3.18 \(\int \frac{\cot ^7(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=122 \[ -\frac{\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac{a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \csc (x)}{b^5}+\frac{\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}+\frac{a \csc ^4(x)}{4 b^2}-\frac{\log (\sin (x))}{a}-\frac{\csc ^5(x)}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.123151, antiderivative size = 122, normalized size of antiderivative = 1., 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} \[ -\frac{\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac{a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac{\left (-3 a^2 b^2+a^4+3 b^4\right ) \csc (x)}{b^5}+\frac{\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}+\frac{a \csc ^4(x)}{4 b^2}-\frac{\log (\sin (x))}{a}-\frac{\csc ^5(x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^7/(a + b*Csc[x]),x]

[Out]

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

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]

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]))

Rubi steps

\begin{align*} \int \frac{\cot ^7(x)}{a+b \csc (x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^3}{x (a+x)} \, dx,x,b \csc (x)\right )}{b^6}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a^4 \left (1+\frac{3 b^2 \left (-a^2+b^2\right )}{a^4}\right )+\frac{b^6}{a x}+a \left (a^2-3 b^2\right ) x-\left (a^2-3 b^2\right ) x^2+a x^3-x^4+\frac{\left (a^2-b^2\right )^3}{a (a+x)}\right ) \, dx,x,b \csc (x)\right )}{b^6}\\ &=-\frac{\left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)}{b^5}+\frac{a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac{\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac{a \csc ^4(x)}{4 b^2}-\frac{\csc ^5(x)}{5 b}+\frac{\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}-\frac{\log (\sin (x))}{a}\\ \end{align*}

Mathematica [A]  time = 0.272671, size = 132, normalized size = 1.08 \[ \frac{-20 b^3 \left (a^2-3 b^2\right ) \csc ^3(x)+30 a b^2 \left (a^2-3 b^2\right ) \csc ^2(x)-60 b \left (-3 a^2 b^2+a^4+3 b^4\right ) \csc (x)-60 a \left (-3 a^2 b^2+a^4+3 b^4\right ) \log (\sin (x))+\frac{60 \left (a^2-b^2\right )^3 \log (a \sin (x)+b)}{a}+15 a b^4 \csc ^4(x)-12 b^5 \csc ^5(x)}{60 b^6} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^7/(a + b*Csc[x]),x]

[Out]

(-60*b*(a^4 - 3*a^2*b^2 + 3*b^4)*Csc[x] + 30*a*b^2*(a^2 - 3*b^2)*Csc[x]^2 - 20*b^3*(a^2 - 3*b^2)*Csc[x]^3 + 15
*a*b^4*Csc[x]^4 - 12*b^5*Csc[x]^5 - 60*a*(a^4 - 3*a^2*b^2 + 3*b^4)*Log[Sin[x]] + (60*(a^2 - b^2)^3*Log[b + a*S
in[x]])/a)/(60*b^6)

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Maple [A]  time = 0.053, size = 181, normalized size = 1.5 \begin{align*}{\frac{{a}^{5}\ln \left ( b+a\sin \left ( x \right ) \right ) }{{b}^{6}}}-3\,{\frac{{a}^{3}\ln \left ( b+a\sin \left ( x \right ) \right ) }{{b}^{4}}}+3\,{\frac{a\ln \left ( b+a\sin \left ( x \right ) \right ) }{{b}^{2}}}-{\frac{\ln \left ( b+a\sin \left ( x \right ) \right ) }{a}}-{\frac{1}{5\,b \left ( \sin \left ( x \right ) \right ) ^{5}}}-{\frac{{a}^{2}}{3\,{b}^{3} \left ( \sin \left ( x \right ) \right ) ^{3}}}+{\frac{1}{b \left ( \sin \left ( x \right ) \right ) ^{3}}}-{\frac{{a}^{4}}{{b}^{5}\sin \left ( x \right ) }}+3\,{\frac{{a}^{2}}{{b}^{3}\sin \left ( x \right ) }}-3\,{\frac{1}{b\sin \left ( x \right ) }}+{\frac{a}{4\,{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{4}}}+{\frac{{a}^{3}}{2\,{b}^{4} \left ( \sin \left ( x \right ) \right ) ^{2}}}-{\frac{3\,a}{2\,{b}^{2} \left ( \sin \left ( x \right ) \right ) ^{2}}}-{\frac{{a}^{5}\ln \left ( \sin \left ( x \right ) \right ) }{{b}^{6}}}+3\,{\frac{{a}^{3}\ln \left ( \sin \left ( x \right ) \right ) }{{b}^{4}}}-3\,{\frac{a\ln \left ( \sin \left ( x \right ) \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^7/(a+b*csc(x)),x)

[Out]

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

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Maxima [A]  time = 0.972714, size = 201, normalized size = 1.65 \begin{align*} -\frac{{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (\sin \left (x\right )\right )}{b^{6}} + \frac{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \sin \left (x\right ) + b\right )}{a b^{6}} + \frac{15 \, a b^{3} \sin \left (x\right ) - 60 \,{\left (a^{4} - 3 \, a^{2} b^{2} + 3 \, b^{4}\right )} \sin \left (x\right )^{4} - 12 \, b^{4} + 30 \,{\left (a^{3} b - 3 \, a b^{3}\right )} \sin \left (x\right )^{3} - 20 \,{\left (a^{2} b^{2} - 3 \, b^{4}\right )} \sin \left (x\right )^{2}}{60 \, b^{5} \sin \left (x\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^7/(a+b*csc(x)),x, algorithm="maxima")

[Out]

-(a^5 - 3*a^3*b^2 + 3*a*b^4)*log(sin(x))/b^6 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*log(a*sin(x) + b)/(a*b^6) +
1/60*(15*a*b^3*sin(x) - 60*(a^4 - 3*a^2*b^2 + 3*b^4)*sin(x)^4 - 12*b^4 + 30*(a^3*b - 3*a*b^3)*sin(x)^3 - 20*(
a^2*b^2 - 3*b^4)*sin(x)^2)/(b^5*sin(x)^5)

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Fricas [B]  time = 0.684086, size = 763, normalized size = 6.25 \begin{align*} -\frac{60 \, a^{5} b - 160 \, a^{3} b^{3} + 132 \, a b^{5} + 60 \,{\left (a^{5} b - 3 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \left (x\right )^{4} - 20 \,{\left (6 \, a^{5} b - 17 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (x\right )^{2} - 60 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} +{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{4} - 2 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \left (x\right )^{2}\right )} \log \left (a \sin \left (x\right ) + b\right ) \sin \left (x\right ) + 60 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} +{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{4} - 2 \,{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \sin \left (x\right )\right ) \sin \left (x\right ) - 15 \,{\left (2 \, a^{4} b^{2} - 5 \, a^{2} b^{4} - 2 \,{\left (a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{60 \,{\left (a b^{6} \cos \left (x\right )^{4} - 2 \, a b^{6} \cos \left (x\right )^{2} + a b^{6}\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^7/(a+b*csc(x)),x, algorithm="fricas")

[Out]

-1/60*(60*a^5*b - 160*a^3*b^3 + 132*a*b^5 + 60*(a^5*b - 3*a^3*b^3 + 3*a*b^5)*cos(x)^4 - 20*(6*a^5*b - 17*a^3*b
^3 + 15*a*b^5)*cos(x)^2 - 60*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^4
- 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cos(x)^2)*log(a*sin(x) + b)*sin(x) + 60*(a^6 - 3*a^4*b^2 + 3*a^2*b^4
+ (a^6 - 3*a^4*b^2 + 3*a^2*b^4)*cos(x)^4 - 2*(a^6 - 3*a^4*b^2 + 3*a^2*b^4)*cos(x)^2)*log(-1/2*sin(x))*sin(x) -
15*(2*a^4*b^2 - 5*a^2*b^4 - 2*(a^4*b^2 - 3*a^2*b^4)*cos(x)^2)*sin(x))/((a*b^6*cos(x)^4 - 2*a*b^6*cos(x)^2 + a
*b^6)*sin(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(cot(x)**7/(a+b*csc(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.30405, size = 209, normalized size = 1.71 \begin{align*} -\frac{{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left ({\left | \sin \left (x\right ) \right |}\right )}{b^{6}} + \frac{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | a \sin \left (x\right ) + b \right |}\right )}{a b^{6}} + \frac{15 \, a b^{4} \sin \left (x\right ) - 12 \, b^{5} - 60 \,{\left (a^{4} b - 3 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \left (x\right )^{4} + 30 \,{\left (a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \left (x\right )^{3} - 20 \,{\left (a^{2} b^{3} - 3 \, b^{5}\right )} \sin \left (x\right )^{2}}{60 \, b^{6} \sin \left (x\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^7/(a+b*csc(x)),x, algorithm="giac")

[Out]

-(a^5 - 3*a^3*b^2 + 3*a*b^4)*log(abs(sin(x)))/b^6 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*log(abs(a*sin(x) + b))
/(a*b^6) + 1/60*(15*a*b^4*sin(x) - 12*b^5 - 60*(a^4*b - 3*a^2*b^3 + 3*b^5)*sin(x)^4 + 30*(a^3*b^2 - 3*a*b^4)*s
in(x)^3 - 20*(a^2*b^3 - 3*b^5)*sin(x)^2)/(b^6*sin(x)^5)