\(\int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx\) [407]

Optimal result

Integrand size = 13, antiderivative size = 61 \[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=-\frac {1}{16} \arcsin (\cos (x)-\sin (x))-\frac {1}{16} \log \left (\cos (x)+\sin (x)+\sqrt {\sin (2 x)}\right )-\frac {\cos ^5(x)}{5 \sin ^{\frac {5}{2}}(2 x)}+\frac {\cos (x)}{4 \sqrt {\sin (2 x)}} \] Output:

-1/16*arcsin(cos(x)-sin(x))-1/16*ln(cos(x)+sin(x)+sin(2*x)^(1/2))-1/5*cos( 
x)^5/sin(2*x)^(5/2)+1/4*cos(x)/sin(2*x)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.92 \[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\frac {1}{16} \left (-\arcsin (\cos (x)-\sin (x))-\log \left (\cos (x)+\sin (x)+\sqrt {\sin (2 x)}\right )\right )+\left (\frac {3 \csc (x)}{20}-\frac {\csc ^3(x)}{40}\right ) \sqrt {\sin (2 x)} \] Input:

Integrate[Cos[x]^7/Sin[2*x]^(7/2),x]
 

Output:

(-ArcSin[Cos[x] - Sin[x]] - Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]])/16 + (( 
3*Csc[x])/20 - Csc[x]^3/40)*Sqrt[Sin[2*x]]
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4781, 3042, 4781, 3042, 4795, 3042, 4794}

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.

\(\displaystyle \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (x)^7}{\sin (2 x)^{7/2}}dx\)

\(\Big \downarrow \) 4781

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

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {1}{4} \int \frac {\cos (x)^3}{\sin (2 x)^{3/2}}dx-\frac {\cos ^5(x)}{5 \sin ^{\frac {5}{2}}(2 x)}\)

\(\Big \downarrow \) 4781

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4795

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4794

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

Input:

Int[Cos[x]^7/Sin[2*x]^(7/2),x]
 

Output:

((-1/2*ArcSin[Cos[x] - Sin[x]] - Log[Cos[x] + Sin[x] + Sqrt[Sin[2*x]]]/2)/ 
2 + Cos[x]/Sqrt[Sin[2*x]])/4 - Cos[x]^5/(5*Sin[2*x]^(5/2))
 

Defintions of rubi rules used

rule 3042

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

rule 4781

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p 
_), x_Symbol] :> Simp[e^2*(e*Cos[a + b*x])^(m - 2)*((g*Sin[c + d*x])^(p + 1 
)/(2*b*g*(p + 1))), x] + Simp[e^4*((m + p - 1)/(4*g^2*(p + 1)))   Int[(e*Co 
s[a + b*x])^(m - 4)*(g*Sin[c + d*x])^(p + 2), x], x] /; FreeQ[{a, b, c, d, 
e, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[m, 2] 
 && LtQ[p, -1] && (GtQ[m, 3] || EqQ[p, -3/2]) && IntegersQ[2*m, 2*p]
 

rule 4794

Int[sin[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim 
p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] - Simp[Log[Cos[a + b*x] + Sin[ 
a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 
 a*d, 0] && EqQ[d/b, 2]
 

rule 4795

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/cos[(a_.) + (b_.)*(x_)], x_Symbol] 
 :> Simp[2*g   Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ 
a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] 
&& IntegerQ[2*p]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 1.59 (sec) , antiderivative size = 1108, normalized size of antiderivative = 18.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1108\)

Input:

int(cos(x)^7/sin(2*x)^(7/2),x,method=_RETURNVERBOSE)
 

Output:

1/160*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(192*(tan(1/2*x)*( 
tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan( 
1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((tan(1/2*x)-1)* 
(1+tan(1/2*x))*tan(1/2*x))^(1/2)*tan(1/2*x)^6-96*(tan(1/2*x)*(tan(1/2*x)^2 
-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2 
)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((tan(1/2*x)-1)*(1+tan(1/2*x 
))*tan(1/2*x))^(1/2)*tan(1/2*x)^6-(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*((ta 
n(1/2*x)-1)*(1+tan(1/2*x))*tan(1/2*x))^(1/2)*tan(1/2*x)^10+96*(tan(1/2*x)* 
(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*tan(1/2*x)^8-384*( 
tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^ 
(1/2)*(-tan(1/2*x))^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((ta 
n(1/2*x)-1)*(1+tan(1/2*x))*tan(1/2*x))^(1/2)*tan(1/2*x)^4+192*(tan(1/2*x)* 
(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan 
(1/2*x))^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((tan(1/2*x)-1) 
*(1+tan(1/2*x))*tan(1/2*x))^(1/2)*tan(1/2*x)^4+3*(tan(1/2*x)*(tan(1/2*x)^2 
-1))^(1/2)*((tan(1/2*x)-1)*(1+tan(1/2*x))*tan(1/2*x))^(1/2)*tan(1/2*x)^8+4 
8*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((tan(1/2*x)-1)*(1+tan(1/2*x))*tan(1/2*x 
))^(1/2)*tan(1/2*x)^8-192*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^ 
3-tan(1/2*x))^(1/2)*tan(1/2*x)^6+192*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*( 
1+tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(1/2)*Ellipti...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (47) = 94\).

Time = 0.09 (sec) , antiderivative size = 205, normalized size of antiderivative = 3.36 \[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\frac {10 \, {\left (\cos \left (x\right )^{2} - 1\right )} \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} {\left (\cos \left (x\right ) - \sin \left (x\right )\right )} + \cos \left (x\right ) \sin \left (x\right )}{\cos \left (x\right )^{2} + 2 \, \cos \left (x\right ) \sin \left (x\right ) - 1}\right ) \sin \left (x\right ) - 10 \, {\left (\cos \left (x\right )^{2} - 1\right )} \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} - \cos \left (x\right ) - \sin \left (x\right )}{\cos \left (x\right ) - \sin \left (x\right )}\right ) \sin \left (x\right ) + 5 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-32 \, \cos \left (x\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (x\right )^{3} - {\left (4 \, \cos \left (x\right )^{2} + 1\right )} \sin \left (x\right ) - 5 \, \cos \left (x\right )\right )} \sqrt {\cos \left (x\right ) \sin \left (x\right )} + 32 \, \cos \left (x\right )^{2} + 16 \, \cos \left (x\right ) \sin \left (x\right ) + 1\right ) \sin \left (x\right ) + 8 \, \sqrt {2} {\left (6 \, \cos \left (x\right )^{2} - 5\right )} \sqrt {\cos \left (x\right ) \sin \left (x\right )} + 48 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{320 \, {\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \] Input:

integrate(cos(x)^7/sin(2*x)^(7/2),x, algorithm="fricas")
 

Output:

1/320*(10*(cos(x)^2 - 1)*arctan(-(sqrt(2)*sqrt(cos(x)*sin(x))*(cos(x) - si 
n(x)) + cos(x)*sin(x))/(cos(x)^2 + 2*cos(x)*sin(x) - 1))*sin(x) - 10*(cos( 
x)^2 - 1)*arctan(-(2*sqrt(2)*sqrt(cos(x)*sin(x)) - cos(x) - sin(x))/(cos(x 
) - sin(x)))*sin(x) + 5*(cos(x)^2 - 1)*log(-32*cos(x)^4 + 4*sqrt(2)*(4*cos 
(x)^3 - (4*cos(x)^2 + 1)*sin(x) - 5*cos(x))*sqrt(cos(x)*sin(x)) + 32*cos(x 
)^2 + 16*cos(x)*sin(x) + 1)*sin(x) + 8*sqrt(2)*(6*cos(x)^2 - 5)*sqrt(cos(x 
)*sin(x)) + 48*(cos(x)^2 - 1)*sin(x))/((cos(x)^2 - 1)*sin(x))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\text {Timed out} \] Input:

integrate(cos(x)**7/sin(2*x)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\int { \frac {\cos \left (x\right )^{7}}{\sin \left (2 \, x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(cos(x)^7/sin(2*x)^(7/2),x, algorithm="maxima")
 

Output:

integrate(cos(x)^7/sin(2*x)^(7/2), x)
 

Giac [F]

\[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\int { \frac {\cos \left (x\right )^{7}}{\sin \left (2 \, x\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(cos(x)^7/sin(2*x)^(7/2),x, algorithm="giac")
 

Output:

integrate(cos(x)^7/sin(2*x)^(7/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\int \frac {{\cos \left (x\right )}^7}{{\sin \left (2\,x\right )}^{7/2}} \,d x \] Input:

int(cos(x)^7/sin(2*x)^(7/2),x)
 

Output:

int(cos(x)^7/sin(2*x)^(7/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^7(x)}{\sin ^{\frac {7}{2}}(2 x)} \, dx=\int \frac {\sqrt {\sin \left (2 x \right )}\, \cos \left (x \right )^{7}}{\sin \left (2 x \right )^{4}}d x \] Input:

int(cos(x)^7/sin(2*x)^(7/2),x)
                                                                                    
                                                                                    
 

Output:

int((sqrt(sin(2*x))*cos(x)**7)/sin(2*x)**4,x)