\(\int \frac {\sin ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx\) [506]

Optimal result

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

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

Mathematica [A] (verified)

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

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (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, 4784, 3042, 4792, 3042, 4791, 3042, 4780}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4792

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4791

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4780

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

Input:

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

Output:

((4*(-1/3*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(3/2)) + (2*Sin[a + b*x])/(3*b* 
Sqrt[Sin[2*a + 2*b*x]])))/5 + Sin[a + b*x]/(5*b*Sin[2*a + 2*b*x]^(5/2)))/3 
 + Sin[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 4780

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^( 
p_), x_Symbol] :> Simp[(e*Sin[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 4784

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p 
_), x_Symbol] :> Simp[(-(e*Sin[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*Sin[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] && L 
tQ[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.

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\frac {128 \, \cos \left (b x + a\right )^{7} - 128 \, \cos \left (b x + a\right )^{5} + \sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 96 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{1440 \, {\left (b \cos \left (b x + a\right )^{7} - b \cos \left (b x + a\right )^{5}\right )}} \] Input:

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

Output:

1/1440*(128*cos(b*x + a)^7 - 128*cos(b*x + a)^5 + sqrt(2)*(128*cos(b*x + a 
)^6 - 96*cos(b*x + a)^4 - 12*cos(b*x + a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x 
 + a)))/(b*cos(b*x + a)^7 - b*cos(b*x + a)^5)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 26.88 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.58 \[ \int \frac {\sin ^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}}\,1{}\mathrm {i}}{60\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^3}-\frac {2\,{\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}+1{}\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}}\,1{}\mathrm {i}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}+1{}\mathrm {i}\right )}^5}+\frac {{\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}}\,8{}\mathrm {i}}{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}+1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {49}{180\,b}-\frac {19\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\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}+1{}\mathrm {i}\right )}^2} \] Input:

int(sin(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)*1i)/(9*b*(exp(a*2i + b*x*2i)*1i + 1i)^5) - (2*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*(ex 
p(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)*1i)/(60*b*(exp(a*2i + b*x*2i)*1i 
+ 1i)^3) + (exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + 
b*x*2i)*1i)/2)^(1/2)*8i)/(45*b*(exp(a*2i + b*x*2i) - 1)*(exp(a*2i + b*x*2i 
)*1i + 1i)) - (exp(a*1i + b*x*1i)*(49/(180*b) - (19*exp(a*2i + b*x*2i))/(1 
80*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]

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

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

Output:

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