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

Optimal. Leaf size=14 \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

[Out]

ArcTanh[Sin[a + b*x]]/(2*b)

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Rubi [A]  time = 0.0169135, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4288, 3770} \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[2*a + 2*b*x]*Sin[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/(2*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 3770

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

Rubi steps

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

Mathematica [A]  time = 0.0045653, size = 14, normalized size = 1. \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(2*b)

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Maple [A]  time = 0.036, size = 20, normalized size = 1.4 \begin{align*}{\frac{\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*ln(sec(b*x+a)+tan(b*x+a))

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Maxima [B]  time = 1.78992, size = 155, normalized size = 11.07 \begin{align*} -\frac{\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 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*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))/b

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Fricas [B]  time = 0.484673, size = 76, normalized size = 5.43 \begin{align*} \frac{\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.51967, size = 235, normalized size = 16.79 \begin{align*} \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{3} + 3 \, \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{2} - \tan \left (\frac{1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right ) + 3 \, \tan \left (\frac{1}{2} \, a\right )^{2} - \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) + 3 \, \tan \left (\frac{1}{2} \, a\right ) - 1 \right |}\right ) - \log \left ({\left | \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{2} + \tan \left (\frac{1}{2} \, a\right )^{3} - 3 \, \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) \tan \left (\frac{1}{2} \, a\right ) + 3 \, \tan \left (\frac{1}{2} \, a\right )^{2} + \tan \left (\frac{1}{2} \, b x + 2 \, a\right ) - 3 \, \tan \left (\frac{1}{2} \, a\right ) - 1 \right |}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(log(abs(tan(1/2*b*x + 2*a)*tan(1/2*a)^3 + 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 3*tan(1/2*b*
x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)^2 - tan(1/2*b*x + 2*a) + 3*tan(1/2*a) - 1)) - log(abs(tan(1/2*b*x + 2*a)*ta
n(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 + tan(1/2*a)^3 - 3*tan(1/2*b*x + 2*a)*tan(1/2*a) + 3*tan(1/2*a)
^2 + tan(1/2*b*x + 2*a) - 3*tan(1/2*a) - 1)))/b