\(\int \frac {-2+x}{(17-18 x+5 x^2) \sqrt {13-22 x+10 x^2}} \, dx\) [249]

Optimal result

Integrand size = 30, antiderivative size = 38 \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \]

[Out]

1/70*arctanh(1/2*(1-x)*35^(1/2)/(10*x^2-22*x+13)^(1/2))*35^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1043, 212} \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {10 x^2-22 x+13}}\right )}{2 \sqrt {35}} \]

[In]

Int[(-2 + x)/((17 - 18*x + 5*x^2)*Sqrt[13 - 22*x + 10*x^2]),x]

[Out]

ArcTanh[(Sqrt[35]*(1 - x))/(2*Sqrt[13 - 22*x + 10*x^2])]/(2*Sqrt[35])

Rule 212

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

Rule 1043

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> Dist[-2*g*(g*b - 2*a*h), Subst[Int[1/Simp[g*(g*b - 2*a*h)*(b^2 - 4*a*c) - (b*d - a*e)*x^2, x], x], x, S
imp[g*b - 2*a*h - (b*h - 2*g*c)*x, x]/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && NeQ[b*d - a*e, 0] && EqQ[h^2*(b*d - a*e) - 2*g*h*(c*d - a*f) + g^2*(
c*e - b*f), 0]

Rubi steps \begin{align*} \text {integral}& = 8 \text {Subst}\left (\int \frac {1}{64-140 x^2} \, dx,x,\frac {2-2 x}{\sqrt {13-22 x+10 x^2}}\right ) \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(85\) vs. \(2(38)=76\).

Time = 0.57 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.24 \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {-135+145 x-50 x^2+\sqrt {10} (-9+5 x) \sqrt {13-22 x+10 x^2}}{-20 \sqrt {14}+10 \sqrt {14} x-2 \sqrt {35} \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \]

[In]

Integrate[(-2 + x)/((17 - 18*x + 5*x^2)*Sqrt[13 - 22*x + 10*x^2]),x]

[Out]

-1/2*ArcTanh[(-135 + 145*x - 50*x^2 + Sqrt[10]*(-9 + 5*x)*Sqrt[13 - 22*x + 10*x^2])/(-20*Sqrt[14] + 10*Sqrt[14
]*x - 2*Sqrt[35]*Sqrt[13 - 22*x + 10*x^2])]/Sqrt[35]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.54 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.16

method	result	size
trager	\(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) \ln \left (-\frac {75 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) x^{2}-158 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right ) x +140 \sqrt {10 x^{2}-22 x +13}\, x +87 \operatorname {RootOf}\left (\textit {\_Z}^{2}-35\right )-140 \sqrt {10 x^{2}-22 x +13}}{5 x^{2}-18 x +17}\right )}{140}\)	\(82\)
default	\(-\frac {\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}\, \operatorname {arctanh}\left (\frac {2 \sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}}{35}\right )}{70 \sqrt {\frac {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}{\left (\frac {-2+x}{1-x}+1\right )^{2}}}\, \left (\frac {-2+x}{1-x}+1\right )}\)	\(94\)

[In]

int((-2+x)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/140*RootOf(_Z^2-35)*ln(-(75*RootOf(_Z^2-35)*x^2-158*RootOf(_Z^2-35)*x+140*(10*x^2-22*x+13)^(1/2)*x+87*RootO
f(_Z^2-35)-140*(10*x^2-22*x+13)^(1/2))/(5*x^2-18*x+17))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (26) = 52\).

Time = 0.23 (sec) , antiderivative size = 81, normalized size of antiderivative = 2.13 \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\frac {1}{280} \, \sqrt {35} \log \left (\frac {11225 \, x^{4} - 47220 \, x^{3} - 8 \, \sqrt {35} {\left (75 \, x^{3} - 233 \, x^{2} + 245 \, x - 87\right )} \sqrt {10 \, x^{2} - 22 \, x + 13} + 75534 \, x^{2} - 54372 \, x + 14849}{25 \, x^{4} - 180 \, x^{3} + 494 \, x^{2} - 612 \, x + 289}\right ) \]

[In]

integrate((-2+x)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="fricas")

[Out]

1/280*sqrt(35)*log((11225*x^4 - 47220*x^3 - 8*sqrt(35)*(75*x^3 - 233*x^2 + 245*x - 87)*sqrt(10*x^2 - 22*x + 13
) + 75534*x^2 - 54372*x + 14849)/(25*x^4 - 180*x^3 + 494*x^2 - 612*x + 289))

Sympy [F]

\[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\int \frac {x - 2}{\left (5 x^{2} - 18 x + 17\right ) \sqrt {10 x^{2} - 22 x + 13}}\, dx \]

[In]

integrate((-2+x)/(5*x**2-18*x+17)/(10*x**2-22*x+13)**(1/2),x)

[Out]

Integral((x - 2)/((5*x**2 - 18*x + 17)*sqrt(10*x**2 - 22*x + 13)), x)

Maxima [F]

\[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\int { \frac {x - 2}{\sqrt {10 \, x^{2} - 22 \, x + 13} {\left (5 \, x^{2} - 18 \, x + 17\right )}} \,d x } \]

[In]

integrate((-2+x)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="maxima")

[Out]

integrate((x - 2)/(sqrt(10*x^2 - 22*x + 13)*(5*x^2 - 18*x + 17)), x)

Giac [F(-1)]

Timed out. \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\text {Timed out} \]

[In]

integrate((-2+x)/(5*x^2-18*x+17)/(10*x^2-22*x+13)^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {-2+x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\int \frac {x-2}{\left (5\,x^2-18\,x+17\right )\,\sqrt {10\,x^2-22\,x+13}} \,d x \]

[In]

int((x - 2)/((5*x^2 - 18*x + 17)*(10*x^2 - 22*x + 13)^(1/2)),x)

[Out]

int((x - 2)/((5*x^2 - 18*x + 17)*(10*x^2 - 22*x + 13)^(1/2)), x)