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

Optimal. Leaf size=49 \[ \frac{3 \sec (a+b x)}{16 b}-\frac{3 \tanh ^{-1}(\cos (a+b x))}{16 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{16 b} \]

[Out]

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

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Rubi [A]  time = 0.0549493, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4287, 2622, 288, 321, 207} \[ \frac{3 \sec (a+b x)}{16 b}-\frac{3 \tanh ^{-1}(\cos (a+b x))}{16 b}-\frac{\csc ^2(a+b x) \sec (a+b x)}{16 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2622

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

Rule 288

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 321

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 207

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

Rubi steps

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

Mathematica [B]  time = 0.254762, size = 143, normalized size = 2.92 \[ \frac{\csc ^4(a+b x) \left (-6 \cos (2 (a+b x))+2 \cos (3 (a+b x))+3 \cos (3 (a+b x)) \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-3 \cos (3 (a+b x)) \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )+\cos (a+b x) \left (3 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )-3 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )-2\right )+2\right )}{16 b \left (\csc ^2\left (\frac{1}{2} (a+b x)\right )-\sec ^2\left (\frac{1}{2} (a+b x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[a + b*x]^4*(2 - 6*Cos[2*(a + b*x)] + 2*Cos[3*(a + b*x)] + 3*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 3*Co
s[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + Cos[a + b*x]*(-2 - 3*Log[Cos[(a + b*x)/2]] + 3*Log[Sin[(a + b*x)/2]])))
/(16*b*(Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2))

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Maple [A]  time = 0.027, size = 57, normalized size = 1.2 \begin{align*} -{\frac{1}{16\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}\cos \left ( bx+a \right ) }}+{\frac{3}{16\,b\cos \left ( bx+a \right ) }}+{\frac{3\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{16\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/b/sin(b*x+a)^2/cos(b*x+a)+3/16/b/cos(b*x+a)+3/16/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [B]  time = 1.2554, size = 1315, normalized size = 26.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(4*(3*cos(5*b*x + 5*a) - 2*cos(3*b*x + 3*a) + 3*cos(b*x + a))*cos(6*b*x + 6*a) - 12*(cos(4*b*x + 4*a) + c
os(2*b*x + 2*a) - 1)*cos(5*b*x + 5*a) + 4*(2*cos(3*b*x + 3*a) - 3*cos(b*x + a))*cos(4*b*x + 4*a) + 8*(cos(2*b*
x + 2*a) - 1)*cos(3*b*x + 3*a) - 12*cos(2*b*x + 2*a)*cos(b*x + 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
+ 2*cos(b*x)*cos(a) + cos(a)^2 + sin(b*x)^2 - 2*sin(b*x)*sin(a) + sin(a)^2) - 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(c
os(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*(3*sin(5*b*x + 5*a)
- 2*sin(3*b*x + 3*a) + 3*sin(b*x + a))*sin(6*b*x + 6*a) - 12*(sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*sin(5*b*x +
5*a) + 4*(2*sin(3*b*x + 3*a) - 3*sin(b*x + a))*sin(4*b*x + 4*a) + 8*sin(3*b*x + 3*a)*sin(2*b*x + 2*a) - 12*si
n(2*b*x + 2*a)*sin(b*x + a) + 12*cos(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)

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Fricas [B]  time = 0.498507, size = 262, normalized size = 5.35 \begin{align*} \frac{6 \, \cos \left (b x + a\right )^{2} - 3 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) + 3 \,{\left (\cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - 4}{32 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(6*cos(b*x + a)^2 - 3*(cos(b*x + a)^3 - cos(b*x + a))*log(1/2*cos(b*x + a) + 1/2) + 3*(cos(b*x + a)^3 - c
os(b*x + a))*log(-1/2*cos(b*x + a) + 1/2) - 4)/(b*cos(b*x + a)^3 - b*cos(b*x + a))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.27516, size = 189, normalized size = 3.86 \begin{align*} \frac{\frac{\frac{14 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}} - \frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 6 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*((14*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)/((cos(b*x +
a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2) - (cos(b*x + a) - 1)/(cos(b*x + a) +
1) + 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b