\(\int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx\) [30]

Optimal result

Integrand size = 20, antiderivative size = 34 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{16 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b} \]

[Out]

1/16*arctanh(sin(b*x+a))/b+1/16*sec(b*x+a)*tan(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4373, 3853, 3855} \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{16 b}+\frac {\tan (a+b x) \sec (a+b x)}{16 b} \]

[In]

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

[Out]

ArcTanh[Sin[a + b*x]]/(16*b) + (Sec[a + b*x]*Tan[a + b*x])/(16*b)

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

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

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 \sec ^3(a+b x) \, dx \\ & = \frac {\sec (a+b x) \tan (a+b x)}{16 b}+\frac {1}{16} \int \sec (a+b x) \, dx \\ & = \frac {\text {arctanh}(\sin (a+b x))}{16 b}+\frac {\sec (a+b x) \tan (a+b x)}{16 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {1}{8} \left (\frac {\text {arctanh}(\sin (a+b x))}{2 b}+\frac {\sec (a+b x) \tan (a+b x)}{2 b}\right ) \]

[In]

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

[Out]

(ArcTanh[Sin[a + b*x]]/(2*b) + (Sec[a + b*x]*Tan[a + b*x])/(2*b))/8

Maple [A] (verified)

Time = 2.89 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09

method	result	size
default	\(\frac {\frac {\sec \left (x b +a \right ) \tan \left (x b +a \right )}{2}+\frac {\ln \left (\sec \left (x b +a \right )+\tan \left (x b +a \right )\right )}{2}}{8 b}\)	\(37\)
risch	\(-\frac {i \left ({\mathrm e}^{3 i \left (x b +a \right )}-{\mathrm e}^{i \left (x b +a \right )}\right )}{8 b \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )^{2}}+\frac {\ln \left (i+{\mathrm e}^{i \left (x b +a \right )}\right )}{16 b}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-i\right )}{16 b}\)	\(78\)

[In]

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

[Out]

1/8/b*(1/2*sec(b*x+a)*tan(b*x+a)+1/2*ln(sec(b*x+a)+tan(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).

Time = 0.24 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) + 2 \, \sin \left (b x + a\right )}{32 \, b \cos \left (b x + a\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

integrate(csc(2*b*x+2*a)**3*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. 480 vs. \(2 (30) = 60\).

Time = 0.34 (sec) , antiderivative size = 480, normalized size of antiderivative = 14.12 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {4 \, {\left (\sin \left (3 \, b x + 3 \, a\right ) - \sin \left (b x + a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) - {\left (2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) + \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \sin \left (2 \, b x + 2 \, a\right )^{2} + 4 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \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 \, {\left (\cos \left (3 \, b x + 3 \, a\right ) - \cos \left (b x + a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right )} \sin \left (3 \, b x + 3 \, a\right ) - 8 \, \cos \left (3 \, b x + 3 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 8 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 8 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) - 4 \, \sin \left (b x + a\right )}{32 \, {\left (b \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, b \sin \left (4 \, b x + 4 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, {\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (4 \, b x + 4 \, a\right ) + 4 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

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

[Out]

1/32*(4*(sin(3*b*x + 3*a) - sin(b*x + a))*cos(4*b*x + 4*a) - (2*(2*cos(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) + co
s(4*b*x + 4*a)^2 + 4*cos(2*b*x + 2*a)^2 + sin(4*b*x + 4*a)^2 + 4*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*sin(2*b
*x + 2*a)^2 + 4*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)*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)) - 4*(cos(3*b*x + 3*a) - cos(b*x + a))*sin(4*b*x + 4*a) + 4*(2*c
os(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) + 8*cos(b*x + a)*sin(2*b*x + 2*a)
- 8*cos(2*b*x + 2*a)*sin(b*x + a) - 4*sin(b*x + a))/(b*cos(4*b*x + 4*a)^2 + 4*b*cos(2*b*x + 2*a)^2 + b*sin(4*b
*x + 4*a)^2 + 4*b*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + 4*b*sin(2*b*x + 2*a)^2 + 2*(2*b*cos(2*b*x + 2*a) + b)*co
s(4*b*x + 4*a) + 4*b*cos(2*b*x + 2*a) + b)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (-\sin \left (b x + a\right ) + 1\right )}{32 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \csc ^3(2 a+2 b x) \sin ^3(a+b x) \, dx=\frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{16\,b}-\frac {\sin \left (a+b\,x\right )}{16\,b\,\left ({\sin \left (a+b\,x\right )}^2-1\right )} \]

[In]

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

[Out]

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