\(\int (4-5 \sin ^2(x))^{5/2} \, dx\) [125]

Optimal result

Integrand size = 12, antiderivative size = 54 \[ \int \left (4-5 \sin ^2(x)\right )^{5/2} \, dx=\frac {72 E\left (x\left |\frac {5}{4}\right .\right )}{5}+\frac {8 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{5}+4 \cos (x) \sin (x) \sqrt {4-5 \sin ^2(x)}+\cos (x) \sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \] Output:

72/5*EllipticE(sin(x),1/2*5^(1/2))+8/5*InverseJacobiAM(x,1/2*5^(1/2))+4*co 
s(x)*sin(x)*(4-5*sin(x)^2)^(1/2)+cos(x)*sin(x)*(4-5*sin(x)^2)^(3/2)
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \left (4-5 \sin ^2(x)\right )^{5/2} \, dx=\frac {72 E\left (x\left |\frac {5}{4}\right .\right )}{5}+\frac {8 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{5}+\frac {\sqrt {3+5 \cos (2 x)} (22 \sin (2 x)+5 \sin (4 x))}{8 \sqrt {2}} \] Input:

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

Output:

(72*EllipticE[x, 5/4])/5 + (8*EllipticF[x, 5/4])/5 + (Sqrt[3 + 5*Cos[2*x]] 
*(22*Sin[2*x] + 5*Sin[4*x]))/(8*Sqrt[2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {3042, 3659, 27, 3042, 3649, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3659

\(\displaystyle \frac {1}{5} \int 60 \sqrt {4-5 \sin ^2(x)} \left (1-\sin ^2(x)\right )dx+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3649

\(\displaystyle 12 \left (\frac {1}{3} \int \frac {8-9 \sin ^2(x)}{\sqrt {4-5 \sin ^2(x)}}dx+\frac {1}{3} \sqrt {4-5 \sin ^2(x)} \sin (x) \cos (x)\right )+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 3042

\(\displaystyle 12 \left (\frac {1}{3} \int \frac {8-9 \sin (x)^2}{\sqrt {4-5 \sin (x)^2}}dx+\frac {1}{3} \sqrt {4-5 \sin ^2(x)} \sin (x) \cos (x)\right )+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 3651

\(\displaystyle 12 \left (\frac {1}{3} \left (\frac {4}{5} \int \frac {1}{\sqrt {4-5 \sin ^2(x)}}dx+\frac {9}{5} \int \sqrt {4-5 \sin ^2(x)}dx\right )+\frac {1}{3} \sqrt {4-5 \sin ^2(x)} \sin (x) \cos (x)\right )+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 3042

\(\displaystyle 12 \left (\frac {1}{3} \left (\frac {4}{5} \int \frac {1}{\sqrt {4-5 \sin (x)^2}}dx+\frac {9}{5} \int \sqrt {4-5 \sin (x)^2}dx\right )+\frac {1}{3} \sqrt {4-5 \sin ^2(x)} \sin (x) \cos (x)\right )+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 3656

\(\displaystyle 12 \left (\frac {1}{3} \left (\frac {4}{5} \int \frac {1}{\sqrt {4-5 \sin (x)^2}}dx+\frac {18 E\left (x\left |\frac {5}{4}\right .\right )}{5}\right )+\frac {1}{3} \sqrt {4-5 \sin ^2(x)} \sin (x) \cos (x)\right )+\sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)\)

\(\Big \downarrow \) 3661

\(\displaystyle \sin (x) \left (4-5 \sin ^2(x)\right )^{3/2} \cos (x)+12 \left (\frac {1}{3} \sin (x) \sqrt {4-5 \sin ^2(x)} \cos (x)+\frac {1}{3} \left (\frac {2 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{5}+\frac {18 E\left (x\left |\frac {5}{4}\right .\right )}{5}\right )\right )\)

Input:

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

Output:

Cos[x]*Sin[x]*(4 - 5*Sin[x]^2)^(3/2) + 12*(((18*EllipticE[x, 5/4])/5 + (2* 
EllipticF[x, 5/4])/5)/3 + (Cos[x]*Sin[x]*Sqrt[4 - 5*Sin[x]^2])/3)
 

Defintions of rubi rules used

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 3649

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]^2), x_Symbol] :> Simp[(-B)*Cos[e + f*x]*Sin[e + f*x]*((a + b* 
Sin[e + f*x]^2)^p/(2*f*(p + 1))), x] + Simp[1/(2*(p + 1))   Int[(a + b*Sin[ 
e + f*x]^2)^(p - 1)*Simp[a*B + 2*a*A*(p + 1) + (2*A*b*(p + 1) + B*(b + 2*a* 
p + 2*b*p))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && G 
tQ[p, 0]
 

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 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 3659

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*f*p)), x] + Sim 
p[1/(2*p)   Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(2*a 
+ b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[ 
a + b, 0] && GtQ[p, 1]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(124\) vs. \(2(51)=102\).

Time = 4.01 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.31

method	result	size
default	\(\frac {\sqrt {-\left (-4+5 \sin \left (x \right )^{2}\right ) \cos \left (x \right )^{2}}\, \left (125 \cos \left (x \right )^{6} \sin \left (x \right )+50 \cos \left (x \right )^{4} \sin \left (x \right )+8 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )+72 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )-15 \cos \left (x \right )^{2} \sin \left (x \right )\right )}{5 \sqrt {5 \cos \left (x \right )^{4}-\cos \left (x \right )^{2}}\, \cos \left (x \right ) \sqrt {4-5 \sin \left (x \right )^{2}}}\)	\(125\)

Input:

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

Output:

1/5*(-(-4+5*sin(x)^2)*cos(x)^2)^(1/2)*(125*cos(x)^6*sin(x)+50*cos(x)^4*sin 
(x)+8*(cos(x)^2)^(1/2)*(5*cos(x)^2-1)^(1/2)*EllipticF(sin(x),1/2*5^(1/2))+ 
72*(cos(x)^2)^(1/2)*(5*cos(x)^2-1)^(1/2)*EllipticE(sin(x),1/2*5^(1/2))-15* 
cos(x)^2*sin(x))/(5*cos(x)^4-cos(x)^2)^(1/2)/cos(x)/(4-5*sin(x)^2)^(1/2)
 

Fricas [F]

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

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

Output:

integral((25*cos(x)^4 - 10*cos(x)^2 + 1)*sqrt(5*cos(x)^2 - 1), x)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

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

Output:

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

Reduce [F]

\[ \int \left (4-5 \sin ^2(x)\right )^{5/2} \, dx=16 \left (\int \sqrt {-5 \sin \left (x \right )^{2}+4}d x \right )+25 \left (\int \sqrt {-5 \sin \left (x \right )^{2}+4}\, \sin \left (x \right )^{4}d x \right )-40 \left (\int \sqrt {-5 \sin \left (x \right )^{2}+4}\, \sin \left (x \right )^{2}d x \right ) \] Input:

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

Output:

16*int(sqrt( - 5*sin(x)**2 + 4),x) + 25*int(sqrt( - 5*sin(x)**2 + 4)*sin(x 
)**4,x) - 40*int(sqrt( - 5*sin(x)**2 + 4)*sin(x)**2,x)