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

Optimal. Leaf size=184 \[ \frac {3 b \tanh ^{-1}(\cos (x))}{a^4}-\frac {2 b}{a^3 (a \cos (x)+b \sin (x))}-\frac {\csc (x)}{a^3}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{2 a^2 \sqrt {a^2+b^2}}-\frac {b \cos (x)-a \sin (x)}{2 a^2 (a \cos (x)+b \sin (x))^2}-\frac {2 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^4 \sqrt {a^2+b^2}}-\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^4} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {3105, 3076, 3074, 206, 3103, 3770, 3093} \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^4 \sqrt {a^2+b^2}}-\frac {\sqrt {a^2+b^2} \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{a^4}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{2 a^2 \sqrt {a^2+b^2}}-\frac {2 b}{a^3 (a \cos (x)+b \sin (x))}-\frac {b \cos (x)-a \sin (x)}{2 a^2 (a \cos (x)+b \sin (x))^2}+\frac {3 b \tanh ^{-1}(\cos (x))}{a^4}-\frac {\csc (x)}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 3076

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

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 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]

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 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.80, size = 193, normalized size = 1.05 \[ \frac {\csc ^3(x) (a \cos (x)+b \sin (x)) \left (a \left (a^2+b^2\right ) \sin (x)+\frac {6 \left (a^2+2 b^2\right ) (a \cos (x)+b \sin (x))^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-5 a b (a \cos (x)+b \sin (x))+6 b \log \left (\cos \left (\frac {x}{2}\right )\right ) (a \cos (x)+b \sin (x))^2-6 b \log \left (\sin \left (\frac {x}{2}\right )\right ) (a \cos (x)+b \sin (x))^2-a \tan \left (\frac {x}{2}\right ) (a \cos (x)+b \sin (x))^2-a \cot \left (\frac {x}{2}\right ) (a \cos (x)+b \sin (x))^2\right )}{2 a^4 (a \cot (x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.92, size = 463, normalized size = 2.52 \[ -\frac {2 \, a^{5} - 10 \, a^{3} b^{2} - 12 \, a b^{4} - 6 \, {\left (a^{5} - a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \relax (x)^{2} - 18 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \relax (x) \sin \relax (x) - 3 \, {\left (2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \relax (x)^{3} - 2 \, {\left (a^{3} b + 2 \, a b^{3}\right )} \cos \relax (x) - {\left (a^{2} b^{2} + 2 \, b^{4} + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 6 \, {\left (2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{3} - 2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \relax (x) - {\left (a^{2} b^{3} + b^{5} + {\left (a^{4} b - b^{5}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + 6 \, {\left (2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{3} - 2 \, {\left (a^{3} b^{2} + a b^{4}\right )} \cos \relax (x) - {\left (a^{2} b^{3} + b^{5} + {\left (a^{4} b - b^{5}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (2 \, {\left (a^{7} b + a^{5} b^{3}\right )} \cos \relax (x)^{3} - 2 \, {\left (a^{7} b + a^{5} b^{3}\right )} \cos \relax (x) - {\left (a^{6} b^{2} + a^{4} b^{4} + {\left (a^{8} - a^{4} b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*a^5 - 10*a^3*b^2 - 12*a*b^4 - 6*(a^5 - a^3*b^2 - 2*a*b^4)*cos(x)^2 - 18*(a^4*b + a^2*b^3)*cos(x)*sin(x
) - 3*(2*(a^3*b + 2*a*b^3)*cos(x)^3 - 2*(a^3*b + 2*a*b^3)*cos(x) - (a^2*b^2 + 2*b^4 + (a^4 + a^2*b^2 - 2*b^4)*
cos(x)^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)) - 6*(2*(a^3*b^2 + a*b^4)*cos
(x)^3 - 2*(a^3*b^2 + a*b^4)*cos(x) - (a^2*b^3 + b^5 + (a^4*b - b^5)*cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2) +
6*(2*(a^3*b^2 + a*b^4)*cos(x)^3 - 2*(a^3*b^2 + a*b^4)*cos(x) - (a^2*b^3 + b^5 + (a^4*b - b^5)*cos(x)^2)*sin(x)
)*log(-1/2*cos(x) + 1/2))/(2*(a^7*b + a^5*b^3)*cos(x)^3 - 2*(a^7*b + a^5*b^3)*cos(x) - (a^6*b^2 + a^4*b^4 + (a
^8 - a^4*b^4)*cos(x)^2)*sin(x))

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giac [A]  time = 3.96, size = 212, normalized size = 1.15 \[ -\frac {3 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{4}} - \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a^{3}} - \frac {3 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{2 \, \sqrt {a^{2} + b^{2}} a^{4}} + \frac {6 \, b \tan \left (\frac {1}{2} \, x\right ) - a}{2 \, a^{4} \tan \left (\frac {1}{2} \, x\right )} + \frac {a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 5 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} - 10 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, x\right ) - 14 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 5 \, a^{2} b}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )}^{2} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*b*log(abs(tan(1/2*x)))/a^4 - 1/2*tan(1/2*x)/a^3 - 3/2*(a^2 + 2*b^2)*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) + 1/2*(6*b*tan(1/2*x) - a)/(a^
4*tan(1/2*x)) + (a^3*tan(1/2*x)^3 + 6*a*b^2*tan(1/2*x)^3 + 5*a^2*b*tan(1/2*x)^2 - 10*b^3*tan(1/2*x)^2 + a^3*ta
n(1/2*x) - 14*a*b^2*tan(1/2*x) - 5*a^2*b)/((a*tan(1/2*x)^2 - 2*b*tan(1/2*x) - a)^2*a^4)

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maple [A]  time = 0.75, size = 333, normalized size = 1.81 \[ -\frac {\tan \left (\frac {x}{2}\right )}{2 a^{3}}-\frac {1}{2 a^{3} \tan \left (\frac {x}{2}\right )}-\frac {3 b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{4}}+\frac {\tan ^{3}\left (\frac {x}{2}\right )}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}+\frac {6 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}+\frac {5 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}-\frac {10 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b^{3}}{a^{4} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}+\frac {\tan \left (\frac {x}{2}\right )}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}-\frac {14 \tan \left (\frac {x}{2}\right ) b^{2}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}-\frac {5 b}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}+\frac {3 \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{2} \sqrt {a^{2}+b^{2}}}+\frac {6 \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right ) b^{2}}{a^{4} \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a^3*tan(1/2*x)-1/2/a^3/tan(1/2*x)-3/a^4*b*ln(tan(1/2*x))+1/a/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2*tan(1/2*
x)^3+6/a^3/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2*tan(1/2*x)^3*b^2+5/a^2/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2*tan(
1/2*x)^2*b-10/a^4/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2*tan(1/2*x)^2*b^3+1/a/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2
*tan(1/2*x)-14/a^3/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^2*tan(1/2*x)*b^2-5/a^2/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)^
2*b+3/a^2/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))+6/a^4/(a^2+b^2)^(1/2)*arctanh(1/2*
(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))*b^2

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maxima [A]  time = 0.44, size = 276, normalized size = 1.50 \[ -\frac {a^{3} + \frac {14 \, a^{2} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {4 \, {\left (a^{3} - 8 \, a b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {2 \, {\left (7 \, a^{2} b - 10 \, b^{3}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {{\left (a^{3} + 12 \, a b^{2}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}}{2 \, {\left (\frac {a^{6} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {4 \, a^{5} b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {4 \, a^{5} b \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {a^{6} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} - \frac {2 \, {\left (a^{6} - 2 \, a^{4} b^{2}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}} - \frac {3 \, b \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{4}} - \frac {3 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{2 \, \sqrt {a^{2} + b^{2}} a^{4}} - \frac {\sin \relax (x)}{2 \, a^{3} {\left (\cos \relax (x) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^3 + 14*a^2*b*sin(x)/(cos(x) + 1) - 4*(a^3 - 8*a*b^2)*sin(x)^2/(cos(x) + 1)^2 - 2*(7*a^2*b - 10*b^3)*si
n(x)^3/(cos(x) + 1)^3 - (a^3 + 12*a*b^2)*sin(x)^4/(cos(x) + 1)^4)/(a^6*sin(x)/(cos(x) + 1) + 4*a^5*b*sin(x)^2/
(cos(x) + 1)^2 - 4*a^5*b*sin(x)^4/(cos(x) + 1)^4 + a^6*sin(x)^5/(cos(x) + 1)^5 - 2*(a^6 - 2*a^4*b^2)*sin(x)^3/
(cos(x) + 1)^3) - 3*b*log(sin(x)/(cos(x) + 1))/a^4 - 3/2*(a^2 + 2*b^2)*log((b - a*sin(x)/(cos(x) + 1) + sqrt(a
^2 + b^2))/(b - a*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/(sqrt(a^2 + b^2)*a^4) - 1/2*sin(x)/(a^3*(cos(x) + 1)
)

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mupad [B]  time = 1.01, size = 813, normalized size = 4.42 \[ \frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^4\,\left (a^2+12\,b^2\right )+{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (4\,a^2-32\,b^2\right )-a^2-14\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (7\,a^2\,b-10\,b^3\right )}{a}}{2\,a^5\,\mathrm {tan}\left (\frac {x}{2}\right )-{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (4\,a^5-8\,a^3\,b^2\right )+2\,a^5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5+8\,a^4\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2-8\,a^4\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{2\,a^3}-\frac {3\,b\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^4}-\frac {\mathrm {atan}\left (\frac {\frac {\left (a^2+2\,b^2\right )\,\sqrt {a^2+b^2}\,\left (\frac {3\,a^6+12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a^4\,b+24\,a^2\,b^3\right )}{a^5}-\frac {3\,\left (a^2+2\,b^2\right )\,\left (2\,a^2\,b+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (6\,a^8+8\,a^6\,b^2\right )}{a^5}\right )\,\sqrt {a^2+b^2}}{2\,\left (a^6+a^4\,b^2\right )}\right )\,3{}\mathrm {i}}{2\,\left (a^6+a^4\,b^2\right )}+\frac {\left (a^2+2\,b^2\right )\,\sqrt {a^2+b^2}\,\left (\frac {3\,a^6+12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a^4\,b+24\,a^2\,b^3\right )}{a^5}+\frac {3\,\left (a^2+2\,b^2\right )\,\left (2\,a^2\,b+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (6\,a^8+8\,a^6\,b^2\right )}{a^5}\right )\,\sqrt {a^2+b^2}}{2\,\left (a^6+a^4\,b^2\right )}\right )\,3{}\mathrm {i}}{2\,\left (a^6+a^4\,b^2\right )}}{\frac {2\,\left (9\,a^2\,b+18\,b^3\right )}{a^6}-\frac {2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (9\,a^2+18\,b^2\right )}{a^5}-\frac {3\,\left (a^2+2\,b^2\right )\,\sqrt {a^2+b^2}\,\left (\frac {3\,a^6+12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a^4\,b+24\,a^2\,b^3\right )}{a^5}-\frac {3\,\left (a^2+2\,b^2\right )\,\left (2\,a^2\,b+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (6\,a^8+8\,a^6\,b^2\right )}{a^5}\right )\,\sqrt {a^2+b^2}}{2\,\left (a^6+a^4\,b^2\right )}\right )}{2\,\left (a^6+a^4\,b^2\right )}+\frac {3\,\left (a^2+2\,b^2\right )\,\sqrt {a^2+b^2}\,\left (\frac {3\,a^6+12\,a^4\,b^2}{a^6}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (12\,a^4\,b+24\,a^2\,b^3\right )}{a^5}+\frac {3\,\left (a^2+2\,b^2\right )\,\left (2\,a^2\,b+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (6\,a^8+8\,a^6\,b^2\right )}{a^5}\right )\,\sqrt {a^2+b^2}}{2\,\left (a^6+a^4\,b^2\right )}\right )}{2\,\left (a^6+a^4\,b^2\right )}}\right )\,\left (a^2+2\,b^2\right )\,\sqrt {a^2+b^2}\,3{}\mathrm {i}}{a^6+a^4\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a*cos(x) + b*sin(x))^3),x)

[Out]

(tan(x/2)^4*(a^2 + 12*b^2) + tan(x/2)^2*(4*a^2 - 32*b^2) - a^2 - 14*a*b*tan(x/2) + (2*tan(x/2)^3*(7*a^2*b - 10
*b^3))/a)/(2*a^5*tan(x/2) - tan(x/2)^3*(4*a^5 - 8*a^3*b^2) + 2*a^5*tan(x/2)^5 + 8*a^4*b*tan(x/2)^2 - 8*a^4*b*t
an(x/2)^4) - tan(x/2)/(2*a^3) - (3*b*log(tan(x/2)))/a^4 - (atan((((a^2 + 2*b^2)*(a^2 + b^2)^(1/2)*((3*a^6 + 12
*a^4*b^2)/a^6 + (tan(x/2)*(12*a^4*b + 24*a^2*b^3))/a^5 - (3*(a^2 + 2*b^2)*(2*a^2*b + (tan(x/2)*(6*a^8 + 8*a^6*
b^2))/a^5)*(a^2 + b^2)^(1/2))/(2*(a^6 + a^4*b^2)))*3i)/(2*(a^6 + a^4*b^2)) + ((a^2 + 2*b^2)*(a^2 + b^2)^(1/2)*
((3*a^6 + 12*a^4*b^2)/a^6 + (tan(x/2)*(12*a^4*b + 24*a^2*b^3))/a^5 + (3*(a^2 + 2*b^2)*(2*a^2*b + (tan(x/2)*(6*
a^8 + 8*a^6*b^2))/a^5)*(a^2 + b^2)^(1/2))/(2*(a^6 + a^4*b^2)))*3i)/(2*(a^6 + a^4*b^2)))/((2*(9*a^2*b + 18*b^3)
)/a^6 - (2*tan(x/2)*(9*a^2 + 18*b^2))/a^5 - (3*(a^2 + 2*b^2)*(a^2 + b^2)^(1/2)*((3*a^6 + 12*a^4*b^2)/a^6 + (ta
n(x/2)*(12*a^4*b + 24*a^2*b^3))/a^5 - (3*(a^2 + 2*b^2)*(2*a^2*b + (tan(x/2)*(6*a^8 + 8*a^6*b^2))/a^5)*(a^2 + b
^2)^(1/2))/(2*(a^6 + a^4*b^2))))/(2*(a^6 + a^4*b^2)) + (3*(a^2 + 2*b^2)*(a^2 + b^2)^(1/2)*((3*a^6 + 12*a^4*b^2
)/a^6 + (tan(x/2)*(12*a^4*b + 24*a^2*b^3))/a^5 + (3*(a^2 + 2*b^2)*(2*a^2*b + (tan(x/2)*(6*a^8 + 8*a^6*b^2))/a^
5)*(a^2 + b^2)^(1/2))/(2*(a^6 + a^4*b^2))))/(2*(a^6 + a^4*b^2))))*(a^2 + 2*b^2)*(a^2 + b^2)^(1/2)*3i)/(a^6 + a
^4*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\relax (x )}}{\left (a \cos {\relax (x )} + b \sin {\relax (x )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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