3.17 \(\int \frac {\cot ^5(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=72 \[ \frac {\left (a^2-b^2\right )^2 \log (a+b \csc (x))}{a b^4}-\frac {\left (a^2-2 b^2\right ) \csc (x)}{b^3}+\frac {a \csc ^2(x)}{2 b^2}+\frac {\log (\sin (x))}{a}-\frac {\csc ^3(x)}{3 b} \]

[Out]

-(a^2-2*b^2)*csc(x)/b^3+1/2*a*csc(x)^2/b^2-1/3*csc(x)^3/b+(a^2-b^2)^2*ln(a+b*csc(x))/a/b^4+ln(sin(x))/a

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3885, 894} \[ -\frac {\left (a^2-2 b^2\right ) \csc (x)}{b^3}+\frac {\left (a^2-b^2\right )^2 \log (a+b \csc (x))}{a b^4}+\frac {a \csc ^2(x)}{2 b^2}+\frac {\log (\sin (x))}{a}-\frac {\csc ^3(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^5/(a + b*Csc[x]),x]

[Out]

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 85, normalized size = 1.18 \[ \frac {3 a^2 b^2 \csc ^2(x)-6 a b \left (a^2-2 b^2\right ) \csc (x)-6 a^2 \left (a^2-2 b^2\right ) \log (\sin (x))+6 \left (a^2-b^2\right )^2 \log (a \sin (x)+b)-2 a b^3 \csc ^3(x)}{6 a b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^5/(a + b*Csc[x]),x]

[Out]

(-6*a*b*(a^2 - 2*b^2)*Csc[x] + 3*a^2*b^2*Csc[x]^2 - 2*a*b^3*Csc[x]^3 - 6*a^2*(a^2 - 2*b^2)*Log[Sin[x]] + 6*(a^
2 - b^2)^2*Log[b + a*Sin[x]])/(6*a*b^4)

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fricas [B]  time = 0.51, size = 151, normalized size = 2.10 \[ -\frac {3 \, a^{2} b^{2} \sin \relax (x) - 6 \, a^{3} b + 10 \, a b^{3} + 6 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \relax (x)^{2} + 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (a \sin \relax (x) + b\right ) \sin \relax (x) - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} - {\left (a^{4} - 2 \, a^{2} b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \sin \relax (x)\right ) \sin \relax (x)}{6 \, {\left (a b^{4} \cos \relax (x)^{2} - a b^{4}\right )} \sin \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5/(a+b*csc(x)),x, algorithm="fricas")

[Out]

-1/6*(3*a^2*b^2*sin(x) - 6*a^3*b + 10*a*b^3 + 6*(a^3*b - 2*a*b^3)*cos(x)^2 + 6*(a^4 - 2*a^2*b^2 + b^4 - (a^4 -
2*a^2*b^2 + b^4)*cos(x)^2)*log(a*sin(x) + b)*sin(x) - 6*(a^4 - 2*a^2*b^2 - (a^4 - 2*a^2*b^2)*cos(x)^2)*log(1/
2*sin(x))*sin(x))/((a*b^4*cos(x)^2 - a*b^4)*sin(x))

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giac [A]  time = 0.39, size = 90, normalized size = 1.25 \[ -\frac {{\left (a^{3} - 2 \, a b^{2}\right )} \log \left ({\left | \sin \relax (x) \right |}\right )}{b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | a \sin \relax (x) + b \right |}\right )}{a b^{4}} + \frac {3 \, a b^{2} \sin \relax (x) - 2 \, b^{3} - 6 \, {\left (a^{2} b - 2 \, b^{3}\right )} \sin \relax (x)^{2}}{6 \, b^{4} \sin \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5/(a+b*csc(x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.32, size = 100, normalized size = 1.39 \[ \frac {a^{3} \ln \left (b +a \sin \relax (x )\right )}{b^{4}}-\frac {2 a \ln \left (b +a \sin \relax (x )\right )}{b^{2}}+\frac {\ln \left (b +a \sin \relax (x )\right )}{a}-\frac {1}{3 b \sin \relax (x )^{3}}-\frac {a^{2}}{b^{3} \sin \relax (x )}+\frac {2}{b \sin \relax (x )}+\frac {a}{2 b^{2} \sin \relax (x )^{2}}-\frac {a^{3} \ln \left (\sin \relax (x )\right )}{b^{4}}+\frac {2 a \ln \left (\sin \relax (x )\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^5/(a+b*csc(x)),x)

[Out]

1/b^4*a^3*ln(b+a*sin(x))-2/b^2*a*ln(b+a*sin(x))+1/a*ln(b+a*sin(x))-1/3/b/sin(x)^3-1/b^3/sin(x)*a^2+2/b/sin(x)+
1/2*a/b^2/sin(x)^2-1/b^4*a^3*ln(sin(x))+2*a/b^2*ln(sin(x))

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maxima [A]  time = 0.31, size = 84, normalized size = 1.17 \[ -\frac {{\left (a^{3} - 2 \, a b^{2}\right )} \log \left (\sin \relax (x)\right )}{b^{4}} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (a \sin \relax (x) + b\right )}{a b^{4}} + \frac {3 \, a b \sin \relax (x) - 6 \, {\left (a^{2} - 2 \, b^{2}\right )} \sin \relax (x)^{2} - 2 \, b^{2}}{6 \, b^{3} \sin \relax (x)^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^5/(a+b*csc(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.57, size = 157, normalized size = 2.18 \[ \mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {7}{8\,b}-\frac {a^2}{2\,b^3}\right )-\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{24\,b}+\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,b^2}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^2-7\,b^2\right )+\frac {b^2}{3}-a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (2\,a\,b^2-a^3\right )}{b^4}+\frac {\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )\,{\left (a^2-b^2\right )}^2}{a\,b^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^5/(a + b/sin(x)),x)

[Out]

tan(x/2)*(7/(8*b) - a^2/(2*b^3)) - log(tan(x/2)^2 + 1)/a - tan(x/2)^3/(24*b) + (a*tan(x/2)^2)/(8*b^2) - (tan(x
/2)^2*(4*a^2 - 7*b^2) + b^2/3 - a*b*tan(x/2))/(8*b^3*tan(x/2)^3) + (log(tan(x/2))*(2*a*b^2 - a^3))/b^4 + (log(
b + 2*a*tan(x/2) + b*tan(x/2)^2)*(a^2 - b^2)^2)/(a*b^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{5}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**5/(a+b*csc(x)),x)

[Out]

Integral(cot(x)**5/(a + b*csc(x)), x)

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