3.183 \(\int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=81 \[ \frac {4 \sin (a+b x)}{21 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-1/7*cos(b*x+a)^3/b/sin(2*b*x+2*a)^(7/2)-2/21*cos(b*x+a)/b/sin(2*b*x+2*a)^(3/2)+4/21*sin(b*x+a)/b/sin(2*b*x+2*
a)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4295, 4303, 4292} \[ \frac {4 \sin (a+b x)}{21 b \sqrt {\sin (2 a+2 b x)}}-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

-Cos[a + b*x]^3/(7*b*Sin[2*a + 2*b*x]^(7/2)) - (2*Cos[a + b*x])/(21*b*Sin[2*a + 2*b*x]^(3/2)) + (4*Sin[a + b*x
])/(21*b*Sqrt[Sin[2*a + 2*b*x]])

Rule 4292

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 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]

Rubi steps

\begin {align*} \int \frac {\cos ^3(a+b x)}{\sin ^{\frac {9}{2}}(2 a+2 b x)} \, dx &=-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}+\frac {2}{7} \int \frac {\cos (a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4}{21} \int \frac {\sin (a+b x)}{\sin ^{\frac {3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {\cos ^3(a+b x)}{7 b \sin ^{\frac {7}{2}}(2 a+2 b x)}-\frac {2 \cos (a+b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {4 \sin (a+b x)}{21 b \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 55, normalized size = 0.68 \[ \frac {\sqrt {\sin (2 (a+b x))} (-12 \cos (2 (a+b x))+4 \cos (4 (a+b x))+5) \csc ^4(a+b x) \sec (a+b x)}{336 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(9/2),x]

[Out]

((5 - 12*Cos[2*(a + b*x)] + 4*Cos[4*(a + b*x)])*Csc[a + b*x]^4*Sec[a + b*x]*Sqrt[Sin[2*(a + b*x)]])/(336*b)

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fricas [A]  time = 0.52, size = 104, normalized size = 1.28 \[ \frac {32 \, \cos \left (b x + a\right )^{5} - 64 \, \cos \left (b x + a\right )^{3} + \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 56 \, \cos \left (b x + a\right )^{2} + 21\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )}{336 \, {\left (b \cos \left (b x + a\right )^{5} - 2 \, b \cos \left (b x + a\right )^{3} + b \cos \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(9/2),x, algorithm="fricas")

[Out]

1/336*(32*cos(b*x + a)^5 - 64*cos(b*x + a)^3 + sqrt(2)*(32*cos(b*x + a)^4 - 56*cos(b*x + a)^2 + 21)*sqrt(cos(b
*x + a)*sin(b*x + a)) + 32*cos(b*x + a))/(b*cos(b*x + a)^5 - 2*b*cos(b*x + a)^3 + b*cos(b*x + a))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(9/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(9/2), x)

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{3}\left (b x +a \right )}{\sin \left (2 b x +2 a \right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(9/2),x)

[Out]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(9/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(9/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(9/2), x)

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mupad [B]  time = 3.66, size = 302, normalized size = 3.73 \[ -\frac {5\,{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{84\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^2}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}\,3{}\mathrm {i}}{14\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^3}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{7\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}^4}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left (\frac {5{}\mathrm {i}}{84\,b}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}}{21\,b}\right )\,\sqrt {\frac {{\mathrm {e}}^{-a\,2{}\mathrm {i}-b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}}}{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^(9/2),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)*3i)/(14*b*(exp(a*2i + b*x*
2i)*1i - 1i)^3) - (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))/(84*b
*(exp(a*2i + b*x*2i)*1i - 1i)^2) - (exp(a*1i + b*x*1i)*((exp(- a*2i - b*x*2i)*1i)/2 - (exp(a*2i + b*x*2i)*1i)/
2)^(1/2))/(7*b*(exp(a*2i + b*x*2i)*1i - 1i)^4) - (exp(a*1i + b*x*1i)*(5i/(84*b) - (exp(a*2i + b*x*2i)*4i)/(21*
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)*(exp(a*2i + b*x
*2i)*1i - 1i))

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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)**(9/2),x)

[Out]

Timed out

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