3.290 \(\int \frac {1}{(a \cot (x)+b \csc (x))^3} \, dx\)

Optimal. Leaf size=50 \[ \frac {2 b}{a^3 (a \cos (x)+b)}+\frac {\log (a \cos (x)+b)}{a^3}+\frac {a^2-b^2}{2 a^3 (a \cos (x)+b)^2} \]

[Out]

1/2*(a^2-b^2)/a^3/(b+a*cos(x))^2+2*b/a^3/(b+a*cos(x))+ln(b+a*cos(x))/a^3

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Rubi [A]  time = 0.08, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4392, 2668, 697} \[ \frac {a^2-b^2}{2 a^3 (a \cos (x)+b)^2}+\frac {2 b}{a^3 (a \cos (x)+b)}+\frac {\log (a \cos (x)+b)}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a*Cot[x] + b*Csc[x])^(-3),x]

[Out]

(a^2 - b^2)/(2*a^3*(b + a*Cos[x])^2) + (2*b)/(a^3*(b + a*Cos[x])) + Log[b + a*Cos[x]]/a^3

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 77, normalized size = 1.54 \[ \frac {a^2 \cos (2 x) \log (a \cos (x)+b)+a^2 \log (a \cos (x)+b)+a^2+2 b^2 \log (a \cos (x)+b)+4 a b \cos (x) (\log (a \cos (x)+b)+1)+3 b^2}{2 a^3 (a \cos (x)+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*Cot[x] + b*Csc[x])^(-3),x]

[Out]

(a^2 + 3*b^2 + a^2*Log[b + a*Cos[x]] + 2*b^2*Log[b + a*Cos[x]] + a^2*Cos[2*x]*Log[b + a*Cos[x]] + 4*a*b*Cos[x]
*(1 + Log[b + a*Cos[x]]))/(2*a^3*(b + a*Cos[x])^2)

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fricas [A]  time = 0.84, size = 70, normalized size = 1.40 \[ \frac {4 \, a b \cos \relax (x) + a^{2} + 3 \, b^{2} + 2 \, {\left (a^{2} \cos \relax (x)^{2} + 2 \, a b \cos \relax (x) + b^{2}\right )} \log \left (a \cos \relax (x) + b\right )}{2 \, {\left (a^{5} \cos \relax (x)^{2} + 2 \, a^{4} b \cos \relax (x) + a^{3} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 45, normalized size = 0.90 \[ \frac {\log \left ({\left | a \cos \relax (x) + b \right |}\right )}{a^{3}} + \frac {4 \, b \cos \relax (x) + \frac {a^{2} + 3 \, b^{2}}{a}}{2 \, {\left (a \cos \relax (x) + b\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^3,x, algorithm="giac")

[Out]

log(abs(a*cos(x) + b))/a^3 + 1/2*(4*b*cos(x) + (a^2 + 3*b^2)/a)/((a*cos(x) + b)^2*a^2)

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maple [A]  time = 0.12, size = 56, normalized size = 1.12 \[ \frac {\ln \left (b +a \cos \relax (x )\right )}{a^{3}}+\frac {1}{2 a \left (b +a \cos \relax (x )\right )^{2}}-\frac {b^{2}}{2 a^{3} \left (b +a \cos \relax (x )\right )^{2}}+\frac {2 b}{a^{3} \left (b +a \cos \relax (x )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*cot(x)+b*csc(x))^3,x)

[Out]

ln(b+a*cos(x))/a^3+1/2/a/(b+a*cos(x))^2-1/2/a^3/(b+a*cos(x))^2*b^2+2*b/a^3/(b+a*cos(x))

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maxima [B]  time = 0.43, size = 177, normalized size = 3.54 \[ \frac {2 \, {\left (a b + b^{2} + \frac {{\left (a^{2} - 2 \, a b + b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}}{a^{5} + a^{4} b - a^{3} b^{2} - a^{2} b^{3} - \frac {2 \, {\left (a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (a^{5} - 3 \, a^{4} b + 3 \, a^{3} b^{2} - a^{2} b^{3}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}} + \frac {\log \left (a + b - \frac {{\left (a - b\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{3}} - \frac {\log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.77, size = 311, normalized size = 6.22 \[ \frac {\frac {2\,\left (b^2+a\,b\right )}{a^2\,\left (a-b\right )}+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a-b\right )}{a^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+2\,a\,b-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+a^2+b^2}-\frac {2\,\mathrm {atanh}\left (\frac {32\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{\frac {32\,b^3}{a^3}-\frac {32\,b^2}{a^2}-\frac {32\,b}{a}+\frac {32\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}-\frac {64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^3}+32}-\frac {64\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32\,a-32\,b+32\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-\frac {32\,b^2}{a}+\frac {32\,b^3}{a^2}-\frac {64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}+\frac {32\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{32\,a^2-32\,a\,b-32\,b^2-64\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+\frac {32\,b^3}{a}+\frac {32\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a}+32\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}\right )}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/sin(x) + a*cot(x))^3,x)

[Out]

((2*(a*b + b^2))/(a^2*(a - b)) + (2*tan(x/2)^2*(a - b))/a^2)/(tan(x/2)^4*(a^2 - 2*a*b + b^2) + 2*a*b - tan(x/2
)^2*(2*a^2 - 2*b^2) + a^2 + b^2) - (2*atanh((32*tan(x/2)^2)/((32*b^3)/a^3 - (32*b^2)/a^2 - (32*b)/a + (32*b*ta
n(x/2)^2)/a - (64*b^2*tan(x/2)^2)/a^2 + (32*b^3*tan(x/2)^2)/a^3 + 32) - (64*b*tan(x/2)^2)/(32*a - 32*b + 32*b*
tan(x/2)^2 - (32*b^2)/a + (32*b^3)/a^2 - (64*b^2*tan(x/2)^2)/a + (32*b^3*tan(x/2)^2)/a^2) + (32*b^2*tan(x/2)^2
)/(32*a^2 - 32*a*b - 32*b^2 - 64*b^2*tan(x/2)^2 + (32*b^3)/a + (32*b^3*tan(x/2)^2)/a + 32*a*b*tan(x/2)^2)))/a^
3

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a \cot {\relax (x )} + b \csc {\relax (x )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*cot(x)+b*csc(x))**3,x)

[Out]

Integral((a*cot(x) + b*csc(x))**(-3), x)

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