\(\int \frac {\sin ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [498]

Optimal result

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

3/10*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b+1/5*sin(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 = 0.76 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=-\frac {12 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\frac {4 (1+6 \cos (2 (a+b x))+3 \cos (4 (a+b x))) \sin ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 (a+b x))}}{40 b} \] Input:

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

Output:

-1/40*(12*EllipticE[a - Pi/4 + b*x, 2] + (4*(1 + 6*Cos[2*(a + b*x)] + 3*Co 
s[4*(a + b*x)])*Sin[a + b*x]^2)/Sin[2*(a + b*x)]^(5/2))/b
 

Rubi [A] (verified)

Time = 0.34 (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, 4784, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

Input:

Int[Sin[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 + Sin[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 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]
 

Maple [B] (verified)

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

Time = 147.19 (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(sin(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 = 212, normalized size of antiderivative = 2.75 \[ \int \frac {\sin ^2(a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {-6 i \, \sqrt {2 i} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 i \, \sqrt {-2 i} \cos \left (b x + a\right )^{3} E(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) + 6 i \, \sqrt {2 i} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - 6 i \, \sqrt {-2 i} \cos \left (b x + a\right )^{3} F(\arcsin \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\,|\,-1) \sin \left (b x + a\right ) - \sqrt {2} {\left (12 \, \cos \left (b x + a\right )^{4} - 6 \, \cos \left (b x + a\right )^{2} - 1\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{40 \, b \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )} \] Input:

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

Output:

1/40*(-6*I*sqrt(2*I)*cos(b*x + a)^3*elliptic_e(arcsin(cos(b*x + a) + I*sin 
(b*x + a)), -1)*sin(b*x + a) + 6*I*sqrt(-2*I)*cos(b*x + a)^3*elliptic_e(ar 
csin(cos(b*x + a) - I*sin(b*x + a)), -1)*sin(b*x + a) + 6*I*sqrt(2*I)*cos( 
b*x + a)^3*elliptic_f(arcsin(cos(b*x + a) + I*sin(b*x + a)), -1)*sin(b*x + 
 a) - 6*I*sqrt(-2*I)*cos(b*x + a)^3*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 - 6*cos(b*x + a) 
^2 - 1)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^3*sin(b*x + a))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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