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

Optimal result

Integrand size = 35, antiderivative size = 19 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=-\log (1+\cos (x)-2 \sin (x))+\log (3+\cos (x)+\sin (x)) \]

[Out]

-ln(1+cos(x)-2*sin(x))+ln(3+cos(x)+sin(x))

Rubi [A] (verified)

Time = 1.56 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63, number of steps used = 32, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {4486, 2736, 12, 6857, 648, 632, 210, 642, 209} \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=\log \left (\tan ^2\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right ) \]

[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 12

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

Rule 209

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

Rule 210

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

Rule 632

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 642

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 648

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 2736

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

Rule 4486

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

Rule 6857

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]

Rubi steps \begin{align*} \text {integral}& = \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}} \arctan \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}} \arctan \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}} \arctan \left (\frac {\sin (x)}{5+2 \sqrt {6}+\cos (x)}\right )+26 \text {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 \text {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 \text {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}} \arctan \left (\frac {\sin (x)}{5+2 \sqrt {6}+\cos (x)}\right )+\frac {13}{4} \text {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} \text {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} \text {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}} \arctan \left (\frac {\sin (x)}{5+2 \sqrt {6}+\cos (x)}\right )+\frac {13}{4} \text {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} \text {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} \text {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}} \arctan \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} \text {Subst}\left (\int \frac {-75-17 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {17}{308} \text {Subst}\left (\int \frac {-3-13 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {23}{154} \text {Subst}\left (\int \frac {29+23 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {17}{56} \text {Subst}\left (\int \frac {1}{3+2 x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {115}{28} \text {Subst}\left (\int \frac {1}{3+2 x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {325}{56} \text {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} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-2 \left (\frac {221}{616} \text {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )\right )+\frac {529}{308} \text {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {115}{44} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {247}{88} \text {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} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )-\frac {115}{22} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )+\frac {247}{44} \text {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] (verified)

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=-\log (1+\cos (x)-2 \sin (x))+\log (3+\cos (x)+\sin (x)) \]

[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]]

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26

method	result	size
parallelrisch	\(-\ln \left (\tan \left (\frac {x}{2}\right )-\frac {1}{2}\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(24\)
default	\(-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(26\)
norman	\(-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(26\)
risch	\(\ln \left ({\mathrm e}^{2 i x}+\left (3+3 i\right ) {\mathrm e}^{i x}+i\right )-\ln \left ({\mathrm e}^{2 i x}+\left (\frac {2}{5}-\frac {4 i}{5}\right ) {\mathrm e}^{i x}-\frac {3}{5}-\frac {4 i}{5}\right )\)	\(41\)

[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,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=-\frac {1}{2} \, \log \left (-\frac {3}{4} \, \cos \left (x\right )^{2} - {\left (\cos \left (x\right ) + 1\right )} \sin \left (x\right ) + \frac {1}{2} \, \cos \left (x\right ) + \frac {5}{4}\right ) + \frac {1}{2} \, \log \left (\frac {1}{2} \, {\left (\cos \left (x\right ) + 3\right )} \sin \left (x\right ) + \frac {3}{2} \, \cos \left (x\right ) + \frac {5}{2}\right ) \]

[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)

Sympy [F]

\[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=\int \frac {2 \sin {\left (x \right )} + 7 \cos {\left (x \right )} + 3}{- \sin {\left (x \right )} \cos {\left (x \right )} - 5 \sin {\left (x \right )} + 3 \cos ^{2}{\left (x \right )} + 4 \cos {\left (x \right )} + 1}\, dx \]

[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]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.29 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.05 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=-\log \left (\frac {2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 2\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=\log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + \tan \left (\frac {1}{2} \, x\right ) + 2\right ) - \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {3+7 \cos (x)+2 \sin (x)}{1+4 \cos (x)+3 \cos ^2(x)-5 \sin (x)-\cos (x) \sin (x)} \, dx=\ln \left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+\mathrm {tan}\left (\frac {x}{2}\right )+2\right )-\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )-\frac {1}{2}\right ) \]

[In]

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

[Out]

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