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

Optimal. Leaf size=28 \[ \frac{\cos (a+b x)}{2 b}-\frac{\tanh ^{-1}(\cos (a+b x))}{2 b} \]

[Out]

-ArcTanh[Cos[a + b*x]]/(2*b) + Cos[a + b*x]/(2*b)

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Rubi [A]  time = 0.0352463, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4287, 2592, 321, 206} \[ \frac{\cos (a+b x)}{2 b}-\frac{\tanh ^{-1}(\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTanh[Cos[a + b*x]]/(2*b) + Cos[a + b*x]/(2*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 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0192191, size = 46, normalized size = 1.64 \[ \frac{1}{2} \left (\frac{\cos (a+b x)}{b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.022, size = 34, normalized size = 1.2 \begin{align*}{\frac{\cos \left ( bx+a \right ) }{2\,b}}+{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{2\,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),x)

[Out]

1/2*cos(b*x+a)/b+1/2/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [B]  time = 1.12094, size = 124, normalized size = 4.43 \begin{align*} \frac{2 \, \cos \left (b x + a\right ) - \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right ) + \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right )}{4 \, 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),x, algorithm="maxima")

[Out]

1/4*(2*cos(b*x + a) - 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) + 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))/b

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Fricas [A]  time = 0.500098, size = 115, normalized size = 4.11 \begin{align*} \frac{2 \, \cos \left (b x + a\right ) - \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right )}{4 \, 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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.24362, size = 77, normalized size = 2.75 \begin{align*} -\frac{\frac{4}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{4 \, 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),x, algorithm="giac")

[Out]

-1/4*(4/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1) - log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b