3.174 \(\int \cos (a+b x) \cot ^5(a+b x) \, dx\)

Optimal. Leaf size=70 \[ \frac{15 \cos (a+b x)}{8 b}-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}+\frac{5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b} \]

[Out]

(-15*ArcTanh[Cos[a + b*x]])/(8*b) + (15*Cos[a + b*x])/(8*b) + (5*Cos[a + b*x]*Cot[a + b*x]^2)/(8*b) - (Cos[a +
b*x]*Cot[a + b*x]^4)/(4*b)

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Rubi [A]  time = 0.0361286, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {2592, 288, 321, 206} \[ \frac{15 \cos (a+b x)}{8 b}-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}+\frac{5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Cot[a + b*x]^5,x]

[Out]

(-15*ArcTanh[Cos[a + b*x]])/(8*b) + (15*Cos[a + b*x])/(8*b) + (5*Cos[a + b*x]*Cot[a + b*x]^2)/(8*b) - (Cos[a +
b*x]*Cot[a + b*x]^4)/(4*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 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 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 (a+b x) \cot ^5(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}+\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\cos (a+b x)\right )}{4 b}\\ &=\frac{5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}-\frac{15 \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\cos (a+b x)\right )}{8 b}\\ &=\frac{15 \cos (a+b x)}{8 b}+\frac{5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}-\frac{15 \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\cos (a+b x)\right )}{8 b}\\ &=-\frac{15 \tanh ^{-1}(\cos (a+b x))}{8 b}+\frac{15 \cos (a+b x)}{8 b}+\frac{5 \cos (a+b x) \cot ^2(a+b x)}{8 b}-\frac{\cos (a+b x) \cot ^4(a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0306889, size = 123, normalized size = 1.76 \[ \frac{\cos (a+b x)}{b}-\frac{\csc ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}+\frac{9 \csc ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{\sec ^4\left (\frac{1}{2} (a+b x)\right )}{64 b}-\frac{9 \sec ^2\left (\frac{1}{2} (a+b x)\right )}{32 b}+\frac{15 \log \left (\sin \left (\frac{1}{2} (a+b x)\right )\right )}{8 b}-\frac{15 \log \left (\cos \left (\frac{1}{2} (a+b x)\right )\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Cot[a + b*x]^5,x]

[Out]

Cos[a + b*x]/b + (9*Csc[(a + b*x)/2]^2)/(32*b) - Csc[(a + b*x)/2]^4/(64*b) - (15*Log[Cos[(a + b*x)/2]])/(8*b)
+ (15*Log[Sin[(a + b*x)/2]])/(8*b) - (9*Sec[(a + b*x)/2]^2)/(32*b) + Sec[(a + b*x)/2]^4/(64*b)

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Maple [A]  time = 0.012, size = 102, normalized size = 1.5 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{8\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{8\,b}}+{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{8\,b}}+{\frac{15\,\cos \left ( bx+a \right ) }{8\,b}}+{\frac{15\,\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{8\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^6/sin(b*x+a)^5,x)

[Out]

-1/4/b*cos(b*x+a)^7/sin(b*x+a)^4+3/8/b*cos(b*x+a)^7/sin(b*x+a)^2+3/8*cos(b*x+a)^5/b+5/8*cos(b*x+a)^3/b+15/8*co
s(b*x+a)/b+15/8/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [A]  time = 0.977546, size = 107, normalized size = 1.53 \begin{align*} -\frac{\frac{2 \,{\left (9 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )}}{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1} - 16 \, \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 )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

-1/16*(2*(9*cos(b*x + a)^3 - 7*cos(b*x + a))/(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1) - 16*cos(b*x + a) + 15*lo
g(cos(b*x + a) + 1) - 15*log(cos(b*x + a) - 1))/b

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Fricas [A]  time = 2.30116, size = 344, normalized size = 4.91 \begin{align*} \frac{16 \, \cos \left (b x + a\right )^{5} - 50 \, \cos \left (b x + a\right )^{3} - 15 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \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 )^{4} - 2 \, \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 )}{16 \,{\left (b \cos \left (b x + a\right )^{4} - 2 \, 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)^5,x, algorithm="fricas")

[Out]

1/16*(16*cos(b*x + a)^5 - 50*cos(b*x + a)^3 - 15*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*log(1/2*cos(b*x + a)
+ 1/2) + 15*(cos(b*x + a)^4 - 2*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)^4 - 2*b*cos(b*x + a)^2 + b)

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Sympy [A]  time = 9.93402, size = 330, normalized size = 4.71 \begin{align*} \begin{cases} \frac{120 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} + \frac{120 \log{\left (\tan{\left (\frac{a}{2} + \frac{b x}{2} \right )} \right )} \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} + \frac{\tan ^{10}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} - \frac{15 \tan ^{8}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} + \frac{160 \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} + \frac{15 \tan ^{2}{\left (\frac{a}{2} + \frac{b x}{2} \right )}}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} - \frac{1}{64 b \tan ^{6}{\left (\frac{a}{2} + \frac{b x}{2} \right )} + 64 b \tan ^{4}{\left (\frac{a}{2} + \frac{b x}{2} \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{6}{\left (a \right )}}{\sin ^{5}{\left (a \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**6/sin(b*x+a)**5,x)

[Out]

Piecewise((120*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4)
+ 120*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + tan(a
/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 15*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 +
b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 160*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*
x/2)**4) + 15*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 1/(64*b*tan(a/2 + b*
x/2)**6 + 64*b*tan(a/2 + b*x/2)**4), Ne(b, 0)), (x*cos(a)**6/sin(a)**5, True))

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Giac [B]  time = 1.19111, size = 221, normalized size = 3.16 \begin{align*} -\frac{\frac{{\left (\frac{16 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{90 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 1\right )}{\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac{16 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\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{128}{\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1} - 60 \, \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)^6/sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/64*((16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 90*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 1)*(cos(b*x
+ a) + 1)^2/(cos(b*x + a) - 1)^2 - 16*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b*x + a) - 1)^2/(cos(b*x +
a) + 1)^2 + 128/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1) - 60*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) +
1)))/b