3.22 \(\int \frac {\cot ^4(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=91 \[ -\frac {\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac {2 \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a b^3}+\frac {a \cot (x)}{b^2}+\frac {x}{a}-\frac {\cot (x) \csc (x)}{2 b} \]

[Out]

x/a-1/2*(2*a^2-3*b^2)*arctanh(cos(x))/b^3+2*(a^2-b^2)^(3/2)*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a/b^3+a*
cot(x)/b^2-1/2*cot(x)*csc(x)/b

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3898, 2893, 3057, 2660, 618, 206, 3770} \[ -\frac {\left (2 a^2-3 b^2\right ) \tanh ^{-1}(\cos (x))}{2 b^3}+\frac {2 \left (a^2-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a b^3}+\frac {a \cot (x)}{b^2}+\frac {x}{a}-\frac {\cot (x) \csc (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]^4/(a + b*Csc[x]),x]

[Out]

x/a - ((2*a^2 - 3*b^2)*ArcTanh[Cos[x]])/(2*b^3) + (2*(a^2 - b^2)^(3/2)*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2
]])/(a*b^3) + (a*Cot[x])/b^2 - (Cot[x]*Csc[x])/(2*b)

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2893

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 1))/(a*d*f*(n + 1)), x] +
(-Dist[1/(a^2*d^2*(n + 1)*(n + 2)), Int[(a + b*Sin[e + f*x])^m*(d*Sin[e + f*x])^(n + 2)*Simp[a^2*n*(n + 2) -
b^2*(m + n + 2)*(m + n + 3) + a*b*m*Sin[e + f*x] - (a^2*(n + 1)*(n + 2) - b^2*(m + n + 2)*(m + n + 4))*Sin[e +
f*x]^2, x], x], x] - Simp[(b*(m + n + 2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(d*Sin[e + f*x])^(n + 2))/
(a^2*d^2*f*(n + 1)*(n + 2)), x]) /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Intege
rsQ[2*m, 2*n]) &&  !m < -1 && LtQ[n, -1] && (LtQ[n, -2] || EqQ[m + n + 4, 0])

Rule 3057

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(C*x)/(b*d), x] + (Dist[(A*b^2 - a*b*B + a
^2*C)/(b*(b*c - a*d)), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)), Int[
1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2
, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3898

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.41, size = 158, normalized size = 1.74 \[ \frac {8 a^3 \log \left (\sin \left (\frac {x}{2}\right )\right )-8 a^3 \log \left (\cos \left (\frac {x}{2}\right )\right )-16 \left (b^2-a^2\right )^{3/2} \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )-4 a^2 b \tan \left (\frac {x}{2}\right )+4 a^2 b \cot \left (\frac {x}{2}\right )-a b^2 \csc ^2\left (\frac {x}{2}\right )+a b^2 \sec ^2\left (\frac {x}{2}\right )-12 a b^2 \log \left (\sin \left (\frac {x}{2}\right )\right )+12 a b^2 \log \left (\cos \left (\frac {x}{2}\right )\right )+8 b^3 x}{8 a b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^4/(a + b*Csc[x]),x]

[Out]

(8*b^3*x - 16*(-a^2 + b^2)^(3/2)*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + 4*a^2*b*Cot[x/2] - a*b^2*Csc[x/2]
^2 - 8*a^3*Log[Cos[x/2]] + 12*a*b^2*Log[Cos[x/2]] + 8*a^3*Log[Sin[x/2]] - 12*a*b^2*Log[Sin[x/2]] + a*b^2*Sec[x
/2]^2 - 4*a^2*b*Tan[x/2])/(8*a*b^3)

________________________________________________________________________________________

fricas [B]  time = 0.63, size = 453, normalized size = 4.98 \[ \left [\frac {4 \, b^{3} x \cos \relax (x)^{2} - 4 \, a^{2} b \cos \relax (x) \sin \relax (x) - 4 \, b^{3} x + 2 \, a b^{2} \cos \relax (x) - 2 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + {\left (2 \, a^{3} - 3 \, a b^{2} - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2} - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a b^{3} \cos \relax (x)^{2} - a b^{3}\right )}}, \frac {4 \, b^{3} x \cos \relax (x)^{2} - 4 \, a^{2} b \cos \relax (x) \sin \relax (x) - 4 \, b^{3} x + 2 \, a b^{2} \cos \relax (x) + 4 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) + {\left (2 \, a^{3} - 3 \, a b^{2} - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2} - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \cos \relax (x)^{2}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a b^{3} \cos \relax (x)^{2} - a b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/4*(4*b^3*x*cos(x)^2 - 4*a^2*b*cos(x)*sin(x) - 4*b^3*x + 2*a*b^2*cos(x) - 2*((a^2 - b^2)*cos(x)^2 - a^2 + b^
2)*sqrt(a^2 - b^2)*log(-((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 - 2*(b*cos(x)*sin(x) + a*cos(x))*sq
rt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log
(1/2*cos(x) + 1/2) - (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a*b^3*cos(x)^2 -
a*b^3), 1/4*(4*b^3*x*cos(x)^2 - 4*a^2*b*cos(x)*sin(x) - 4*b^3*x + 2*a*b^2*cos(x) + 4*((a^2 - b^2)*cos(x)^2 - a
^2 + b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) + (2*a^3 - 3*a*b^2 -
(2*a^3 - 3*a*b^2)*cos(x)^2)*log(1/2*cos(x) + 1/2) - (2*a^3 - 3*a*b^2 - (2*a^3 - 3*a*b^2)*cos(x)^2)*log(-1/2*co
s(x) + 1/2))/(a*b^3*cos(x)^2 - a*b^3)]

________________________________________________________________________________________

giac [B]  time = 0.38, size = 163, normalized size = 1.79 \[ \frac {x}{a} + \frac {b \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a \tan \left (\frac {1}{2} \, x\right )}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, b^{3}} - \frac {2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a b^{3}} - \frac {12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 18 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 4 \, a b \tan \left (\frac {1}{2} \, x\right ) + b^{2}}{8 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4/(a+b*csc(x)),x, algorithm="giac")

[Out]

x/a + 1/8*(b*tan(1/2*x)^2 - 4*a*tan(1/2*x))/b^2 + 1/2*(2*a^2 - 3*b^2)*log(abs(tan(1/2*x)))/b^3 - 2*(a^4 - 2*a^
2*b^2 + b^4)*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)
*a*b^3) - 1/8*(12*a^2*tan(1/2*x)^2 - 18*b^2*tan(1/2*x)^2 - 4*a*b*tan(1/2*x) + b^2)/(b^3*tan(1/2*x)^2)

________________________________________________________________________________________

maple [B]  time = 0.33, size = 206, normalized size = 2.26 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 b}-\frac {a \tan \left (\frac {x}{2}\right )}{2 b^{2}}-\frac {2 a^{3} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b^{3} \sqrt {-a^{2}+b^{2}}}+\frac {4 a \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \sqrt {-a^{2}+b^{2}}}-\frac {2 b \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}-\frac {1}{8 b \tan \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right ) a^{2}}{b^{3}}-\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 b}+\frac {a}{2 b^{2} \tan \left (\frac {x}{2}\right )}+\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4/(a+b*csc(x)),x)

[Out]

1/8/b*tan(1/2*x)^2-1/2/b^2*a*tan(1/2*x)-2/b^3*a^3/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^
(1/2))+4/b*a/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-2*b/a/(-a^2+b^2)^(1/2)*arctan(
1/2*(2*tan(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-1/8/b/tan(1/2*x)^2+1/b^3*ln(tan(1/2*x))*a^2-3/2/b*ln(tan(1/2*x))+1/
2*a/b^2/tan(1/2*x)+2/a*arctan(tan(1/2*x))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)^4/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 1.19, size = 2163, normalized size = 23.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^4/(a + b/sin(x)),x)

[Out]

(a^3*log(sin(x/2)/cos(x/2)))/(a*b^3 - a*b^3*cos(2*x)) - (2*b^3*atan((2*b^3*cos(x/2) + 2*a^3*sin(x/2) - 3*a*b^2
*sin(x/2))/(2*b^3*sin(x/2) - 2*a^3*cos(x/2) + 3*a*b^2*cos(x/2))))/(a*b^3 - a*b^3*cos(2*x)) - (atan((32*a^6*sin
(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 14*b^6*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) -
14*b^12*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 19*a*b^5*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*
b^2)^(3/2) + 16*a^5*b*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 13*a*b^11*cos(x/2)*(a^6 - b^6 + 3*a
^2*b^4 - 3*a^4*b^2)^(1/2) - 36*a^3*b^3*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 24*a^3*b^9*cos(x/2
)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 8*a^5*b^7*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 2
*a^7*b^5*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 63*a^2*b^4*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a
^4*b^2)^(3/2) - 82*a^4*b^2*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 72*a^2*b^10*sin(x/2)*(a^6 - b^
6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 106*a^4*b^8*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 56*a^6*b^6
*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 11*a^8*b^4*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^
(1/2) + 2*a^10*b^2*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(a^15*sin(x/2)*32i + a^2*b^13*cos(x/2)*
27i - a^4*b^11*cos(x/2)*165i + a^6*b^9*cos(x/2)*390i - a^8*b^7*cos(x/2)*469i + a^10*b^5*cos(x/2)*309i - a^12*b
^3*cos(x/2)*108i + a^3*b^12*sin(x/2)*108i - a^5*b^10*sin(x/2)*482i + a^7*b^8*sin(x/2)*982i - a^9*b^6*sin(x/2)*
1080i + a^11*b^4*sin(x/2)*670i - a^13*b^2*sin(x/2)*224i + a^14*b*cos(x/2)*16i - a*b^14*sin(x/2)*6i))*(a^6 - b^
6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2)*2i)/(a*b^3 - a*b^3*cos(2*x)) - (a^3*cos(2*x)*log(sin(x/2)/cos(x/2)))/(a*b^3 -
a*b^3*cos(2*x)) + (atan((32*a^6*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 14*b^6*sin(x/2)*(a^6 - b
^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 14*b^12*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 19*a*b^5*cos(
x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) + 16*a^5*b*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) +
13*a*b^11*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 36*a^3*b^3*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3
*a^4*b^2)^(3/2) - 24*a^3*b^9*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 8*a^5*b^7*cos(x/2)*(a^6 - b^
6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 2*a^7*b^5*cos(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 63*a^2*b^4*s
in(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3/2) - 82*a^4*b^2*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(3
/2) + 72*a^2*b^10*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 106*a^4*b^8*sin(x/2)*(a^6 - b^6 + 3*a^2
*b^4 - 3*a^4*b^2)^(1/2) + 56*a^6*b^6*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) - 11*a^8*b^4*sin(x/2)*
(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2) + 2*a^10*b^2*sin(x/2)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/(a^
15*sin(x/2)*32i + a^2*b^13*cos(x/2)*27i - a^4*b^11*cos(x/2)*165i + a^6*b^9*cos(x/2)*390i - a^8*b^7*cos(x/2)*46
9i + a^10*b^5*cos(x/2)*309i - a^12*b^3*cos(x/2)*108i + a^3*b^12*sin(x/2)*108i - a^5*b^10*sin(x/2)*482i + a^7*b
^8*sin(x/2)*982i - a^9*b^6*sin(x/2)*1080i + a^11*b^4*sin(x/2)*670i - a^13*b^2*sin(x/2)*224i + a^14*b*cos(x/2)*
16i - a*b^14*sin(x/2)*6i))*cos(2*x)*(a^6 - b^6 + 3*a^2*b^4 - 3*a^4*b^2)^(1/2)*2i)/(a*b^3 - a*b^3*cos(2*x)) - (
3*a*b^2*log(sin(x/2)/cos(x/2)))/(2*(a*b^3 - a*b^3*cos(2*x))) + (2*b^3*cos(2*x)*atan((2*b^3*cos(x/2) + 2*a^3*si
n(x/2) - 3*a*b^2*sin(x/2))/(2*b^3*sin(x/2) - 2*a^3*cos(x/2) + 3*a*b^2*cos(x/2))))/(a*b^3 - a*b^3*cos(2*x)) - (
a*b^2*cos(x))/(a*b^3 - a*b^3*cos(2*x)) + (a^2*b*sin(2*x))/(a*b^3 - a*b^3*cos(2*x)) + (3*a*b^2*cos(2*x)*log(sin
(x/2)/cos(x/2)))/(2*(a*b^3 - a*b^3*cos(2*x)))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{4}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)**4/(a+b*csc(x)),x)

[Out]

Integral(cot(x)**4/(a + b*csc(x)), x)

________________________________________________________________________________________