\(\int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [585]

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {3 \cos (2 a+2 b x)}{10 b \sqrt {\sin (2 a+2 b x)}} \] Output:

3/10*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/5*cos(b*x+a)^2/b/sin(2*b*x+2 
*a)^(5/2)-3/10*cos(2*b*x+2*a)/b/sin(2*b*x+2*a)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {-12 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\frac {2 (1-6 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \cot (a+b x)}{\sin ^{\frac {3}{2}}(2 (a+b x))}}{40 b} \] Input:

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

Output:

(-12*EllipticE[a - Pi/4 + b*x, 2] + (2*(1 - 6*Cos[2*(a + b*x)] + 3*Cos[4*( 
a + b*x)])*Cot[a + b*x])/Sin[2*(a + b*x)]^(3/2))/(40*b)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4783, 3042, 3116, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4783

\(\displaystyle \frac {3}{10} \int \frac {1}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{10} \int \frac {1}{\sin (2 a+2 b x)^{3/2}}dx-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {3}{10} \left (-\int \sqrt {\sin (2 a+2 b x)}dx-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3}{10} \left (-\int \sqrt {\sin (2 a+2 b x)}dx-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {3}{10} \left (-\frac {E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b}-\frac {\cos (2 a+2 b x)}{b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos ^2(a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

Input:

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

Output:

(3*(-(EllipticE[a - Pi/4 + b*x, 2]/b) - Cos[2*a + 2*b*x]/(b*Sqrt[Sin[2*a + 
 2*b*x]])))/10 - Cos[a + b*x]^2/(5*b*Sin[2*a + 2*b*x]^(5/2))
 

Defintions of rubi rules used

rule 3042

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

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4783

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(68)=136\).

Time = 134.03 (sec) , antiderivative size = 227, normalized size of antiderivative = 2.95

method	result	size
default	\(\frac {\sqrt {2}\, \left (-\frac {8 \sqrt {2}}{5 \sin \left (2 b x +2 a \right )^{\frac {5}{2}}}+\frac {4 \sqrt {2}\, \left (6 \sqrt {\sin \left (2 b x +2 a \right )+1}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \sin \left (2 b x +2 a \right )^{2} \operatorname {EllipticE}\left (\sqrt {\sin \left (2 b x +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {\sin \left (2 b x +2 a \right )+1}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \sin \left (2 b x +2 a \right )^{2} \operatorname {EllipticF}\left (\sqrt {\sin \left (2 b x +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+6 \sin \left (2 b x +2 a \right )^{4}-4 \sin \left (2 b x +2 a \right )^{2}-2\right )}{5 \sin \left (2 b x +2 a \right )^{\frac {5}{2}} \cos \left (2 b x +2 a \right )}\right )}{32 b}\)	\(227\)

Input:

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

Output:

1/32*2^(1/2)*(-8/5*2^(1/2)/sin(2*b*x+2*a)^(5/2)+4/5*2^(1/2)/sin(2*b*x+2*a) 
^(5/2)*(6*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x 
+2*a))^(1/2)*sin(2*b*x+2*a)^2*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/ 
2))-3*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a 
))^(1/2)*sin(2*b*x+2*a)^2*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+ 
6*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-2)/cos(2*b*x+2*a))/b
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.45 \[ \int \frac {\cos ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {6 \, \sqrt {2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {2 i} {\left (-i \, \cos \left (b x + a\right )^{3} + i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 \, \sqrt {-2 i} {\left (i \, \cos \left (b x + a\right )^{3} - i \, \cos \left (b x + a\right )\right )} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + \sqrt {2} {\left (12 \, \cos \left (b x + a\right )^{4} - 18 \, \cos \left (b x + a\right )^{2} + 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{40 \, {\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \] Input:

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

Output:

-1/40*(6*sqrt(2*I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_e(arcsin(c 
os(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + a) + 6*sqrt(-2*I)*(-I*cos(b*x 
 + a)^3 + I*cos(b*x + a))*elliptic_e(arcsin(cos(b*x + a) - I*sin(b*x + a)) 
, -1)*sin(b*x + a) + 6*sqrt(2*I)*(-I*cos(b*x + a)^3 + I*cos(b*x + a))*elli 
ptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + a) + 6*sqrt(-2 
*I)*(I*cos(b*x + a)^3 - I*cos(b*x + a))*elliptic_f(arcsin(cos(b*x + a) - I 
*sin(b*x + a)), -1)*sin(b*x + a) + sqrt(2)*(12*cos(b*x + a)^4 - 18*cos(b*x 
 + a)^2 + 5)*sqrt(cos(b*x + a)*sin(b*x + a)))/((b*cos(b*x + a)^3 - b*cos(b 
*x + a))*sin(b*x + a))
 

Sympy [F(-1)]

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

integrate(cos(b*x+a)**2/sin(2*b*x+2*a)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(sin(2*a + 2*b*x))*cos(a + b*x)**2)/sin(2*a + 2*b*x)**4,x)