3.121 \(\int \cos ^4(a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=40 \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{\sin ^2(a+b x)}{b}+\frac{\log (\sin (a+b x))}{b} \]

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/b + Sin[a + b*x]^4/(4*b)

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Rubi [A]  time = 0.0273101, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2590, 266, 43} \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{\sin ^2(a+b x)}{b}+\frac{\log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^4*Cot[a + b*x],x]

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/b + Sin[a + b*x]^4/(4*b)

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0148495, size = 40, normalized size = 1. \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{\sin ^2(a+b x)}{b}+\frac{\log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Log[Sin[a + b*x]]/b - Sin[a + b*x]^2/b + Sin[a + b*x]^4/(4*b)

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Maple [A]  time = 0.016, size = 39, normalized size = 1. \begin{align*}{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{4\,b}}+{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+{\frac{\ln \left ( \sin \left ( bx+a \right ) \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*cos(b*x+a)^4/b+1/2*cos(b*x+a)^2/b+ln(sin(b*x+a))/b

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Maxima [A]  time = 0.973472, size = 47, normalized size = 1.18 \begin{align*} \frac{\sin \left (b x + a\right )^{4} - 4 \, \sin \left (b x + a\right )^{2} + 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sin(b*x + a)^4 - 4*sin(b*x + a)^2 + 2*log(sin(b*x + a)^2))/b

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Fricas [A]  time = 1.72162, size = 93, normalized size = 2.32 \begin{align*} \frac{\cos \left (b x + a\right )^{4} + 2 \, \cos \left (b x + a\right )^{2} + 4 \, \log \left (\frac{1}{2} \, \sin \left (b x + a\right )\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(cos(b*x + a)^4 + 2*cos(b*x + a)^2 + 4*log(1/2*sin(b*x + a)))/b

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Sympy [A]  time = 6.07945, size = 1086, normalized size = 27.15 \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)**5/sin(b*x+a),x)

[Out]

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

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Giac [B]  time = 1.16468, size = 230, normalized size = 5.75 \begin{align*} -\frac{\frac{\frac{52 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{102 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{52 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac{25 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} - 25}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} - 6 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 12 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*((52*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 102*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 52*(cos(b*x
+ a) - 1)^3/(cos(b*x + a) + 1)^3 - 25*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 - 25)/((cos(b*x + a) - 1)/(co
s(b*x + a) + 1) - 1)^4 - 6*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 12*log(abs(-(cos(b*x + a) - 1)/
(cos(b*x + a) + 1) + 1)))/b