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

Optimal. Leaf size=107 \[ \frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \]

[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]  time = 0.09, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4295, 4303, 4304, 4291} \[ \frac {4 \sin (a+b x)}{45 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\cos ^3(a+b x)}{9 b \sin ^{\frac {9}{2}}(2 a+2 b x)}-\frac {\cos (a+b x)}{15 b \sin ^{\frac {5}{2}}(2 a+2 b x)}-\frac {8 \cos (a+b x)}{45 b \sqrt {\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cos[a + b*x]^3/(9*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 4291

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

Rule 4295

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 4303

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 4304

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] && Inte
gerQ[2*p]

Rubi steps

\begin {align*} \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 {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]  time = 0.09, size = 62, normalized size = 0.58 \[ -\frac {\sqrt {\sin (2 (a+b x))} \left (5 \csc ^5(a+b x)+17 \csc ^3(a+b x)+113 \csc (a+b x)-15 \tan (a+b x) \sec (a+b x)\right )}{1440 b} \]

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 131, normalized size = 1.22 \[ -\frac {\sqrt {2} {\left (128 \, \cos \left (b x + a\right )^{6} - 288 \, \cos \left (b x + a\right )^{4} + 180 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 128 \, {\left (\cos \left (b x + a\right )^{6} - 2 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )}{1440 \, {\left (b \cos \left (b x + a\right )^{6} - 2 \, b \cos \left (b x + a\right )^{4} + b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{3}\left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {11}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 4.88, size = 383, normalized size = 3.58 \[ -\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} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________