3.182 \(\int \frac{\cos ^3(a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=55 \[ -\frac{\cos ^3(a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{5 b \sqrt{\sin (2 a+2 b x)}} \]

[Out]

-Cos[a + b*x]^3/(5*b*Sin[2*a + 2*b*x]^(5/2)) - Cos[a + b*x]/(5*b*Sqrt[Sin[2*a + 2*b*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0471591, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {4295, 4291} \[ -\frac{\cos ^3(a+b x)}{5 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{\cos (a+b x)}{5 b \sqrt{\sin (2 a+2 b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-Cos[a + b*x]^3/(5*b*Sin[2*a + 2*b*x]^(5/2)) - Cos[a + b*x]/(5*b*Sqrt[Sin[2*a + 2*b*x]])

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0934569, size = 35, normalized size = 0.64 \[ -\frac{\sqrt{\sin (2 (a+b x))} \csc (a+b x) \left (\csc ^2(a+b x)+4\right )}{40 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-(Csc[a + b*x]*(4 + Csc[a + b*x]^2)*Sqrt[Sin[2*(a + b*x)]])/(40*b)

________________________________________________________________________________________

Maple [C]  time = 213.665, size = 482, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x)

[Out]

1/160*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)/tan(1/2*b*x+1/2*a)^3*(16*(tan(1/2*b*x+1/2*a)*(tan(1
/2*b*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(
1/2)*EllipticE((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-8*(tan(1/2*b*x+1/2*a)*(tan(1/2*b
*x+1/2*a)^2-1))^(1/2)*(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)
*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)^2-(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2
*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^6+(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^4+
8*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)*tan(1/2*b*x+1/2*a)^4+(tan(1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)
^2-1))^(1/2)*tan(1/2*b*x+1/2*a)^2-8*(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)*tan(1/2*b*x+1/2*a)^2-(tan(
1/2*b*x+1/2*a)*(tan(1/2*b*x+1/2*a)^2-1))^(1/2))/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)/b

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(7/2), x)

________________________________________________________________________________________

Fricas [A]  time = 0.50141, size = 200, normalized size = 3.64 \begin{align*} -\frac{\sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{2} - 5\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 4 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sin \left (b x + a\right )}{40 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")

[Out]

-1/40*(sqrt(2)*(4*cos(b*x + a)^2 - 5)*sqrt(cos(b*x + a)*sin(b*x + a)) + 4*(cos(b*x + a)^2 - 1)*sin(b*x + a))/(
(b*cos(b*x + a)^2 - b)*sin(b*x + a))

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3/sin(2*b*x+2*a)**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )^{3}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^(7/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^3/sin(2*b*x + 2*a)^(7/2), x)