3.158 \(\int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx\)

Optimal. Leaf size=34 \[ -\frac {\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc (a+b x)}{16 b} \]

[Out]

-1/16*arctanh(cos(b*x+a))/b-1/16*cot(b*x+a)*csc(b*x+a)/b

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Rubi [A]  time = 0.04, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4287, 3768, 3770} \[ -\frac {\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3768

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 3770

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

Rule 4287

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*} \int \cos ^3(a+b x) \csc ^3(2 a+2 b x) \, dx &=\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 {\tanh ^{-1}(\cos (a+b x))}{16 b}-\frac {\cot (a+b x) \csc (a+b x)}{16 b}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 79, normalized size = 2.32 \[ \frac {1}{8} \left (-\frac {\csc ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {\sec ^2\left (\frac {1}{2} (a+b x)\right )}{8 b}+\frac {\log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}-\frac {\log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{2 b}\right ) \]

Antiderivative was successfully verified.

[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

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fricas [B]  time = 0.48, size = 72, normalized size = 2.12 \[ -\frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - {\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - 2 \, \cos \left (b x + a\right )}{32 \, {\left (b \cos \left (b x + a\right )^{2} - b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 0.66, size = 92, normalized size = 2.71 \[ -\frac {\frac {{\left (\frac {2 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}}{\cos \left (b x + a\right ) - 1} + \frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 2 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*((2*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) + (cos(b*x + a) - 1
)/(cos(b*x + a) + 1) - 2*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b

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maple [A]  time = 0.86, size = 40, normalized size = 1.18 \[ -\frac {\cot \left (b x +a \right ) \csc \left (b x +a \right )}{16 b}+\frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*cot(b*x+a)*csc(b*x+a)/b+1/16/b*ln(csc(b*x+a)-cot(b*x+a))

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maxima [B]  time = 0.35, size = 558, normalized size = 16.41 \[ \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 \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{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 \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 36, normalized size = 1.06 \[ \frac {\cos \left (a+b\,x\right )}{16\,b\,\left ({\cos \left (a+b\,x\right )}^2-1\right )}-\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{16\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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