3.145 \(\int \cos ^4(a+b x) \cot ^3(a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.04388, 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} \[ -\frac{\sin ^4(a+b x)}{4 b}+\frac{3 \sin ^2(a+b x)}{2 b}-\frac{\csc ^2(a+b x)}{2 b}-\frac{3 \log (\sin (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^4*Cot[a + b*x]^3,x]

[Out]

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

Mathematica [A]  time = 0.102288, size = 45, normalized size = 0.78 \[ -\frac{\sin ^4(a+b x)-6 \sin ^2(a+b x)+2 \csc ^2(a+b x)+12 \log (\sin (a+b x))}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^4*Cot[a + b*x]^3,x]

[Out]

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

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Maple [A]  time = 0.014, size = 74, normalized size = 1.3 \begin{align*} -{\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)^3,x)

[Out]

-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 = 0.971924, size = 61, normalized size = 1.05 \begin{align*} -\frac{\sin \left (b x + a\right )^{4} - 6 \, \sin \left (b x + a\right )^{2} + \frac{2}{\sin \left (b x + a\right )^{2}} + 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)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.02392, size = 190, normalized size = 3.28 \begin{align*} -\frac{8 \, \cos \left (b x + a\right )^{6} + 24 \, \cos \left (b x + a\right )^{4} - 51 \, \cos \left (b x + a\right )^{2} + 96 \,{\left (\cos \left (b x + a\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \sin \left (b x + a\right )\right ) + 3}{32 \,{\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)^7/sin(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/32*(8*cos(b*x + a)^6 + 24*cos(b*x + a)^4 - 51*cos(b*x + a)^2 + 96*(cos(b*x + a)^2 - 1)*log(1/2*sin(b*x + a)
) + 3)/(b*cos(b*x + a)^2 - b)

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Sympy [A]  time = 22.4741, size = 1484, normalized size = 25.59 \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)**3,x)

[Out]

Piecewise((24*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**10/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x
/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 96*log(tan(a/2 + b*x
/2)**2 + 1)*tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**
6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 144*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**6
/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 +
8*b*tan(a/2 + b*x/2)**2) + 96*log(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**10 + 32*
b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 24*lo
g(tan(a/2 + b*x/2)**2 + 1)*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan
(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 24*log(tan(a/2 + b*x/2))*tan(a/2 + b*
x/2)**10/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/
2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 96*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)**10 + 32
*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 144*
log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**6/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2
+ b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - 96*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)*
*4/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4
+ 8*b*tan(a/2 + b*x/2)**2) - 24*log(tan(a/2 + b*x/2))*tan(a/2 + b*x/2)**2/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan
(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) - tan(a/2 +
b*x/2)**12/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*
x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 57*tan(a/2 + b*x/2)**8/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)*
*8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2) + 80*tan(a/2 + b*x/2)**6/(
8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*
b*tan(a/2 + b*x/2)**2) + 57*tan(a/2 + b*x/2)**4/(8*b*tan(a/2 + b*x/2)**10 + 32*b*tan(a/2 + b*x/2)**8 + 48*b*ta
n(a/2 + b*x/2)**6 + 32*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)**10 + 32*b*t
an(a/2 + b*x/2)**8 + 48*b*tan(a/2 + b*x/2)**6 + 32*b*tan(a/2 + b*x/2)**4 + 8*b*tan(a/2 + b*x/2)**2), Ne(b, 0))
, (x*cos(a)**7/sin(a)**3, True))

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Giac [B]  time = 1.24494, size = 312, normalized size = 5.38 \begin{align*} \frac{\frac{{\left (\frac{12 \,{\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{2 \,{\left (\frac{76 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac{118 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{76 \,{\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\right )}}{{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} - 12 \, \log \left (\frac{{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right ) + 24 \, \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)^7/sin(b*x+a)^3,x, algorithm="giac")

[Out]

1/8*((12*(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) + 2*(76*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 118*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1
)^2 + 76*(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)/(cos(b*x + a) + 1) - 1)^4 - 12*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)) + 24*log(abs(-(
cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 1)))/b