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

[Out]

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

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Rubi [A]  time = 0.123304, 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 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]

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 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 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])

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 = 0.114487, 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])

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Maple [B]  time = 0.02, size = 192, normalized size = 2.4 \begin{align*}{\frac{\sqrt{35}}{70}\sqrt{{\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9} \left ({\it Artanh} \left ({\frac{2\,\sqrt{35}}{35}\sqrt{{\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9}} \right ) A-4\,\arctan \left ({\frac{\sqrt{35} \left ( -2+x \right ) }{1-x}{\frac{1}{\sqrt{{\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9}}}} \right ) A+{\it Artanh} \left ({\frac{2\,\sqrt{35}}{35}\sqrt{{\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9}} \right ) B-2\,\arctan \left ({\frac{\sqrt{35} \left ( -2+x \right ) }{1-x}{\frac{1}{\sqrt{{\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9}}}} \right ) B \right ){\frac{1}{\sqrt{{ \left ({\frac{ \left ( -2+x \right ) ^{2}}{ \left ( 1-x \right ) ^{2}}}+9 \right ) \left ( 1+{\frac{-2+x}{1-x}} \right ) ^{-2}}}}} \left ( 1+{\frac{-2+x}{1-x}} \right ) ^{-1}} \end{align*}

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)

[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))

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

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)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

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

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)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

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]

Exception raised: TypeError