3.1.31 \(\int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx\) [31]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {2747, 755, 815} \begin {gather*} \frac {\csc ^2(x) (b-a \cos (x))}{2 \left (a^2-b^2\right )}-\frac {b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac {(a+2 b) \log (1-\cos (x))}{4 (a+b)^2}-\frac {(a-2 b) \log (\cos (x)+1)}{4 (a-b)^2} \end {gather*}

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 755

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 815

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]

Rule 2747

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]

Rubi steps

\begin {align*} \int \frac {\csc ^3(x)}{a+b \cos (x)} \, dx &=-\left (b^3 \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.46, size = 99, normalized size = 1.08 \begin {gather*} \frac {1}{8} \left (-\frac {\csc ^2\left (\frac {x}{2}\right )}{a+b}-\frac {4 (a-2 b) \log \left (\cos \left (\frac {x}{2}\right )\right )}{(a-b)^2}-\frac {8 b^3 \log (a+b \cos (x))}{\left (a^2-b^2\right )^2}+\frac {4 (a+2 b) \log \left (\sin \left (\frac {x}{2}\right )\right )}{(a+b)^2}+\frac {\sec ^2\left (\frac {x}{2}\right )}{a-b}\right ) \end {gather*}

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 = 0.16, size = 96, normalized size = 1.04

method	result	size
default	\(-\frac {b^{3} \ln \left (a +b \cos \left (x \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (x \right )\right )}+\frac {\left (a +2 b \right ) \ln \left (-1+\cos \left (x \right )\right )}{4 \left (a +b \right )^{2}}+\frac {1}{\left (4 a -4 b \right ) \left (\cos \left (x \right )+1\right )}+\frac {\left (-a +2 b \right ) \ln \left (\cos \left (x \right )+1\right )}{4 \left (a -b \right )^{2}}\)	\(96\)
norman	\(\frac {-\frac {1}{8 \left (a +b \right )}+\frac {\tan ^{4}\left (\frac {x}{2}\right )}{8 a -8 b}}{\tan \left (\frac {x}{2}\right )^{2}}-\frac {b^{3} \ln \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+a +b \right )}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {\left (a +2 b \right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}+4 a b +2 b^{2}}\)	\(100\)
risch	\(-\frac {i x a}{2 \left (a^{2}+2 a b +b^{2}\right )}-\frac {i x b}{a^{2}+2 a b +b^{2}}+\frac {i x a}{2 a^{2}-4 a b +2 b^{2}}-\frac {i x b}{a^{2}-2 a b +b^{2}}+\frac {2 i x \,b^{3}}{a^{4}-2 a^{2} b^{2}+b^{4}}-\frac {a \,{\mathrm e}^{3 i x}-2 b \,{\mathrm e}^{2 i x}+a \,{\mathrm e}^{i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2} \left (-a^{2}+b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) a}{2 a^{2}+4 a b +2 b^{2}}+\frac {\ln \left ({\mathrm e}^{i x}-1\right ) b}{a^{2}+2 a b +b^{2}}-\frac {\ln \left ({\mathrm e}^{i x}+1\right ) a}{2 \left (a^{2}-2 a b +b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{i x}+1\right ) b}{a^{2}-2 a b +b^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i x}+\frac {2 a \,{\mathrm e}^{i x}}{b}+1\right )}{a^{4}-2 a^{2} b^{2}+b^{4}}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [A]
time = 0.37, size = 115, normalized size = 1.25 \begin {gather*} -\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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (87) = 174\).
time = 0.42, size = 181, normalized size = 1.97 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{3}{\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)**3/(a+b*cos(x)),x)

[Out]

Integral(csc(x)**3/(a + b*cos(x)), x)

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Giac [A]
time = 0.47, size = 136, normalized size = 1.48 \begin {gather*} -\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 {gather*}

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))

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Mupad [B]
time = 0.51, size = 112, normalized size = 1.22 \begin {gather*} \ln \left (\cos \left (x\right )-1\right )\,\left (\frac {b}{4\,{\left (a+b\right )}^2}+\frac {1}{4\,\left (a+b\right )}\right )+\frac {\frac {b}{2\,\left (a^2-b^2\right )}-\frac {a\,\cos \left (x\right )}{2\,\left (a^2-b^2\right )}}{{\sin \left (x\right )}^2}-\frac {b^3\,\ln \left (a+b\,\cos \left (x\right )\right )}{a^4-2\,a^2\,b^2+b^4}-\frac {\ln \left (\cos \left (x\right )+1\right )\,\left (a-2\,b\right )}{4\,{\left (a-b\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^3*(a + b*cos(x))),x)

[Out]

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

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