3.173 \(\int \cos ^2(a+b x) \cot ^5(a+b x) \, dx\)

Optimal. Leaf size=58 \[ -\frac{\sin ^2(a+b x)}{2 b}-\frac{\csc ^4(a+b x)}{4 b}+\frac{3 \csc ^2(a+b x)}{2 b}+\frac{3 \log (\sin (a+b x))}{b} \]

[Out]

(3*Csc[a + b*x]^2)/(2*b) - Csc[a + b*x]^4/(4*b) + (3*Log[Sin[a + b*x]])/b - Sin[a + b*x]^2/(2*b)

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Rubi [A]  time = 0.0426495, antiderivative size = 58, 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} \[ -\frac{\sin ^2(a+b x)}{2 b}-\frac{\csc ^4(a+b x)}{4 b}+\frac{3 \csc ^2(a+b x)}{2 b}+\frac{3 \log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x]^5,x]

[Out]

(3*Csc[a + b*x]^2)/(2*b) - Csc[a + b*x]^4/(4*b) + (3*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 ^5(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^5} \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{(1-x)^3}{x^3} \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^3}-\frac{3}{x^2}+\frac{3}{x}\right ) \, dx,x,\sin ^2(a+b x)\right )}{2 b}\\ &=\frac{3 \csc ^2(a+b x)}{2 b}-\frac{\csc ^4(a+b x)}{4 b}+\frac{3 \log (\sin (a+b x))}{b}-\frac{\sin ^2(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.154481, size = 47, normalized size = 0.81 \[ \frac{-2 \sin ^2(a+b x)-\csc ^4(a+b x)+6 \csc ^2(a+b x)+12 \log (\sin (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Cot[a + b*x]^5,x]

[Out]

(6*Csc[a + b*x]^2 - Csc[a + b*x]^4 + 12*Log[Sin[a + b*x]] - 2*Sin[a + b*x]^2)/(4*b)

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Maple [A]  time = 0.013, size = 95, normalized size = 1.6 \begin{align*} -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{4\,b \left ( \sin \left ( bx+a \right ) \right ) ^{4}}}+{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{8}}{2\,b \left ( \sin \left ( bx+a \right ) \right ) ^{2}}}+{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{2\,b}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{4\,b}}+{\frac{3\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}}{2\,b}}+3\,{\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)^7/sin(b*x+a)^5,x)

[Out]

-1/4/b*cos(b*x+a)^8/sin(b*x+a)^4+1/2/b/sin(b*x+a)^2*cos(b*x+a)^8+1/2*cos(b*x+a)^6/b+3/4*cos(b*x+a)^4/b+3/2*cos
(b*x+a)^2/b+3*ln(sin(b*x+a))/b

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Maxima [A]  time = 0.961055, size = 66, normalized size = 1.14 \begin{align*} -\frac{2 \, \sin \left (b x + a\right )^{2} - \frac{6 \, \sin \left (b x + a\right )^{2} - 1}{\sin \left (b x + a\right )^{4}} - 6 \, \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)^7/sin(b*x+a)^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.24891, size = 239, normalized size = 4.12 \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{6} - 5 \, \cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 12 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \log \left (\frac{1}{2} \, \sin \left (b x + a\right )\right ) + 4}{4 \,{\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)^7/sin(b*x+a)^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 18.9652, size = 733, normalized size = 12.64 \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)**7/sin(b*x+a)**5,x)

[Out]

Piecewise((-192*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b
*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 384*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*
x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 192*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b
*x/2)**4/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 192*log(tan(a/2 +
b*x/2))*tan(a/2 + b*x/2)**8/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4)
+ 384*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*
tan(a/2 + b*x/2)**4) + 192*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**4/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2
+ b*x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) - tan(a/2 + b*x/2)**12/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*
x/2)**6 + 64*b*tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**10/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/
2)**6 + 64*b*tan(a/2 + b*x/2)**4) - 166*tan(a/2 + b*x/2)**6/(64*b*tan(a/2 + b*x/2)**8 + 128*b*tan(a/2 + b*x/2)
**6 + 64*b*tan(a/2 + b*x/2)**4) + 18*tan(a/2 + b*x/2)**2/(64*b*tan(a/2 + b*x/2)**8 + 128*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)**8 + 128*b*tan(a/2 + b*x/2)**6 + 64*b*tan(a/2 + b*x/2)
**4), Ne(b, 0)), (x*cos(a)**7/sin(a)**5, True))

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Giac [B]  time = 1.20473, size = 313, normalized size = 5.4 \begin{align*} -\frac{\frac{20 \,{\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{\frac{18 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{111 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{36 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} - \frac{72 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\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}}\right )}^{2}} - 96 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 192 \, \log \left ({\left | -\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1 \right |}\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^7/sin(b*x+a)^5,x, algorithm="giac")

[Out]

-1/64*(20*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + (18*(cos(b*x + a
) - 1)/(cos(b*x + a) + 1) - 111*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 36*(cos(b*x + a) - 1)^3/(cos(b*x +
a) + 1)^3 - 72*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - (cos(b
*x + a) - 1)^2/(cos(b*x + a) + 1)^2)^2 - 96*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 192*log(abs(-(
cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b