\(\int \frac {\sin ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\) [502]

Optimal result

Integrand size = 22, antiderivative size = 81 \[ \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{4 b}-\frac {\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{4 b}+\frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}} \] Output:

1/4*arcsin(cos(b*x+a)-sin(b*x+a))/b-1/4*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x 
+2*a)^(1/2))/b+sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=\frac {\arcsin (\cos (a+b x)-\sin (a+b x))-\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )+2 \sec (a+b x) \sqrt {\sin (2 (a+b x))}}{4 b} \] Input:

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

Output:

(ArcSin[Cos[a + b*x] - Sin[a + b*x]] - Log[Cos[a + b*x] + Sin[a + b*x] + S 
qrt[Sin[2*(a + b*x)]]] + 2*Sec[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/(4*b)
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4782, 3042, 4796, 3042, 4793}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4782

\(\displaystyle \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{4} \int \csc (a+b x) \sqrt {\sin (2 a+2 b x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{4} \int \frac {\sqrt {\sin (2 a+2 b x)}}{\sin (a+b x)}dx\)

\(\Big \downarrow \) 4796

\(\displaystyle \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}-\frac {1}{2} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx\)

\(\Big \downarrow \) 4793

\(\displaystyle \frac {1}{2} \left (\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x)}{b \sqrt {\sin (2 a+2 b x)}}\)

Input:

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

Output:

(ArcSin[Cos[a + b*x] - Sin[a + b*x]]/(2*b) - Log[Cos[a + b*x] + Sin[a + b* 
x] + Sqrt[Sin[2*a + 2*b*x]]]/(2*b))/2 + Sin[a + b*x]/(b*Sqrt[Sin[2*a + 2*b 
*x]])
 

Defintions of rubi rules used

rule 3042

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

rule 4782

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p 
_), x_Symbol] :> Simp[(-e^2)*(e*Sin[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 
*Sin[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 4793

Int[cos[(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 4796

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] 
 :> Simp[2*g   Int[Cos[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 [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (73) = 146\).

Time = 0.09 (sec) , antiderivative size = 296, normalized size of antiderivative = 3.65 \[ \int \frac {\sin ^3(a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx=-\frac {2 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) \cos \left (b x + a\right ) - 2 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) - \cos \left (b x + a\right ) \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right ) - 8 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 8 \, \cos \left (b x + a\right )}{16 \, b \cos \left (b x + a\right )} \] Input:

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

Output:

-1/16*(2*arctan(-(sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a))*(cos(b*x + a) - 
sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*cos(b*x + a 
)*sin(b*x + a) - 1))*cos(b*x + a) - 2*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a) 
*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a) 
))*cos(b*x + a) - cos(b*x + a)*log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b 
*x + a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos 
(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) 
 + 1) - 8*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - 8*cos(b*x + a))/(b*cos 
(b*x + a))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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