\(\int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\) [168]

Optimal result

Integrand size = 20, antiderivative size = 105 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {16 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \]

[Out]

-1/7*cos(b*x+a)/b/sin(2*b*x+2*a)^(7/2)+6/35*sin(b*x+a)/b/sin(2*b*x+2*a)^(5/2)-8/35*cos(b*x+a)/b/sin(2*b*x+2*a)
^(3/2)+16/35*sin(b*x+a)/b/sin(2*b*x+2*a)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4388, 4389, 4377} \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\frac {6 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {16 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}}-\frac {8 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)} \]

[In]

Int[Cos[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

-1/7*Cos[a + b*x]/(b*Sin[2*a + 2*b*x]^(7/2)) + (6*Sin[a + b*x])/(35*b*Sin[2*a + 2*b*x]^(5/2)) - (8*Cos[a + b*x
])/(35*b*Sin[2*a + 2*b*x]^(3/2)) + (16*Sin[a + b*x])/(35*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4377

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] && Eq
Q[d/b, 2] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rule 4388

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] + Dist[(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] &&  !IntegerQ[p] && LtQ[p, -1] && Integ
erQ[2*p]

Rule 4389

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] + Dist[(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] && In
tegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6}{7} \int \frac {\sin (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {24}{35} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {16}{35} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos (a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {6 \sin (a+b x)}{35 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{35 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {16 \sin (a+b x)}{35 b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.64 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\frac {(5-10 \cos (2 (a+b x))-4 \cos (4 (a+b x))+4 \cos (6 (a+b x))) \csc ^4(a+b x) \sec ^3(a+b x) \sqrt {\sin (2 (a+b x))}}{560 b} \]

[In]

Integrate[Cos[a + b*x]/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

((5 - 10*Cos[2*(a + b*x)] - 4*Cos[4*(a + b*x)] + 4*Cos[6*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x]^3*Sqrt[Sin[2*
(a + b*x)]])/(560*b)

Maple [C] (verified)

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

Time = 157.90 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.11

\[\frac {\sqrt {-\frac {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}{\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1}}\, \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right ) \left (3 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{8}+40 \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )+2}\, \sqrt {-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-26 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{6}+26 \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-3\right )}{2688 b \tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3} \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right ) \left (\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {a}{2}+\frac {x b}{2}\right )^{3}-\tan \left (\frac {a}{2}+\frac {x b}{2}\right )}}\]

[In]

int(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x)

[Out]

1/2688/b*(-tan(1/2*a+1/2*x*b)/(tan(1/2*a+1/2*x*b)^2-1))^(1/2)*(tan(1/2*a+1/2*x*b)^2-1)/tan(1/2*a+1/2*x*b)^3*(3
*tan(1/2*a+1/2*x*b)^8+40*(tan(1/2*a+1/2*x*b)+1)^(1/2)*(-2*tan(1/2*a+1/2*x*b)+2)^(1/2)*(-tan(1/2*a+1/2*x*b))^(1
/2)*EllipticF((tan(1/2*a+1/2*x*b)+1)^(1/2),1/2*2^(1/2))*tan(1/2*a+1/2*x*b)^3-26*tan(1/2*a+1/2*x*b)^6+26*tan(1/
2*a+1/2*x*b)^2-3)/(tan(1/2*a+1/2*x*b)*(tan(1/2*a+1/2*x*b)^2-1))^(1/2)/(tan(1/2*a+1/2*x*b)^3-tan(1/2*a+1/2*x*b)
)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\frac {128 \, \cos \left (b x + a\right )^{7} - 256 \, \cos \left (b x + a\right )^{5} + 128 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 224 \, \cos \left (b x + a\right )^{4} + 84 \, \cos \left (b x + a\right )^{2} + 7\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )}}{560 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \]

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")

[Out]

1/560*(128*cos(b*x + a)^7 - 256*cos(b*x + a)^5 + 128*cos(b*x + a)^3 + sqrt(2)*(128*cos(b*x + a)^6 - 224*cos(b*
x + a)^4 + 84*cos(b*x + a)^2 + 7)*sqrt(cos(b*x + a)*sin(b*x + a)))/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*
cos(b*x + a)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\int { \frac {\cos \left (b x + a\right )}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)/sin(2*b*x + 2*a)^(9/2), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)/sin(2*b*x+2*a)^(9/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 24.20 (sec) , antiderivative size = 350, normalized size of antiderivative = 3.33 \[ \int \frac {\cos (a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx=-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}+\frac {{\mathrm {e}}^{a\,3{}\mathrm {i}+b\,x\,3{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,16{}\mathrm {i}}{35\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {1}{7\,b}-\frac {8\,{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}}{35\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {16{}\mathrm {i}}{35\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,44{}\mathrm {i}}{35\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3} \]

[In]

int(cos(a + b*x)/sin(2*a + 2*b*x)^(9/2),x)

[Out]

(exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*16i)/(35*b*(exp(a*2i + b*x
*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i)) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2
i)*1i)/2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(1/(7*b) - (8*exp(a*2i + b*x*2i))/
(35*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) + (exp(a*1i + b*x*1i)*(16i/(35*b) + (exp(a*2i + b*x*2i)*44i)/(35*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)^3*(exp(a*2i + b*x*2i)*1i - 1i)^3)