3.11 \(\int \frac{\cos ^4(x)}{a+b \cot (x)} \, dx\)

Optimal. Leaf size=126 \[ \frac{a x \left (-6 a^2 b^2+3 a^4-b^4\right )}{8 \left (a^2+b^2\right )^3}-\frac{\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}+\frac{\sin ^2(x) \left (a \left (5 a^2+b^2\right ) \cot (x)+4 b \left (2 a^2+b^2\right )\right )}{8 \left (a^2+b^2\right )^2}-\frac{a^4 b \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3} \]

[Out]

(a*(3*a^4 - 6*a^2*b^2 - b^4)*x)/(8*(a^2 + b^2)^3) - (a^4*b*Log[b*Cos[x] + a*Sin[x]])/(a^2 + b^2)^3 + ((4*b*(2*
a^2 + b^2) + a*(5*a^2 + b^2)*Cot[x])*Sin[x]^2)/(8*(a^2 + b^2)^2) - ((b + a*Cot[x])*Sin[x]^4)/(4*(a^2 + b^2))

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Rubi [A]  time = 0.303197, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3516, 1647, 801, 635, 203, 260} \[ \frac{a x \left (-6 a^2 b^2+3 a^4-b^4\right )}{8 \left (a^2+b^2\right )^3}-\frac{\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}+\frac{\sin ^2(x) \left (a \left (5 a^2+b^2\right ) \cot (x)+4 b \left (2 a^2+b^2\right )\right )}{8 \left (a^2+b^2\right )^2}-\frac{a^4 b \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^4/(a + b*Cot[x]),x]

[Out]

(a*(3*a^4 - 6*a^2*b^2 - b^4)*x)/(8*(a^2 + b^2)^3) - (a^4*b*Log[b*Cos[x] + a*Sin[x]])/(a^2 + b^2)^3 + ((4*b*(2*
a^2 + b^2) + a*(5*a^2 + b^2)*Cot[x])*Sin[x]^2)/(8*(a^2 + b^2)^2) - ((b + a*Cot[x])*Sin[x]^4)/(4*(a^2 + b^2))

Rule 3516

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

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

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])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\cos ^4(x)}{a+b \cot (x)} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{x^4}{(a+x) \left (b^2+x^2\right )^3} \, dx,x,b \cot (x)\right )\right )\\ &=-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^4}{a^2+b^2}-\frac{3 a b^4 x}{a^2+b^2}-4 b^2 x^2}{(a+x) \left (b^2+x^2\right )^2} \, dx,x,b \cot (x)\right )}{4 b}\\ &=\frac{\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^4 \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^2}-\frac{a b^4 \left (5 a^2+b^2\right ) x}{\left (a^2+b^2\right )^2}}{(a+x) \left (b^2+x^2\right )} \, dx,x,b \cot (x)\right )}{8 b^3}\\ &=\frac{\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac{\operatorname{Subst}\left (\int \left (\frac{8 a^4 b^4}{\left (a^2+b^2\right )^3 (a+x)}+\frac{a b^4 \left (3 a^4-6 a^2 b^2-b^4-8 a^3 x\right )}{\left (a^2+b^2\right )^3 \left (b^2+x^2\right )}\right ) \, dx,x,b \cot (x)\right )}{8 b^3}\\ &=-\frac{a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}+\frac{\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac{(a b) \operatorname{Subst}\left (\int \frac{3 a^4-6 a^2 b^2-b^4-8 a^3 x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=-\frac{a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}+\frac{\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac{\left (a^4 b\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{\left (a^2+b^2\right )^3}-\frac{\left (a b \left (3 a^4-6 a^2 b^2-b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3}\\ &=\frac{a \left (3 a^4-6 a^2 b^2-b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac{a^4 b \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac{a^4 b \log (\sin (x))}{\left (a^2+b^2\right )^3}+\frac{\left (4 b \left (2 a^2+b^2\right )+a \left (5 a^2+b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac{(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}\\ \end{align*}

Mathematica [C]  time = 0.55147, size = 179, normalized size = 1.42 \[ \frac{-24 a^3 b^2 x+8 a^3 b^2 \sin (2 x)+2 a^3 b^2 \sin (4 x)-2 a^2 b^3 \cos (4 x)-4 b \left (4 a^2 b^2+3 a^4+b^4\right ) \cos (2 x)-32 i a^4 b x-a^4 b \cos (4 x)+32 i a^4 b \tan ^{-1}(\tan (x))-16 a^4 b \log \left ((a \sin (x)+b \cos (x))^2\right )+12 a^5 x+8 a^5 \sin (2 x)+a^5 \sin (4 x)-4 a b^4 x+a b^4 \sin (4 x)-b^5 \cos (4 x)}{32 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^4/(a + b*Cot[x]),x]

[Out]

(12*a^5*x - (32*I)*a^4*b*x - 24*a^3*b^2*x - 4*a*b^4*x + (32*I)*a^4*b*ArcTan[Tan[x]] - 4*b*(3*a^4 + 4*a^2*b^2 +
b^4)*Cos[2*x] - a^4*b*Cos[4*x] - 2*a^2*b^3*Cos[4*x] - b^5*Cos[4*x] - 16*a^4*b*Log[(b*Cos[x] + a*Sin[x])^2] +
8*a^5*Sin[2*x] + 8*a^3*b^2*Sin[2*x] + a^5*Sin[4*x] + 2*a^3*b^2*Sin[4*x] + a*b^4*Sin[4*x])/(32*(a^2 + b^2)^3)

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Maple [B]  time = 0.125, size = 385, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*a^4/(a^2+b^2)^3*ln(a*tan(x)+b)+3/8/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)^3*a^5+1/4/(a^2+b^2)^3/(tan(x)^2+1)^2*t
an(x)^3*a^3*b^2-1/8/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)^3*a*b^4-1/2/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)^2*a^4*b-1/
2/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)^2*a^2*b^3+5/8/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)*a^5+3/4/(a^2+b^2)^3/(tan(x
)^2+1)^2*tan(x)*a^3*b^2+1/8/(a^2+b^2)^3/(tan(x)^2+1)^2*tan(x)*a*b^4-3/4/(a^2+b^2)^3/(tan(x)^2+1)^2*a^4*b-1/(a^
2+b^2)^3/(tan(x)^2+1)^2*a^2*b^3-1/4/(a^2+b^2)^3/(tan(x)^2+1)^2*b^5+1/2/(a^2+b^2)^3*ln(tan(x)^2+1)*a^4*b+3/8/(a
^2+b^2)^3*arctan(tan(x))*a^5-3/4/(a^2+b^2)^3*arctan(tan(x))*a^3*b^2-1/8/(a^2+b^2)^3*arctan(tan(x))*a*b^4

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Maxima [A]  time = 1.80643, size = 324, normalized size = 2.57 \begin{align*} -\frac{a^{4} b \log \left (a \tan \left (x\right ) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{a^{4} b \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac{{\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x}{8 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac{4 \, a^{2} b \tan \left (x\right )^{2} -{\left (3 \, a^{3} - a b^{2}\right )} \tan \left (x\right )^{3} + 6 \, a^{2} b + 2 \, b^{3} -{\left (5 \, a^{3} + a b^{2}\right )} \tan \left (x\right )}{8 \,{\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^4*b*log(a*tan(x) + b)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 1/2*a^4*b*log(tan(x)^2 + 1)/(a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6) + 1/8*(3*a^5 - 6*a^3*b^2 - a*b^4)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/8*(4*a^2*b*tan(x)
^2 - (3*a^3 - a*b^2)*tan(x)^3 + 6*a^2*b + 2*b^3 - (5*a^3 + a*b^2)*tan(x))/((a^4 + 2*a^2*b^2 + b^4)*tan(x)^4 +
a^4 + 2*a^2*b^2 + b^4 + 2*(a^4 + 2*a^2*b^2 + b^4)*tan(x)^2)

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Fricas [A]  time = 2.24849, size = 396, normalized size = 3.14 \begin{align*} -\frac{4 \, a^{4} b \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) -{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}\right ) + 2 \,{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{4} + 4 \,{\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right )^{2} -{\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x -{\left (2 \,{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} +{\left (3 \, a^{5} + 2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*a^4*b*log(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2) + 2*(a^4*b + 2*a^2*b^3 + b^5)*cos(x)^4 + 4
*(a^4*b + a^2*b^3)*cos(x)^2 - (3*a^5 - 6*a^3*b^2 - a*b^4)*x - (2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x)^3 + (3*a^5 +
2*a^3*b^2 - a*b^4)*cos(x))*sin(x))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{4}{\left (x \right )}}{a + b \cot{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**4/(a+b*cot(x)),x)

[Out]

Integral(cos(x)**4/(a + b*cot(x)), x)

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Giac [B]  time = 1.26137, size = 365, normalized size = 2.9 \begin{align*} -\frac{a^{5} b \log \left ({\left | a \tan \left (x\right ) + b \right |}\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}} + \frac{a^{4} b \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac{{\left (3 \, a^{5} - 6 \, a^{3} b^{2} - a b^{4}\right )} x}{8 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac{6 \, a^{4} b \tan \left (x\right )^{4} - 3 \, a^{5} \tan \left (x\right )^{3} - 2 \, a^{3} b^{2} \tan \left (x\right )^{3} + a b^{4} \tan \left (x\right )^{3} + 16 \, a^{4} b \tan \left (x\right )^{2} + 4 \, a^{2} b^{3} \tan \left (x\right )^{2} - 5 \, a^{5} \tan \left (x\right ) - 6 \, a^{3} b^{2} \tan \left (x\right ) - a b^{4} \tan \left (x\right ) + 12 \, a^{4} b + 8 \, a^{2} b^{3} + 2 \, b^{5}}{8 \,{\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}{\left (\tan \left (x\right )^{2} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^5*b*log(abs(a*tan(x) + b))/(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6) + 1/2*a^4*b*log(tan(x)^2 + 1)/(a^6 + 3*a^4
*b^2 + 3*a^2*b^4 + b^6) + 1/8*(3*a^5 - 6*a^3*b^2 - a*b^4)*x/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 1/8*(6*a^4*b
*tan(x)^4 - 3*a^5*tan(x)^3 - 2*a^3*b^2*tan(x)^3 + a*b^4*tan(x)^3 + 16*a^4*b*tan(x)^2 + 4*a^2*b^3*tan(x)^2 - 5*
a^5*tan(x) - 6*a^3*b^2*tan(x) - a*b^4*tan(x) + 12*a^4*b + 8*a^2*b^3 + 2*b^5)/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b
^6)*(tan(x)^2 + 1)^2)