3.67 \(\int \csc ^3(a+b x) \sin ^2(2 a+2 b x) \, dx\)

Optimal. Leaf size=24 \[ \frac{4 \cos (a+b x)}{b}-\frac{4 \tanh ^{-1}(\cos (a+b x))}{b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0430228, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4288, 2592, 321, 206} \[ \frac{4 \cos (a+b x)}{b}-\frac{4 \tanh ^{-1}(\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4288

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, 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 \csc ^3(a+b x) \sin ^2(2 a+2 b x) \, dx &=4 \int \cos (a+b x) \cot (a+b x) \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=\frac{4 \cos (a+b x)}{b}-\frac{4 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{4 \tanh ^{-1}(\cos (a+b x))}{b}+\frac{4 \cos (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0256678, size = 44, normalized size = 1.83 \[ 4 \left (\frac{\cos (a+b x)}{b}+\frac{\log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 34, normalized size = 1.4 \begin{align*} 4\,{\frac{\cos \left ( bx+a \right ) }{b}}+4\,{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*cos(b*x+a)/b+4/b*ln(csc(b*x+a)-cot(b*x+a))

________________________________________________________________________________________

Maxima [B]  time = 1.05241, size = 124, normalized size = 5.17 \begin{align*} \frac{2 \,{\left (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 )\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(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

________________________________________________________________________________________

Fricas [A]  time = 0.503847, size = 112, normalized size = 4.67 \begin{align*} \frac{2 \,{\left (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 )\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(2*cos(b*x + a) - log(1/2*cos(b*x + a) + 1/2) + log(-1/2*cos(b*x + a) + 1/2))/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(csc(b*x+a)**3*sin(2*b*x+2*a)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.24985, size = 77, normalized size = 3.21 \begin{align*} -\frac{2 \,{\left (\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 )\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(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