\(\int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx\) [16]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=-\frac {\csc (x)}{b}-\frac {\left (1-\frac {a^2}{b^2}\right ) \log (a+b \csc (x))}{a}-\frac {\log (\sin (x))}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3970, 908} \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=-\frac {\left (1-\frac {a^2}{b^2}\right ) \log (a+b \csc (x))}{a}-\frac {\log (\sin (x))}{a}-\frac {\csc (x)}{b} \]

[In]

Int[Cot[x]^3/(a + b*Csc[x]),x]

[Out]

-(Csc[x]/b) - ((1 - a^2/b^2)*Log[a + b*Csc[x]])/a - Log[Sin[x]]/a

Rule 908

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 3970

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=\frac {-a b \csc (x)-a^2 \log (\sin (x))+\left (a^2-b^2\right ) \log (b+a \sin (x))}{a b^2} \]

[In]

Integrate[Cot[x]^3/(a + b*Csc[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {1}{b \sin \left (x \right )}-\frac {a \ln \left (\sin \left (x \right )\right )}{b^{2}}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (a \sin \left (x \right )+b \right )}{b^{2} a}\)	\(43\)
risch	\(\frac {i x}{a}-\frac {2 i {\mathrm e}^{i x}}{b \left ({\mathrm e}^{2 i x}-1\right )}-\frac {a \ln \left ({\mathrm e}^{2 i x}-1\right )}{b^{2}}+\frac {a \ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{b^{2}}-\frac {\ln \left ({\mathrm e}^{2 i x}-1+\frac {2 i b \,{\mathrm e}^{i x}}{a}\right )}{a}\)	\(93\)

[In]

int(cot(x)^3/(a+b*csc(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=-\frac {a^{2} \log \left (-\frac {1}{2} \, \sin \left (x\right )\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \log \left (a \sin \left (x\right ) + b\right ) \sin \left (x\right ) + a b}{a b^{2} \sin \left (x\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=\int \frac {\cot ^{3}{\left (x \right )}}{a + b \csc {\left (x \right )}}\, dx \]

[In]

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

[Out]

Integral(cot(x)**3/(a + b*csc(x)), x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.11 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=-\frac {a \log \left (\sin \left (x\right )\right )}{b^{2}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (a \sin \left (x\right ) + b\right )}{a b^{2}} - \frac {1}{b \sin \left (x\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=-\frac {a \log \left ({\left | \sin \left (x\right ) \right |}\right )}{b^{2}} + \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a b^{2}} - \frac {1}{b \sin \left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.97 \[ \int \frac {\cot ^3(x)}{a+b \csc (x)} \, dx=\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )\,\left (\frac {a}{b^2}-\frac {1}{a}\right )-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,b}+\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {1}{2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {a\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{b^2} \]

[In]

int(cot(x)^3/(a + b/sin(x)),x)

[Out]

log(b + 2*a*tan(x/2) + b*tan(x/2)^2)*(a/b^2 - 1/a) - tan(x/2)/(2*b) + log(tan(x/2)^2 + 1)/a - 1/(2*b*tan(x/2))
- (a*log(tan(x/2)))/b^2