3.4 \(\int \frac{3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx\)

Optimal. Leaf size=19 \[ \log (\sin (x)+\cos (x)+3)-\log (-2 \sin (x)+\cos (x)+1) \]

[Out]

-Log[1 + Cos[x] - 2*Sin[x]] + Log[3 + Cos[x] + Sin[x]]

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Rubi [A]  time = 2.06192, antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 32, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used = {4401, 2657, 12, 6725, 634, 618, 204, 628, 203} \[ \log \left (\tan ^2\left (\frac{x}{2}\right )+\tan \left (\frac{x}{2}\right )+2\right )-\log \left (1-2 \tan \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[(3 + 7*Cos[x] + 2*Sin[x])/(1 + 4*Cos[x] + 3*Cos[x]^2 - 5*Sin[x] - Cos[x]*Sin[x]),x]

[Out]

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

Rule 4401

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

Rule 2657

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2*ArcTan[(b*Cos[c + d*x])/(a + q + b*Sin[c + d*x])])/(d*q), x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rule 12

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 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 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 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 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 203

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

Rubi steps

\begin{align*} \int \frac{3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx &=\int \left (-\frac{2}{5+\cos (x)}+\frac{17+46 \cos (x)+13 \cos ^2(x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac{1}{5+\cos (x)} \, dx\right )+\int \frac{17+46 \cos (x)+13 \cos ^2(x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )} \, dx\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )+\int \left (\frac{17}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )}+\frac{46 \cos (x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )}+\frac{13 \cos ^2(x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )}\right ) \, dx\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )+13 \int \frac{\cos ^2(x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )} \, dx+17 \int \frac{1}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )} \, dx+46 \int \frac{\cos (x)}{(5+\cos (x)) \left (1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)\right )} \, dx\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )+26 \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{8 \left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+34 \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{8 \left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+92 \operatorname{Subst}\left (\int \frac{1-x^4}{8 \left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )+\frac{13}{4} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{\left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{17}{4} \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{\left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{23}{2} \operatorname{Subst}\left (\int \frac{1-x^4}{\left (3+2 x^2\right ) \left (2-3 x-x^2-2 x^3\right )} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )+\frac{13}{4} \operatorname{Subst}\left (\int \left (-\frac{9}{154 (-1+2 x)}+\frac{-75-17 x}{77 \left (2+x+x^2\right )}+\frac{25}{14 \left (3+2 x^2\right )}\right ) \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{17}{4} \operatorname{Subst}\left (\int \left (-\frac{25}{154 (-1+2 x)}+\frac{-3-13 x}{77 \left (2+x+x^2\right )}+\frac{1}{14 \left (3+2 x^2\right )}\right ) \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{23}{2} \operatorname{Subst}\left (\int \left (-\frac{15}{154 (-1+2 x)}+\frac{29+23 x}{77 \left (2+x+x^2\right )}-\frac{5}{14 \left (3+2 x^2\right )}\right ) \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\frac{x}{\sqrt{6}}+\sqrt{\frac{2}{3}} \tan ^{-1}\left (\frac{\sin (x)}{5+2 \sqrt{6}+\cos (x)}\right )-\log \left (1-2 \tan \left (\frac{x}{2}\right )\right )+\frac{13}{308} \operatorname{Subst}\left (\int \frac{-75-17 x}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{17}{308} \operatorname{Subst}\left (\int \frac{-3-13 x}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{23}{154} \operatorname{Subst}\left (\int \frac{29+23 x}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{17}{56} \operatorname{Subst}\left (\int \frac{1}{3+2 x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )-\frac{115}{28} \operatorname{Subst}\left (\int \frac{1}{3+2 x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{325}{56} \operatorname{Subst}\left (\int \frac{1}{3+2 x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\\ &=-\log \left (1-2 \tan \left (\frac{x}{2}\right )\right )+\frac{17}{88} \operatorname{Subst}\left (\int \frac{1}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )-2 \left (\frac{221}{616} \operatorname{Subst}\left (\int \frac{1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )\right )+\frac{529}{308} \operatorname{Subst}\left (\int \frac{1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )+\frac{115}{44} \operatorname{Subst}\left (\int \frac{1}{2+x+x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )-\frac{247}{88} \operatorname{Subst}\left (\int \frac{1}{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 (2+\tan \left (\frac{x}{2}\right )+\tan ^2\left (\frac{x}{2}\right )\right )-\frac{17}{44} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac{x}{2}\right )\right )-\frac{115}{22} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac{x}{2}\right )\right )+\frac{247}{44} \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 (2+\tan \left (\frac{x}{2}\right )+\tan ^2\left (\frac{x}{2}\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.178451, size = 19, normalized size = 1. \[ \log (\sin (x)+\cos (x)+3)-\log (-2 \sin (x)+\cos (x)+1) \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 7*Cos[x] + 2*Sin[x])/(1 + 4*Cos[x] + 3*Cos[x]^2 - 5*Sin[x] - Cos[x]*Sin[x]),x]

[Out]

-Log[1 + Cos[x] - 2*Sin[x]] + Log[3 + Cos[x] + Sin[x]]

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Maple [A]  time = 0.092, size = 26, normalized size = 1.4 \begin{align*} \ln \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+\tan \left ({\frac{x}{2}} \right ) +2 \right ) -\ln \left ( 2\,\tan \left ( x/2 \right ) -1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+7*cos(x)+2*sin(x))/(1+4*cos(x)+3*cos(x)^2-5*sin(x)-cos(x)*sin(x)),x)

[Out]

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 = 1.46038, size = 53, normalized size = 2.79 \begin{align*} -\log \left (\frac{2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \log \left (\frac{\sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac{\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+7*cos(x)+2*sin(x))/(1+4*cos(x)+3*cos(x)^2-5*sin(x)-cos(x)*sin(x)),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)

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Fricas [B]  time = 1.79597, size = 162, normalized size = 8.53 \begin{align*} -\frac{1}{2} \, \log \left (-\frac{3}{4} \, \cos \left (x\right )^{2} -{\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) + \frac{1}{2} \, \cos \left (x\right ) + \frac{5}{4}\right ) + \frac{1}{2} \, \log \left (\frac{1}{2} \,{\left (\cos \left (x\right ) + 3\right )} \sin \left (x\right ) + \frac{3}{2} \, \cos \left (x\right ) + \frac{5}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+7*cos(x)+2*sin(x))/(1+4*cos(x)+3*cos(x)^2-5*sin(x)-cos(x)*sin(x)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.34972, size = 24, normalized size = 1.26 \begin{align*} - \log{\left (\tan{\left (\frac{x}{2} \right )} - \frac{1}{2} \right )} + \log{\left (\tan ^{2}{\left (\frac{x}{2} \right )} + \tan{\left (\frac{x}{2} \right )} + 2 \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+7*cos(x)+2*sin(x))/(1+4*cos(x)+3*cos(x)**2-5*sin(x)-cos(x)*sin(x)),x)

[Out]

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

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Giac [A]  time = 1.11713, size = 35, normalized size = 1.84 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+7*cos(x)+2*sin(x))/(1+4*cos(x)+3*cos(x)^2-5*sin(x)-cos(x)*sin(x)),x, algorithm="giac")

[Out]

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