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

Optimal. Leaf size=28 \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{\sin (a+b x)}{2 b} \]

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) - Sin[a + b*x]/(2*b)

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Rubi [A]  time = 0.0372884, 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 = {4288, 2592, 321, 206} \[ \frac{\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{\sin (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(2*b) - 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 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 (2 a+2 b x) \sin ^3(a+b x) \, dx &=\frac{1}{2} \int \sin (a+b x) \tan (a+b x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b}\\ &=-\frac{\sin (a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b}\\ &=\frac{\tanh ^{-1}(\sin (a+b x))}{2 b}-\frac{\sin (a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0120244, size = 27, normalized size = 0.96 \[ \frac{1}{2} \left (\frac{\tanh ^{-1}(\sin (a+b x))}{b}-\frac{\sin (a+b x)}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[a + b*x]]/b - Sin[a + b*x]/b)/2

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Maple [A]  time = 0.026, size = 32, normalized size = 1.1 \begin{align*} -{\frac{\sin \left ( bx+a \right ) }{2\,b}}+{\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)^3,x)

[Out]

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

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Maxima [B]  time = 1.75747, size = 167, normalized size = 5.96 \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 ) + 2 \, \sin \left (b x + a\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)^3,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)) + 2*sin(b*x + a))/b

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Fricas [A]  time = 0.502638, size = 99, normalized size = 3.54 \begin{align*} \frac{\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, \sin \left (b x + a\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)^3,x, algorithm="fricas")

[Out]

1/4*(log(sin(b*x + a) + 1) - log(-sin(b*x + a) + 1) - 2*sin(b*x + a))/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)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.61057, size = 834, normalized size = 29.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((tan(1/2*a)^15 + 3*tan(1/2*a)^14 + 3*tan(1/2*a)^13 + 17*tan(1/2*a)^12 - 3*tan(1/2*a)^11 + 39*tan(1/2*a)^1
0 - 25*tan(1/2*a)^9 + 45*tan(1/2*a)^8 - 45*tan(1/2*a)^7 + 25*tan(1/2*a)^6 - 39*tan(1/2*a)^5 + 3*tan(1/2*a)^4 -
17*tan(1/2*a)^3 - 3*tan(1/2*a)^2 - 3*tan(1/2*a) - 1)*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))/(tan(1/2*a)^3 + 3*tan(1/2*a)^2 - 3*tan(1/2*a) - 1) - (tan(1/2*a)^15 - 3*tan(1/2*a)^14 + 3*ta
n(1/2*a)^13 - 17*tan(1/2*a)^12 - 3*tan(1/2*a)^11 - 39*tan(1/2*a)^10 - 25*tan(1/2*a)^9 - 45*tan(1/2*a)^8 - 45*t
an(1/2*a)^7 - 25*tan(1/2*a)^6 - 39*tan(1/2*a)^5 - 3*tan(1/2*a)^4 - 17*tan(1/2*a)^3 + 3*tan(1/2*a)^2 - 3*tan(1/
2*a) + 1)*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))/(tan(1/2*a)^3 - 3*tan(1/2*
a)^2 - 3*tan(1/2*a) + 1) + 2*(tan(1/2*b*x + 2*a)*tan(1/2*a)^12 - 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^10 + 6*tan(1
/2*a)^11 - 27*tan(1/2*b*x + 2*a)*tan(1/2*a)^8 - 2*tan(1/2*a)^9 - 36*tan(1/2*a)^7 + 27*tan(1/2*b*x + 2*a)*tan(1
/2*a)^4 - 36*tan(1/2*a)^5 + 12*tan(1/2*b*x + 2*a)*tan(1/2*a)^2 - 2*tan(1/2*a)^3 - tan(1/2*b*x + 2*a) + 6*tan(1
/2*a))/(tan(1/2*b*x + 2*a)^2 + 1))/b