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

Optimal. Leaf size=12 \[ \frac {2 \log (\sin (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 12, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4288, 3475} \[ \frac {2 \log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Log[Sin[a + b*x]])/b

Rule 3475

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

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 ^2(a+b x) \sin (2 a+2 b x) \, dx &=2 \int \cot (a+b x) \, dx\\ &=\frac {2 \log (\sin (a+b x))}{b}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 1.67 \[ \frac {2 (\log (\tan (a+b x))+\log (\cos (a+b x)))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 14, normalized size = 1.17 \[ \frac {2 \, \log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(1/2*sin(b*x + a))/b

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giac [B]  time = 0.99, size = 55, normalized size = 4.58 \[ \frac {\log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) - 2 \, \log \left ({\left | -\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.28, size = 13, normalized size = 1.08 \[ \frac {2 \ln \left (\sin \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.35, size = 81, normalized size = 6.75 \[ \frac {\log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) + \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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 = 0.13, size = 13, normalized size = 1.08 \[ \frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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