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

Optimal. Leaf size=80 \[ -\frac {(2 A+B) \tan ^{-1}\left (\frac {\sqrt {35} (2-x)}{\sqrt {10 x^2-22 x+13}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {35} (1-x)}{2 \sqrt {10 x^2-22 x+13}}\right )}{2 \sqrt {35}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1035, 1029, 206, 204} \[ -\frac {(2 A+B) \tan ^{-1}\left (\frac {\sqrt {35} (2-x)}{\sqrt {10 x^2-22 x+13}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\left (\frac {\sqrt {35} (-x (A+B)+A+B)}{2 \sqrt {10 x^2-22 x+13} (A+B)}\right )}{2 \sqrt {35}} \]

Antiderivative was successfully verified.

[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 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 206

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

Rule 1029

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 1035

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*} \int \frac {B+A x}{\left (17-18 x+5 x^2\right ) \sqrt {13-22 x+10 x^2}} \, dx &=\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 ) \operatorname {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 ) \operatorname {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) \tan ^{-1}\left (\frac {\sqrt {35} (2-x)}{\sqrt {13-22 x+10 x^2}}\right )}{\sqrt {35}}-\frac {(A+B) \tanh ^{-1}\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]  time = 0.12, size = 94, normalized size = 1.18 \[ \frac {((4-i) A+(2-i) B) \tan ^{-1}\left (\frac {(2-18 i)-(1-18 i) x}{\sqrt {35} \sqrt {10 x^2-22 x+13}}\right )+((1-4 i) A+(1-2 i) B) \tanh ^{-1}\left (\frac {(18-i) x-(18-2 i)}{\sqrt {35} \sqrt {10 x^2-22 x+13}}\right )}{4 \sqrt {35}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4 - I)*A + (2 - I)*B)*ArcTan[((2 - 18*I) - (1 - 18*I)*x)/(Sqrt[35]*Sqrt[13 - 22*x + 10*x^2])] + ((1 - 4*I)*
A + (1 - 2*I)*B)*ArcTanh[((-18 + 2*I) + (18 - I)*x)/(Sqrt[35]*Sqrt[13 - 22*x + 10*x^2])])/(4*Sqrt[35])

________________________________________________________________________________________

IntegrateAlgebraic [C]  time = 1.13, size = 143, normalized size = 1.79 \[ \frac {1}{70} \left ((1+4 i) \sqrt {35} A+(1+2 i) \sqrt {35} B\right ) \tanh ^{-1}\left (\frac {(2-i) \sqrt {10 x^2-22 x+13}+(-2+i) \sqrt {10} x+(4-i) \sqrt {10}}{\sqrt {35}}\right )-\frac {1}{70} i \left ((4+i) \sqrt {35} A+(2+i) \sqrt {35} B\right ) \tanh ^{-1}\left (\frac {(2+i) \sqrt {10 x^2-22 x+13}+(-2-i) \sqrt {10} x+(4+i) \sqrt {10}}{\sqrt {35}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((1 + 4*I)*Sqrt[35]*A + (1 + 2*I)*Sqrt[35]*B)*ArcTanh[((4 - I)*Sqrt[10] - (2 - I)*Sqrt[10]*x + (2 - I)*Sqrt[1
3 - 22*x + 10*x^2])/Sqrt[35]])/70 - (I/70)*((4 + I)*Sqrt[35]*A + (2 + I)*Sqrt[35]*B)*ArcTanh[((4 + I)*Sqrt[10]
- (2 + I)*Sqrt[10]*x + (2 + I)*Sqrt[13 - 22*x + 10*x^2])/Sqrt[35]]

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [B]  time = 0.99, size = 629, normalized size = 7.86 \[ \frac {2 \, \sqrt {35} {\left (2 \, A^{2} + 3 \, A B + B^{2}\right )} \sqrt {A^{2} + 2 \, A B + B^{2}} {\left (\arctan \relax (3) + \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 ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [B]  time = 0.46, size = 192, normalized size = 2.40

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

Verification of antiderivative is not currently implemented for this CAS.

[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)/(1+(-2+x)/(1-x))^2)^(1/2)/(1+(-2+x)/(1-
x))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A x + B}{\sqrt {10 \, x^{2} - 22 \, x + 13} {\left (5 \, x^{2} - 18 \, x + 17\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {B+A\,x}{\left (5\,x^2-18\,x+17\right )\,\sqrt {10\,x^2-22\,x+13}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A x + B}{\left (5 x^{2} - 18 x + 17\right ) \sqrt {10 x^{2} - 22 x + 13}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________