\(\int (1+\cos ^2(x))^{5/2} \, dx\) [47]

Optimal result

Integrand size = 10, antiderivative size = 63 \[ \int \left (1+\cos ^2(x)\right )^{5/2} \, dx=\frac {18}{5} E\left (\left .\frac {\pi }{2}+x\right |-1\right )-\frac {8}{5} \operatorname {EllipticF}\left (\frac {\pi }{2}+x,-1\right )+\frac {4}{5} \cos (x) \sqrt {1+\cos ^2(x)} \sin (x)+\frac {1}{5} \cos (x) \left (1+\cos ^2(x)\right )^{3/2} \sin (x) \] Output:

18/5*EllipticE(cos(x),I)-8/5*InverseJacobiAM(1/2*Pi+x,I)+4/5*cos(x)*(1+cos 
(x)^2)^(1/2)*sin(x)+1/5*cos(x)*(1+cos(x)^2)^(3/2)*sin(x)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int \left (1+\cos ^2(x)\right )^{5/2} \, dx=\frac {288 E\left (x\left |\frac {1}{2}\right .\right )-64 \operatorname {EllipticF}\left (x,\frac {1}{2}\right )+\sqrt {3+\cos (2 x)} (22 \sin (2 x)+\sin (4 x))}{40 \sqrt {2}} \] Input:

Integrate[(1 + Cos[x]^2)^(5/2),x]
 

Output:

(288*EllipticE[x, 1/2] - 64*EllipticF[x, 1/2] + Sqrt[3 + Cos[2*x]]*(22*Sin 
[2*x] + Sin[4*x]))/(40*Sqrt[2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3659

\(\displaystyle \frac {1}{5} \int 6 \sqrt {\cos ^2(x)+1} \left (2 \cos ^2(x)+1\right )dx+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {6}{5} \int \sqrt {\cos ^2(x)+1} \left (2 \cos ^2(x)+1\right )dx+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{5} \int \sqrt {\sin \left (x+\frac {\pi }{2}\right )^2+1} \left (2 \sin \left (x+\frac {\pi }{2}\right )^2+1\right )dx+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3649

\(\displaystyle \frac {6}{5} \left (\frac {1}{3} \int \frac {9 \cos ^2(x)+5}{\sqrt {\cos ^2(x)+1}}dx+\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}\right )+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{5} \left (\frac {1}{3} \int \frac {9 \sin \left (x+\frac {\pi }{2}\right )^2+5}{\sqrt {\sin \left (x+\frac {\pi }{2}\right )^2+1}}dx+\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}\right )+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3651

\(\displaystyle \frac {6}{5} \left (\frac {1}{3} \left (9 \int \sqrt {\cos ^2(x)+1}dx-4 \int \frac {1}{\sqrt {\cos ^2(x)+1}}dx\right )+\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}\right )+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {6}{5} \left (\frac {1}{3} \left (9 \int \sqrt {\sin \left (x+\frac {\pi }{2}\right )^2+1}dx-4 \int \frac {1}{\sqrt {\sin \left (x+\frac {\pi }{2}\right )^2+1}}dx\right )+\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}\right )+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3656

\(\displaystyle \frac {6}{5} \left (\frac {1}{3} \left (9 E\left (\left .x+\frac {\pi }{2}\right |-1\right )-4 \int \frac {1}{\sqrt {\sin \left (x+\frac {\pi }{2}\right )^2+1}}dx\right )+\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}\right )+\frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}\)

\(\Big \downarrow \) 3661

\(\displaystyle \frac {1}{5} \sin (x) \cos (x) \left (\cos ^2(x)+1\right )^{3/2}+\frac {6}{5} \left (\frac {2}{3} \sin (x) \cos (x) \sqrt {\cos ^2(x)+1}+\frac {1}{3} \left (9 E\left (\left .x+\frac {\pi }{2}\right |-1\right )-4 \operatorname {EllipticF}\left (x+\frac {\pi }{2},-1\right )\right )\right )\)

Input:

Int[(1 + Cos[x]^2)^(5/2),x]
 

Output:

(Cos[x]*(1 + Cos[x]^2)^(3/2)*Sin[x])/5 + (6*((9*EllipticE[Pi/2 + x, -1] - 
4*EllipticF[Pi/2 + x, -1])/3 + (2*Cos[x]*Sqrt[1 + Cos[x]^2]*Sin[x])/3))/5
 

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)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (46 ) = 92\).

Time = 1.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {\sqrt {\left (1+\cos \left (x \right )^{2}\right ) \sin \left (x \right )^{2}}\, \left (\sin \left (x \right )^{6} \cos \left (x \right )-8 \sin \left (x \right )^{4} \cos \left (x \right )+8 \sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {-\sin \left (x \right )^{2}+2}\, \operatorname {EllipticF}\left (\cos \left (x \right ), i\right )-18 \sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {-\sin \left (x \right )^{2}+2}\, \operatorname {EllipticE}\left (\cos \left (x \right ), i\right )+12 \sin \left (x \right )^{2} \cos \left (x \right )\right )}{5 \sqrt {1-\cos \left (x \right )^{4}}\, \sin \left (x \right ) \sqrt {1+\cos \left (x \right )^{2}}}\)	\(108\)

Input:

int((1+cos(x)^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/5*((1+cos(x)^2)*sin(x)^2)^(1/2)*(sin(x)^6*cos(x)-8*sin(x)^4*cos(x)+8*(si 
n(x)^2)^(1/2)*(-sin(x)^2+2)^(1/2)*EllipticF(cos(x),I)-18*(sin(x)^2)^(1/2)* 
(-sin(x)^2+2)^(1/2)*EllipticE(cos(x),I)+12*sin(x)^2*cos(x))/(1-cos(x)^4)^( 
1/2)/sin(x)/(1+cos(x)^2)^(1/2)
 

Fricas [F]

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

integrate((1+cos(x)^2)^(5/2),x, algorithm="fricas")
 

Output:

integral((cos(x)^4 + 2*cos(x)^2 + 1)*sqrt(cos(x)^2 + 1), x)
 

Sympy [F(-1)]

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

integrate((1+cos(x)**2)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((1+cos(x)^2)^(5/2),x, algorithm="maxima")
 

Output:

integrate((cos(x)^2 + 1)^(5/2), x)
 

Giac [F]

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

integrate((1+cos(x)^2)^(5/2),x, algorithm="giac")
 

Output:

integrate((cos(x)^2 + 1)^(5/2), x)
 

Mupad [F(-1)]

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

int((cos(x)^2 + 1)^(5/2),x)
 

Output:

int((cos(x)^2 + 1)^(5/2), x)
 

Reduce [F]

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

int((1+cos(x)^2)^(5/2),x)
 

Output:

int(sqrt(cos(x)**2 + 1),x) + int(sqrt(cos(x)**2 + 1)*cos(x)**4,x) + 2*int( 
sqrt(cos(x)**2 + 1)*cos(x)**2,x)