\(\int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx\) [455]

Optimal result

Integrand size = 15, antiderivative size = 102 \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\frac {\arctan \left (\frac {1-\sqrt {\cos (2 x)}}{\sqrt {2} \sqrt [4]{\cos (2 x)}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {1+\sqrt {\cos (2 x)}}{\sqrt {2} \sqrt [4]{\cos (2 x)}}\right )}{\sqrt {2}}+\frac {7}{4} \sqrt [4]{\cos (2 x)}-\frac {1}{5} \cos ^{\frac {5}{4}}(2 x)+\frac {1}{36} \cos ^{\frac {9}{4}}(2 x) \] Output:

7/4*cos(2*x)^(1/4)-1/5*cos(2*x)^(5/4)+1/36*cos(2*x)^(9/4)+1/2*arctan(1/2*( 
1-cos(2*x)^(1/2))/cos(2*x)^(1/4)*2^(1/2))*2^(1/2)-1/2*arctanh(1/2*(1+cos(2 
*x)^(1/2))/cos(2*x)^(1/4)*2^(1/2))*2^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.50 \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\frac {1}{360} \left (180 \sqrt {2} \arctan \left (1-\sqrt {2} \sqrt [4]{\cos (2 x)}\right )-180 \sqrt {2} \arctan \left (1+\sqrt {2} \sqrt [4]{\cos (2 x)}\right )+635 \sqrt [4]{\cos (2 x)}-72 \cos ^{\frac {5}{4}}(2 x)+5 \sqrt [4]{\cos (2 x)} \cos (4 x)+90 \sqrt {2} \log \left (1-\sqrt {2} \sqrt [4]{\cos (2 x)}+\sqrt {\cos (2 x)}\right )-90 \sqrt {2} \log \left (1+\sqrt {2} \sqrt [4]{\cos (2 x)}+\sqrt {\cos (2 x)}\right )\right ) \] Input:

Integrate[(Sin[x]^6*Tan[x])/Cos[2*x]^(3/4),x]
 

Output:

(180*Sqrt[2]*ArcTan[1 - Sqrt[2]*Cos[2*x]^(1/4)] - 180*Sqrt[2]*ArcTan[1 + S 
qrt[2]*Cos[2*x]^(1/4)] + 635*Cos[2*x]^(1/4) - 72*Cos[2*x]^(5/4) + 5*Cos[2* 
x]^(1/4)*Cos[4*x] + 90*Sqrt[2]*Log[1 - Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos[2 
*x]]] - 90*Sqrt[2]*Log[1 + Sqrt[2]*Cos[2*x]^(1/4) + Sqrt[Cos[2*x]]])/360
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.85, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4861, 354, 99, 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.

\(\displaystyle \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (x)^6 \tan (x)}{\cos (2 x)^{3/4}}dx\)

\(\Big \downarrow \) 4861

\(\displaystyle -\int \frac {\left (1-\cos ^2(x)\right )^3 \sec (x)}{\left (2 \cos ^2(x)-1\right )^{3/4}}d\cos (x)\)

\(\Big \downarrow \) 354

\(\displaystyle -\frac {1}{2} \int \frac {\left (1-\cos ^2(x)\right )^3 \sec (x)}{\left (2 \cos ^2(x)-1\right )^{3/4}}d\cos ^2(x)\)

\(\Big \downarrow \) 99

\(\displaystyle -\frac {1}{2} \int \left (-\frac {1}{4} \left (2 \cos ^2(x)-1\right )^{5/4}+\sqrt [4]{2 \cos ^2(x)-1}+\frac {\sec (x)}{\left (2 \cos ^2(x)-1\right )^{3/4}}-\frac {7}{4 \left (2 \cos ^2(x)-1\right )^{3/4}}\right )d\cos ^2(x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\sqrt {2} \arctan \left (1-\sqrt {2} \sqrt [4]{2 \cos ^2(x)-1}\right )-\sqrt {2} \arctan \left (\sqrt {2} \sqrt [4]{2 \cos ^2(x)-1}+1\right )+\frac {1}{18} \left (2 \cos ^2(x)-1\right )^{9/4}-\frac {2}{5} \left (2 \cos ^2(x)-1\right )^{5/4}+\frac {7}{2} \sqrt [4]{2 \cos ^2(x)-1}+\frac {\log \left (\sqrt {2 \cos ^2(x)-1}-\sqrt {2} \sqrt [4]{2 \cos ^2(x)-1}+1\right )}{\sqrt {2}}-\frac {\log \left (\sqrt {2 \cos ^2(x)-1}+\sqrt {2} \sqrt [4]{2 \cos ^2(x)-1}+1\right )}{\sqrt {2}}\right )\)

Input:

Int[(Sin[x]^6*Tan[x])/Cos[2*x]^(3/4),x]
 

Output:

(Sqrt[2]*ArcTan[1 - Sqrt[2]*(-1 + 2*Cos[x]^2)^(1/4)] - Sqrt[2]*ArcTan[1 + 
Sqrt[2]*(-1 + 2*Cos[x]^2)^(1/4)] + (7*(-1 + 2*Cos[x]^2)^(1/4))/2 - (2*(-1 
+ 2*Cos[x]^2)^(5/4))/5 + (-1 + 2*Cos[x]^2)^(9/4)/18 + Log[1 - Sqrt[2]*(-1 
+ 2*Cos[x]^2)^(1/4) + Sqrt[-1 + 2*Cos[x]^2]]/Sqrt[2] - Log[1 + Sqrt[2]*(-1 
 + 2*Cos[x]^2)^(1/4) + Sqrt[-1 + 2*Cos[x]^2]]/Sqrt[2])/2
 

Defintions of rubi rules used

rule 99

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | 
| (GtQ[m, 0] && GeQ[n, -1]))
 

rule 354

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S 
ymbol] :> Simp[1/2   Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x 
, x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ 
[(m - 1)/2]
 

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])
 

Maple [F]

\[\int \frac {\sin \left (x \right )^{6} \tan \left (x \right )}{\cos \left (2 x \right )^{\frac {3}{4}}}d x\]

Input:

int(sin(x)^6*tan(x)/cos(2*x)^(3/4),x)
 

Output:

int(sin(x)^6*tan(x)/cos(2*x)^(3/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\text {Timed out} \] Input:

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\text {Timed out} \] Input:

integrate(sin(x)**6*tan(x)/cos(2*x)**(3/4),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="maxima")
 

Output:

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.18 \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\frac {1}{36} \, \cos \left (2 \, x\right )^{\frac {9}{4}} - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \cos \left (2 \, x\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \cos \left (2 \, x\right )^{\frac {1}{4}}\right )}\right ) - \frac {1}{4} \, \sqrt {2} \log \left (\sqrt {2} \cos \left (2 \, x\right )^{\frac {1}{4}} + \sqrt {\cos \left (2 \, x\right )} + 1\right ) + \frac {1}{4} \, \sqrt {2} \log \left (-\sqrt {2} \cos \left (2 \, x\right )^{\frac {1}{4}} + \sqrt {\cos \left (2 \, x\right )} + 1\right ) - \frac {1}{5} \, \cos \left (2 \, x\right )^{\frac {5}{4}} + \frac {7}{4} \, \cos \left (2 \, x\right )^{\frac {1}{4}} \] Input:

integrate(sin(x)^6*tan(x)/cos(2*x)^(3/4),x, algorithm="giac")
 

Output:

1/36*cos(2*x)^(9/4) - 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*cos(2*x) 
^(1/4))) - 1/2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*cos(2*x)^(1/4))) - 
 1/4*sqrt(2)*log(sqrt(2)*cos(2*x)^(1/4) + sqrt(cos(2*x)) + 1) + 1/4*sqrt(2 
)*log(-sqrt(2)*cos(2*x)^(1/4) + sqrt(cos(2*x)) + 1) - 1/5*cos(2*x)^(5/4) + 
 7/4*cos(2*x)^(1/4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^6(x) \tan (x)}{\cos ^{\frac {3}{4}}(2 x)} \, dx=\int \frac {{\sin \left (x\right )}^6\,\mathrm {tan}\left (x\right )}{{\cos \left (2\,x\right )}^{3/4}} \,d x \] Input:

int((sin(x)^6*tan(x))/cos(2*x)^(3/4),x)
 

Output:

int((sin(x)^6*tan(x))/cos(2*x)^(3/4), x)
 

Reduce [F]

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

int(sin(x)^6*tan(x)/cos(2*x)^(3/4),x)
 

Output:

int((sin(x)**6*tan(x))/cos(2*x)**(3/4),x)