\(\int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx\) [158]

Optimal result

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

[Out]

-1/16*arctanh(cos(b*x+a))/b-1/16*cot(b*x+a)*csc(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 = {4372, 3853, 3855} \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {\text {arctanh}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc (a+b x)}{16 b} \]

[In]

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

[Out]

-1/16*ArcTanh[Cos[a + b*x]]/b - (Cot[a + b*x]*Csc[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 4372

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{8} \int \csc ^3(a+b x) \, dx \\ & = -\frac {\cot (a+b x) \csc (a+b x)}{16 b}+\frac {1}{16} \int \csc (a+b x) \, dx \\ & = -\frac {\text {arctanh}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc (a+b x)}{16 b} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(34)=68\).

Time = 0.01 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.32 \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {1}{8} \left (-\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}\right ) \]

[In]

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

[Out]

(-1/8*Csc[(a + b*x)/2]^2/b - Log[Cos[(a + b*x)/2]]/(2*b) + Log[Sin[(a + b*x)/2]]/(2*b) + Sec[(a + b*x)/2]^2/(8
*b))/8

Maple [A] (verified)

Time = 5.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15

method	result	size
default	\(\frac {-\frac {\csc \left (x b +a \right ) \cot \left (x b +a \right )}{2}+\frac {\ln \left (\csc \left (x b +a \right )-\cot \left (x b +a \right )\right )}{2}}{8 b}\)	\(39\)
risch	\(\frac {{\mathrm e}^{3 i \left (x b +a \right )}+{\mathrm e}^{i \left (x b +a \right )}}{8 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}+1\right )}{16 b}+\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-1\right )}{16 b}\)	\(73\)

[In]

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

[Out]

1/8/b*(-1/2*csc(b*x+a)*cot(b*x+a)+1/2*ln(csc(b*x+a)-cot(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (30) = 60\).

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.12 \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=-\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, \cos \left (b x + a\right )}{32 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 558 vs. \(2 (30) = 60\).

Time = 0.21 (sec) , antiderivative size = 558, normalized size of antiderivative = 16.41 \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {4 \, {\left (\cos \left (3 \, b x + 3 \, a\right ) + \cos \left (b x + a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) - 4 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \cos \left (3 \, b x + 3 \, a\right ) - 8 \, \cos \left (2 \, b x + 2 \, a\right ) \cos \left (b x + 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 (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\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 (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right ) + 4 \, {\left (\sin \left (3 \, b x + 3 \, a\right ) + \sin \left (b x + a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) - 8 \, \sin \left (3 \, b x + 3 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) - 8 \, \sin \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 4 \, \cos \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(cos(b*x+a)^3/sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.41 \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {\frac {2 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} - \log \left (\cos \left (b x + a\right ) + 1\right ) + \log \left (-\cos \left (b x + a\right ) + 1\right )}{32 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.82 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx=\frac {\cos \left (a+b\,x\right )}{16\,b\,\left ({\cos \left (a+b\,x\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{16\,b} \]

[In]

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

[Out]

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