3.160 \(\int \cos ^2(a+b x) \cot ^4(a+b x) \, dx\)

Optimal. Leaf size=57 \[ -\frac{5 \cot ^3(a+b x)}{6 b}+\frac{5 \cot (a+b x)}{2 b}+\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}+\frac{5 x}{2} \]

[Out]

(5*x)/2 + (5*Cot[a + b*x])/(2*b) - (5*Cot[a + b*x]^3)/(6*b) + (Cos[a + b*x]^2*Cot[a + b*x]^3)/(2*b)

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Rubi [A]  time = 0.0410902, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {2591, 288, 302, 203} \[ -\frac{5 \cot ^3(a+b x)}{6 b}+\frac{5 \cot (a+b x)}{2 b}+\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}+\frac{5 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Cot[a + b*x]^4,x]

[Out]

(5*x)/2 + (5*Cot[a + b*x])/(2*b) - (5*Cot[a + b*x]^3)/(6*b) + (Cos[a + b*x]^2*Cot[a + b*x]^3)/(2*b)

Rule 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \cot ^4(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^2} \, dx,x,\cot (a+b x)\right )}{b}\\ &=\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \frac{x^4}{1+x^2} \, dx,x,\cot (a+b x)\right )}{2 b}\\ &=\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \left (-1+x^2+\frac{1}{1+x^2}\right ) \, dx,x,\cot (a+b x)\right )}{2 b}\\ &=\frac{5 \cot (a+b x)}{2 b}-\frac{5 \cot ^3(a+b x)}{6 b}+\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\cot (a+b x)\right )}{2 b}\\ &=\frac{5 x}{2}+\frac{5 \cot (a+b x)}{2 b}-\frac{5 \cot ^3(a+b x)}{6 b}+\frac{\cos ^2(a+b x) \cot ^3(a+b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.177181, size = 43, normalized size = 0.75 \[ \frac{30 (a+b x)+3 \sin (2 (a+b x))-4 \cot (a+b x) \left (\csc ^2(a+b x)-7\right )}{12 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Cot[a + b*x]^4,x]

[Out]

(30*(a + b*x) - 4*Cot[a + b*x]*(-7 + Csc[a + b*x]^2) + 3*Sin[2*(a + b*x)])/(12*b)

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Maple [A]  time = 0.012, size = 84, normalized size = 1.5 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}}}+{\frac{4\, \left ( \cos \left ( bx+a \right ) \right ) ^{7}}{3\,\sin \left ( bx+a \right ) }}+{\frac{4\,\sin \left ( bx+a \right ) }{3} \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( bx+a \right ) }{8}} \right ) }+{\frac{5\,bx}{2}}+{\frac{5\,a}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^6/sin(b*x+a)^4,x)

[Out]

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

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Maxima [A]  time = 1.48475, size = 74, normalized size = 1.3 \begin{align*} \frac{15 \, b x + 15 \, a + \frac{15 \, \tan \left (b x + a\right )^{4} + 10 \, \tan \left (b x + a\right )^{2} - 2}{\tan \left (b x + a\right )^{5} + \tan \left (b x + a\right )^{3}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="maxima")

[Out]

1/6*(15*b*x + 15*a + (15*tan(b*x + a)^4 + 10*tan(b*x + a)^2 - 2)/(tan(b*x + a)^5 + tan(b*x + a)^3))/b

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Fricas [A]  time = 2.23404, size = 197, normalized size = 3.46 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{5} - 20 \, \cos \left (b x + a\right )^{3} - 15 \,{\left (b x \cos \left (b x + a\right )^{2} - b x\right )} \sin \left (b x + a\right ) + 15 \, \cos \left (b x + a\right )}{6 \,{\left (b \cos \left (b x + a\right )^{2} - b\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="fricas")

[Out]

-1/6*(3*cos(b*x + a)^5 - 20*cos(b*x + a)^3 - 15*(b*x*cos(b*x + a)^2 - b*x)*sin(b*x + a) + 15*cos(b*x + a))/((b
*cos(b*x + a)^2 - b)*sin(b*x + a))

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Sympy [A]  time = 4.73463, size = 97, normalized size = 1.7 \begin{align*} \begin{cases} \frac{5 x \sin ^{2}{\left (a + b x \right )}}{2} + \frac{5 x \cos ^{2}{\left (a + b x \right )}}{2} + \frac{5 \sin{\left (a + b x \right )} \cos{\left (a + b x \right )}}{2 b} + \frac{5 \cos ^{3}{\left (a + b x \right )}}{3 b \sin{\left (a + b x \right )}} - \frac{\cos ^{5}{\left (a + b x \right )}}{3 b \sin ^{3}{\left (a + b x \right )}} & \text{for}\: b \neq 0 \\\frac{x \cos ^{6}{\left (a \right )}}{\sin ^{4}{\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)**4,x)

[Out]

Piecewise((5*x*sin(a + b*x)**2/2 + 5*x*cos(a + b*x)**2/2 + 5*sin(a + b*x)*cos(a + b*x)/(2*b) + 5*cos(a + b*x)*
*3/(3*b*sin(a + b*x)) - cos(a + b*x)**5/(3*b*sin(a + b*x)**3), Ne(b, 0)), (x*cos(a)**6/sin(a)**4, True))

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Giac [A]  time = 1.15451, size = 74, normalized size = 1.3 \begin{align*} \frac{15 \, b x + 15 \, a + \frac{3 \, \tan \left (b x + a\right )}{\tan \left (b x + a\right )^{2} + 1} + \frac{2 \,{\left (6 \, \tan \left (b x + a\right )^{2} - 1\right )}}{\tan \left (b x + a\right )^{3}}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^6/sin(b*x+a)^4,x, algorithm="giac")

[Out]

1/6*(15*b*x + 15*a + 3*tan(b*x + a)/(tan(b*x + a)^2 + 1) + 2*(6*tan(b*x + a)^2 - 1)/tan(b*x + a)^3)/b