\(\int \frac {1}{(1+\cos ^2(x))^{5/2}} \, dx\) [52]

Optimal result

Integrand size = 10, antiderivative size = 63 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^{5/2}} \, dx=\frac {1}{2} E\left (\left .\frac {\pi }{2}+x\right |-1\right )-\frac {1}{6} \operatorname {EllipticF}\left (\frac {\pi }{2}+x,-1\right )-\frac {\cos (x) \sin (x)}{6 \left (1+\cos ^2(x)\right )^{3/2}}-\frac {\cos (x) \sin (x)}{2 \sqrt {1+\cos ^2(x)}} \] Output:

1/2*EllipticE(cos(x),I)-1/6*InverseJacobiAM(1/2*Pi+x,I)-1/6*cos(x)*sin(x)/ 
(1+cos(x)^2)^(3/2)-1/2*cos(x)*sin(x)/(1+cos(x)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^{5/2}} \, dx=\frac {12 E\left (x\left |\frac {1}{2}\right .\right )-2 \operatorname {EllipticF}\left (x,\frac {1}{2}\right )-\frac {22 \sin (2 x)+3 \sin (4 x)}{(3+\cos (2 x))^{3/2}}}{12 \sqrt {2}} \] Input:

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

Output:

(12*EllipticE[x, 1/2] - 2*EllipticF[x, 1/2] - (22*Sin[2*x] + 3*Sin[4*x])/( 
3 + Cos[2*x])^(3/2))/(12*Sqrt[2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.100, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3663

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3652

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3651

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3656

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

\(\Big \downarrow \) 3661

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

Input:

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

Output:

-1/6*(Cos[x]*Sin[x])/(1 + Cos[x]^2)^(3/2) + (3*EllipticE[Pi/2 + x, -1] - E 
llipticF[Pi/2 + x, -1] - (3*Cos[x]*Sin[x])/Sqrt[1 + Cos[x]^2])/6
 

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)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (46 ) = 92\).

Time = 0.21 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.35

method	result	size
default	\(-\frac {\sqrt {-\left (-1-\cos \left (x \right )^{2}\right ) \sin \left (x \right )^{2}}\, \left (\frac {\cos \left (x \right ) \sqrt {1-\cos \left (x \right )^{4}}}{6 \left (1+\cos \left (x \right )^{2}\right )^{2}}+\frac {\sin \left (x \right )^{2} \cos \left (x \right )}{2 \sqrt {-\left (-1-\cos \left (x \right )^{2}\right ) \sin \left (x \right )^{2}}}+\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {1+\cos \left (x \right )^{2}}\, \operatorname {EllipticF}\left (\cos \left (x \right ), i\right )}{3 \sqrt {1-\cos \left (x \right )^{4}}}-\frac {\sqrt {\frac {1}{2}-\frac {\cos \left (2 x \right )}{2}}\, \sqrt {1+\cos \left (x \right )^{2}}\, \left (\operatorname {EllipticF}\left (\cos \left (x \right ), i\right )-\operatorname {EllipticE}\left (\cos \left (x \right ), i\right )\right )}{2 \sqrt {1-\cos \left (x \right )^{4}}}\right )}{\sin \left (x \right ) \sqrt {1+\cos \left (x \right )^{2}}}\)	\(148\)

Input:

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

Output:

-(-(-1-cos(x)^2)*sin(x)^2)^(1/2)*(1/6*cos(x)*(1-cos(x)^4)^(1/2)/(1+cos(x)^ 
2)^2+1/2*sin(x)^2*cos(x)/(-(-1-cos(x)^2)*sin(x)^2)^(1/2)+1/3*(sin(x)^2)^(1 
/2)*(1+cos(x)^2)^(1/2)/(1-cos(x)^4)^(1/2)*EllipticF(cos(x),I)-1/2*(sin(x)^ 
2)^(1/2)*(1+cos(x)^2)^(1/2)/(1-cos(x)^4)^(1/2)*(EllipticF(cos(x),I)-Ellipt 
icE(cos(x),I)))/sin(x)/(1+cos(x)^2)^(1/2)
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (44) = 88\).

Time = 0.15 (sec) , antiderivative size = 316, normalized size of antiderivative = 5.02 \[ \int \frac {1}{\left (1+\cos ^2(x)\right )^{5/2}} \, dx=-\frac {3 \, {\left ({\left (-2 i \, \sqrt {2} + 3 i\right )} \cos \left (x\right )^{4} + 2 \, {\left (-2 i \, \sqrt {2} + 3 i\right )} \cos \left (x\right )^{2} - 2 i \, \sqrt {2} + 3 i\right )} \sqrt {2 \, \sqrt {2} - 3} E(\arcsin \left (\sqrt {2 \, \sqrt {2} - 3} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,12 \, \sqrt {2} + 17) + 3 \, {\left ({\left (2 i \, \sqrt {2} - 3 i\right )} \cos \left (x\right )^{4} + 2 \, {\left (2 i \, \sqrt {2} - 3 i\right )} \cos \left (x\right )^{2} + 2 i \, \sqrt {2} - 3 i\right )} \sqrt {2 \, \sqrt {2} - 3} E(\arcsin \left (\sqrt {2 \, \sqrt {2} - 3} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,12 \, \sqrt {2} + 17) + 2 \, {\left ({\left (-4 i \, \sqrt {2} - 15 i\right )} \cos \left (x\right )^{4} + 2 \, {\left (-4 i \, \sqrt {2} - 15 i\right )} \cos \left (x\right )^{2} - 4 i \, \sqrt {2} - 15 i\right )} \sqrt {2 \, \sqrt {2} - 3} F(\arcsin \left (\sqrt {2 \, \sqrt {2} - 3} {\left (\cos \left (x\right ) + i \, \sin \left (x\right )\right )}\right )\,|\,12 \, \sqrt {2} + 17) + 2 \, {\left ({\left (4 i \, \sqrt {2} + 15 i\right )} \cos \left (x\right )^{4} + 2 \, {\left (4 i \, \sqrt {2} + 15 i\right )} \cos \left (x\right )^{2} + 4 i \, \sqrt {2} + 15 i\right )} \sqrt {2 \, \sqrt {2} - 3} F(\arcsin \left (\sqrt {2 \, \sqrt {2} - 3} {\left (\cos \left (x\right ) - i \, \sin \left (x\right )\right )}\right )\,|\,12 \, \sqrt {2} + 17) + 2 \, {\left (3 \, \cos \left (x\right )^{3} + 4 \, \cos \left (x\right )\right )} \sqrt {\cos \left (x\right )^{2} + 1} \sin \left (x\right )}{12 \, {\left (\cos \left (x\right )^{4} + 2 \, \cos \left (x\right )^{2} + 1\right )}} \] Input:

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

Output:

-1/12*(3*((-2*I*sqrt(2) + 3*I)*cos(x)^4 + 2*(-2*I*sqrt(2) + 3*I)*cos(x)^2 
- 2*I*sqrt(2) + 3*I)*sqrt(2*sqrt(2) - 3)*elliptic_e(arcsin(sqrt(2*sqrt(2) 
- 3)*(cos(x) + I*sin(x))), 12*sqrt(2) + 17) + 3*((2*I*sqrt(2) - 3*I)*cos(x 
)^4 + 2*(2*I*sqrt(2) - 3*I)*cos(x)^2 + 2*I*sqrt(2) - 3*I)*sqrt(2*sqrt(2) - 
 3)*elliptic_e(arcsin(sqrt(2*sqrt(2) - 3)*(cos(x) - I*sin(x))), 12*sqrt(2) 
 + 17) + 2*((-4*I*sqrt(2) - 15*I)*cos(x)^4 + 2*(-4*I*sqrt(2) - 15*I)*cos(x 
)^2 - 4*I*sqrt(2) - 15*I)*sqrt(2*sqrt(2) - 3)*elliptic_f(arcsin(sqrt(2*sqr 
t(2) - 3)*(cos(x) + I*sin(x))), 12*sqrt(2) + 17) + 2*((4*I*sqrt(2) + 15*I) 
*cos(x)^4 + 2*(4*I*sqrt(2) + 15*I)*cos(x)^2 + 4*I*sqrt(2) + 15*I)*sqrt(2*s 
qrt(2) - 3)*elliptic_f(arcsin(sqrt(2*sqrt(2) - 3)*(cos(x) - I*sin(x))), 12 
*sqrt(2) + 17) + 2*(3*cos(x)^3 + 4*cos(x))*sqrt(cos(x)^2 + 1)*sin(x))/(cos 
(x)^4 + 2*cos(x)^2 + 1)
 

Sympy [F]

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

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

Output:

Integral((cos(x)**2 + 1)**(-5/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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