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

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 122, 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} \[ -\frac {\left (a^2-3 b^2\right ) \csc ^3(x)}{3 b^3}+\frac {a \left (a^2-3 b^2\right ) \csc ^2(x)}{2 b^4}-\frac {\left (-3 a^2 b^2+a^4+3 b^4\right ) \csc (x)}{b^5}+\frac {\left (a^2-b^2\right )^3 \log (a+b \csc (x))}{a b^6}+\frac {a \csc ^4(x)}{4 b^2}-\frac {\log (\sin (x))}{a}-\frac {\csc ^5(x)}{5 b} \]

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 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 ^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*}

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Mathematica [A]  time = 0.28, size = 132, normalized size = 1.08 \[ \frac {30 a b^2 \left (a^2-3 b^2\right ) \csc ^2(x)+\frac {60 \left (a^2-b^2\right )^3 \log (a \sin (x)+b)}{a}-20 b^3 \left (a^2-3 b^2\right ) \csc ^3(x)-60 b \left (a^4-3 a^2 b^2+3 b^4\right ) \csc (x)-60 a \left (a^4-3 a^2 b^2+3 b^4\right ) \log (\sin (x))+15 a b^4 \csc ^4(x)-12 b^5 \csc ^5(x)}{60 b^6} \]

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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fricas [B]  time = 0.56, size = 327, normalized size = 2.68 \[ -\frac {60 \, a^{5} b - 160 \, a^{3} b^{3} + 132 \, a b^{5} + 60 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 3 \, a b^{5}\right )} \cos \relax (x)^{4} - 20 \, {\left (6 \, a^{5} b - 17 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \relax (x)^{2} - 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6} + {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{4} - 2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \cos \relax (x)^{2}\right )} \log \left (a \sin \relax (x) + b\right ) \sin \relax (x) + 60 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \relax (x)^{4} - 2 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \sin \relax (x)\right ) \sin \relax (x) - 15 \, {\left (2 \, a^{4} b^{2} - 5 \, a^{2} b^{4} - 2 \, {\left (a^{4} b^{2} - 3 \, a^{2} b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{60 \, {\left (a b^{6} \cos \relax (x)^{4} - 2 \, a b^{6} \cos \relax (x)^{2} + a b^{6}\right )} \sin \relax (x)} \]

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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giac [A]  time = 0.41, size = 155, normalized size = 1.27 \[ -\frac {{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left ({\left | \sin \relax (x) \right |}\right )}{b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left ({\left | a \sin \relax (x) + b \right |}\right )}{a b^{6}} + \frac {15 \, a b^{4} \sin \relax (x) - 12 \, b^{5} - 60 \, {\left (a^{4} b - 3 \, a^{2} b^{3} + 3 \, b^{5}\right )} \sin \relax (x)^{4} + 30 \, {\left (a^{3} b^{2} - 3 \, a b^{4}\right )} \sin \relax (x)^{3} - 20 \, {\left (a^{2} b^{3} - 3 \, b^{5}\right )} \sin \relax (x)^{2}}{60 \, b^{6} \sin \relax (x)^{5}} \]

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)

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maple [A]  time = 0.33, size = 181, normalized size = 1.48 \[ \frac {a^{5} \ln \left (b +a \sin \relax (x )\right )}{b^{6}}-\frac {3 a^{3} \ln \left (b +a \sin \relax (x )\right )}{b^{4}}+\frac {3 a \ln \left (b +a \sin \relax (x )\right )}{b^{2}}-\frac {\ln \left (b +a \sin \relax (x )\right )}{a}-\frac {1}{5 b \sin \relax (x )^{5}}-\frac {a^{2}}{3 b^{3} \sin \relax (x )^{3}}+\frac {1}{b \sin \relax (x )^{3}}-\frac {a^{4}}{b^{5} \sin \relax (x )}+\frac {3 a^{2}}{b^{3} \sin \relax (x )}-\frac {3}{b \sin \relax (x )}+\frac {a}{4 b^{2} \sin \relax (x )^{4}}+\frac {a^{3}}{2 b^{4} \sin \relax (x )^{2}}-\frac {3 a}{2 b^{2} \sin \relax (x )^{2}}-\frac {a^{5} \ln \left (\sin \relax (x )\right )}{b^{6}}+\frac {3 a^{3} \ln \left (\sin \relax (x )\right )}{b^{4}}-\frac {3 a \ln \left (\sin \relax (x )\right )}{b^{2}} \]

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 = 0.32, size = 149, normalized size = 1.22 \[ -\frac {{\left (a^{5} - 3 \, a^{3} b^{2} + 3 \, a b^{4}\right )} \log \left (\sin \relax (x)\right )}{b^{6}} + \frac {{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} \log \left (a \sin \relax (x) + b\right )}{a b^{6}} + \frac {15 \, a b^{3} \sin \relax (x) - 60 \, {\left (a^{4} - 3 \, a^{2} b^{2} + 3 \, b^{4}\right )} \sin \relax (x)^{4} - 12 \, b^{4} + 30 \, {\left (a^{3} b - 3 \, a b^{3}\right )} \sin \relax (x)^{3} - 20 \, {\left (a^{2} b^{2} - 3 \, b^{4}\right )} \sin \relax (x)^{2}}{60 \, b^{5} \sin \relax (x)^{5}} \]

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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mupad [B]  time = 0.87, size = 284, normalized size = 2.33 \[ {\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (\frac {3}{32\,b}-\frac {a^2}{24\,b^3}\right )-\frac {19\,\mathrm {tan}\left (\frac {x}{2}\right )}{16\,b}+\frac {\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}{a}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{160\,b}-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (\frac {a}{32\,b^2}+\frac {a\,\left (\frac {9}{32\,b}-\frac {a^2}{8\,b^3}\right )}{b}\right )+\frac {11\,a^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,b^3}+\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{64\,b^2}-\frac {a^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{2\,b^5}-\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (a^5-3\,a^3\,b^2+3\,a\,b^4\right )}{b^6}-\frac {{\mathrm {cot}\left (\frac {x}{2}\right )}^5\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (10\,a\,b^3-4\,a^3\,b\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (3\,b^4-\frac {4\,a^2\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (16\,a^4-44\,a^2\,b^2+38\,b^4\right )+\frac {b^4}{5}-\frac {a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{2}\right )}{32\,b^5}+\frac {\ln \left (b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )+b\right )\,{\left (a^2-b^2\right )}^3}{a\,b^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^7/(a + b/sin(x)),x)

[Out]

tan(x/2)^3*(3/(32*b) - a^2/(24*b^3)) - (19*tan(x/2))/(16*b) + log(tan(x/2)^2 + 1)/a - tan(x/2)^5/(160*b) - tan
(x/2)^2*(a/(32*b^2) + (a*(9/(32*b) - a^2/(8*b^3)))/b) + (11*a^2*tan(x/2))/(8*b^3) + (a*tan(x/2)^4)/(64*b^2) -
(a^4*tan(x/2))/(2*b^5) - (log(tan(x/2))*(3*a*b^4 + a^5 - 3*a^3*b^2))/b^6 - (cot(x/2)^5*(tan(x/2)^3*(10*a*b^3 -
4*a^3*b) - tan(x/2)^2*(3*b^4 - (4*a^2*b^2)/3) + tan(x/2)^4*(16*a^4 + 38*b^4 - 44*a^2*b^2) + b^4/5 - (a*b^3*ta
n(x/2))/2))/(32*b^5) + (log(b + 2*a*tan(x/2) + b*tan(x/2)^2)*(a^2 - b^2)^3)/(a*b^6)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{7}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(x)**7/(a + b*csc(x)), x)

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