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

Optimal. Leaf size=110 \[ \frac{\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}-\frac{\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{3 \left (a^2-b^2\right )^2}+\frac{2 b^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

[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))

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Rubi [A]  time = 0.270887, antiderivative size = 110, normalized size of antiderivative = 1., 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 = {2696, 2866, 12, 2659, 205} \[ \frac{\csc ^3(x) (b-a \cos (x))}{3 \left (a^2-b^2\right )}-\frac{\csc (x) \left (a \left (2 a^2-5 b^2\right ) \cos (x)+3 b^3\right )}{3 \left (a^2-b^2\right )^2}+\frac{2 b^4 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2}} \]

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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[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 2866

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]

Rule 12

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

Rule 2659

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 205

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

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 ) \operatorname{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*}

Mathematica [A]  time = 0.627142, size = 112, normalized size = 1.02 \[ \frac{\csc ^3(x) \left (\left (9 a b^2-6 a^3\right ) \cos (x)+\left (2 a^2-5 b^2\right ) (a \cos (3 x)+2 b)+6 b^3 \cos (2 x)\right )}{12 (a-b)^2 (a+b)^2}-\frac{2 b^4 \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}} \]

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 = 0.088, size = 153, normalized size = 1.4 \begin{align*}{\frac{a}{24\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{24\, \left ( a-b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{3\,a}{8\, \left ( a-b \right ) ^{2}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{5\,b}{8\, \left ( a-b \right ) ^{2}}\tan \left ({\frac{x}{2}} \right ) }-{\frac{1}{24\,a+24\,b} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{3\,a}{8\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{5\,b}{8\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+2\,{\frac{{b}^{4}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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Fricas [B]  time = 1.89326, size = 1027, normalized size = 9.34 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (x \right )}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

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]  time = 1.20125, size = 278, normalized size = 2.53 \begin{align*} -\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{align*}

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)