3.291 \(\int \frac{1}{(a \cot (x)+b \csc (x))^4} \, dx\)

Optimal. Leaf size=159 \[ \frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (a \cos (x)+b)^2}-\frac{\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 a^3 \left (a^2-b^2\right ) (a \cos (x)+b)}-\frac{b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}+\frac{x}{a^4}+\frac{\sin ^3(x)}{3 a (a \cos (x)+b)^3} \]

[Out]

x/a^4 - (b*(3*a^2 - 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^(3/2)) - ((
2*(a^2 - b^2) - a*b*Cos[x])*Sin[x])/(2*a^3*(a^2 - b^2)*(b + a*Cos[x])) + Sin[x]^3/(3*a*(b + a*Cos[x])^3) + (b*
Sin[x]^3)/(2*a*(a^2 - b^2)*(b + a*Cos[x])^2)

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Rubi [A]  time = 0.338455, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {4392, 2693, 2864, 2863, 2735, 2659, 208} \[ \frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (a \cos (x)+b)^2}-\frac{\sin (x) \left (2 \left (a^2-b^2\right )-a b \cos (x)\right )}{2 a^3 \left (a^2-b^2\right ) (a \cos (x)+b)}-\frac{b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}+\frac{x}{a^4}+\frac{\sin ^3(x)}{3 a (a \cos (x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^(-4),x]

[Out]

x/a^4 - (b*(3*a^2 - 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/(a^4*(a - b)^(3/2)*(a + b)^(3/2)) - ((
2*(a^2 - b^2) - a*b*Cos[x])*Sin[x])/(2*a^3*(a^2 - b^2)*(b + a*Cos[x])) + Sin[x]^3/(3*a*(b + a*Cos[x])^3) + (b*
Sin[x]^3)/(2*a*(a^2 - b^2)*(b + a*Cos[x])^2)

Rule 4392

Int[(cot[(c_.) + (d_.)*(x_)]^(n_.)*(a_.) + csc[(c_.) + (d_.)*(x_)]^(n_.)*(b_.))^(p_)*(u_.), x_Symbol] :> Int[A
ctivateTrig[u]*Csc[c + d*x]^(n*p)*(b + a*Cos[c + d*x]^n)^p, x] /; FreeQ[{a, b, c, d}, x] && IntegersQ[n, p]

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2864

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a
^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*Sim
p[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + p + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2863

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + 1)*Sin[e + f*x]))/(b^2*f*(m + 1)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + 1)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Simp[b*d*(m + 1) + (b*c*(m + p + 1) - a*d*p)*Si
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && N
eQ[m + p + 1, 0] && IntegerQ[2*m]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{(a \cot (x)+b \csc (x))^4} \, dx &=\int \frac{\sin ^4(x)}{(b+a \cos (x))^4} \, dx\\ &=\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}-\frac{\int \frac{\cos (x) \sin ^2(x)}{(b+a \cos (x))^3} \, dx}{a}\\ &=\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2}-\frac{\int \frac{(2 a+b \cos (x)) \sin ^2(x)}{(b+a \cos (x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (x))}+\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2}+\frac{\int \frac{-a b+2 \left (a^2-b^2\right ) \cos (x)}{b+a \cos (x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac{x}{a^4}-\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (x))}+\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2}-\frac{\left (b \left (3 a^2-2 b^2\right )\right ) \int \frac{1}{b+a \cos (x)} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac{x}{a^4}-\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (x))}+\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2}-\frac{\left (b \left (3 a^2-2 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^4 \left (a^2-b^2\right )}\\ &=\frac{x}{a^4}-\frac{b \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2}}-\frac{\left (2 \left (a^2-b^2\right )-a b \cos (x)\right ) \sin (x)}{2 a^3 \left (a^2-b^2\right ) (b+a \cos (x))}+\frac{\sin ^3(x)}{3 a (b+a \cos (x))^3}+\frac{b \sin ^3(x)}{2 a \left (a^2-b^2\right ) (b+a \cos (x))^2}\\ \end{align*}

Mathematica [A]  time = 0.466912, size = 150, normalized size = 0.94 \[ \frac{\sin (x) \left (-\frac{a \left (8 a^2-11 b^2\right ) (a \cos (x)+b)^2}{(a-b) (a+b)}-\frac{6 b \left (2 b^2-3 a^2\right ) \csc (x) (a \cos (x)+b)^3 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+2 a \left (a^2-b^2\right )+7 a b (a \cos (x)+b)+6 x \csc (x) (a \cos (x)+b)^3\right )}{6 a^4 (a \cos (x)+b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cot[x] + b*Csc[x])^(-4),x]

[Out]

((2*a*(a^2 - b^2) + 7*a*b*(b + a*Cos[x]) - (a*(8*a^2 - 11*b^2)*(b + a*Cos[x])^2)/((a - b)*(a + b)) + 6*x*(b +
a*Cos[x])^3*Csc[x] - (6*b*(-3*a^2 + 2*b^2)*ArcTanh[((-a + b)*Tan[x/2])/Sqrt[a^2 - b^2]]*(b + a*Cos[x])^3*Csc[x
])/(a^2 - b^2)^(3/2))*Sin[x])/(6*a^4*(b + a*Cos[x])^3)

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Maple [B]  time = 0.062, size = 534, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cot(x)+b*csc(x))^4,x)

[Out]

2/a^4*arctan(tan(1/2*x))+2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5-1/a/(a*tan(1/2*x)^2-b*tan(
1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5*b-3/a^2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5*b^2+2/a^3/
(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a+b)*tan(1/2*x)^5*b^3-20/3/a/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3*tan(
1/2*x)^3+4/a^3/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3*tan(1/2*x)^3*b^2+2/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/
(a-b)*tan(1/2*x)+1/a/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2*x)*b-3/a^2/(a*tan(1/2*x)^2-b*tan(1/2*
x)^2-a-b)^3/(a-b)*tan(1/2*x)*b^2-2/a^3/(a*tan(1/2*x)^2-b*tan(1/2*x)^2-a-b)^3/(a-b)*tan(1/2*x)*b^3-3/a^2*b/(a^2
-b^2)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))+2/a^4*b^3/(a^2-b^2)/((a-b)*(a+b))^(1/2
)*arctanh(tan(1/2*x)*(a-b)/((a-b)*(a+b))^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.93663, size = 1945, normalized size = 12.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="fricas")

[Out]

[1/12*(12*(a^7 - 2*a^5*b^2 + a^3*b^4)*x*cos(x)^3 + 36*(a^6*b - 2*a^4*b^3 + a^2*b^5)*x*cos(x)^2 + 36*(a^5*b^2 -
2*a^3*b^4 + a*b^6)*x*cos(x) + 3*(3*a^2*b^4 - 2*b^6 + (3*a^5*b - 2*a^3*b^3)*cos(x)^3 + 3*(3*a^4*b^2 - 2*a^2*b^
4)*cos(x)^2 + 3*(3*a^3*b^3 - 2*a*b^5)*cos(x))*sqrt(a^2 - b^2)*log((2*a*b*cos(x) - (a^2 - 2*b^2)*cos(x)^2 - 2*s
qrt(a^2 - b^2)*(b*cos(x) + a)*sin(x) + 2*a^2 - b^2)/(a^2*cos(x)^2 + 2*a*b*cos(x) + b^2)) + 12*(a^4*b^3 - 2*a^2
*b^5 + b^7)*x + 2*(2*a^7 - 7*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6 - (8*a^7 - 19*a^5*b^2 + 11*a^3*b^4)*cos(x)^2 - 3*(
3*a^6*b - 8*a^4*b^3 + 5*a^2*b^5)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + a^4*b^7 + (a^11 - 2*a^9*b^2 + a^7*b^4)
*cos(x)^3 + 3*(a^10*b - 2*a^8*b^3 + a^6*b^5)*cos(x)^2 + 3*(a^9*b^2 - 2*a^7*b^4 + a^5*b^6)*cos(x)), 1/6*(6*(a^7
- 2*a^5*b^2 + a^3*b^4)*x*cos(x)^3 + 18*(a^6*b - 2*a^4*b^3 + a^2*b^5)*x*cos(x)^2 + 18*(a^5*b^2 - 2*a^3*b^4 + a
*b^6)*x*cos(x) - 3*(3*a^2*b^4 - 2*b^6 + (3*a^5*b - 2*a^3*b^3)*cos(x)^3 + 3*(3*a^4*b^2 - 2*a^2*b^4)*cos(x)^2 +
3*(3*a^3*b^3 - 2*a*b^5)*cos(x))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(x) + a)/((a^2 - b^2)*sin(x)))
+ 6*(a^4*b^3 - 2*a^2*b^5 + b^7)*x + (2*a^7 - 7*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6 - (8*a^7 - 19*a^5*b^2 + 11*a^3*
b^4)*cos(x)^2 - 3*(3*a^6*b - 8*a^4*b^3 + 5*a^2*b^5)*cos(x))*sin(x))/(a^8*b^3 - 2*a^6*b^5 + a^4*b^7 + (a^11 - 2
*a^9*b^2 + a^7*b^4)*cos(x)^3 + 3*(a^10*b - 2*a^8*b^3 + a^6*b^5)*cos(x)^2 + 3*(a^9*b^2 - 2*a^7*b^4 + a^5*b^6)*c
os(x))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.14053, size = 381, normalized size = 2.4 \begin{align*} -\frac{{\left (3 \, a^{2} b - 2 \, b^{3}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{6 \, a^{4} \tan \left (\frac{1}{2} \, x\right )^{5} - 9 \, a^{3} b \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, x\right )^{5} + 15 \, a b^{3} \tan \left (\frac{1}{2} \, x\right )^{5} - 6 \, b^{4} \tan \left (\frac{1}{2} \, x\right )^{5} - 20 \, a^{4} \tan \left (\frac{1}{2} \, x\right )^{3} + 32 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, x\right )^{3} - 12 \, b^{4} \tan \left (\frac{1}{2} \, x\right )^{3} + 6 \, a^{4} \tan \left (\frac{1}{2} \, x\right ) + 9 \, a^{3} b \tan \left (\frac{1}{2} \, x\right ) - 6 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, x\right ) - 15 \, a b^{3} \tan \left (\frac{1}{2} \, x\right ) - 6 \, b^{4} \tan \left (\frac{1}{2} \, x\right )}{3 \,{\left (a^{5} - a^{3} b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - b \tan \left (\frac{1}{2} \, x\right )^{2} - a - b\right )}^{3}} + \frac{x}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^4,x, algorithm="giac")

[Out]

-(3*a^2*b - 2*b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(-a^2
+ b^2)))/((a^6 - a^4*b^2)*sqrt(-a^2 + b^2)) + 1/3*(6*a^4*tan(1/2*x)^5 - 9*a^3*b*tan(1/2*x)^5 - 6*a^2*b^2*tan(
1/2*x)^5 + 15*a*b^3*tan(1/2*x)^5 - 6*b^4*tan(1/2*x)^5 - 20*a^4*tan(1/2*x)^3 + 32*a^2*b^2*tan(1/2*x)^3 - 12*b^4
*tan(1/2*x)^3 + 6*a^4*tan(1/2*x) + 9*a^3*b*tan(1/2*x) - 6*a^2*b^2*tan(1/2*x) - 15*a*b^3*tan(1/2*x) - 6*b^4*tan
(1/2*x))/((a^5 - a^3*b^2)*(a*tan(1/2*x)^2 - b*tan(1/2*x)^2 - a - b)^3) + x/a^4