\(\int (4-5 \sin ^2(x))^{3/2} \, dx\) [126]

Optimal result

Integrand size = 12, antiderivative size = 37 \[ \int \left (4-5 \sin ^2(x)\right )^{3/2} \, dx=4 E\left (x\left |\frac {5}{4}\right .\right )+\frac {2 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{3}+\frac {5}{3} \cos (x) \sin (x) \sqrt {4-5 \sin ^2(x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \left (4-5 \sin ^2(x)\right )^{3/2} \, dx=4 E\left (x\left |\frac {5}{4}\right .\right )+\frac {2 \operatorname {EllipticF}\left (x,\frac {5}{4}\right )}{3}+\frac {5}{6} \cos (x) \sqrt {6+10 \cos (2 x)} \sin (x) \] Input:

Integrate[(4 - 5*Sin[x]^2)^(3/2),x]
 

Output:

4*EllipticE[x, 5/4] + (2*EllipticF[x, 5/4])/3 + (5*Cos[x]*Sqrt[6 + 10*Cos[ 
2*x]]*Sin[x])/6
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3659, 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.

\(\displaystyle \int \left (4-5 \sin ^2(x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (4-5 \sin (x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 3659

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3651

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3656

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

\(\Big \downarrow \) 3661

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

Input:

Int[(4 - 5*Sin[x]^2)^(3/2),x]
 

Output:

(2*(6*EllipticE[x, 5/4] + EllipticF[x, 5/4]))/3 + (5*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 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. \(116\) vs. \(2(36)=72\).

Time = 1.70 (sec) , antiderivative size = 117, normalized size of antiderivative = 3.16

method	result	size
default	\(\frac {\sqrt {-\left (-4+5 \sin \left (x \right )^{2}\right ) \cos \left (x \right )^{2}}\, \left (25 \cos \left (x \right )^{4} \sin \left (x \right )+2 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticF}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )+12 \sqrt {\frac {\cos \left (2 x \right )}{2}+\frac {1}{2}}\, \sqrt {5 \cos \left (x \right )^{2}-1}\, \operatorname {EllipticE}\left (\sin \left (x \right ), \frac {\sqrt {5}}{2}\right )-5 \cos \left (x \right )^{2} \sin \left (x \right )\right )}{3 \sqrt {5 \cos \left (x \right )^{4}-\cos \left (x \right )^{2}}\, \cos \left (x \right ) \sqrt {4-5 \sin \left (x \right )^{2}}}\)	\(117\)

Input:

int((4-5*sin(x)^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*(-(-4+5*sin(x)^2)*cos(x)^2)^(1/2)*(25*cos(x)^4*sin(x)+2*(cos(x)^2)^(1/ 
2)*(5*cos(x)^2-1)^(1/2)*EllipticF(sin(x),1/2*5^(1/2))+12*(cos(x)^2)^(1/2)* 
(5*cos(x)^2-1)^(1/2)*EllipticE(sin(x),1/2*5^(1/2))-5*cos(x)^2*sin(x))/(5*c 
os(x)^4-cos(x)^2)^(1/2)/cos(x)/(4-5*sin(x)^2)^(1/2)
 

Fricas [F]

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

integrate((4-5*sin(x)^2)^(3/2),x, algorithm="fricas")
 

Output:

integral((5*cos(x)^2 - 1)^(3/2), x)
 

Sympy [F]

\[ \int \left (4-5 \sin ^2(x)\right )^{3/2} \, dx=\int \left (4 - 5 \sin ^{2}{\left (x \right )}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((4-5*sin(x)**2)**(3/2),x)
 

Output:

Integral((4 - 5*sin(x)**2)**(3/2), x)
 

Maxima [F]

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

integrate((4-5*sin(x)^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate((-5*sin(x)^2 + 4)^(3/2), x)
 

Giac [F]

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

integrate((4-5*sin(x)^2)^(3/2),x, algorithm="giac")
 

Output:

integrate((-5*sin(x)^2 + 4)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (4-5 \sin ^2(x)\right )^{3/2} \, dx=\int {\left (4-5\,{\sin \left (x\right )}^2\right )}^{3/2} \,d x \] Input:

int((4 - 5*sin(x)^2)^(3/2),x)
 

Output:

int((4 - 5*sin(x)^2)^(3/2), x)
 

Reduce [F]

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

int((4-5*sin(x)^2)^(3/2),x)
 

Output:

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