3.31 \(\int \frac{\csc ^3(x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=92 \[ -\frac{b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac{\csc ^2(x) (b-a \cos (x))}{2 \left (a^2-b^2\right )}+\frac{(a+2 b) \log (1-\cos (x))}{4 (a+b)^2}-\frac{(a-2 b) \log (\cos (x)+1)}{4 (a-b)^2} \]

[Out]

((b - a*Cos[x])*Csc[x]^2)/(2*(a^2 - b^2)) + ((a + 2*b)*Log[1 - Cos[x]])/(4*(a + b)^2) - ((a - 2*b)*Log[1 + Cos
[x]])/(4*(a - b)^2) - (b^3*Log[a + b*Cos[x]])/(a^2 - b^2)^2

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Rubi [A]  time = 0.154131, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2668, 741, 801} \[ -\frac{b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac{\csc ^2(x) (b-a \cos (x))}{2 \left (a^2-b^2\right )}+\frac{(a+2 b) \log (1-\cos (x))}{4 (a+b)^2}-\frac{(a-2 b) \log (\cos (x)+1)}{4 (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a + b*Cos[x]),x]

[Out]

((b - a*Cos[x])*Csc[x]^2)/(2*(a^2 - b^2)) + ((a + 2*b)*Log[1 - Cos[x]])/(4*(a + b)^2) - ((a - 2*b)*Log[1 + Cos
[x]])/(4*(a - b)^2) - (b^3*Log[a + b*Cos[x]])/(a^2 - b^2)^2

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.42895, size = 99, normalized size = 1.08 \[ \frac{1}{8} \left (-\frac{8 b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}-\frac{\csc ^2\left (\frac{x}{2}\right )}{a+b}+\frac{\sec ^2\left (\frac{x}{2}\right )}{a-b}+\frac{4 (a+2 b) \log \left (\sin \left (\frac{x}{2}\right )\right )}{(a+b)^2}-\frac{4 (a-2 b) \log \left (\cos \left (\frac{x}{2}\right )\right )}{(a-b)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a + b*Cos[x]),x]

[Out]

(-(Csc[x/2]^2/(a + b)) - (4*(a - 2*b)*Log[Cos[x/2]])/(a - b)^2 - (8*b^3*Log[a + b*Cos[x]])/(a^2 - b^2)^2 + (4*
(a + 2*b)*Log[Sin[x/2]])/(a + b)^2 + Sec[x/2]^2/(a - b))/8

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Maple [A]  time = 0.052, size = 114, normalized size = 1.2 \begin{align*} -{\frac{{b}^{3}\ln \left ( a+b\cos \left ( x \right ) \right ) }{ \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}+{\frac{1}{ \left ( 4\,a+4\,b \right ) \left ( -1+\cos \left ( x \right ) \right ) }}+{\frac{\ln \left ( -1+\cos \left ( x \right ) \right ) a}{4\, \left ( a+b \right ) ^{2}}}+{\frac{\ln \left ( -1+\cos \left ( x \right ) \right ) b}{2\, \left ( a+b \right ) ^{2}}}+{\frac{1}{ \left ( 4\,a-4\,b \right ) \left ( \cos \left ( x \right ) +1 \right ) }}-{\frac{\ln \left ( \cos \left ( x \right ) +1 \right ) a}{4\, \left ( a-b \right ) ^{2}}}+{\frac{\ln \left ( \cos \left ( x \right ) +1 \right ) b}{2\, \left ( a-b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^3/(a+b*cos(x)),x)

[Out]

-b^3/(a+b)^2/(a-b)^2*ln(a+b*cos(x))+1/(4*a+4*b)/(-1+cos(x))+1/4/(a+b)^2*ln(-1+cos(x))*a+1/2/(a+b)^2*ln(-1+cos(
x))*b+1/(4*a-4*b)/(cos(x)+1)-1/4/(a-b)^2*ln(cos(x)+1)*a+1/2/(a-b)^2*ln(cos(x)+1)*b

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Maxima [A]  time = 1.50236, size = 155, normalized size = 1.68 \begin{align*} -\frac{b^{3} \log \left (b \cos \left (x\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (a - 2 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (a + 2 \, b\right )} \log \left (\cos \left (x\right ) - 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{a \cos \left (x\right ) - b}{2 \,{\left ({\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="maxima")

[Out]

-b^3*log(b*cos(x) + a)/(a^4 - 2*a^2*b^2 + b^4) - 1/4*(a - 2*b)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2) + 1/4*(a +
2*b)*log(cos(x) - 1)/(a^2 + 2*a*b + b^2) + 1/2*(a*cos(x) - b)/((a^2 - b^2)*cos(x)^2 - a^2 + b^2)

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Fricas [B]  time = 2.02705, size = 429, normalized size = 4.66 \begin{align*} \frac{2 \, a^{2} b - 2 \, b^{3} - 2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (x\right ) + 4 \,{\left (b^{3} \cos \left (x\right )^{2} - b^{3}\right )} \log \left (-b \cos \left (x\right ) - a\right ) -{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3} -{\left (a^{3} - 3 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3} -{\left (a^{3} - 3 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (x\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="fricas")

[Out]

1/4*(2*a^2*b - 2*b^3 - 2*(a^3 - a*b^2)*cos(x) + 4*(b^3*cos(x)^2 - b^3)*log(-b*cos(x) - a) - (a^3 - 3*a*b^2 - 2
*b^3 - (a^3 - 3*a*b^2 - 2*b^3)*cos(x)^2)*log(1/2*cos(x) + 1/2) + (a^3 - 3*a*b^2 + 2*b^3 - (a^3 - 3*a*b^2 + 2*b
^3)*cos(x)^2)*log(-1/2*cos(x) + 1/2))/(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(x)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\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)**3/(a+b*cos(x)),x)

[Out]

Integral(csc(x)**3/(a + b*cos(x)), x)

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Giac [A]  time = 1.16905, size = 184, normalized size = 2. \begin{align*} -\frac{b^{4} \log \left ({\left | b \cos \left (x\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac{{\left (a - 2 \, b\right )} \log \left (\cos \left (x\right ) + 1\right )}{4 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac{{\left (a + 2 \, b\right )} \log \left (-\cos \left (x\right ) + 1\right )}{4 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{2} b - b^{3} -{\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \,{\left (a + b\right )}^{2}{\left (a - b\right )}^{2}{\left (\cos \left (x\right ) + 1\right )}{\left (\cos \left (x\right ) - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*cos(x)),x, algorithm="giac")

[Out]

-b^4*log(abs(b*cos(x) + a))/(a^4*b - 2*a^2*b^3 + b^5) - 1/4*(a - 2*b)*log(cos(x) + 1)/(a^2 - 2*a*b + b^2) + 1/
4*(a + 2*b)*log(-cos(x) + 1)/(a^2 + 2*a*b + b^2) - 1/2*(a^2*b - b^3 - (a^3 - a*b^2)*cos(x))/((a + b)^2*(a - b)
^2*(cos(x) + 1)*(cos(x) - 1))