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

Optimal result

Integrand size = 31, antiderivative size = 19 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=-\log (1-3 \cos (x)+\sin (x))+\log (3+\cos (x)+\sin (x)) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(42\) vs. \(2(19)=38\).

Time = 0.54 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21, number of steps used = 25, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4486, 12, 2099, 648, 632, 210, 642} \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=\log \left (\tan ^2\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )-\log \left (1-2 \tan \left (\frac {x}{2}\right )\right )-\log \left (\tan \left (\frac {x}{2}\right )+1\right ) \]

[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 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 2099

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 4486

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

Rubi steps \begin{align*} \text {integral}& = \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 \text {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 \text {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 \text {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 \text {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 \text {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 )+\text {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} \text {Subst}\left (\int \frac {-2-5 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-2 \text {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 )+\text {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} \text {Subst}\left (\int \frac {7+x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {2}{11} \text {Subst}\left (\int \frac {3+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {5}{22} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {25}{22} \text {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} \text {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {2}{11} \text {Subst}\left (\int \frac {1+2 x}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )-\frac {4}{11} \text {Subst}\left (\int \frac {1}{2+x+x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )+\frac {5}{11} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )+\frac {13}{22} \text {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 \arctan \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} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,1+2 \tan \left (\frac {x}{2}\right )\right )-\frac {13}{11} \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 (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*}

Mathematica [A] (verified)

Time = 2.59 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=-\log (1-3 \cos (x)+\sin (x))+\log (3+\cos (x)+\sin (x)) \]

[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]]

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84

method	result	size
default	\(-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )-\ln \left (1+\tan \left (\frac {x}{2}\right )\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(35\)
norman	\(-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )-\ln \left (1+\tan \left (\frac {x}{2}\right )\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(35\)
parallelrisch	\(-\ln \left (\frac {1}{2}+\frac {\tan \left (\frac {x}{2}\right )}{2}\right )-\ln \left (2 \tan \left (\frac {x}{2}\right )-1\right )+\ln \left (\tan ^{2}\left (\frac {x}{2}\right )+\tan \left (\frac {x}{2}\right )+2\right )\)	\(37\)
risch	\(\ln \left ({\mathrm e}^{2 i x}+\left (3+3 i\right ) {\mathrm e}^{i x}+i\right )-\ln \left ({\mathrm e}^{2 i x}+\left (-\frac {3}{5}+\frac {i}{5}\right ) {\mathrm e}^{i x}+\frac {4}{5}-\frac {3 i}{5}\right )\)	\(41\)

[In]

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

[Out]

-ln(2*tan(1/2*x)-1)-ln(1+tan(1/2*x))+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. 43 vs. \(2 (19) = 38\).

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.26 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=-\frac {1}{2} \, \log \left (2 \, \cos \left (x\right )^{2} - \frac {1}{2} \, {\left (3 \, \cos \left (x\right ) - 1\right )} \sin \left (x\right ) - \frac {3}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) + \frac {1}{2} \, \log \left (\frac {1}{2} \, {\left (\cos \left (x\right ) + 3\right )} \sin \left (x\right ) + \frac {3}{2} \, \cos \left (x\right ) + \frac {5}{2}\right ) \]

[In]

integrate((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)

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=- \log {\left (\tan {\left (\frac {x}{2} \right )} + 1 \right )} - \log {\left (2 \tan {\left (\frac {x}{2} \right )} - 1 \right )} + \log {\left (\tan ^{2}{\left (\frac {x}{2} \right )} + \tan {\left (\frac {x}{2} \right )} + 2 \right )} \]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (19) = 38\).

Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.79 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=-\log \left (\frac {2 \, \sin \left (x\right )}{\cos \left (x\right ) + 1} - 1\right ) + \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {\sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + 2\right ) - \log \left (\frac {\sin \left (x\right )}{\cos \left (x\right ) + 1} + 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.89 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=\log \left (\tan \left (\frac {1}{2} \, x\right )^{2} + \tan \left (\frac {1}{2} \, x\right ) + 2\right ) - \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, x\right ) - 1 \right |}\right ) - \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) + 1 \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.68 \[ \int \frac {2+\cos (x)+5 \sin (x)}{4 \cos (x)-2 \sin (x)+\cos (x) \sin (x)-2 \sin ^2(x)} \, dx=-2\,\mathrm {atanh}\left (\frac {\frac {252\,\mathrm {tan}\left (\frac {x}{2}\right )}{19}+\frac {1260}{19}}{19\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+5\,\mathrm {tan}\left (\frac {x}{2}\right )-32}+\frac {37}{19}\right ) \]

[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)