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

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

[Out]

-1/2*arctanh(cos(x))/a^2-2*b^2*arctanh(cos(x))/a^4-(a^2+b^2)*arctanh(cos(x))/a^4+2*b*csc(x)/a^3-1/2*cot(x)*csc
(x)/a^2+(a^2+b^2)/a^3/(a*cos(x)+b*sin(x))+3*b*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 = 0.18, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 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 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 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 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 ^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*}

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Mathematica [B]  time = 1.91, size = 270, normalized size = 2.29 \[ \frac {8 a^3 \csc (x)+a^3 \cot (x) \sec ^2\left (\frac {x}{2}\right )-12 a^3 \cot (x) \log \left (\cos \left (\frac {x}{2}\right )\right )+12 a^3 \cot (x) \log \left (\sin \left (\frac {x}{2}\right )\right )-48 b \sqrt {a^2+b^2} (a \cot (x)+b) \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )+a^2 b \sec ^2\left (\frac {x}{2}\right )+12 a^2 b \log \left (\sin \left (\frac {x}{2}\right )\right )-12 a^2 b \log \left (\cos \left (\frac {x}{2}\right )\right )+8 a^2 b \tan \left (\frac {x}{2}\right ) \cot (x)-a \csc ^2\left (\frac {x}{2}\right ) \left (a^2 \cot (x)+b (a-4 b \sin (x))-4 a b \cos (x)\right )+8 a b^2 \tan \left (\frac {x}{2}\right )+8 a b^2 \csc (x)-24 a b^2 \cot (x) \log \left (\cos \left (\frac {x}{2}\right )\right )+24 a b^2 \cot (x) \log \left (\sin \left (\frac {x}{2}\right )\right )+24 b^3 \log \left (\sin \left (\frac {x}{2}\right )\right )-24 b^3 \log \left (\cos \left (\frac {x}{2}\right )\right )}{8 a^4 (a \cot (x)+b)} \]

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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fricas [B]  time = 0.60, size = 345, normalized size = 2.92 \[ -\frac {6 \, a^{2} b \cos \relax (x) \sin \relax (x) + 4 \, a^{3} + 12 \, a b^{2} - 6 \, {\left (a^{3} + 2 \, a b^{2}\right )} \cos \relax (x)^{2} - 6 \, {\left (a b \cos \relax (x)^{3} - a b \cos \relax (x) + {\left (b^{2} \cos \relax (x)^{2} - b^{2}\right )} \sin \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) + 3 \, {\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \relax (x)^{3} - {\left (a^{3} + 2 \, a b^{2}\right )} \cos \relax (x) - {\left (a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 2 \, b^{3}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) - 3 \, {\left ({\left (a^{3} + 2 \, a b^{2}\right )} \cos \relax (x)^{3} - {\left (a^{3} + 2 \, a b^{2}\right )} \cos \relax (x) - {\left (a^{2} b + 2 \, b^{3} - {\left (a^{2} b + 2 \, b^{3}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right )}{4 \, {\left (a^{5} \cos \relax (x)^{3} - a^{5} \cos \relax (x) + {\left (a^{4} b \cos \relax (x)^{2} - a^{4} b\right )} \sin \relax (x)\right )}} \]

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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giac [A]  time = 8.78, size = 215, normalized size = 1.82 \[ \frac {3 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{4}} + \frac {3 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 8 \, a b \tan \left (\frac {1}{2} \, x\right )}{8 \, a^{4}} - \frac {2 \, {\left (a^{2} b \tan \left (\frac {1}{2} \, x\right ) + b^{3} \tan \left (\frac {1}{2} \, x\right ) + a^{3} + a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )} a^{4}} - \frac {18 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 36 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, x\right ) + a^{2}}{8 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2}} \]

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)

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maple [B]  time = 0.72, size = 224, normalized size = 1.90 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 a^{2}}+\frac {\tan \left (\frac {x}{2}\right ) b}{a^{3}}-\frac {1}{8 a^{2} \tan \left (\frac {x}{2}\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{4}}+\frac {b}{a^{3} \tan \left (\frac {x}{2}\right )}-\frac {2 \tan \left (\frac {x}{2}\right ) b}{a^{2} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {2 \tan \left (\frac {x}{2}\right ) b^{3}}{a^{4} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {2}{a \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {2 b^{2}}{a^{3} \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )}-\frac {6 b \sqrt {a^{2}+b^{2}}\, \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{a^{4}} \]

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-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)-2/a^2/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)*tan(1/2*x)*b-2/a^4/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)
*tan(1/2*x)*b^3-2/a/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a)-2/a^3/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-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))

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maxima [B]  time = 0.43, size = 242, normalized size = 2.05 \[ -\frac {a^{3} - \frac {6 \, a^{2} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {{\left (17 \, a^{3} + 32 \, a b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {8 \, {\left (a^{2} b + 2 \, b^{3}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{8 \, {\left (\frac {a^{5} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {2 \, a^{4} b \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {a^{5} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}\right )}} + \frac {\frac {8 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}}{8 \, a^{3}} + \frac {3 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{2 \, a^{4}} + \frac {3 \, {\left (a^{2} b + b^{3}\right )} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}} a^{4}} \]

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]

-1/8*(a^3 - 6*a^2*b*sin(x)/(cos(x) + 1) - (17*a^3 + 32*a*b^2)*sin(x)^2/(cos(x) + 1)^2 - 8*(a^2*b + 2*b^3)*sin(
x)^3/(cos(x) + 1)^3)/(a^5*sin(x)^2/(cos(x) + 1)^2 + 2*a^4*b*sin(x)^3/(cos(x) + 1)^3 - a^5*sin(x)^4/(cos(x) + 1
)^4) + 1/8*(8*b*sin(x)/(cos(x) + 1) + a*sin(x)^2/(cos(x) + 1)^2)/a^3 + 3/2*(a^2 + 2*b^2)*log(sin(x)/(cos(x) +
1))/a^4 + 3*(a^2*b + b^3)*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)

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mupad [B]  time = 0.74, size = 511, normalized size = 4.33 \[ \frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (\frac {17\,a^2}{2}+16\,b^2\right )-\frac {a^2}{2}+3\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a^2\,b+2\,b^3\right )}{a}}{-4\,a^4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a^4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+8\,b\,a^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{8\,a^2}+\frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )\,\left (3\,a^2+6\,b^2\right )}{2\,a^4}+\frac {b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^3}-\frac {6\,b\,\mathrm {atanh}\left (\frac {54\,b^2\,\sqrt {a^2+b^2}}{18\,a^2\,b+90\,b^3+\frac {72\,b^5}{a^2}+\frac {216\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+\frac {144\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^3}+72\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {72\,b^4\,\sqrt {a^2+b^2}}{18\,a^4\,b+72\,b^5+90\,a^2\,b^3+72\,a^3\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {144\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+216\,a\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {144\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{216\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )+90\,a\,b^3+18\,a^3\,b+\frac {72\,b^5}{a}+72\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {144\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2}}+\frac {144\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{18\,a^5\,b+72\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^4\,b^2+90\,a^3\,b^3+216\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b^4+72\,a\,b^5+144\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^6}+\frac {18\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2+b^2}}{18\,a\,b+72\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {90\,b^3}{a}+\frac {72\,b^5}{a^3}+\frac {216\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2}+\frac {144\,b^6\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4}}\right )\,\sqrt {a^2+b^2}}{a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(x/2)^2*((17*a^2)/2 + 16*b^2) - a^2/2 + 3*a*b*tan(x/2) + (4*tan(x/2)^3*(a^2*b + 2*b^3))/a)/(4*a^4*tan(x/2)
^2 - 4*a^4*tan(x/2)^4 + 8*a^3*b*tan(x/2)^3) + tan(x/2)^2/(8*a^2) + (log(tan(x/2))*(3*a^2 + 6*b^2))/(2*a^4) + (
b*tan(x/2))/a^3 - (6*b*atanh((54*b^2*(a^2 + b^2)^(1/2))/(18*a^2*b + 90*b^3 + (72*b^5)/a^2 + (216*b^4*tan(x/2))
/a + (144*b^6*tan(x/2))/a^3 + 72*a*b^2*tan(x/2)) + (72*b^4*(a^2 + b^2)^(1/2))/(18*a^4*b + 72*b^5 + 90*a^2*b^3
+ 72*a^3*b^2*tan(x/2) + (144*b^6*tan(x/2))/a + 216*a*b^4*tan(x/2)) + (144*b^3*tan(x/2)*(a^2 + b^2)^(1/2))/(216
*b^4*tan(x/2) + 90*a*b^3 + 18*a^3*b + (72*b^5)/a + 72*a^2*b^2*tan(x/2) + (144*b^6*tan(x/2))/a^2) + (144*b^5*ta
n(x/2)*(a^2 + b^2)^(1/2))/(144*b^6*tan(x/2) + 72*a*b^5 + 18*a^5*b + 90*a^3*b^3 + 216*a^2*b^4*tan(x/2) + 72*a^4
*b^2*tan(x/2)) + (18*b*tan(x/2)*(a^2 + b^2)^(1/2))/(18*a*b + 72*b^2*tan(x/2) + (90*b^3)/a + (72*b^5)/a^3 + (21
6*b^4*tan(x/2))/a^2 + (144*b^6*tan(x/2))/a^4))*(a^2 + b^2)^(1/2))/a^4

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

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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