3.139 \(\int \cos (a+b x) \csc ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac{5 \csc ^3(a+b x)}{96 b}-\frac{5 \csc (a+b x)}{32 b}+\frac{5 \tanh ^{-1}(\sin (a+b x))}{32 b}+\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b} \]

[Out]

(5*ArcTanh[Sin[a + b*x]])/(32*b) - (5*Csc[a + b*x])/(32*b) - (5*Csc[a + b*x]^3)/(96*b) + (Csc[a + b*x]^3*Sec[a
+ b*x]^2)/(32*b)

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Rubi [A]  time = 0.0586615, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.278, Rules used = {4287, 2621, 288, 302, 207} \[ -\frac{5 \csc ^3(a+b x)}{96 b}-\frac{5 \csc (a+b x)}{32 b}+\frac{5 \tanh ^{-1}(\sin (a+b x))}{32 b}+\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Csc[2*a + 2*b*x]^4,x]

[Out]

(5*ArcTanh[Sin[a + b*x]])/(32*b) - (5*Csc[a + b*x])/(32*b) - (5*Csc[a + b*x]^3)/(96*b) + (Csc[a + b*x]^3*Sec[a
+ b*x]^2)/(32*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 2621

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 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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 ^4(2 a+2 b x) \, dx &=\frac{1}{16} \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (-1+x^2\right )^2} \, dx,x,\csc (a+b x)\right )}{16 b}\\ &=\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{32 b}\\ &=\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b}-\frac{5 \operatorname{Subst}\left (\int \left (1+x^2+\frac{1}{-1+x^2}\right ) \, dx,x,\csc (a+b x)\right )}{32 b}\\ &=-\frac{5 \csc (a+b x)}{32 b}-\frac{5 \csc ^3(a+b x)}{96 b}+\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{32 b}\\ &=\frac{5 \tanh ^{-1}(\sin (a+b x))}{32 b}-\frac{5 \csc (a+b x)}{32 b}-\frac{5 \csc ^3(a+b x)}{96 b}+\frac{\csc ^3(a+b x) \sec ^2(a+b x)}{32 b}\\ \end{align*}

Mathematica [C]  time = 0.0239051, size = 31, normalized size = 0.47 \[ -\frac{\csc ^3(a+b x) \text{Hypergeometric2F1}\left (-\frac{3}{2},2,-\frac{1}{2},\sin ^2(a+b x)\right )}{48 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Csc[2*a + 2*b*x]^4,x]

[Out]

-(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, Sin[a + b*x]^2])/(48*b)

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Maple [A]  time = 0.027, size = 76, normalized size = 1.2 \begin{align*} -{\frac{1}{48\,b \left ( \sin \left ( bx+a \right ) \right ) ^{3} \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}+{\frac{5}{96\,b\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}}}-{\frac{5}{32\,b\sin \left ( bx+a \right ) }}+{\frac{5\,\ln \left ( \sec \left ( bx+a \right ) +\tan \left ( bx+a \right ) \right ) }{32\,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)^4,x)

[Out]

-1/48/b/sin(b*x+a)^3/cos(b*x+a)^2+5/96/b/sin(b*x+a)/cos(b*x+a)^2-5/32/b/sin(b*x+a)+5/32/b*ln(sec(b*x+a)+tan(b*
x+a))

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Maxima [B]  time = 2.16879, size = 2403, normalized size = 36.41 \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)^4,x, algorithm="maxima")

[Out]

1/192*(4*(15*sin(9*b*x + 9*a) - 20*sin(7*b*x + 7*a) - 22*sin(5*b*x + 5*a) - 20*sin(3*b*x + 3*a) + 15*sin(b*x +
a))*cos(10*b*x + 10*a) + 60*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*c
os(9*b*x + 9*a) + 4*(20*sin(7*b*x + 7*a) + 22*sin(5*b*x + 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(8*
b*x + 8*a) - 80*(2*sin(6*b*x + 6*a) - 2*sin(4*b*x + 4*a) - sin(2*b*x + 2*a))*cos(7*b*x + 7*a) + 8*(22*sin(5*b*
x + 5*a) + 20*sin(3*b*x + 3*a) - 15*sin(b*x + a))*cos(6*b*x + 6*a) + 88*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a)
)*cos(5*b*x + 5*a) - 40*(4*sin(3*b*x + 3*a) - 3*sin(b*x + a))*cos(4*b*x + 4*a) + 15*(2*(cos(8*b*x + 8*a) + 2*c
os(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(10*b*x + 10*a) - cos(10*b*x + 10*a)^2 - 2*(2*
cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*cos(8*b*x + 8*a) - cos(8*b*x + 8*a)^2 + 4*(2*cos
(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*cos(6*b*x + 6*a) - 4*cos(6*b*x + 6*a)^2 - 4*(cos(2*b*x + 2*a) - 1)*cos(4
*b*x + 4*a) - 4*cos(4*b*x + 4*a)^2 - cos(2*b*x + 2*a)^2 + 2*(sin(8*b*x + 8*a) + 2*sin(6*b*x + 6*a) - 2*sin(4*b
*x + 4*a) - sin(2*b*x + 2*a))*sin(10*b*x + 10*a) - sin(10*b*x + 10*a)^2 - 2*(2*sin(6*b*x + 6*a) - 2*sin(4*b*x
+ 4*a) - sin(2*b*x + 2*a))*sin(8*b*x + 8*a) - sin(8*b*x + 8*a)^2 + 4*(2*sin(4*b*x + 4*a) + sin(2*b*x + 2*a))*s
in(6*b*x + 6*a) - 4*sin(6*b*x + 6*a)^2 - 4*sin(4*b*x + 4*a)^2 - 4*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)*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*(15*cos(9*b*x + 9*a) - 20*cos(7*b*x + 7*a) - 22*cos(5*b*x +
5*a) - 20*cos(3*b*x + 3*a) + 15*cos(b*x + a))*sin(10*b*x + 10*a) - 60*(cos(8*b*x + 8*a) + 2*cos(6*b*x + 6*a) -
2*cos(4*b*x + 4*a) - cos(2*b*x + 2*a) + 1)*sin(9*b*x + 9*a) - 4*(20*cos(7*b*x + 7*a) + 22*cos(5*b*x + 5*a) +
20*cos(3*b*x + 3*a) - 15*cos(b*x + a))*sin(8*b*x + 8*a) + 80*(2*cos(6*b*x + 6*a) - 2*cos(4*b*x + 4*a) - cos(2*
b*x + 2*a) + 1)*sin(7*b*x + 7*a) - 8*(22*cos(5*b*x + 5*a) + 20*cos(3*b*x + 3*a) - 15*cos(b*x + a))*sin(6*b*x +
6*a) - 88*(2*cos(4*b*x + 4*a) + cos(2*b*x + 2*a) - 1)*sin(5*b*x + 5*a) + 40*(4*cos(3*b*x + 3*a) - 3*cos(b*x +
a))*sin(4*b*x + 4*a) - 80*(cos(2*b*x + 2*a) - 1)*sin(3*b*x + 3*a) + 80*cos(3*b*x + 3*a)*sin(2*b*x + 2*a) - 60
*cos(b*x + a)*sin(2*b*x + 2*a) + 60*cos(2*b*x + 2*a)*sin(b*x + a) - 60*sin(b*x + a))/(b*cos(10*b*x + 10*a)^2 +
b*cos(8*b*x + 8*a)^2 + 4*b*cos(6*b*x + 6*a)^2 + 4*b*cos(4*b*x + 4*a)^2 + b*cos(2*b*x + 2*a)^2 + b*sin(10*b*x
+ 10*a)^2 + b*sin(8*b*x + 8*a)^2 + 4*b*sin(6*b*x + 6*a)^2 + 4*b*sin(4*b*x + 4*a)^2 + 4*b*sin(4*b*x + 4*a)*sin(
2*b*x + 2*a) + b*sin(2*b*x + 2*a)^2 - 2*(b*cos(8*b*x + 8*a) + 2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - b*
cos(2*b*x + 2*a) + b)*cos(10*b*x + 10*a) + 2*(2*b*cos(6*b*x + 6*a) - 2*b*cos(4*b*x + 4*a) - b*cos(2*b*x + 2*a)
+ b)*cos(8*b*x + 8*a) - 4*(2*b*cos(4*b*x + 4*a) + b*cos(2*b*x + 2*a) - b)*cos(6*b*x + 6*a) + 4*(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(8*b*x + 8*a) + 2*b*sin(6*b*x + 6*a) - 2*b*sin(4*
b*x + 4*a) - b*sin(2*b*x + 2*a))*sin(10*b*x + 10*a) + 2*(2*b*sin(6*b*x + 6*a) - 2*b*sin(4*b*x + 4*a) - b*sin(2
*b*x + 2*a))*sin(8*b*x + 8*a) - 4*(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 = 0.514639, size = 343, normalized size = 5.2 \begin{align*} -\frac{30 \, \cos \left (b x + a\right )^{4} - 15 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \,{\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{192 \,{\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\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)^4,x, algorithm="fricas")

[Out]

-1/192*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b*x + a) + 1)*sin(b*x + a) + 15*(cos(
b*x + a)^4 - cos(b*x + a)^2)*log(-sin(b*x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 -
b*cos(b*x + a)^2)*sin(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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.3007, size = 97, normalized size = 1.47 \begin{align*} -\frac{\frac{6 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + \frac{4 \,{\left (6 \, \sin \left (b x + a\right )^{2} + 1\right )}}{\sin \left (b x + a\right )^{3}} - 15 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 15 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{192 \, 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)^4,x, algorithm="giac")

[Out]

-1/192*(6*sin(b*x + a)/(sin(b*x + a)^2 - 1) + 4*(6*sin(b*x + a)^2 + 1)/sin(b*x + a)^3 - 15*log(abs(sin(b*x + a
) + 1)) + 15*log(abs(sin(b*x + a) - 1)))/b