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

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

[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)

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Rubi [A]
time = 0.08, 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 = {1049, 1043, 212, 210} \begin {gather*} -\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}} \end {gather*}

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 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*} \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 ) \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) \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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.83, size = 124, normalized size = 1.55 \begin {gather*} \frac {((1+4 i) A+(1+2 i) B) \tanh ^{-1}\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) \tanh ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[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])

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

cought exception: maximum recursion depth exceeded

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(191\) vs. \(2(64)=128\).
time = 0.21, size = 192, normalized size = 2.40

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

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A x + B}{\left (5 x^{2} - 18 x + 17\right ) \sqrt {10 x^{2} - 22 x + 13}}\, dx \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 629 vs. \(2 (61) = 122\).
time = 0.10, size = 1186, normalized size = 14.82 \begin {gather*} 2 \left (\frac {1}{280} \sqrt {35} \sqrt {A^{2}+B^{2}+2 A B} \ln \left (\left (2730 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}+14035 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+5070 \sqrt {10} \sqrt {14}+26065 \sqrt {10}+1170 \sqrt {35} \sqrt {14}+6015 \sqrt {35}\right ) \left (2730 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}+14035 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+5070 \sqrt {10} \sqrt {14}+26065 \sqrt {10}+1170 \sqrt {35} \sqrt {14}+6015 \sqrt {35}\right )+\left (390 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}+2005 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )-1040 \sqrt {10} \sqrt {14}-3055 \sqrt {10}-240 \sqrt {35} \sqrt {14}-705 \sqrt {35}\right ) \left (390 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}+2005 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )-1040 \sqrt {10} \sqrt {14}-3055 \sqrt {10}-240 \sqrt {35} \sqrt {14}-705 \sqrt {35}\right )\right )-\frac {\sqrt {35} \left (-17920 A^{2}-8960 B^{2}-26880 A B\right ) \sqrt {A^{2}+B^{2}+2 A B} \left (\arctan \left (\frac {1}{7}\right )+\arctan \left (\frac {3163726848000 \sqrt {10}[phone] \sqrt {35}+\left (320286720000 \sqrt {14}[phone]\right ) \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )}{952162304000 \sqrt {10}[phone] \sqrt {35}}\right )\right )}{140 \left (-33600 A^{2}-6720 B^{2}-31360 A B+2240 \sqrt {289 A^{4}+25 B^{4}+180 A B^{3}+494 A^{2} B^{2}+612 A^{3} B}\right )}-\frac {1}{280} \sqrt {35} \sqrt {A^{2}+B^{2}+2 A B} \ln \left (\left (150 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}-625 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+450 \sqrt {10} \sqrt {14}-1875 \sqrt {10}-150 \sqrt {35} \sqrt {14}+625 \sqrt {35}\right ) \left (150 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}-625 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+450 \sqrt {10} \sqrt {14}-1875 \sqrt {10}-150 \sqrt {35} \sqrt {14}+625 \sqrt {35}\right )+\left (450 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}-1875 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+600 \sqrt {10} \sqrt {14}-2775 \sqrt {10}-200 \sqrt {35} \sqrt {14}+925 \sqrt {35}\right ) \left (450 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right ) \sqrt {14}-1875 \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )+600 \sqrt {10} \sqrt {14}-2775 \sqrt {10}-200 \sqrt {35} \sqrt {14}+925 \sqrt {35}\right )\right )+\frac {\sqrt {35} \left (-17920 A^{2}-8960 B^{2}-26880 A B\right ) \sqrt {A^{2}+B^{2}+2 A B} \left (\arctan \left (3\right )+\arctan \left (\frac {73528320000 \sqrt {10}-39208960000 \sqrt {35}+\left (-7680000000 \sqrt {14}[phone]\right ) \left (\sqrt {10 x^{2}-22 x+13}-\sqrt {10} x\right )}{22312960000 \sqrt {10}-11924480000 \sqrt {35}}\right )\right )}{140 \left (-33600 A^{2}-6720 B^{2}-31360 A B+2240 \sqrt {289 A^{4}+25 B^{4}+180 A B^{3}+494 A^{2} B^{2}+612 A^{3} B}\right )}\right ) \end {gather*}

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]

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)

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

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)

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