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

Optimal. Leaf size=59 \[ -\frac {\log (a+b \tan (x))}{a^3}+\frac {\log (\tan (x))}{a^3}+\frac {\frac {1}{a^2}-\frac {1}{b^2}}{a+b \tan (x)}+\frac {\frac {a}{b^2}+\frac {1}{a}}{2 (a+b \tan (x))^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3087, 894} \[ \frac {\frac {1}{a^2}-\frac {1}{b^2}}{a+b \tan (x)}-\frac {\log (a+b \tan (x))}{a^3}+\frac {\log (\tan (x))}{a^3}+\frac {\frac {a}{b^2}+\frac {1}{a}}{2 (a+b \tan (x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 3087

Int[sin[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[1/d, Subst[Int[(x^m*(a + b*x)^n)/(1 + x^2)^((m + n + 2)/2), x], x, Tan[c + d*x]], x] /; FreeQ[{a,
b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 96, normalized size = 1.63 \[ \frac {2 a^2 \cot ^2(x) (\log (\sin (x))-\log (a \cos (x)+b \sin (x)))+a^2 \csc ^2(x)+2 b^2 (-\log (a \cos (x)+b \sin (x))+\log (\sin (x))-1)+2 a b \cot (x) (-2 \log (a \cos (x)+b \sin (x))+2 \log (\sin (x))-1)}{2 a^3 (a \cot (x)+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.76, size = 220, normalized size = 3.73 \[ \frac {4 \, a^{2} b^{2} \cos \relax (x)^{2} + a^{4} - a^{2} b^{2} - 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x) \sin \relax (x) - {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \relax (x) \sin \relax (x)\right )} \log \left (2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}\right ) + {\left (a^{2} b^{2} + b^{4} + {\left (a^{4} - b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{3} b + a b^{3}\right )} \cos \relax (x) \sin \relax (x)\right )} \log \left (-\frac {1}{4} \, \cos \relax (x)^{2} + \frac {1}{4}\right )}{2 \, {\left (a^{5} b^{2} + a^{3} b^{4} + {\left (a^{7} - a^{3} b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{6} b + a^{4} b^{3}\right )} \cos \relax (x) \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 2.02, size = 77, normalized size = 1.31 \[ -\frac {\log \left ({\left | b \tan \relax (x) + a \right |}\right )}{a^{3}} + \frac {\log \left ({\left | \tan \relax (x) \right |}\right )}{a^{3}} + \frac {3 \, b^{4} \tan \relax (x)^{2} - 2 \, a^{3} b \tan \relax (x) + 8 \, a b^{3} \tan \relax (x) - a^{4} + 6 \, a^{2} b^{2}}{2 \, {\left (b \tan \relax (x) + a\right )}^{2} a^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.67, size = 73, normalized size = 1.24 \[ \frac {a}{2 b^{2} \left (a +b \tan \relax (x )\right )^{2}}+\frac {1}{2 a \left (a +b \tan \relax (x )\right )^{2}}-\frac {1}{b^{2} \left (a +b \tan \relax (x )\right )}+\frac {1}{a^{2} \left (a +b \tan \relax (x )\right )}-\frac {\ln \left (a +b \tan \relax (x )\right )}{a^{3}}+\frac {\ln \left (\tan \relax (x )\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 172, normalized size = 2.92 \[ -\frac {2 \, {\left (\frac {2 \, a b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, a b \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} - \frac {{\left (a^{2} - 3 \, b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}}{a^{5} + \frac {4 \, a^{4} b \sin \relax (x)}{\cos \relax (x) + 1} - \frac {4 \, a^{4} b \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {a^{5} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}} - \frac {\log \left (-a - \frac {2 \, b \sin \relax (x)}{\cos \relax (x) + 1} + \frac {a \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )}{a^{3}} + \frac {\log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right )}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.71, size = 131, normalized size = 2.22 \[ \frac {\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{a^3}-\frac {\ln \left (-a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a\right )}{a^3}+\frac {\frac {2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2-3\,b^2\right )}{a^3}+\frac {4\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a^2}-\frac {4\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2}}{a^2-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(tan(x/2))/a^3 - log(a + 2*b*tan(x/2) - a*tan(x/2)^2)/a^3 + ((2*tan(x/2)^2*(a^2 - 3*b^2))/a^3 + (4*b*tan(x/
2)^3)/a^2 - (4*b*tan(x/2))/a^2)/(a^2 - tan(x/2)^2*(2*a^2 - 4*b^2) + a^2*tan(x/2)^4 + 4*a*b*tan(x/2) - 4*a*b*ta
n(x/2)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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