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

Optimal. Leaf size=28 \[ \frac{\tanh ^{-1}(\sin (a+b x))}{4 b}-\frac{\csc (a+b x)}{4 b} \]

[Out]

ArcTanh[Sin[a + b*x]]/(4*b) - Csc[a + b*x]/(4*b)

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Rubi [A]  time = 0.0377685, 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, 2621, 321, 207} \[ \frac{\tanh ^{-1}(\sin (a+b x))}{4 b}-\frac{\csc (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(4*b) - Csc[a + b*x]/(4*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 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

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 207

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

Rubi steps

\begin{align*} \int \cos (a+b x) \csc ^2(2 a+2 b x) \, dx &=\frac{1}{4} \int \csc ^2(a+b x) \sec (a+b x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=-\frac{\csc (a+b x)}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{4 b}\\ &=\frac{\tanh ^{-1}(\sin (a+b x))}{4 b}-\frac{\csc (a+b x)}{4 b}\\ \end{align*}

Mathematica [C]  time = 0.0177062, size = 29, normalized size = 1.04 \[ -\frac{\csc (a+b x) \text{Hypergeometric2F1}\left (-\frac{1}{2},1,\frac{1}{2},\sin ^2(a+b x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[a + b*x]^2])/(4*b)

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Maple [A]  time = 0.023, size = 34, normalized size = 1.2 \begin{align*} -{\frac{1}{4\,b\sin \left ( bx+a \right ) }}+{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b/sin(b*x+a)+1/4/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [B]  time = 1.81805, size = 315, normalized size = 11.25 \begin{align*} -\frac{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \log \left (\frac{\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} - 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} + 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}{\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} + 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} - 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 4 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 4 \, \sin \left (b x + a\right )}{8 \,{\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.498404, size = 136, normalized size = 4.86 \begin{align*} \frac{\log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 2}{8 \, b \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(log(sin(b*x + a) + 1)*sin(b*x + a) - log(-sin(b*x + a) + 1)*sin(b*x + a) - 2)/(b*sin(b*x + a))

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

[Out]

Timed out

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Giac [A]  time = 1.23463, size = 51, normalized size = 1.82 \begin{align*} -\frac{\frac{2}{\sin \left (b x + a\right )} - \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2/sin(b*x + a) - log(abs(sin(b*x + a) + 1)) + log(abs(sin(b*x + a) - 1)))/b