3.250 \(\int \frac {1+3 x+4 x^2}{(1+2 x)^2 (2-x+3 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=87 \[ -\frac {2 (197-837 x)}{3887 \sqrt {3 x^2-x+2}}-\frac {4 \sqrt {3 x^2-x+2}}{169 (2 x+1)}+\frac {2 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{169 \sqrt {13}} \]

[Out]

2/2197*arctanh(1/26*(9-8*x)*13^(1/2)/(3*x^2-x+2)^(1/2))*13^(1/2)-2/3887*(197-837*x)/(3*x^2-x+2)^(1/2)-4/169*(3
*x^2-x+2)^(1/2)/(1+2*x)

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Rubi [A]  time = 0.09, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1646, 806, 724, 206} \[ -\frac {2 (197-837 x)}{3887 \sqrt {3 x^2-x+2}}-\frac {4 \sqrt {3 x^2-x+2}}{169 (2 x+1)}+\frac {2 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{169 \sqrt {13}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*(2 - x + 3*x^2)^(3/2)),x]

[Out]

(-2*(197 - 837*x))/(3887*Sqrt[2 - x + 3*x^2]) - (4*Sqrt[2 - x + 3*x^2])/(169*(1 + 2*x)) + (2*ArcTanh[(9 - 8*x)
/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(169*Sqrt[13])

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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 1646

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = Polynomi
alQuotient[(d + e*x)^m*Pq, a + b*x + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2,
x], x, 0], g = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + b*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2
*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(d
+ e*x)^m*(a + b*x + c*x^2)^(p + 1)*ExpandToSum[((p + 1)*(b^2 - 4*a*c)*Q)/(d + e*x)^m - ((2*p + 3)*(2*c*f - b*
g))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {1+3 x+4 x^2}{(1+2 x)^2 \left (2-x+3 x^2\right )^{3/2}} \, dx &=-\frac {2 (197-837 x)}{3887 \sqrt {2-x+3 x^2}}+\frac {2}{23} \int \frac {\frac {184}{169}-\frac {230 x}{169}}{(1+2 x)^2 \sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {2 (197-837 x)}{3887 \sqrt {2-x+3 x^2}}-\frac {4 \sqrt {2-x+3 x^2}}{169 (1+2 x)}-\frac {2}{169} \int \frac {1}{(1+2 x) \sqrt {2-x+3 x^2}} \, dx\\ &=-\frac {2 (197-837 x)}{3887 \sqrt {2-x+3 x^2}}-\frac {4 \sqrt {2-x+3 x^2}}{169 (1+2 x)}+\frac {4}{169} \operatorname {Subst}\left (\int \frac {1}{52-x^2} \, dx,x,\frac {9-8 x}{\sqrt {2-x+3 x^2}}\right )\\ &=-\frac {2 (197-837 x)}{3887 \sqrt {2-x+3 x^2}}-\frac {4 \sqrt {2-x+3 x^2}}{169 (1+2 x)}+\frac {2 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {2-x+3 x^2}}\right )}{169 \sqrt {13}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 74, normalized size = 0.85 \[ \frac {2 \left (1536 x^2+489 x-289\right )}{3887 (2 x+1) \sqrt {3 x^2-x+2}}+\frac {2 \tanh ^{-1}\left (\frac {9-8 x}{2 \sqrt {13} \sqrt {3 x^2-x+2}}\right )}{169 \sqrt {13}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 3*x + 4*x^2)/((1 + 2*x)^2*(2 - x + 3*x^2)^(3/2)),x]

[Out]

(2*(-289 + 489*x + 1536*x^2))/(3887*(1 + 2*x)*Sqrt[2 - x + 3*x^2]) + (2*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 -
x + 3*x^2])])/(169*Sqrt[13])

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fricas [A]  time = 0.69, size = 106, normalized size = 1.22 \[ \frac {23 \, \sqrt {13} {\left (6 \, x^{3} + x^{2} + 3 \, x + 2\right )} \log \left (\frac {4 \, \sqrt {13} \sqrt {3 \, x^{2} - x + 2} {\left (8 \, x - 9\right )} - 220 \, x^{2} + 196 \, x - 185}{4 \, x^{2} + 4 \, x + 1}\right ) + 26 \, {\left (1536 \, x^{2} + 489 \, x - 289\right )} \sqrt {3 \, x^{2} - x + 2}}{50531 \, {\left (6 \, x^{3} + x^{2} + 3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(3/2),x, algorithm="fricas")

[Out]

1/50531*(23*sqrt(13)*(6*x^3 + x^2 + 3*x + 2)*log((4*sqrt(13)*sqrt(3*x^2 - x + 2)*(8*x - 9) - 220*x^2 + 196*x -
185)/(4*x^2 + 4*x + 1)) + 26*(1536*x^2 + 489*x - 289)*sqrt(3*x^2 - x + 2))/(6*x^3 + x^2 + 3*x + 2)

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giac [B]  time = 0.30, size = 168, normalized size = 1.93 \[ -\frac {2}{50531} \, \sqrt {13} {\left (256 \, \sqrt {13} \sqrt {3} + 23 \, \log \left (\sqrt {13} \sqrt {3} - 4\right )\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right ) - \frac {2 \, {\left (\frac {\frac {1047}{\mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right )} + \frac {299}{{\left (2 \, x + 1\right )} \mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right )}}{2 \, x + 1} - \frac {768}{\mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right )}\right )}}{3887 \, \sqrt {-\frac {8}{2 \, x + 1} + \frac {13}{{\left (2 \, x + 1\right )}^{2}} + 3}} + \frac {2 \, \sqrt {13} \log \left (\sqrt {13} {\left (\sqrt {-\frac {8}{2 \, x + 1} + \frac {13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac {\sqrt {13}}{2 \, x + 1}\right )} - 4\right )}{2197 \, \mathrm {sgn}\left (\frac {1}{2 \, x + 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(3/2),x, algorithm="giac")

[Out]

-2/50531*sqrt(13)*(256*sqrt(13)*sqrt(3) + 23*log(sqrt(13)*sqrt(3) - 4))*sgn(1/(2*x + 1)) - 2/3887*((1047/sgn(1
/(2*x + 1)) + 299/((2*x + 1)*sgn(1/(2*x + 1))))/(2*x + 1) - 768/sgn(1/(2*x + 1)))/sqrt(-8/(2*x + 1) + 13/(2*x
+ 1)^2 + 3) + 2/2197*sqrt(13)*log(sqrt(13)*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1)) - 4)
/sgn(1/(2*x + 1))

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maple [A]  time = 0.01, size = 109, normalized size = 1.25 \[ \frac {2 \sqrt {13}\, \arctanh \left (\frac {2 \left (-4 x +\frac {9}{2}\right ) \sqrt {13}}{13 \sqrt {-16 x +12 \left (x +\frac {1}{2}\right )^{2}+5}}\right )}{2197}+\frac {\frac {12 x}{23}-\frac {2}{23}}{\sqrt {3 x^{2}-x +2}}-\frac {1}{169 \sqrt {-4 x +3 \left (x +\frac {1}{2}\right )^{2}+\frac {5}{4}}}-\frac {82 \left (6 x -1\right )}{3887 \sqrt {-4 x +3 \left (x +\frac {1}{2}\right )^{2}+\frac {5}{4}}}-\frac {1}{26 \left (x +\frac {1}{2}\right ) \sqrt {-4 x +3 \left (x +\frac {1}{2}\right )^{2}+\frac {5}{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^2+3*x+1)/(2*x+1)^2/(3*x^2-x+2)^(3/2),x)

[Out]

2/23*(6*x-1)/(3*x^2-x+2)^(1/2)-1/169/(-4*x+3*(x+1/2)^2+5/4)^(1/2)-82/3887*(6*x-1)/(-4*x+3*(x+1/2)^2+5/4)^(1/2)
+2/2197*13^(1/2)*arctanh(2/13*(-4*x+9/2)*13^(1/2)/(-16*x+12*(x+1/2)^2+5)^(1/2))-1/26/(x+1/2)/(-4*x+3*(x+1/2)^2
+5/4)^(1/2)

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maxima [A]  time = 0.96, size = 96, normalized size = 1.10 \[ -\frac {2}{2197} \, \sqrt {13} \operatorname {arsinh}\left (\frac {8 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 1 \right |}} - \frac {9 \, \sqrt {23}}{23 \, {\left | 2 \, x + 1 \right |}}\right ) + \frac {1536 \, x}{3887 \, \sqrt {3 \, x^{2} - x + 2}} - \frac {279}{3887 \, \sqrt {3 \, x^{2} - x + 2}} - \frac {1}{13 \, {\left (2 \, \sqrt {3 \, x^{2} - x + 2} x + \sqrt {3 \, x^{2} - x + 2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^2+3*x+1)/(1+2*x)^2/(3*x^2-x+2)^(3/2),x, algorithm="maxima")

[Out]

-2/2197*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 1536/3887*x/sqrt(3*x^2 -
x + 2) - 279/3887/sqrt(3*x^2 - x + 2) - 1/13/(2*sqrt(3*x^2 - x + 2)*x + sqrt(3*x^2 - x + 2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {4\,x^2+3\,x+1}{{\left (2\,x+1\right )}^2\,{\left (3\,x^2-x+2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 4*x^2 + 1)/((2*x + 1)^2*(3*x^2 - x + 2)^(3/2)),x)

[Out]

int((3*x + 4*x^2 + 1)/((2*x + 1)^2*(3*x^2 - x + 2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {4 x^{2} + 3 x + 1}{\left (2 x + 1\right )^{2} \left (3 x^{2} - x + 2\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**2+3*x+1)/(1+2*x)**2/(3*x**2-x+2)**(3/2),x)

[Out]

Integral((4*x**2 + 3*x + 1)/((2*x + 1)**2*(3*x**2 - x + 2)**(3/2)), x)

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