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

Optimal. Leaf size=34 \[ -\frac{\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac{\cot (a+b x) \csc (a+b x)}{16 b} \]

[Out]

-ArcTanh[Cos[a + b*x]]/(16*b) - (Cot[a + b*x]*Csc[a + b*x])/(16*b)

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Rubi [A]  time = 0.040758, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4287, 3768, 3770} \[ -\frac{\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac{\cot (a+b x) \csc (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[a + b*x]]/(16*b) - (Cot[a + b*x]*Csc[a + b*x])/(16*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx &=\frac{1}{8} \int \csc ^3(a+b x) \, dx\\ &=-\frac{\cot (a+b x) \csc (a+b x)}{16 b}+\frac{1}{16} \int \csc (a+b x) \, dx\\ &=-\frac{\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac{\cot (a+b x) \csc (a+b x)}{16 b}\\ \end{align*}

Mathematica [B]  time = 0.0151075, size = 79, normalized size = 2.32 \[ \frac{1}{8} \left (-\frac{\csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-Csc[(a + b*x)/2]^2/(8*b) - Log[Cos[(a + b*x)/2]]/(2*b) + Log[Sin[(a + b*x)/2]]/(2*b) + Sec[(a + b*x)/2]^2/(8
*b))/8

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Maple [A]  time = 0.048, size = 40, normalized size = 1.2 \begin{align*} -{\frac{\cot \left ( bx+a \right ) \csc \left ( bx+a \right ) }{16\,b}}+{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{16\,b}} \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)^3,x)

[Out]

-1/16*cot(b*x+a)*csc(b*x+a)/b+1/16/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [B]  time = 1.20577, size = 753, normalized size = 22.15 \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)^3/sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

1/32*(4*(cos(3*b*x + 3*a) + cos(b*x + a))*cos(4*b*x + 4*a) - 4*(2*cos(2*b*x + 2*a) - 1)*cos(3*b*x + 3*a) - 8*c
os(2*b*x + 2*a)*cos(b*x + a) + (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*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*cos(b*x)*cos(
a) + cos(a)^2 + sin(b*x)^2 + 2*sin(b*x)*sin(a) + sin(a)^2) + 4*(sin(3*b*x + 3*a) + sin(b*x + a))*sin(4*b*x + 4
*a) - 8*sin(3*b*x + 3*a)*sin(2*b*x + 2*a) - 8*sin(2*b*x + 2*a)*sin(b*x + a) + 4*cos(b*x + 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)

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Fricas [B]  time = 0.493301, size = 203, normalized size = 5.97 \begin{align*} -\frac{{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) -{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - 2 \, \cos \left (b x + a\right )}{32 \,{\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)^3/sin(2*b*x+2*a)^3,x, algorithm="fricas")

[Out]

-1/32*((cos(b*x + a)^2 - 1)*log(1/2*cos(b*x + a) + 1/2) - (cos(b*x + a)^2 - 1)*log(-1/2*cos(b*x + a) + 1/2) -
2*cos(b*x + a))/(b*cos(b*x + a)^2 - b)

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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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.31529, size = 124, normalized size = 3.65 \begin{align*} -\frac{\frac{{\left (\frac{2 \,{\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} - 2 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \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)^3,x, algorithm="giac")

[Out]

-1/64*((2*(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) - 2*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b