3.1.32 \(\int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx\) [32]

Optimal. Leaf size=110 \[ \frac {2 b^4 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )} \]

[Out]

2*b^4*arctan((a-b)^(1/2)*tan(1/2*x)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)-1/3*(3*b^3+a*(2*a^2-5*b^2)*cos(x))*cs
c(x)/(a^2-b^2)^2+1/3*(b-a*cos(x))*csc(x)^3/(a^2-b^2)

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Rubi [A]
time = 0.18, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2775, 2945, 12, 2738, 211} \begin {gather*} \frac {\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}-\frac {\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{3 \left (a^2-b^2\right )^2}+\frac {2 b^4 \text {ArcTan}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[x]^4/(a + b*Cos[x]),x]

[Out]

(2*b^4*ArcTan[(Sqrt[a - b]*Tan[x/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)) - ((3*b^3 + a*(2*a^2 - 5*b^2)
*Cos[x])*Csc[x])/(3*(a^2 - b^2)^2) + ((b - a*Cos[x])*Csc[x]^3)/(3*(a^2 - b^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[2*(e/d), Subst[Int[1/(a + b + (a - b)*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 2775

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*Cos
[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b - a*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2945

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c -
b*d)*Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {\csc ^4(x)}{a+b \cos (x)} \, dx &=\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\int \frac {\left (-2 a^2+3 b^2-2 a b \cos (x)\right ) \csc ^2(x)}{a+b \cos (x)} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\int \frac {3 b^4}{a+b \cos (x)} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )}+\frac {b^4 \int \frac {1}{a+b \cos (x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )}+\frac {\left (2 b^4\right ) \text {Subst}\left (\int \frac {1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{\left (a^2-b^2\right )^2}\\ &=\frac {2 b^4 \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (3 b^3+a \left (2 a^2-5 b^2\right ) \cos (x)\right ) \csc (x)}{3 \left (a^2-b^2\right )^2}+\frac {(b-a \cos (x)) \csc ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]
time = 0.70, size = 112, normalized size = 1.02 \begin {gather*} -\frac {2 b^4 \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{5/2}}+\frac {\left (\left (-6 a^3+9 a b^2\right ) \cos (x)+6 b^3 \cos (2 x)+\left (2 a^2-5 b^2\right ) (2 b+a \cos (3 x))\right ) \csc ^3(x)}{12 (a-b)^2 (a+b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^4/(a + b*Cos[x]),x]

[Out]

(-2*b^4*ArcTanh[((a - b)*Tan[x/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (((-6*a^3 + 9*a*b^2)*Cos[x] + 6*b^3
*Cos[2*x] + (2*a^2 - 5*b^2)*(2*b + a*Cos[3*x]))*Csc[x]^3)/(12*(a - b)^2*(a + b)^2)

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Maple [A]
time = 0.21, size = 127, normalized size = 1.15

method	result	size
default	\(\frac {\frac {a \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}-\frac {b \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{3}+3 a \tan \left (\frac {x}{2}\right )-5 b \tan \left (\frac {x}{2}\right )}{8 \left (a -b \right )^{2}}+\frac {2 b^{4} \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{24 \left (a +b \right ) \tan \left (\frac {x}{2}\right )^{3}}-\frac {3 a +5 b}{8 \left (a +b \right )^{2} \tan \left (\frac {x}{2}\right )}\)	\(127\)
risch	\(-\frac {2 i \left (3 b^{3} {\mathrm e}^{5 i x}-3 a \,b^{2} {\mathrm e}^{4 i x}+4 a^{2} b \,{\mathrm e}^{3 i x}-10 b^{3} {\mathrm e}^{3 i x}-6 a^{3} {\mathrm e}^{2 i x}+12 a \,b^{2} {\mathrm e}^{2 i x}+3 b^{3} {\mathrm e}^{i x}+2 a^{3}-5 b^{2} a \right )}{3 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left ({\mathrm e}^{2 i x}-1\right )^{3}}-\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {b^{4} \ln \left ({\mathrm e}^{i x}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2}}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^4/(a+b*cos(x)),x,method=_RETURNVERBOSE)

[Out]

1/8/(a-b)^2*(1/3*a*tan(1/2*x)^3-1/3*b*tan(1/2*x)^3+3*a*tan(1/2*x)-5*b*tan(1/2*x))+2/(a-b)^2/(a+b)^2*b^4/((a-b)
*(a+b))^(1/2)*arctan((a-b)*tan(1/2*x)/((a-b)*(a+b))^(1/2))-1/24/(a+b)/tan(1/2*x)^3-1/8*(3*a+5*b)/(a+b)^2/tan(1
/2*x)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^4/(a+b*cos(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*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (97) = 194\).
time = 0.39, size = 459, normalized size = 4.17 \begin {gather*} \left [\frac {2 \, a^{4} b - 10 \, a^{2} b^{3} + 8 \, b^{5} + 2 \, {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (x\right )^{3} + 3 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (x\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) \sin \left (x\right ) + 6 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2} - 6 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (x\right )}{6 \, {\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 )^{2}\right )} \sin \left (x\right )}, \frac {a^{4} b - 5 \, a^{2} b^{3} + 4 \, b^{5} + {\left (2 \, a^{5} - 7 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (x\right )^{3} - 3 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (x\right )}\right ) \sin \left (x\right ) + 3 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (x\right )^{2} - 3 \, {\left (a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (x\right )}{3 \, {\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 )^{2}\right )} \sin \left (x\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*a^4*b - 10*a^2*b^3 + 8*b^5 + 2*(2*a^5 - 7*a^3*b^2 + 5*a*b^4)*cos(x)^3 + 3*(b^4*cos(x)^2 - b^4)*sqrt(-a
^2 + b^2)*log((2*a*b*cos(x) + (2*a^2 - b^2)*cos(x)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(x) + b)*sin(x) - a^2 + 2*b^2)
/(b^2*cos(x)^2 + 2*a*b*cos(x) + a^2))*sin(x) + 6*(a^2*b^3 - b^5)*cos(x)^2 - 6*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cos(
x))/((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)^2)*sin(x)), 1/3*(a^4*b -
5*a^2*b^3 + 4*b^5 + (2*a^5 - 7*a^3*b^2 + 5*a*b^4)*cos(x)^3 - 3*(b^4*cos(x)^2 - b^4)*sqrt(a^2 - b^2)*arctan(-(a
*cos(x) + b)/(sqrt(a^2 - b^2)*sin(x)))*sin(x) + 3*(a^2*b^3 - b^5)*cos(x)^2 - 3*(a^5 - 3*a^3*b^2 + 2*a*b^4)*cos
(x))/((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)^2)*sin(x))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{4}{\left (x \right )}}{a + b \cos {\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**4/(a + b*cos(x)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (97) = 194\).
time = 0.44, size = 206, normalized size = 1.87 \begin {gather*} -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, x\right ) - b \tan \left (\frac {1}{2} \, x\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b \tan \left (\frac {1}{2} \, x\right )^{3} + b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) - 24 \, a b \tan \left (\frac {1}{2} \, x\right ) + 15 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac {9 \, a \tan \left (\frac {1}{2} \, x\right )^{2} + 15 \, b \tan \left (\frac {1}{2} \, x\right )^{2} + a + b}{24 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (\frac {1}{2} \, x\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*b^4/((a
^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + 1/24*(a^2*tan(1/2*x)^3 - 2*a*b*tan(1/2*x)^3 + b^2*tan(1/2*x)^3 + 9*a^
2*tan(1/2*x) - 24*a*b*tan(1/2*x) + 15*b^2*tan(1/2*x))/(a^3 - 3*a^2*b + 3*a*b^2 - b^3) - 1/24*(9*a*tan(1/2*x)^2
+ 15*b*tan(1/2*x)^2 + a + b)/((a^2 + 2*a*b + b^2)*tan(1/2*x)^3)

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Mupad [B]
time = 0.56, size = 184, normalized size = 1.67 \begin {gather*} \mathrm {tan}\left (\frac {x}{2}\right )\,\left (\frac {4}{8\,a-8\,b}-\frac {8\,a+8\,b}{{\left (8\,a-8\,b\right )}^2}\right )+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{3\,\left (8\,a-8\,b\right )}-\frac {\frac {a^2-2\,a\,b+b^2}{3\,\left (a+b\right )}-\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (-3\,a^3+a^2\,b+7\,a\,b^2-5\,b^3\right )}{{\left (a+b\right )}^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (8\,a^2-16\,a\,b+8\,b^2\right )}+\frac {2\,b^4\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^4*(a + b*cos(x))),x)

[Out]

tan(x/2)*(4/(8*a - 8*b) - (8*a + 8*b)/(8*a - 8*b)^2) + tan(x/2)^3/(3*(8*a - 8*b)) - ((a^2 - 2*a*b + b^2)/(3*(a
+ b)) - (tan(x/2)^2*(7*a*b^2 + a^2*b - 3*a^3 - 5*b^3))/(a + b)^2)/(tan(x/2)^3*(8*a^2 - 16*a*b + 8*b^2)) + (2*
b^4*atan((tan(x/2)*(a^4 + b^4 - 2*a^2*b^2))/((a + b)^(5/2)*(a - b)^(3/2))))/((a + b)^(5/2)*(a - b)^(5/2))

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