3.3 \(\int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx\)

Optimal. Leaf size=19 \[ \log (\sin (x)+\cos (x)+3)-\log (\sin (x)-3 \cos (x)+1) \]

[Out]

-ln(1-3*cos(x)+sin(x))+ln(3+cos(x)+sin(x))

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Rubi [B]  time = 0.63, antiderivative size = 42, normalized size of antiderivative = 2.21, number of steps used = 25, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4401, 12, 2074, 634, 618, 204, 628} \[ \log \left (\tan ^2\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (\tan \left (\frac {x}{2}\right )+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 + Cos[x] + 5*Sin[x])/(4*Cos[x] - 2*Sin[x] + Cos[x]*Sin[x] - 2*Sin[x]^2),x]

[Out]

-Log[1 - 2*Tan[x/2]] - Log[1 + Tan[x/2]] + Log[2 + Tan[x/2] + Tan[x/2]^2]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx &=\int \left (\frac {\cos (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)}-\frac {2}{-4 \cos (x)+2 \sin (x)-\cos (x) \sin (x)+2 \sin ^2(x)}-\frac {5 \sin (x)}{-4 \cos (x)+2 \sin (x)-\cos (x) \sin (x)+2 \sin ^2(x)}\right ) \, dx\\ &=-\left (2 \int \frac {1}{-4 \cos (x)+2 \sin (x)-\cos (x) \sin (x)+2 \sin ^2(x)} \, dx\right )-5 \int \frac {\sin (x)}{-4 \cos (x)+2 \sin (x)-\cos (x) \sin (x)+2 \sin ^2(x)} \, dx+\int \frac {\cos (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {1-x}{2 \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-4 \operatorname {Subst}\left (\int \frac {-1-x^2}{2 \left (2-x-4 x^2-3 x^3-2 x^4\right )} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-10 \operatorname {Subst}\left (\int \frac {x}{-2+x+4 x^2+3 x^3+2 x^4} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {-1-x^2}{2-x-4 x^2-3 x^3-2 x^4} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\right )-10 \operatorname {Subst}\left (\int \left (\frac {1}{6 (1+x)}+\frac {4}{33 (-1+2 x)}+\frac {-2-5 x}{22 \left (2+x+x^2\right )}\right ) \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\operatorname {Subst}\left (\int \frac {1-x}{2-3 x-x^2-2 x^3} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\frac {20}{33} \log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\frac {5}{3} \log \left (1+\tan \left (\frac {x}{2}\right )\right )-\frac {5}{11} \operatorname {Subst}\left (\int \frac {-2-5 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-2 \operatorname {Subst}\left (\int \left (-\frac {1}{3 (1+x)}+\frac {10}{33 (-1+2 x)}+\frac {3+2 x}{11 \left (2+x+x^2\right )}\right ) \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\operatorname {Subst}\left (\int \left (-\frac {2}{11 (-1+2 x)}+\frac {7+x}{11 \left (2+x+x^2\right )}\right ) \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (1+\tan \left (\frac {x}{2}\right )\right )+\frac {1}{11} \operatorname {Subst}\left (\int \frac {7+x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {2}{11} \operatorname {Subst}\left (\int \frac {3+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {5}{22} \operatorname {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {25}{22} \operatorname {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (1+\tan \left (\frac {x}{2}\right )\right )+\frac {25}{22} \log \left (2+\tan \left (\frac {x}{2}\right )+\tan ^2\left (\frac {x}{2}\right )\right )+\frac {1}{22} \operatorname {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {2}{11} \operatorname {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {4}{11} \operatorname {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {5}{11} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )+\frac {13}{22} \operatorname {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\\ &=-\frac {5 x}{22 \sqrt {7}}-\frac {5 \tan ^{-1}\left (\frac {\cos (x)-\sin (x)}{3+\sqrt {7}+\cos (x)+\sin (x)}\right )}{11 \sqrt {7}}-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (1+\tan \left (\frac {x}{2}\right )\right )+\log \left (2+\tan \left (\frac {x}{2}\right )+\tan ^2\left (\frac {x}{2}\right )\right )+\frac {8}{11} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )-\frac {13}{11} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )\\ &=-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (1+\tan \left (\frac {x}{2}\right )\right )+\log \left (2+\tan \left (\frac {x}{2}\right )+\tan ^2\left (\frac {x}{2}\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 19, normalized size = 1.00 \[ \log (\sin (x)+\cos (x)+3)-\log (\sin (x)-3 \cos (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(2 + Cos[x] + 5*Sin[x])/(4*Cos[x] - 2*Sin[x] + Cos[x]*Sin[x] - 2*Sin[x]^2),x]

[Out]

-Log[1 - 3*Cos[x] + Sin[x]] + Log[3 + Cos[x] + Sin[x]]

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fricas [B]  time = 0.47, size = 43, normalized size = 2.26 \[ -\frac {1}{2} \, \log \left (2 \, \cos \relax (x)^{2} - \frac {1}{2} \, {\left (3 \, \cos \relax (x) - 1\right )} \sin \relax (x) - \frac {3}{2} \, \cos \relax (x) + \frac {1}{2}\right ) + \frac {1}{2} \, \log \left (\frac {1}{2} \, {\left (\cos \relax (x) + 3\right )} \sin \relax (x) + \frac {3}{2} \, \cos \relax (x) + \frac {5}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+cos(x)+5*sin(x))/(4*cos(x)-2*sin(x)+cos(x)*sin(x)-2*sin(x)^2),x, algorithm="fricas")

[Out]

-1/2*log(2*cos(x)^2 - 1/2*(3*cos(x) - 1)*sin(x) - 3/2*cos(x) + 1/2) + 1/2*log(1/2*(cos(x) + 3)*sin(x) + 3/2*co
s(x) + 5/2)

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giac [A]  time = 1.26, size = 36, normalized size = 1.89 \[ \log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + \tan \left (\frac {1}{2} \, x\right ) + 2\right ) - \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) - \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+cos(x)+5*sin(x))/(4*cos(x)-2*sin(x)+cos(x)*sin(x)-2*sin(x)^2),x, algorithm="giac")

[Out]

log(tan(1/2*x)^2 + tan(1/2*x) + 2) - log(abs(2*tan(1/2*x) - 1)) - log(abs(tan(1/2*x) + 1))

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maple [A]  time = 0.13, size = 35, normalized size = 1.84 \[ -\ln \left (\tan \left (\frac {x}{2}\right )+1\right )-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+cos(x)+5*sin(x))/(4*cos(x)-2*sin(x)+cos(x)*sin(x)-2*sin(x)^2),x)

[Out]

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

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maxima [B]  time = 1.48, size = 53, normalized size = 2.79 \[ -\log \left (\frac {2 \, \sin \relax (x)}{\cos \relax (x) + 1} - 1\right ) + \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + \frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 2\right ) - \log \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+cos(x)+5*sin(x))/(4*cos(x)-2*sin(x)+cos(x)*sin(x)-2*sin(x)^2),x, algorithm="maxima")

[Out]

-log(2*sin(x)/(cos(x) + 1) - 1) + log(sin(x)/(cos(x) + 1) + sin(x)^2/(cos(x) + 1)^2 + 2) - log(sin(x)/(cos(x)
+ 1) + 1)

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mupad [B]  time = 0.85, size = 32, normalized size = 1.68 \[ -2\,\mathrm {atanh}\left (\frac {\frac {252\,\mathrm {tan}\left (\frac {x}{2}\right )}{19}+\frac {1260}{19}}{19\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+5\,\mathrm {tan}\left (\frac {x}{2}\right )-32}+\frac {37}{19}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(x) + 5*sin(x) + 2)/(4*cos(x) - 2*sin(x) + cos(x)*sin(x) - 2*sin(x)^2),x)

[Out]

-2*atanh(((252*tan(x/2))/19 + 1260/19)/(5*tan(x/2) + 19*tan(x/2)^2 - 32) + 37/19)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {5 \sin {\relax (x )} + \cos {\relax (x )} + 2}{- 2 \sin ^{2}{\relax (x )} + \sin {\relax (x )} \cos {\relax (x )} - 2 \sin {\relax (x )} + 4 \cos {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+cos(x)+5*sin(x))/(4*cos(x)-2*sin(x)+cos(x)*sin(x)-2*sin(x)**2),x)

[Out]

Integral((5*sin(x) + cos(x) + 2)/(-2*sin(x)**2 + sin(x)*cos(x) - 2*sin(x) + 4*cos(x)), x)

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