\(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx\) [184]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4380, 4388, 4389, 4376} \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=\frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \]

[In]

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

[Out]

-1/9*Cos[a + b*x]^3/(b*Sin[2*a + 2*b*x]^(9/2)) - Cos[a + b*x]/(15*b*Sin[2*a + 2*b*x]^(5/2)) + (4*Sin[a + b*x])
/(45*b*Sin[2*a + 2*b*x]^(3/2)) - (8*Cos[a + b*x])/(45*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4376

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-(e*Cos[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 4380

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e*Cos[a + b*
x])^m*((g*Sin[c + d*x])^(p + 1)/(2*b*g*(p + 1))), x] + Dist[e^2*((m + 2*p + 2)/(4*g^2*(p + 1))), Int[(e*Cos[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] && LtQ[p, -1] && NeQ[m + 2*p + 2, 0] && (LtQ[p, -2] || EqQ[m, 2]) && Inte
gersQ[2*m, 2*p]

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 ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}+\frac {1}{3} \int \frac {\cos (a+b x)}{\sin ^{\frac {7}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4}{15} \int \frac {\sin (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {8}{45} \int \frac {\cos (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx \\ & = -\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.58 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {\sin (2 (a+b x))} \left (113 \csc (a+b x)+17 \csc ^3(a+b x)+5 \csc ^5(a+b x)-15 \sec (a+b x) \tan (a+b x)\right )}{1440 b} \]

[In]

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

[Out]

-1/1440*(Sqrt[Sin[2*(a + b*x)]]*(113*Csc[a + b*x] + 17*Csc[a + b*x]^3 + 5*Csc[a + b*x]^5 - 15*Sec[a + b*x]*Tan
[a + b*x]))/b

Maple [F(-1)]

Timed out.

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22 \[ \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {11}{2}}(2 a+2 b x)} \, dx=-\frac {\sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 288 \, \cos \left (b x + a\right )^{4} + 180 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{1440 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]

[In]

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

[Out]

-1/1440*(sqrt(2)*(128*cos(b*x + a)^6 - 288*cos(b*x + a)^4 + 180*cos(b*x + a)^2 - 15)*sqrt(cos(b*x + a)*sin(b*x
+ a)) + 128*(cos(b*x + a)^6 - 2*cos(b*x + a)^4 + cos(b*x + a)^2)*sin(b*x + a))/((b*cos(b*x + a)^6 - 2*b*cos(b
*x + a)^4 + b*cos(b*x + a)^2)*sin(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 25.36 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.58 \[ \int \frac {\cos ^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}}}{60\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\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}}\,2{}\mathrm {i}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\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}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^5}+\frac {8\,{\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}}}{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}-\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {49{}\mathrm {i}}{180\,b}+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,19{}\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}-\mathrm {i}\right )}^2} \]

[In]

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

[Out]

(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*(exp(a*2i + b*x*2i)*
1i - 1i)^5) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/2)^(1/2)*2i)/(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))/(60*b*(exp(a*2i + b*x*2i)*1i - 1i)^3) + (8*exp(a*3i + b*x*3i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i +
b*x*2i)*1i)/2)^(1/2))/(45*b*(exp(a*2i + b*x*2i) + 1)*(exp(a*2i + b*x*2i)*1i - 1i)) - (exp(a*1i + b*x*1i)*(49i
/(180*b) + (exp(a*2i + b*x*2i)*19i)/(180*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)