3.21 \(\int \frac{\csc ^3(x)}{(a \cos (x)+b \sin (x))^2} \, dx\)

Optimal. Leaf size=118 \[ \frac{a^2+b^2}{a^3 (a \cos (x)+b \sin (x))}-\frac{2 b^2 \tanh ^{-1}(\cos (x))}{a^4}-\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{a^4}+\frac{3 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{a^4}+\frac{2 b \csc (x)}{a^3}-\frac{\tanh ^{-1}(\cos (x))}{2 a^2}-\frac{\cot (x) \csc (x)}{2 a^2} \]

[Out]

-ArcTanh[Cos[x]]/(2*a^2) - (2*b^2*ArcTanh[Cos[x]])/a^4 - ((a^2 + b^2)*ArcTanh[Cos[x]])/a^4 + (3*b*Sqrt[a^2 + b
^2]*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/a^4 + (2*b*Csc[x])/a^3 - (Cot[x]*Csc[x])/(2*a^2) + (a^2 +
b^2)/(a^3*(a*Cos[x] + b*Sin[x]))

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Rubi [A]  time = 0.180463, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3105, 3093, 3770, 3074, 206, 3768, 3103} \[ \frac{a^2+b^2}{a^3 (a \cos (x)+b \sin (x))}-\frac{2 b^2 \tanh ^{-1}(\cos (x))}{a^4}-\frac{\left (a^2+b^2\right ) \tanh ^{-1}(\cos (x))}{a^4}+\frac{3 b \sqrt{a^2+b^2} \tanh ^{-1}\left (\frac{b \cos (x)-a \sin (x)}{\sqrt{a^2+b^2}}\right )}{a^4}+\frac{2 b \csc (x)}{a^3}-\frac{\tanh ^{-1}(\cos (x))}{2 a^2}-\frac{\cot (x) \csc (x)}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

-ArcTanh[Cos[x]]/(2*a^2) - (2*b^2*ArcTanh[Cos[x]])/a^4 - ((a^2 + b^2)*ArcTanh[Cos[x]])/a^4 + (3*b*Sqrt[a^2 + b
^2]*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/a^4 + (2*b*Csc[x])/a^3 - (Cot[x]*Csc[x])/(2*a^2) + (a^2 +
b^2)/(a^3*(a*Cos[x] + b*Sin[x]))

Rule 3105

Int[sin[(c_.) + (d_.)*(x_)]^(m_)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbo
l] :> Dist[(a^2 + b^2)/a^2, Int[Sin[c + d*x]^(m + 2)*(a*Cos[c + d*x] + b*Sin[c + d*x])^n, x], x] + (Dist[1/a^2
, Int[Sin[c + d*x]^m*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 2), x], x] - Dist[(2*b)/a^2, Int[Sin[c + d*x]^(m +
1)*(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n
, -1] && LtQ[m, -1]

Rule 3093

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_)/sin[(c_.) + (d_.)*(x_)], x_Symbol] :>
-Simp[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1)/(a*d*(n + 1)), x] + (Dist[1/a^2, Int[(a*Cos[c + d*x] + b*Sin[
c + d*x])^(n + 2)/Sin[c + d*x], x], x] - Dist[b/a^2, Int[(a*Cos[c + d*x] + b*Sin[c + d*x])^(n + 1), x], x]) /;
FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3770

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

Rule 3074

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3103

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

Rubi steps

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

Mathematica [B]  time = 1.73063, size = 270, normalized size = 2.29 \[ \frac{-48 b \sqrt{a^2+b^2} (a \cot (x)+b) \tanh ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )-b}{\sqrt{a^2+b^2}}\right )+a^2 b \sec ^2\left (\frac{x}{2}\right )+12 a^2 b \log \left (\sin \left (\frac{x}{2}\right )\right )-12 a^2 b \log \left (\cos \left (\frac{x}{2}\right )\right )+8 a^2 b \tan \left (\frac{x}{2}\right ) \cot (x)-a \csc ^2\left (\frac{x}{2}\right ) \left (a^2 \cot (x)+b (a-4 b \sin (x))-4 a b \cos (x)\right )+8 a^3 \csc (x)+a^3 \cot (x) \sec ^2\left (\frac{x}{2}\right )-12 a^3 \cot (x) \log \left (\cos \left (\frac{x}{2}\right )\right )+12 a^3 \cot (x) \log \left (\sin \left (\frac{x}{2}\right )\right )+8 a b^2 \tan \left (\frac{x}{2}\right )+8 a b^2 \csc (x)-24 a b^2 \cot (x) \log \left (\cos \left (\frac{x}{2}\right )\right )+24 a b^2 \cot (x) \log \left (\sin \left (\frac{x}{2}\right )\right )+24 b^3 \log \left (\sin \left (\frac{x}{2}\right )\right )-24 b^3 \log \left (\cos \left (\frac{x}{2}\right )\right )}{8 a^4 (a \cot (x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a*Cos[x] + b*Sin[x])^2,x]

[Out]

(-48*b*Sqrt[a^2 + b^2]*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]]*(b + a*Cot[x]) + 8*a^3*Csc[x] + 8*a*b^2*Csc[
x] - 12*a^2*b*Log[Cos[x/2]] - 24*b^3*Log[Cos[x/2]] - 12*a^3*Cot[x]*Log[Cos[x/2]] - 24*a*b^2*Cot[x]*Log[Cos[x/2
]] + 12*a^2*b*Log[Sin[x/2]] + 24*b^3*Log[Sin[x/2]] + 12*a^3*Cot[x]*Log[Sin[x/2]] + 24*a*b^2*Cot[x]*Log[Sin[x/2
]] + a^2*b*Sec[x/2]^2 + a^3*Cot[x]*Sec[x/2]^2 - a*Csc[x/2]^2*(-4*a*b*Cos[x] + a^2*Cot[x] + b*(a - 4*b*Sin[x]))
+ 8*a*b^2*Tan[x/2] + 8*a^2*b*Cot[x]*Tan[x/2])/(8*a^4*(b + a*Cot[x]))

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Maple [B]  time = 0.144, size = 224, normalized size = 1.9 \begin{align*}{\frac{1}{8\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{b}{{a}^{3}}\tan \left ({\frac{x}{2}} \right ) }-2\,{\frac{b\tan \left ( x/2 \right ) }{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) }}-2\,{\frac{\tan \left ( x/2 \right ){b}^{3}}{{a}^{4} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) }}-2\,{\frac{1}{a \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) }}-2\,{\frac{{b}^{2}}{{a}^{3} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a-2\,b\tan \left ( x/2 \right ) -a \right ) }}-6\,{\frac{b\sqrt{{a}^{2}+{b}^{2}}}{{a}^{4}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) }-{\frac{1}{8\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-2}}+{\frac{3}{2\,{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }+3\,{\frac{\ln \left ( \tan \left ( x/2 \right ) \right ){b}^{2}}{{a}^{4}}}+{\frac{b}{{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^3/(a*cos(x)+b*sin(x))^2,x)

[Out]

1/8/a^2*tan(1/2*x)^2+1/a^3*tan(1/2*x)*b-2/a^2/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)*tan(1/2*x)*b-2/a^4/(tan(1/2*x)
^2*a-2*b*tan(1/2*x)-a)*tan(1/2*x)*b^3-2/a/(tan(1/2*x)^2*a-2*b*tan(1/2*x)-a)-2/a^3/(tan(1/2*x)^2*a-2*b*tan(1/2*
x)-a)*b^2-6/a^4*b*(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))-1/8/a^2/tan(1/2*x)^2+3/2/a
^2*ln(tan(1/2*x))+3/a^4*ln(tan(1/2*x))*b^2+1/a^3*b/tan(1/2*x)

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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)^3/(a*cos(x)+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.708885, size = 859, normalized size = 7.28 \begin{align*} -\frac{6 \, a^{2} b \cos \left (x\right ) \sin \left (x\right ) + 4 \, a^{3} + 12 \, a b^{2} - 6 \,{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (x\right )^{2} - 6 \,{\left (a b \cos \left (x\right )^{3} - a b \cos \left (x\right ) +{\left (b^{2} \cos \left (x\right )^{2} - b^{2}\right )} \sin \left (x\right )\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) + 3 \,{\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (x\right )^{3} -{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (x\right ) -{\left (a^{2} b + 2 \, b^{3} -{\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) - 3 \,{\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (x\right )^{3} -{\left (a^{3} + 2 \, a b^{2}\right )} \cos \left (x\right ) -{\left (a^{2} b + 2 \, b^{3} -{\left (a^{2} b + 2 \, b^{3}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{4 \,{\left (a^{5} \cos \left (x\right )^{3} - a^{5} \cos \left (x\right ) +{\left (a^{4} b \cos \left (x\right )^{2} - a^{4} b\right )} \sin \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a*cos(x)+b*sin(x))^2,x, algorithm="fricas")

[Out]

-1/4*(6*a^2*b*cos(x)*sin(x) + 4*a^3 + 12*a*b^2 - 6*(a^3 + 2*a*b^2)*cos(x)^2 - 6*(a*b*cos(x)^3 - a*b*cos(x) + (
b^2*cos(x)^2 - b^2)*sin(x))*sqrt(a^2 + b^2)*log((2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 - 2*a^2 - b^2 - 2*
sqrt(a^2 + b^2)*(b*cos(x) - a*sin(x)))/(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2)) + 3*((a^3 + 2*a*b^2
)*cos(x)^3 - (a^3 + 2*a*b^2)*cos(x) - (a^2*b + 2*b^3 - (a^2*b + 2*b^3)*cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2)
- 3*((a^3 + 2*a*b^2)*cos(x)^3 - (a^3 + 2*a*b^2)*cos(x) - (a^2*b + 2*b^3 - (a^2*b + 2*b^3)*cos(x)^2)*sin(x))*l
og(-1/2*cos(x) + 1/2))/(a^5*cos(x)^3 - a^5*cos(x) + (a^4*b*cos(x)^2 - a^4*b)*sin(x))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (x \right )}}{\left (a \cos{\left (x \right )} + b \sin{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a*cos(x)+b*sin(x))**2,x)

[Out]

Integral(csc(x)**3/(a*cos(x) + b*sin(x))**2, x)

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Giac [A]  time = 1.25371, size = 290, normalized size = 2.46 \begin{align*} \frac{3 \,{\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{2 \, a^{4}} + \frac{3 \,{\left (a^{2} b + b^{3}\right )} \log \left (\frac{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt{a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac{1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt{a^{2} + b^{2}} \right |}}\right )}{\sqrt{a^{2} + b^{2}} a^{4}} + \frac{a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 8 \, a b \tan \left (\frac{1}{2} \, x\right )}{8 \, a^{4}} - \frac{2 \,{\left (a^{2} b \tan \left (\frac{1}{2} \, x\right ) + b^{3} \tan \left (\frac{1}{2} \, x\right ) + a^{3} + a b^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac{1}{2} \, x\right ) - a\right )} a^{4}} - \frac{18 \, a^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 36 \, b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} - 8 \, a b \tan \left (\frac{1}{2} \, x\right ) + a^{2}}{8 \, a^{4} \tan \left (\frac{1}{2} \, x\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a*cos(x)+b*sin(x))^2,x, algorithm="giac")

[Out]

3/2*(a^2 + 2*b^2)*log(abs(tan(1/2*x)))/a^4 + 3*(a^2*b + b^3)*log(abs(2*a*tan(1/2*x) - 2*b - 2*sqrt(a^2 + b^2))
/abs(2*a*tan(1/2*x) - 2*b + 2*sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^4) + 1/8*(a^2*tan(1/2*x)^2 + 8*a*b*tan(1/2*
x))/a^4 - 2*(a^2*b*tan(1/2*x) + b^3*tan(1/2*x) + a^3 + a*b^2)/((a*tan(1/2*x)^2 - 2*b*tan(1/2*x) - a)*a^4) - 1/
8*(18*a^2*tan(1/2*x)^2 + 36*b^2*tan(1/2*x)^2 - 8*a*b*tan(1/2*x) + a^2)/(a^4*tan(1/2*x)^2)