\(\int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx\) [28]

Optimal result

Integrand size = 18, antiderivative size = 28 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b} \]

[Out]

1/2*arctanh(sin(b*x+a))/b-1/2*sin(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4373, 2672, 327, 212} \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 327

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 2672

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 4373

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*} \text {integral}& = \frac {1}{2} \int \sin (a+b x) \tan (a+b x) \, dx \\ & = \frac {\text {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 {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = \frac {\text {arctanh}(\sin (a+b x))}{2 b}-\frac {\sin (a+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {1}{2} \left (\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}\right ) \]

[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

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {-\sin \left (x b +a \right )+\ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{2 b}\)	\(29\)
risch	\(\frac {i {\mathrm e}^{i \left (x b +a \right )}}{4 b}-\frac {i {\mathrm e}^{-i \left (x b +a \right )}}{4 b}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{2 b}+\frac {\ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{2 b}\)	\(68\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, \sin \left (b x + a\right )}{4 \, b} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 124 vs. \(2 (24) = 48\).

Time = 0.32 (sec) , antiderivative size = 124, normalized size of antiderivative = 4.43 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\log \left (\frac {\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} - 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} + 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}{\cos \left (b x + 2 \, a\right )^{2} + \cos \left (a\right )^{2} + 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, a\right )^{2} - 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}}\right ) + 2 \, \sin \left (b x + a\right )}{4 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\log \left (\sin \left (b x + a\right ) + 1\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, \sin \left (b x + a\right )}{4 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \csc (2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\frac {\sin \left (a+b\,x\right )}{2}-\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2}}{b} \]

[In]

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

[Out]

-(sin(a + b*x)/2 - atanh(sin(a + b*x))/2)/b