\(\int \frac {B+A x}{(17-18 x+5 x^2) \sqrt {13-22 x+10 x^2}} \, dx\) [248]

Optimal result

Integrand size = 32, antiderivative size = 80 \[ \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=-\frac {(2 A+B) \arctan \left (\frac {\sqrt {35} (2-x)}{\sqrt {13-22 x+10 x^2}}\right )}{\sqrt {35}}-\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1049, 1043, 212, 210} \[ \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=-\frac {(2 A+B) \arctan \left (\frac {\sqrt {35} (2-x)}{\sqrt {10 x^2-22 x+13}}\right )}{\sqrt {35}}-\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {35} (-x (A+B)+A+B)}{2 \sqrt {10 x^2-22 x+13} (A+B)}\right )}{2 \sqrt {35}} \]

[In]

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

[Out]

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

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 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]

Rule 1049

Int[((g_.) + (h_.)*(x_))/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symb
ol] :> With[{q = Rt[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*
d - a*f - q) - (g*(c*e - b*f) - h*(c*d - a*f + q))*x, x]/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - D
ist[1/(2*q), Int[Simp[h*(b*d - a*e) - g*(c*d - a*f + q) - (g*(c*e - b*f) - h*(c*d - a*f - q))*x, x]/((a + b*x
+ c*x^2)*Sqrt[d + e*x + f*x^2]), x], 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] && NegQ[b^2 - 4*a*c]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{70} \int \frac {140 (A+B)-70 (A+B) x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx-\frac {1}{70} \int \frac {70 (2 A+B)-70 (2 A+B) x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx \\ & = \left (560 (A+B)^2\right ) \text {Subst}\left (\int \frac {1}{313600 (A+B)^2-140 x^2} \, dx,x,\frac {-140 (A+B)+140 (A+B) x}{\sqrt {13-22 x+10 x^2}}\right )+\left (2240 (2 A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-1254400 (2 A+B)^2-140 x^2} \, dx,x,\frac {1120 (2 A+B)-560 (2 A+B) x}{\sqrt {13-22 x+10 x^2}}\right ) \\ & = -\frac {(2 A+B) \arctan \left (\frac {\sqrt {35} (2-x)}{\sqrt {13-22 x+10 x^2}}\right )}{\sqrt {35}}-\frac {(A+B) \text {arctanh}\left (\frac {\sqrt {35} (A+B-(A+B) x)}{2 (A+B) \sqrt {13-22 x+10 x^2}}\right )}{2 \sqrt {35}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.31 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.55 \[ \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\frac {((1+4 i) A+(1+2 i) B) \text {arctanh}\left (\frac {(4-i) \sqrt {10}-(2-i) \sqrt {10} x+(2-i) \sqrt {13-22 x+10 x^2}}{\sqrt {35}}\right )+((1-4 i) A+(1-2 i) B) \text {arctanh}\left (\frac {(4+i) \sqrt {10}-(2+i) \sqrt {10} x+(2+i) \sqrt {13-22 x+10 x^2}}{\sqrt {35}}\right )}{2 \sqrt {35}} \]

[In]

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

[Out]

(((1 + 4*I)*A + (1 + 2*I)*B)*ArcTanh[((4 - I)*Sqrt[10] - (2 - I)*Sqrt[10]*x + (2 - I)*Sqrt[13 - 22*x + 10*x^2]
)/Sqrt[35]] + ((1 - 4*I)*A + (1 - 2*I)*B)*ArcTanh[((4 + I)*Sqrt[10] - (2 + I)*Sqrt[10]*x + (2 + I)*Sqrt[13 - 2
2*x + 10*x^2])/Sqrt[35]])/(2*Sqrt[35])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(64)=128\).

Time = 0.94 (sec) , antiderivative size = 192, normalized size of antiderivative = 2.40

method	result	size
default	\(\frac {\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}\, \left (\operatorname {arctanh}\left (\frac {2 \sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}}{35}\right ) A -4 \arctan \left (\frac {\sqrt {35}\, \left (-2+x \right )}{\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \left (1-x \right )}\right ) A +\operatorname {arctanh}\left (\frac {2 \sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \sqrt {35}}{35}\right ) B -2 \arctan \left (\frac {\sqrt {35}\, \left (-2+x \right )}{\sqrt {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}\, \left (1-x \right )}\right ) B \right )}{70 \sqrt {\frac {\frac {\left (-2+x \right )^{2}}{\left (1-x \right )^{2}}+9}{\left (\frac {-2+x}{1-x}+1\right )^{2}}}\, \left (\frac {-2+x}{1-x}+1\right )}\)	\(192\)

[In]

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

[Out]

1/70*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2)*(arctanh(2/35*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2))*A-4*arctan(35^(1/2
)/((-2+x)^2/(1-x)^2+9)^(1/2)*(-2+x)/(1-x))*A+arctanh(2/35*((-2+x)^2/(1-x)^2+9)^(1/2)*35^(1/2))*B-2*arctan(35^(
1/2)/((-2+x)^2/(1-x)^2+9)^(1/2)*(-2+x)/(1-x))*B)/(((-2+x)^2/(1-x)^2+9)/((-2+x)/(1-x)+1)^2)^(1/2)/((-2+x)/(1-x)
+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1339 vs. \(2 (61) = 122\).

Time = 0.34 (sec) , antiderivative size = 1339, normalized size of antiderivative = 16.74 \[ \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/280*sqrt(35)*sqrt(-15*A^2 - 14*A*B - 3*B^2 + 4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*log(-(2
380*A^4 + 6090*A^3*B + 5670*A^2*B^2 + 2310*A*B^3 + 350*B^4 + (sqrt(35)*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6
*A*B^3 - B^4)*(15*A + 7*B) - 2*sqrt(35)*(8*A^3 + 18*A^2*B + 13*A*B^2 + 3*B^3))*sqrt(-15*A^2 - 14*A*B - 3*B^2 +
4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*sqrt(10*x^2 - 22*x + 13) - 71*(34*A^4 + 87*A^3*B + 81
*A^2*B^2 + 33*A*B^3 + 5*B^4)*x + 2*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4)*(221*A^2 + 234*A*B + 6
5*B^2 - 11*(17*A^2 + 18*A*B + 5*B^2)*x))/x) - 1/280*sqrt(35)*sqrt(-15*A^2 - 14*A*B - 3*B^2 + 4*sqrt(-4*A^4 - 1
2*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*log(-(2380*A^4 + 6090*A^3*B + 5670*A^2*B^2 + 2310*A*B^3 + 350*B^4 - (sq
rt(35)*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4)*(15*A + 7*B) - 2*sqrt(35)*(8*A^3 + 18*A^2*B + 13*A
*B^2 + 3*B^3))*sqrt(-15*A^2 - 14*A*B - 3*B^2 + 4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*sqrt(10
*x^2 - 22*x + 13) - 71*(34*A^4 + 87*A^3*B + 81*A^2*B^2 + 33*A*B^3 + 5*B^4)*x + 2*sqrt(-4*A^4 - 12*A^3*B - 13*A
^2*B^2 - 6*A*B^3 - B^4)*(221*A^2 + 234*A*B + 65*B^2 - 11*(17*A^2 + 18*A*B + 5*B^2)*x))/x) - 1/280*sqrt(35)*sqr
t(-15*A^2 - 14*A*B - 3*B^2 - 4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*log(-(2380*A^4 + 6090*A^3
*B + 5670*A^2*B^2 + 2310*A*B^3 + 350*B^4 + (sqrt(35)*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4)*(15*
A + 7*B) + 2*sqrt(35)*(8*A^3 + 18*A^2*B + 13*A*B^2 + 3*B^3))*sqrt(-15*A^2 - 14*A*B - 3*B^2 - 4*sqrt(-4*A^4 - 1
2*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*sqrt(10*x^2 - 22*x + 13) - 71*(34*A^4 + 87*A^3*B + 81*A^2*B^2 + 33*A*B^
3 + 5*B^4)*x - 2*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4)*(221*A^2 + 234*A*B + 65*B^2 - 11*(17*A^2
+ 18*A*B + 5*B^2)*x))/x) + 1/280*sqrt(35)*sqrt(-15*A^2 - 14*A*B - 3*B^2 - 4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B
^2 - 6*A*B^3 - B^4))*log(-(2380*A^4 + 6090*A^3*B + 5670*A^2*B^2 + 2310*A*B^3 + 350*B^4 - (sqrt(35)*sqrt(-4*A^4
- 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4)*(15*A + 7*B) + 2*sqrt(35)*(8*A^3 + 18*A^2*B + 13*A*B^2 + 3*B^3))*sqr
t(-15*A^2 - 14*A*B - 3*B^2 - 4*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 - B^4))*sqrt(10*x^2 - 22*x + 13)
- 71*(34*A^4 + 87*A^3*B + 81*A^2*B^2 + 33*A*B^3 + 5*B^4)*x - 2*sqrt(-4*A^4 - 12*A^3*B - 13*A^2*B^2 - 6*A*B^3 -
B^4)*(221*A^2 + 234*A*B + 65*B^2 - 11*(17*A^2 + 18*A*B + 5*B^2)*x))/x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (61) = 122\).

Time = 0.42 (sec) , antiderivative size = 629, normalized size of antiderivative = 7.86 \[ \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx=\frac {2 \, \sqrt {35} {\left (2 \, A^{2} + 3 \, A B + B^{2}\right )} \sqrt {A^{2} + 2 \, A B + B^{2}} {\left (\arctan \left (3\right ) + \arctan \left (-\frac {5 \, {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} {\left (300 \, \sqrt {14} - 1129\right )} - 7658 \, \sqrt {35} + 14361 \, \sqrt {10}}{2329 \, \sqrt {35} - 4358 \, \sqrt {10}}\right )\right )}}{35 \, {\left (15 \, A^{2} + 14 \, A B + 3 \, B^{2} - \sqrt {289 \, A^{4} + 612 \, A^{3} B + 494 \, A^{2} B^{2} + 180 \, A B^{3} + 25 \, B^{4}}\right )}} - \frac {2 \, \sqrt {35} {\left (2 \, A^{2} + 3 \, A B + B^{2}\right )} \sqrt {A^{2} + 2 \, A B + B^{2}} {\left (\arctan \left (\frac {1}{7}\right ) + \arctan \left (-\frac {5 \, {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} {\left (62556 \, \sqrt {14} + 245977\right )} - 1617962 \, \sqrt {35} - 3089577 \, \sqrt {10}}{496201 \, \sqrt {35} + 929846 \, \sqrt {10}}\right )\right )}}{35 \, {\left (15 \, A^{2} + 14 \, A B + 3 \, B^{2} - \sqrt {289 \, A^{4} + 612 \, A^{3} B + 494 \, A^{2} B^{2} + 180 \, A B^{3} + 25 \, B^{4}}\right )}} + \frac {1}{140} \, \sqrt {35} \sqrt {A^{2} + 2 \, A B + B^{2}} \log \left (25 \, {\left (546 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} + 2807 \, \sqrt {10} x - 234 \, \sqrt {35} \sqrt {14} - 1014 \, \sqrt {14} \sqrt {10} - 1203 \, \sqrt {35} - 5213 \, \sqrt {10} - 2807 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2} + 25 \, {\left (78 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} + 401 \, \sqrt {10} x + 48 \, \sqrt {35} \sqrt {14} + 208 \, \sqrt {14} \sqrt {10} + 141 \, \sqrt {35} + 611 \, \sqrt {10} - 401 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2}\right ) - \frac {1}{140} \, \sqrt {35} \sqrt {A^{2} + 2 \, A B + B^{2}} \log \left (625 \, {\left (18 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} - 75 \, \sqrt {10} x + 8 \, \sqrt {35} \sqrt {14} - 24 \, \sqrt {14} \sqrt {10} - 37 \, \sqrt {35} + 111 \, \sqrt {10} + 75 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2} + 625 \, {\left (6 \, \sqrt {14} {\left (\sqrt {10} x - \sqrt {10 \, x^{2} - 22 \, x + 13}\right )} - 25 \, \sqrt {10} x + 6 \, \sqrt {35} \sqrt {14} - 18 \, \sqrt {14} \sqrt {10} - 25 \, \sqrt {35} + 75 \, \sqrt {10} + 25 \, \sqrt {10 \, x^{2} - 22 \, x + 13}\right )}^{2}\right ) \]

[In]

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

[Out]

2/35*sqrt(35)*(2*A^2 + 3*A*B + B^2)*sqrt(A^2 + 2*A*B + B^2)*(arctan(3) + arctan(-(5*(sqrt(10)*x - sqrt(10*x^2
- 22*x + 13))*(300*sqrt(14) - 1129) - 7658*sqrt(35) + 14361*sqrt(10))/(2329*sqrt(35) - 4358*sqrt(10))))/(15*A^
2 + 14*A*B + 3*B^2 - sqrt(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)) - 2/35*sqrt(35)*(2*A^2 + 3*
A*B + B^2)*sqrt(A^2 + 2*A*B + B^2)*(arctan(1/7) + arctan(-(5*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13))*(62556*sq
rt(14) + 245977) - 1617962*sqrt(35) - 3089577*sqrt(10))/(496201*sqrt(35) + 929846*sqrt(10))))/(15*A^2 + 14*A*B
+ 3*B^2 - sqrt(289*A^4 + 612*A^3*B + 494*A^2*B^2 + 180*A*B^3 + 25*B^4)) + 1/140*sqrt(35)*sqrt(A^2 + 2*A*B + B
^2)*log(25*(546*sqrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13)) + 2807*sqrt(10)*x - 234*sqrt(35)*sqrt(14) - 1
014*sqrt(14)*sqrt(10) - 1203*sqrt(35) - 5213*sqrt(10) - 2807*sqrt(10*x^2 - 22*x + 13))^2 + 25*(78*sqrt(14)*(sq
rt(10)*x - sqrt(10*x^2 - 22*x + 13)) + 401*sqrt(10)*x + 48*sqrt(35)*sqrt(14) + 208*sqrt(14)*sqrt(10) + 141*sqr
t(35) + 611*sqrt(10) - 401*sqrt(10*x^2 - 22*x + 13))^2) - 1/140*sqrt(35)*sqrt(A^2 + 2*A*B + B^2)*log(625*(18*s
qrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*x + 13)) - 75*sqrt(10)*x + 8*sqrt(35)*sqrt(14) - 24*sqrt(14)*sqrt(10) -
37*sqrt(35) + 111*sqrt(10) + 75*sqrt(10*x^2 - 22*x + 13))^2 + 625*(6*sqrt(14)*(sqrt(10)*x - sqrt(10*x^2 - 22*
x + 13)) - 25*sqrt(10)*x + 6*sqrt(35)*sqrt(14) - 18*sqrt(14)*sqrt(10) - 25*sqrt(35) + 75*sqrt(10) + 25*sqrt(10
*x^2 - 22*x + 13))^2)

Mupad [F(-1)]

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

[In]

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

[Out]

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