3.146 \(\int \cos ^3(a+b x) \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=66 \[ -\frac{5 \cos ^3(a+b x)}{6 b}-\frac{5 \cos (a+b x)}{2 b}-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}+\frac{5 \tanh ^{-1}(\cos (a+b x))}{2 b} \]

[Out]

(5*ArcTanh[Cos[a + b*x]])/(2*b) - (5*Cos[a + b*x])/(2*b) - (5*Cos[a + b*x]^3)/(6*b) - (Cos[a + b*x]^3*Cot[a +
b*x]^2)/(2*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0449874, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2592, 288, 302, 206} \[ -\frac{5 \cos ^3(a+b x)}{6 b}-\frac{5 \cos (a+b x)}{2 b}-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}+\frac{5 \tanh ^{-1}(\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Cot[a + b*x]^3,x]

[Out]

(5*ArcTanh[Cos[a + b*x]])/(2*b) - (5*Cos[a + b*x])/(2*b) - (5*Cos[a + b*x]^3)/(6*b) - (Cos[a + b*x]^3*Cot[a +
b*x]^2)/(2*b)

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

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 206

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

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \cot ^3(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^2} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1-x^2} \, dx,x,\cos (a+b x)\right )}{2 b}\\ &=-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}+\frac{5 \operatorname{Subst}\left (\int \left (-1-x^2+\frac{1}{1-x^2}\right ) \, dx,x,\cos (a+b x)\right )}{2 b}\\ &=-\frac{5 \cos (a+b x)}{2 b}-\frac{5 \cos ^3(a+b x)}{6 b}-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{2 b}\\ &=\frac{5 \tanh ^{-1}(\cos (a+b x))}{2 b}-\frac{5 \cos (a+b x)}{2 b}-\frac{5 \cos ^3(a+b x)}{6 b}-\frac{\cos ^3(a+b x) \cot ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.038896, size = 103, normalized size = 1.56 \[ -\frac{9 \cos (a+b x)}{4 b}-\frac{\cos (3 (a+b x))}{12 b}-\frac{\csc ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}+\frac{\sec ^2\left (\frac{1}{2} (a+b x)\right )}{8 b}-\frac{5 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{2 b}+\frac{5 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*Cos[a + b*x])/(4*b) - Cos[3*(a + b*x)]/(12*b) - Csc[(a + b*x)/2]^2/(8*b) + (5*Log[Cos[(a + b*x)/2]])/(2*b)
- (5*Log[Sin[(a + b*x)/2]])/(2*b) + Sec[(a + b*x)/2]^2/(8*b)

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 81, normalized size = 1.2 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{2\,b}}-{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{6\,b}}-{\frac{5\,\cos \left ( bx+a \right ) }{2\,b}}-{\frac{5\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^6/sin(b*x+a)^3,x)

[Out]

-1/2/b*cos(b*x+a)^7/sin(b*x+a)^2-1/2*cos(b*x+a)^5/b-5/6*cos(b*x+a)^3/b-5/2*cos(b*x+a)/b-5/2/b*ln(csc(b*x+a)-co
t(b*x+a))

________________________________________________________________________________________

Maxima [A]  time = 0.977276, size = 89, normalized size = 1.35 \begin{align*} -\frac{4 \, \cos \left (b x + a\right )^{3} - \frac{6 \, \cos \left (b x + a\right )}{\cos \left (b x + a\right )^{2} - 1} + 24 \, \cos \left (b x + a\right ) - 15 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/12*(4*cos(b*x + a)^3 - 6*cos(b*x + a)/(cos(b*x + a)^2 - 1) + 24*cos(b*x + a) - 15*log(cos(b*x + a) + 1) + 1
5*log(cos(b*x + a) - 1))/b

________________________________________________________________________________________

Fricas [A]  time = 2.04597, size = 265, normalized size = 4.02 \begin{align*} -\frac{4 \, \cos \left (b x + a\right )^{5} + 20 \, \cos \left (b x + a\right )^{3} - 15 \,{\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 ) + 15 \,{\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 ) - 30 \, \cos \left (b x + a\right )}{12 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/12*(4*cos(b*x + a)^5 + 20*cos(b*x + a)^3 - 15*(cos(b*x + a)^2 - 1)*log(1/2*cos(b*x + a) + 1/2) + 15*(cos(b*
x + a)^2 - 1)*log(-1/2*cos(b*x + a) + 1/2) - 30*cos(b*x + a))/(b*cos(b*x + a)^2 - b)

________________________________________________________________________________________

Sympy [A]  time = 11.7231, size = 719, normalized size = 10.89 \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)**6/sin(b*x+a)**3,x)

[Out]

Piecewise((-60*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6
+ 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 180*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(24*b*t
an(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 180*log
(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a/2 + b
*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 60*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(24*b*tan(a/2 + b*x/2)**8
+ 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) + 3*tan(a/2 + b*x/2)**10/(24
*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 165
*tan(a/2 + b*x/2)**6/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*ta
n(a/2 + b*x/2)**2) - 225*tan(a/2 + b*x/2)**4/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2 + b*x/2)**6 + 72*b*tan(a
/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 130*tan(a/2 + b*x/2)**2/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/2
+ b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2) - 3/(24*b*tan(a/2 + b*x/2)**8 + 72*b*tan(a/
2 + b*x/2)**6 + 72*b*tan(a/2 + b*x/2)**4 + 24*b*tan(a/2 + b*x/2)**2), Ne(b, 0)), (x*cos(a)**6/sin(a)**3, True)
)

________________________________________________________________________________________

Giac [B]  time = 1.16818, size = 220, normalized size = 3.33 \begin{align*} \frac{\frac{3 \,{\left (\frac{10 \,{\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{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{16 \,{\left (\frac{12 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{9 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 7\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{3}} - 30 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{24 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/24*(3*(10*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) - 3*(cos(b*x + a)
- 1)/(cos(b*x + a) + 1) - 16*(12*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 9*(cos(b*x + a) - 1)^2/(cos(b*x + a)
+ 1)^2 - 7)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^3 - 30*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) +
1)))/b