\(\int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx\) [489]

Optimal result

Integrand size = 20, antiderivative size = 79 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {4 \cos (a+b x)}{15 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8 \sin (a+b x)}{15 b \sqrt {\sin (2 a+2 b x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.66 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {\left (-5 \cot (a+b x) \csc (a+b x)+3 \sec (a+b x) \left (9+\sec ^2(a+b x)\right )\right ) \sqrt {\sin (2 (a+b x))}}{120 b} \] Input:

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

Output:

((-5*Cot[a + b*x]*Csc[a + b*x] + 3*Sec[a + b*x]*(9 + Sec[a + b*x]^2))*Sqrt 
[Sin[2*(a + b*x)]])/(120*b)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4792

\(\displaystyle \frac {4}{5} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)}dx+\frac {\sin (a+b x)}{5 b \sin ^{\frac {5}{2}}(2 a+2 b x)}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4791

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4780

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

Input:

Int[Sin[a + b*x]/Sin[2*a + 2*b*x]^(7/2),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*S 
qrt[Sin[2*a + 2*b*x]])))/5 + Sin[a + b*x]/(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 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 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 [C] (verified)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 97.64 (sec) , antiderivative size = 308, normalized size of antiderivative = 3.90

\[-\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}{\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right ) \left (5 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{6}+5 \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {b x}{2}\right )-7 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{4}+7 \tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}+1\right )}{48 b \tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {b x}{2}\right )}\, \left (1+\tan \left (\frac {a}{2}+\frac {b x}{2}\right )^{2}\right )}\]

Input:

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

Output:

-1/48/b*(-tan(1/2*a+1/2*b*x)/(tan(1/2*a+1/2*b*x)^2-1))^(1/2)*(tan(1/2*a+1/ 
2*b*x)^2-1)/tan(1/2*a+1/2*b*x)*(5*(tan(1/2*a+1/2*b*x)+1)^(1/2)*(-2*tan(1/2 
*a+1/2*b*x)+2)^(1/2)*(-tan(1/2*a+1/2*b*x))^(1/2)*EllipticF((tan(1/2*a+1/2* 
b*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*b*x)^3-tan(1/2*a+1/2*b*x)^6+5*(ta 
n(1/2*a+1/2*b*x)+1)^(1/2)*(-2*tan(1/2*a+1/2*b*x)+2)^(1/2)*(-tan(1/2*a+1/2* 
b*x))^(1/2)*EllipticF((tan(1/2*a+1/2*b*x)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+ 
1/2*b*x)-7*tan(1/2*a+1/2*b*x)^4+7*tan(1/2*a+1/2*b*x)^2+1)/(tan(1/2*a+1/2*b 
*x)*(tan(1/2*a+1/2*b*x)^2-1))^(1/2)/(tan(1/2*a+1/2*b*x)^3-tan(1/2*a+1/2*b* 
x))^(1/2)/(1+tan(1/2*a+1/2*b*x)^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.11 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {32 \, \cos \left (b x + a\right )^{5} - 32 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 24 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{120 \, {\left (b \cos \left (b x + a\right )^{5} - b \cos \left (b x + a\right )^{3}\right )}} \] Input:

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

Output:

1/120*(32*cos(b*x + a)^5 - 32*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 
- 24*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^ 
5 - b*cos(b*x + a)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18022 vs. \(2 (67) = 134\).

Time = 157.24 (sec) , antiderivative size = 18022, normalized size of antiderivative = 228.13 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/480*sqrt(2)*sqrt(-tan(1/2*b*x)^4*tan(1/2*a)^3 - tan(1/2*b*x)^3*tan(1/2*a 
)^4 + tan(1/2*b*x)^4*tan(1/2*a) + 6*tan(1/2*b*x)^3*tan(1/2*a)^2 + 6*tan(1/ 
2*b*x)^2*tan(1/2*a)^3 + tan(1/2*b*x)*tan(1/2*a)^4 - tan(1/2*b*x)^3 - 6*tan 
(1/2*b*x)^2*tan(1/2*a) - 6*tan(1/2*b*x)*tan(1/2*a)^2 - tan(1/2*a)^3 + tan( 
1/2*b*x) + tan(1/2*a))*((((((((((2*(sqrt(2)*tan(1/2*a)^87 - 92*sqrt(2)*tan 
(1/2*a)^85 - 3213*sqrt(2)*tan(1/2*a)^83 - 48008*sqrt(2)*tan(1/2*a)^81 - 44 
3462*sqrt(2)*tan(1/2*a)^79 - 2875040*sqrt(2)*tan(1/2*a)^77 - 13907802*sqrt 
(2)*tan(1/2*a)^75 - 51781432*sqrt(2)*tan(1/2*a)^73 - 149943911*sqrt(2)*tan 
(1/2*a)^71 - 333456564*sqrt(2)*tan(1/2*a)^69 - 536442973*sqrt(2)*tan(1/2*a 
)^67 - 482080288*sqrt(2)*tan(1/2*a)^65 + 316221080*sqrt(2)*tan(1/2*a)^63 + 
 2190937152*sqrt(2)*tan(1/2*a)^61 + 4607763368*sqrt(2)*tan(1/2*a)^59 + 574 
2984608*sqrt(2)*tan(1/2*a)^57 + 3316624962*sqrt(2)*tan(1/2*a)^55 - 3241815 
576*sqrt(2)*tan(1/2*a)^53 - 11030972730*sqrt(2)*tan(1/2*a)^51 - 1471202712 
0*sqrt(2)*tan(1/2*a)^49 - 10524179460*sqrt(2)*tan(1/2*a)^47 + 10524179460* 
sqrt(2)*tan(1/2*a)^43 + 14712027120*sqrt(2)*tan(1/2*a)^41 + 11030972730*sq 
rt(2)*tan(1/2*a)^39 + 3241815576*sqrt(2)*tan(1/2*a)^37 - 3316624962*sqrt(2 
)*tan(1/2*a)^35 - 5742984608*sqrt(2)*tan(1/2*a)^33 - 4607763368*sqrt(2)*ta 
n(1/2*a)^31 - 2190937152*sqrt(2)*tan(1/2*a)^29 - 316221080*sqrt(2)*tan(1/2 
*a)^27 + 482080288*sqrt(2)*tan(1/2*a)^25 + 536442973*sqrt(2)*tan(1/2*a)^23 
 + 333456564*sqrt(2)*tan(1/2*a)^21 + 149943911*sqrt(2)*tan(1/2*a)^19 + ...
 

Mupad [B] (verification not implemented)

Time = 22.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx=\frac {4\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,\left (2\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-3\,{\mathrm {e}}^{a\,4{}\mathrm {i}+b\,x\,4{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,6{}\mathrm {i}+b\,x\,6{}\mathrm {i}}+2\,{\mathrm {e}}^{a\,8{}\mathrm {i}+b\,x\,8{}\mathrm {i}}+2\right )}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}-1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3} \] Input:

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

Output:

(4*exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1 
i)/2)^(1/2)*(2*exp(a*2i + b*x*2i) - 3*exp(a*4i + b*x*4i) + 2*exp(a*6i + b* 
x*6i) + 2*exp(a*8i + b*x*8i) + 2))/(15*b*(exp(a*2i + b*x*2i) - 1)^2*(exp(a 
*2i + b*x*2i) + 1)^3)
 

Reduce [F]

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

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

Output:

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