\(\int \frac {(5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} (2+\sqrt {-1+5 \sin ^2(x)})} \, dx\) [453]

Optimal result

Integrand size = 52, antiderivative size = 101 \[ \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx=-\frac {3 \arctan \left (\frac {\sqrt [4]{-1+5 \sin ^2(x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{-1+5 \sin ^2(x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}+2 \sqrt [4]{-1+5 \sin ^2(x)}-\frac {\sqrt [4]{-1+5 \sin ^2(x)}}{2 \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \] Output:

2*(-1+5*sin(x)^2)^(1/4)-3/2*arctan(1/2*(-1+5*sin(x)^2)^(1/4)*2^(1/2))*2^(1 
/2)-1/4*arctanh(1/2*(-1+5*sin(x)^2)^(1/4)*2^(1/2))*2^(1/2)-1/2*(-1+5*sin(x 
)^2)^(1/4)/(2+(-1+5*sin(x)^2)^(1/2))
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88 \[ \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx=\frac {1}{4} \left (-6 \sqrt {2} \arctan \left (\frac {\sqrt [4]{3-5 \cos (2 x)}}{2^{3/4}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt [4]{3-5 \cos (2 x)}}{2^{3/4}}\right )-2 \sqrt [4]{4-5 \cos ^2(x)} \left (-4+\frac {1}{2+\sqrt {4-5 \cos ^2(x)}}\right )\right ) \] Input:

Integrate[((5*Cos[x]^2 - Sqrt[-1 + 5*Sin[x]^2])*Tan[x])/((-1 + 5*Sin[x]^2) 
^(1/4)*(2 + Sqrt[-1 + 5*Sin[x]^2])),x]
 

Output:

(-6*Sqrt[2]*ArcTan[(3 - 5*Cos[2*x])^(1/4)/2^(3/4)] - Sqrt[2]*ArcTanh[(3 - 
5*Cos[2*x])^(1/4)/2^(3/4)] - 2*(4 - 5*Cos[x]^2)^(1/4)*(-4 + (2 + Sqrt[4 - 
5*Cos[x]^2])^(-1)))/4
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.25, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3042, 4861, 7293, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4861

\(\displaystyle -\int \frac {5 \cos ^2(x)-\sqrt {4-5 \cos ^2(x)}}{\sqrt [4]{4-5 \cos ^2(x)} \left (\sqrt {4-5 \cos ^2(x)} \cos (x)+2 \cos (x)\right )}d\cos (x)\)

\(\Big \downarrow \) 7293

\(\displaystyle -\int \left (\frac {5 \cos (x)}{\sqrt [4]{4-5 \cos ^2(x)} \left (\sqrt {4-5 \cos ^2(x)}+2\right )}-\frac {\sqrt [4]{4-5 \cos ^2(x)} \sec (x)}{\sqrt {4-5 \cos ^2(x)}+2}\right )d\cos (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \sqrt {2} \arctan \left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )+\frac {\arctan \left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{4-5 \cos ^2(x)}}{\sqrt {2}}\right )}{2 \sqrt {2}}+2 \sqrt [4]{4-5 \cos ^2(x)}-\frac {\sqrt [4]{4-5 \cos ^2(x)}}{2 \left (\sqrt {4-5 \cos ^2(x)}+2\right )}\)

Input:

Int[((5*Cos[x]^2 - Sqrt[-1 + 5*Sin[x]^2])*Tan[x])/((-1 + 5*Sin[x]^2)^(1/4) 
*(2 + Sqrt[-1 + 5*Sin[x]^2])),x]
 

Output:

ArcTan[(4 - 5*Cos[x]^2)^(1/4)/Sqrt[2]]/Sqrt[2] - 2*Sqrt[2]*ArcTan[(4 - 5*C 
os[x]^2)^(1/4)/Sqrt[2]] - ArcTanh[(4 - 5*Cos[x]^2)^(1/4)/Sqrt[2]]/(2*Sqrt[ 
2]) + 2*(4 - 5*Cos[x]^2)^(1/4) - (4 - 5*Cos[x]^2)^(1/4)/(2*(2 + Sqrt[4 - 5 
*Cos[x]^2]))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4861

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto 
rs[Cos[c*(a + b*x)], x]}, Simp[-(b*c)^(-1)   Subst[Int[SubstFor[1/x, Cos[c* 
(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a 
 + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\left (5 \cos \left (x \right )^{2}-\sqrt {-1+5 \sin \left (x \right )^{2}}\right ) \tan \left (x \right )}{\left (-1+5 \sin \left (x \right )^{2}\right )^{\frac {1}{4}} \left (2+\sqrt {-1+5 \sin \left (x \right )^{2}}\right )}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 461 vs. \(2 (81) = 162\).

Time = 35.60 (sec) , antiderivative size = 461, normalized size of antiderivative = 4.56 \[ \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

1/160*(70*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2)*arctan(-2/5*((5*sqrt(2 
)*cos(x)^2 - 4*sqrt(2))*(-5*cos(x)^2 + 4)^(3/4) - 2*sqrt(2)*(-5*cos(x)^2 + 
 4)^(5/4))/(5*cos(x)^4 - 4*cos(x)^2)) - 50*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2) 
*cos(x)^2)*arctan(2/5*(sqrt(2)*(-5*cos(x)^2 + 4)^(3/4) + 2*sqrt(2)*(-5*cos 
(x)^2 + 4)^(1/4))/cos(x)^2) + 35*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2) 
*log(-(125*cos(x)^6 - 1700*cos(x)^4 - 8*(15*sqrt(2)*cos(x)^2 - 16*sqrt(2)) 
*(-5*cos(x)^2 + 4)^(5/4) + 2560*cos(x)^2 + 4*(25*sqrt(2)*cos(x)^4 - 100*sq 
rt(2)*cos(x)^2 + 64*sqrt(2))*(-5*cos(x)^2 + 4)^(3/4) - 16*(25*cos(x)^4 - 6 
0*cos(x)^2 + 32)*sqrt(-5*cos(x)^2 + 4) - 1024)/(5*cos(x)^6 - 4*cos(x)^4)) 
+ 25*(5*sqrt(2)*cos(x)^4 - 4*sqrt(2)*cos(x)^2)*log(-(25*cos(x)^4 - 320*cos 
(x)^2 - 4*(5*sqrt(2)*cos(x)^2 - 16*sqrt(2))*(-5*cos(x)^2 + 4)^(3/4) - 16*( 
5*cos(x)^2 - 8)*sqrt(-5*cos(x)^2 + 4) - 8*(15*sqrt(2)*cos(x)^2 - 16*sqrt(2 
))*(-5*cos(x)^2 + 4)^(1/4) + 256)/cos(x)^4) + 16*(5*cos(x)^2 - 2*(10*cos(x 
)^2 - 1)*sqrt(-5*cos(x)^2 + 4) - 4)*(-5*cos(x)^2 + 4)^(3/4))/(5*cos(x)^4 - 
 4*cos(x)^2)
 

Sympy [F]

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

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

Output:

Integral((-sqrt(5*sin(x)**2 - 1) + 5*cos(x)**2)*tan(x)/((sqrt(5*sin(x)**2 
- 1) + 2)*(5*sin(x)**2 - 1)**(1/4)), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.99 \[ \int \frac {\left (5 \cos ^2(x)-\sqrt {-1+5 \sin ^2(x)}\right ) \tan (x)}{\sqrt [4]{-1+5 \sin ^2(x)} \left (2+\sqrt {-1+5 \sin ^2(x)}\right )} \, dx=-\frac {3}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}{\sqrt {2} + {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}\right ) + 2 \, {\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}} - \frac {{\left (5 \, \sin \left (x\right )^{2} - 1\right )}^{\frac {1}{4}}}{2 \, {\left (\sqrt {5 \, \sin \left (x\right )^{2} - 1} + 2\right )}} \] Input:

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

Output:

-3/2*sqrt(2)*arctan(1/2*sqrt(2)*(5*sin(x)^2 - 1)^(1/4)) + 1/8*sqrt(2)*log( 
-(sqrt(2) - (5*sin(x)^2 - 1)^(1/4))/(sqrt(2) + (5*sin(x)^2 - 1)^(1/4))) + 
2*(5*sin(x)^2 - 1)^(1/4) - 1/2*(5*sin(x)^2 - 1)^(1/4)/(sqrt(5*sin(x)^2 - 1 
) + 2)
 

Giac [F]

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

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

Output:

integrate((5*cos(x)^2 - sqrt(5*sin(x)^2 - 1))*tan(x)/((5*sin(x)^2 - 1)^(1/ 
4)*(sqrt(5*sin(x)^2 - 1) + 2)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 10*int(((5*sin(x)**2 - 1)**(3/4)*cos(x)**2*tan(x))/(5*sin(x)**4 - 6*si 
n(x)**2 + 1),x) - 5*int(((5*sin(x)**2 - 1)**(3/4)*sin(x)**2*tan(x))/(5*sin 
(x)**4 - 6*sin(x)**2 + 1),x) + int(((5*sin(x)**2 - 1)**(3/4)*tan(x))/(5*si 
n(x)**4 - 6*sin(x)**2 + 1),x) + 5*int(((5*sin(x)**2 - 1)**(1/4)*cos(x)**2* 
tan(x))/(sin(x)**2 - 1),x) + 2*int(((5*sin(x)**2 - 1)**(1/4)*tan(x))/(sin( 
x)**2 - 1),x))/5