \right )-\sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )-6 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )+25 \cos \left (x \right )^{2} \sin \left (x \right )\right ) \sqrt {5 \cos \left (x \right )^{4}-\cos \left (x \right )^{2}}}{24 \left (25 \cos \left (x \right )^{4}-10 \cos \left (x \right )^{2}+1\right ) \cos \left (x \right )^{3} \sqrt {4-5 \sin \left (x \right )^{2}}}\) \(193\)

Input:

int(1/(4-5*sin(x)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/24*(-(-4+5*sin(x)^2)*cos(x)^2)^(1/2)/(25*cos(x)^4-10*cos(x)^2+1)/cos(x)^ 
3*(5*(5*cos(x)^2-1)^(1/2)*(cos(x)^2)^(1/2)*EllipticF(sin(x),1/2*5^(1/2))*c 
os(x)^2+30*(5*cos(x)^2-1)^(1/2)*(cos(x)^2)^(1/2)*EllipticE(sin(x),1/2*5^(1 
/2))*cos(x)^2-75*cos(x)^4*sin(x)-(cos(x)^2)^(1/2)*(5*cos(x)^2-1)^(1/2)*Ell 
ipticF(sin(x),1/2*5^(1/2))-6*(cos(x)^2)^(1/2)*(5*cos(x)^2-1)^(1/2)*Ellipti 
cE(sin(x),1/2*5^(1/2))+25*cos(x)^2*sin(x))*(5*cos(x)^4-cos(x)^2)^(1/2)/(4- 
5*sin(x)^2)^(1/2)
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.66 \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=-\frac {15 \, \sqrt {\frac {4}{5} i - \frac {3}{5}} {\left (\left (75 i + 100\right ) \, \sqrt {5} \cos \left (x\right )^{4} - \left (30 i + 40\right ) \, \sqrt {5} \cos \left (x\right )^{2} + \left (3 i + 4\right ) \, \sqrt {5}\right )} E(\arcsin \left (\sqrt {\frac {4}{5} i - \frac {3}{5}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {24}{25} i - \frac {7}{25}) + 15 \, \sqrt {-\frac {4}{5} i - \frac {3}{5}} {\left (-\left (75 i - 100\right ) \, \sqrt {5} \cos \left (x\right )^{4} + \left (30 i - 40\right ) \, \sqrt {5} \cos \left (x\right )^{2} - \left (3 i - 4\right ) \, \sqrt {5}\right )} E(\arcsin \left (\sqrt {-\frac {4}{5} i - \frac {3}{5}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,-\frac {24}{25} i - \frac {7}{25}) + 4 \, \sqrt {\frac {4}{5} i - \frac {3}{5}} {\left (-\left (525 i + 50\right ) \, \sqrt {5} \cos \left (x\right )^{4} + \left (210 i + 20\right ) \, \sqrt {5} \cos \left (x\right )^{2} - \left (21 i + 2\right ) \, \sqrt {5}\right )} F(\arcsin \left (\sqrt {\frac {4}{5} i - \frac {3}{5}} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,\frac {24}{25} i - \frac {7}{25}) + 4 \, \sqrt {-\frac {4}{5} i - \frac {3}{5}} {\left (\left (525 i - 50\right ) \, \sqrt {5} \cos \left (x\right )^{4} - \left (210 i - 20\right ) \, \sqrt {5} \cos \left (x\right )^{2} + \left (21 i - 2\right ) \, \sqrt {5}\right )} F(\arcsin \left (\sqrt {-\frac {4}{5} i - \frac {3}{5}} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,-\frac {24}{25} i - \frac {7}{25}) + 1250 \, {\left (3 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sqrt {5 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right )}{1200 \, {\left (25 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:

integrate(1/(4-5*sin(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

-1/1200*(15*sqrt(4/5*I - 3/5)*((75*I + 100)*sqrt(5)*cos(x)^4 - (30*I + 40) 
*sqrt(5)*cos(x)^2 + (3*I + 4)*sqrt(5))*elliptic_e(arcsin(sqrt(4/5*I - 3/5) 
*(cos(x) + I*sin(x))), 24/25*I - 7/25) + 15*sqrt(-4/5*I - 3/5)*(-(75*I - 1 
00)*sqrt(5)*cos(x)^4 + (30*I - 40)*sqrt(5)*cos(x)^2 - (3*I - 4)*sqrt(5))*e 
lliptic_e(arcsin(sqrt(-4/5*I - 3/5)*(cos(x) - I*sin(x))), -24/25*I - 7/25) 
 + 4*sqrt(4/5*I - 3/5)*(-(525*I + 50)*sqrt(5)*cos(x)^4 + (210*I + 20)*sqrt 
(5)*cos(x)^2 - (21*I + 2)*sqrt(5))*elliptic_f(arcsin(sqrt(4/5*I - 3/5)*(co 
s(x) + I*sin(x))), 24/25*I - 7/25) + 4*sqrt(-4/5*I - 3/5)*((525*I - 50)*sq 
rt(5)*cos(x)^4 - (210*I - 20)*sqrt(5)*cos(x)^2 + (21*I - 2)*sqrt(5))*ellip 
tic_f(arcsin(sqrt(-4/5*I - 3/5)*(cos(x) - I*sin(x))), -24/25*I - 7/25) + 1 
250*(3*cos(x)^3 - cos(x))*sqrt(5*cos(x)^2 - 1)*sin(x))/(25*cos(x)^4 - 10*c 
os(x)^2 + 1)
 

Sympy [F]

\[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{\left (4 - 5 \sin ^{2}{\left (x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(4-5*sin(x)**2)**(5/2),x)
 

Output:

Integral((4 - 5*sin(x)**2)**(-5/2), x)
 

Maxima [F]

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

integrate(1/(4-5*sin(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((-5*sin(x)^2 + 4)^(-5/2), x)
 

Giac [F]

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

integrate(1/(4-5*sin(x)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((-5*sin(x)^2 + 4)^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=\int \frac {1}{{\left (4-5\,{\sin \left (x\right )}^2\right )}^{5/2}} \,d x \] Input:

int(1/(4 - 5*sin(x)^2)^(5/2),x)
 

Output:

int(1/(4 - 5*sin(x)^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (4-5 \sin ^2(x)\right )^{5/2}} \, dx=-\left (\int \frac {\sqrt {-5 \sin \left (x \right )^{2}+4}}{125 \sin \left (x \right )^{6}-300 \sin \left (x \right )^{4}+240 \sin \left (x \right )^{2}-64}d x \right ) \] Input:

int(1/(4-5*sin(x)^2)^(5/2),x)
 

Output:

 - int(sqrt( - 5*sin(x)**2 + 4)/(125*sin(x)**6 - 300*sin(x)**4 + 240*sin(x 
)**2 - 64),x)