\(\int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx\) [10]

Optimal result

Integrand size = 18, antiderivative size = 49 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{16 b}-\frac {3 \csc (a+b x)}{16 b}+\frac {\csc (a+b x) \sec ^2(a+b x)}{16 b} \]

[Out]

3/16*arctanh(sin(b*x+a))/b-3/16*csc(b*x+a)/b+1/16*csc(b*x+a)*sec(b*x+a)^2/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {4373, 2701, 294, 327, 213} \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=\frac {3 \text {arctanh}(\sin (a+b x))}{16 b}-\frac {3 \csc (a+b x)}{16 b}+\frac {\csc (a+b x) \sec ^2(a+b x)}{16 b} \]

[In]

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

[Out]

(3*ArcTanh[Sin[a + b*x]])/(16*b) - (3*Csc[a + b*x])/(16*b) + (Csc[a + b*x]*Sec[a + b*x]^2)/(16*b)

Rule 213

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

Rule 294

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*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

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}{8} \int \csc ^2(a+b x) \sec ^3(a+b x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {x^4}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{8 b} \\ & = \frac {\csc (a+b x) \sec ^2(a+b x)}{16 b}-\frac {3 \text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b} \\ & = -\frac {3 \csc (a+b x)}{16 b}+\frac {\csc (a+b x) \sec ^2(a+b x)}{16 b}-\frac {3 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{16 b} \\ & = \frac {3 \text {arctanh}(\sin (a+b x))}{16 b}-\frac {3 \csc (a+b x)}{16 b}+\frac {\csc (a+b x) \sec ^2(a+b x)}{16 b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.59 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=-\frac {\csc (a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},2,\frac {1}{2},\sin ^2(a+b x)\right )}{8 b} \]

[In]

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

[Out]

-1/8*(Csc[a + b*x]*Hypergeometric2F1[-1/2, 2, 1/2, Sin[a + b*x]^2])/b

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {\frac {1}{2 \sin \left (x b +a \right ) \cos \left (x b +a \right )^{2}}-\frac {3}{2 \sin \left (x b +a \right )}+\frac {3 \ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{2}}{8 b}\)	\(51\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{5 i \left (x b +a \right )}+2 \,{\mathrm e}^{3 i \left (x b +a \right )}+3 \,{\mathrm e}^{i \left (x b +a \right )}\right )}{8 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2} \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}+\frac {3 \ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{16 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{16 b}\)	\(104\)

[In]

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

[Out]

1/8/b*(1/2/sin(b*x+a)/cos(b*x+a)^2-3/2/sin(b*x+a)+3/2*ln(sec(b*x+a)+tan(b*x+a)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.73 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=\frac {3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 6 \, \cos \left (b x + a\right )^{2} + 2}{32 \, b \cos \left (b x + a\right )^{2} \sin \left (b x + a\right )} \]

[In]

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

[Out]

1/32*(3*cos(b*x + a)^2*log(sin(b*x + a) + 1)*sin(b*x + a) - 3*cos(b*x + a)^2*log(-sin(b*x + a) + 1)*sin(b*x +
a) - 6*cos(b*x + a)^2 + 2)/(b*cos(b*x + a)^2*sin(b*x + a))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 808 vs. \(2 (43) = 86\).

Time = 0.36 (sec) , antiderivative size = 808, normalized size of antiderivative = 16.49 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/32*(4*(3*sin(5*b*x + 5*a) + 2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(6*b*x + 6*a) - 12*(sin(4*b*x + 4*a) - s
in(2*b*x + 2*a))*cos(5*b*x + 5*a) + 4*(2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*cos(4*b*x + 4*a) - 3*(2*(cos(4*b*x
+ 4*a) - cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) + cos(6*b*x + 6*a)^2 - 2*(cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4
*a) + cos(4*b*x + 4*a)^2 + cos(2*b*x + 2*a)^2 + 2*(sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*sin(6*b*x + 6*a) + sin
(6*b*x + 6*a)^2 + sin(4*b*x + 4*a)^2 - 2*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x
+ 2*a) + 1)*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)*s
in(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)) - 4*(3*cos(5*b*x + 5*a) + 2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(6*b*x + 6*a) + 12*(cos
(4*b*x + 4*a) - cos(2*b*x + 2*a) - 1)*sin(5*b*x + 5*a) - 4*(2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*sin(4*b*x + 4
*a) - 8*(cos(2*b*x + 2*a) + 1)*sin(3*b*x + 3*a) + 8*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) + 12*cos(b*x + a)*sin(2*
b*x + 2*a) - 12*cos(2*b*x + 2*a)*sin(b*x + a) - 12*sin(b*x + a))/(b*cos(6*b*x + 6*a)^2 + b*cos(4*b*x + 4*a)^2
+ b*cos(2*b*x + 2*a)^2 + b*sin(6*b*x + 6*a)^2 + b*sin(4*b*x + 4*a)^2 - 2*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) +
b*sin(2*b*x + 2*a)^2 + 2*(b*cos(4*b*x + 4*a) - b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) - 2*(b*cos(2*b*x + 2*
a) + b)*cos(4*b*x + 4*a) + 2*b*cos(2*b*x + 2*a) + 2*(b*sin(4*b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(6*b*x + 6*a)
+ b)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.29 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, \sin \left (b x + a\right )^{2} - 2\right )}}{\sin \left (b x + a\right )^{3} - \sin \left (b x + a\right )} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{32 \, b} \]

[In]

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

[Out]

-1/32*(2*(3*sin(b*x + a)^2 - 2)/(sin(b*x + a)^3 - sin(b*x + a)) - 3*log(sin(b*x + a) + 1) + 3*log(-sin(b*x + a
) + 1))/b

Mupad [B] (verification not implemented)

Time = 19.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.98 \[ \int \csc ^3(2 a+2 b x) \sin (a+b x) \, dx=\frac {3\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}+\frac {\frac {3\,{\sin \left (a+b\,x\right )}^2}{16}-\frac {1}{8}}{b\,\left (\sin \left (a+b\,x\right )-{\sin \left (a+b\,x\right )}^3\right )} \]

[In]

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

[Out]

(3*atanh(sin(a + b*x)))/(16*b) + ((3*sin(a + b*x)^2)/16 - 1/8)/(b*(sin(a + b*x) - sin(a + b*x)^3))