3.148 \(\int \cos ^2(a+b x) \csc ^3(2 a+2 b x) \, dx\)

Optimal. Leaf size=30 \[ \frac{\log (\tan (a+b x))}{8 b}-\frac{\cot ^2(a+b x)}{16 b} \]

[Out]

-Cot[a + b*x]^2/(16*b) + Log[Tan[a + b*x]]/(8*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0467969, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4287, 2620, 14} \[ \frac{\log (\tan (a+b x))}{8 b}-\frac{\cot ^2(a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Csc[2*a + 2*b*x]^3,x]

[Out]

-Cot[a + b*x]^2/(16*b) + Log[Tan[a + b*x]]/(8*b)

Rule 4287

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \csc ^3(2 a+2 b x) \, dx &=\frac{1}{8} \int \csc ^3(a+b x) \sec (a+b x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{1+x^2}{x^3} \, dx,x,\tan (a+b x)\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{x^3}+\frac{1}{x}\right ) \, dx,x,\tan (a+b x)\right )}{8 b}\\ &=-\frac{\cot ^2(a+b x)}{16 b}+\frac{\log (\tan (a+b x))}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.0435241, size = 34, normalized size = 1.13 \[ -\frac{\csc ^2(a+b x)-2 \log (\sin (a+b x))+2 \log (\cos (a+b x))}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Csc[2*a + 2*b*x]^3,x]

[Out]

-(Csc[a + b*x]^2 + 2*Log[Cos[a + b*x]] - 2*Log[Sin[a + b*x]])/(16*b)

________________________________________________________________________________________

Maple [A]  time = 0.023, size = 27, normalized size = 0.9 \begin{align*} -{\frac{1}{16\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/b/sin(b*x+a)^2+1/8*ln(tan(b*x+a))/b

________________________________________________________________________________________

Maxima [B]  time = 1.20243, size = 886, normalized size = 29.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) - 8*cos(2*b*x + 2*a)^2 + (2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a
) - cos(4*b*x + 4*a)^2 - 4*cos(2*b*x + 2*a)^2 - sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 4*s
in(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) - 1)*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)
^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2) - (2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 -
4*cos(2*b*x + 2*a)^2 - sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 4*sin(2*b*x + 2*a)^2 + 4*co
s(2*b*x + 2*a) - 1)*log(cos(b*x)^2 + 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2)
- (2*(2*cos(2*b*x + 2*a) - 1)*cos(4*b*x + 4*a) - cos(4*b*x + 4*a)^2 - 4*cos(2*b*x + 2*a)^2 - sin(4*b*x + 4*a)
^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) - 4*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a) - 1)*log(cos(b*x)^2 - 2*c
os(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2) + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) -
8*sin(2*b*x + 2*a)^2 + 4*cos(2*b*x + 2*a))/(b*cos(4*b*x + 4*a)^2 + 4*b*cos(2*b*x + 2*a)^2 + b*sin(4*b*x + 4*a
)^2 - 4*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b*sin(2*b*x + 2*a)^2 - 2*(2*b*cos(2*b*x + 2*a) - b)*cos(4*b*x
+ 4*a) - 4*b*cos(2*b*x + 2*a) + b)

________________________________________________________________________________________

Fricas [B]  time = 0.499714, size = 177, normalized size = 5.9 \begin{align*} -\frac{{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\cos \left (b x + a\right )^{2}\right ) -{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac{1}{4} \, \cos \left (b x + a\right )^{2} + \frac{1}{4}\right ) - 1}{16 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((cos(b*x + a)^2 - 1)*log(cos(b*x + a)^2) - (cos(b*x + a)^2 - 1)*log(-1/4*cos(b*x + a)^2 + 1/4) - 1)/(b*
cos(b*x + a)^2 - b)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.28487, size = 161, normalized size = 5.37 \begin{align*} -\frac{\frac{{\left (\frac{4 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 4 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 8 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1 \right |}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*((4*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - (cos(b*x + a) - 1
)/(cos(b*x + a) + 1) - 4*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 8*log(abs(-(cos(b*x + a) - 1)/(co
s(b*x + a) + 1) - 1)))/b