\(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx\) [593]

Optimal result

Integrand size = 22, antiderivative size = 107 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \] Output:

-1/9*cos(b*x+a)^3/b/sin(2*b*x+2*a)^(9/2)-1/15*cos(b*x+a)/b/sin(2*b*x+2*a)^ 
(5/2)+4/45*sin(b*x+a)/b/sin(2*b*x+2*a)^(3/2)-8/45*cos(b*x+a)/b/sin(2*b*x+2 
*a)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.79 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (113 \csc (a+b x)+17 \csc ^3(a+b x)+5 \csc ^5(a+b x)-15 \sec (a+b x) \tan (a+b x)\right )}{1440 b} \] Input:

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]
 

Output:

-1/1440*(Sqrt[Sin[2*(a + b*x)]]*(113*Csc[a + b*x] + 17*Csc[a + b*x]^3 + 5* 
Csc[a + b*x]^5 - 15*Sec[a + b*x]*Tan[a + b*x]))/b
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4783, 3042, 4791, 3042, 4792, 3042, 4779}

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.

\(\displaystyle \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\cos (a+b x)^3}{\sin (2 a+2 b x)^{11/2}}dx\)

\(\Big \downarrow \) 4783

\(\displaystyle \frac {1}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)}dx-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{7/2}}dx-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 4791

\(\displaystyle \frac {1}{3} \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {4}{5} \int \frac {\sin (a+b x)}{\sin (2 a+2 b x)^{5/2}}dx-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 4792

\(\displaystyle \frac {1}{3} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\cos (a+b x)}{\sin (2 a+2 b x)^{3/2}}dx+\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 4779

\(\displaystyle \frac {1}{3} \left (\frac {4}{5} \left (\frac {\sin (a+b x)}{3 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{3 b \sqrt {\sin (2 a+2 b x)}}\right )-\frac {\cos (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\right )-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}\)

Input:

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(11/2),x]
 

Output:

((4*(Sin[a + b*x]/(3*b*Sin[2*a + 2*b*x]^(3/2)) - (2*Cos[a + b*x])/(3*b*Sqr 
t[Sin[2*a + 2*b*x]])))/5 - Cos[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)))/3 - 
Cos[a + b*x]^3/(9*b*Sin[2*a + 2*b*x]^(9/2))
 

Defintions of rubi rules used

rule 3042

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

rule 4779

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)/(b*g 
*m)), x] /; FreeQ[{a, b, c, d, e, g, m, p}, x] && EqQ[b*c - a*d, 0] && EqQ[ 
d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]
 

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]
 

rule 4791

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[Cos[a + b*x]*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Simp 
[(2*p + 3)/(2*g*(p + 1))   Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p + 1), x], x 
] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !Int 
egerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 4792

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[(-Sin[a + b*x])*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + S 
imp[(2*p + 3)/(2*g*(p + 1))   Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p + 1), x] 
, x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  ! 
IntegerQ[p] && LtQ[p, -1] && IntegerQ[2*p]
 

Maple [F(-1)]

Timed out.

\[\int \frac {\cos \left (b x +a \right )^{3}}{\sin \left (2 b x +2 a \right )^{\frac {11}{2}}}d x\]

Input:

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)
 

Output:

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 288 \, \cos \left (b x + a\right )^{4} + 180 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{1440 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="fricas")
 

Output:

-1/1440*(sqrt(2)*(128*cos(b*x + a)^6 - 288*cos(b*x + a)^4 + 180*cos(b*x + 
a)^2 - 15)*sqrt(cos(b*x + a)*sin(b*x + a)) + 128*(cos(b*x + a)^6 - 2*cos(b 
*x + a)^4 + cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^6 - 2*b*cos(b*x 
 + a)^4 + b*cos(b*x + a)^2)*sin(b*x + a))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \] Input:

integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**(11/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="maxima")
 

Output:

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(11/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \] Input:

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 24.68 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.58 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{60\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,2{}\mathrm {i}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^5}+\frac {8\,{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{45\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {49{}\mathrm {i}}{180\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,19{}\mathrm {i}}{180\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2} \] Input:

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(11/2),x)
 

Output:

(exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i) 
/2)^(1/2))/(9*b*(exp(a*2i + b*x*2i)*1i - 1i)^5) - (exp(a*1i + b*x*1i)*((ex 
p(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*2i)/(9*b*(exp( 
a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i 
)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/(60*b*(exp(a*2i + b*x*2i)*1i - 1i) 
^3) + (8*exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x 
*2i)*1i)/2)^(1/2))/(45*b*(exp(a*2i + b*x*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 
 1i)) - (exp(a*1i + b*x*1i)*(49i/(180*b) + (exp(a*2i + b*x*2i)*19i)/(180*b 
))*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2))/((exp( 
a*2i + b*x*2i) + 1)^2*(exp(a*2i + b*x*2i)*1i - 1i)^2)
 

Reduce [F]

\[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\int \frac {\sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right )^{3}}{\sin \left (2 b x +2 a \right )^{6}}d x \] Input:

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(11/2),x)
 

Output:

int((sqrt(sin(2*a + 2*b*x))*cos(a + b*x)**3)/sin(2*a + 2*b*x)**6,x)