3.147 \(\int \cos ^2(a+b x) \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=43 \[ \frac{\sin ^2(a+b x)}{2 b}-\frac{\csc ^2(a+b x)}{2 b}-\frac{2 \log (\sin (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0388649, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2590, 266, 43} \[ \frac{\sin ^2(a+b x)}{2 b}-\frac{\csc ^2(a+b x)}{2 b}-\frac{2 \log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x]^3,x]

[Out]

-Csc[a + b*x]^2/(2*b) - (2*Log[Sin[a + b*x]])/b + Sin[a + b*x]^2/(2*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 ^2(a+b x) \cot ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^3} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^2}{x^2} \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^2}-\frac{2}{x}\right ) \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=-\frac{\csc ^2(a+b x)}{2 b}-\frac{2 \log (\sin (a+b x))}{b}+\frac{\sin ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0617029, size = 35, normalized size = 0.81 \[ -\frac{-\sin ^2(a+b x)+\csc ^2(a+b x)+4 \log (\sin (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 61, normalized size = 1.4 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{2\,b}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{b}}-2\,{\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)^3,x)

[Out]

-1/2/b*cos(b*x+a)^6/sin(b*x+a)^2-1/2*cos(b*x+a)^4/b-cos(b*x+a)^2/b-2*ln(sin(b*x+a))/b

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Maxima [A]  time = 0.97445, size = 47, normalized size = 1.09 \begin{align*} \frac{\sin \left (b x + a\right )^{2} - \frac{1}{\sin \left (b x + a\right )^{2}} - 2 \, \log \left (\sin \left (b x + a\right )^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.94162, size = 159, normalized size = 3.7 \begin{align*} -\frac{2 \, \cos \left (b x + a\right )^{4} - 3 \, \cos \left (b x + a\right )^{2} + 8 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \sin \left (b x + a\right )\right ) - 1}{4 \,{\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)^5/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/4*(2*cos(b*x + a)^4 - 3*cos(b*x + a)^2 + 8*(cos(b*x + a)^2 - 1)*log(1/2*sin(b*x + a)) - 1)/(b*cos(b*x + a)^
2 - b)

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Sympy [A]  time = 6.34138, size = 614, normalized size = 14.28 \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)**3,x)

[Out]

Piecewise((16*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2
)**4 + 8*b*tan(a/2 + b*x/2)**2) + 32*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6
+ 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 16*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(
8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 16*log(tan(a/2 + b*x/2))*tan(a
/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 32*log(tan(a/2
+ b*x/2))*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2)
- 16*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a
/2 + b*x/2)**2) - tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/
2)**2) + 18*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2)
- 1/(8*b*tan(a/2 + b*x/2)**6 + 16*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2), Ne(b, 0)), (x*cos(a)**5/s
in(a)**3, True))

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Giac [B]  time = 1.21101, size = 252, normalized size = 5.86 \begin{align*} \frac{\frac{{\left (\frac{8 \,{\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{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac{8 \,{\left (\frac{4 \,{\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}} - 3\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{2}} - 8 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 16 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^5/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*((8*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)*(cos(b*x + a) + 1)/(cos(b*x + a) - 1) + (cos(b*x + a) - 1)/
(cos(b*x + a) + 1) + 8*(4*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2
- 3)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^2 - 8*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 16*
log(abs(-(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b