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(b*x+a))/b+4*cos(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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)

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fricas [A]  time = 0.62, size = 38, normalized size = 1.58 \[ \frac {2 \, {\left (2 \, \cos \left (b x + a\right ) - \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right )\right )}}{b} \]

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

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giac [B]  time = 0.57, size = 57, normalized size = 2.38 \[ -\frac {2 \, {\left (\frac {4}{\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )\right )}}{b} \]

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

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maple [A]  time = 0.57, size = 34, normalized size = 1.42 \[ \frac {4 \cos \left (b x +a \right )}{b}+\frac {4 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b} \]

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))

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maxima [B]  time = 0.99, size = 92, normalized size = 3.83 \[ \frac {2 \, {\left (2 \, \cos \left (b x + a\right ) - \log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) + \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right )\right )}}{b} \]

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

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mupad [B]  time = 0.11, size = 22, normalized size = 0.92 \[ \frac {4\,\cos \left (a+b\,x\right )-4\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*cos(a + b*x) - 4*atanh(cos(a + b*x)))/b

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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