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

Optimal. Leaf size=89 \[ \frac{7 \sqrt{3 x^2-x+2}}{169 (2 x+1)}-\frac{\sqrt{3 x^2-x+2}}{26 (2 x+1)^2}-\frac{581 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{676 \sqrt{13}} \]

[Out]

-Sqrt[2 - x + 3*x^2]/(26*(1 + 2*x)^2) + (7*Sqrt[2 - x + 3*x^2])/(169*(1 + 2*x)) - (581*ArcTanh[(9 - 8*x)/(2*Sq
rt[13]*Sqrt[2 - x + 3*x^2])])/(676*Sqrt[13])

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Rubi [A]  time = 0.0882716, antiderivative size = 89, normalized size of antiderivative = 1., 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 = {1650, 806, 724, 206} \[ \frac{7 \sqrt{3 x^2-x+2}}{169 (2 x+1)}-\frac{\sqrt{3 x^2-x+2}}{26 (2 x+1)^2}-\frac{581 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{676 \sqrt{13}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[2 - x + 3*x^2]/(26*(1 + 2*x)^2) + (7*Sqrt[2 - x + 3*x^2])/(169*(1 + 2*x)) - (581*ArcTanh[(9 - 8*x)/(2*Sq
rt[13]*Sqrt[2 - x + 3*x^2])])/(676*Sqrt[13])

Rule 1650

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

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 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 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])

Rubi steps

\begin{align*} \int \frac{1+3 x+4 x^2}{(1+2 x)^3 \sqrt{2-x+3 x^2}} \, dx &=-\frac{\sqrt{2-x+3 x^2}}{26 (1+2 x)^2}-\frac{1}{26} \int \frac{-\frac{35}{2}-49 x}{(1+2 x)^2 \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{\sqrt{2-x+3 x^2}}{26 (1+2 x)^2}+\frac{7 \sqrt{2-x+3 x^2}}{169 (1+2 x)}+\frac{581}{676} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{\sqrt{2-x+3 x^2}}{26 (1+2 x)^2}+\frac{7 \sqrt{2-x+3 x^2}}{169 (1+2 x)}-\frac{581}{338} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )\\ &=-\frac{\sqrt{2-x+3 x^2}}{26 (1+2 x)^2}+\frac{7 \sqrt{2-x+3 x^2}}{169 (1+2 x)}-\frac{581 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{676 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0447765, size = 69, normalized size = 0.78 \[ \frac{\frac{26 (28 x+1) \sqrt{3 x^2-x+2}}{(2 x+1)^2}-581 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{8788} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((26*(1 + 28*x)*Sqrt[2 - x + 3*x^2])/(1 + 2*x)^2 - 581*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x
^2])])/8788

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Maple [A]  time = 0.055, size = 74, normalized size = 0.8 \begin{align*} -{\frac{1}{104}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-2}}+{\frac{7}{338}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}-{\frac{581\,\sqrt{13}}{8788}{\it Artanh} \left ({\frac{2\,\sqrt{13}}{13} \left ({\frac{9}{2}}-4\,x \right ){\frac{1}{\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/104/(x+1/2)^2*(3*(x+1/2)^2-4*x+5/4)^(1/2)+7/338/(x+1/2)*(3*(x+1/2)^2-4*x+5/4)^(1/2)-581/8788*13^(1/2)*arcta
nh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))

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Maxima [A]  time = 1.59528, size = 111, normalized size = 1.25 \begin{align*} \frac{581}{8788} \, \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{\sqrt{3 \, x^{2} - x + 2}}{26 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} + \frac{7 \, \sqrt{3 \, x^{2} - x + 2}}{169 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

581/8788*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) - 1/26*sqrt(3*x^2 - x + 2
)/(4*x^2 + 4*x + 1) + 7/169*sqrt(3*x^2 - x + 2)/(2*x + 1)

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Fricas [A]  time = 1.28669, size = 252, normalized size = 2.83 \begin{align*} \frac{581 \, \sqrt{13}{\left (4 \, x^{2} + 4 \, x + 1\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 ) + 52 \, \sqrt{3 \, x^{2} - x + 2}{\left (28 \, x + 1\right )}}{17576 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/17576*(581*sqrt(13)*(4*x^2 + 4*x + 1)*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)) + 52*sqrt(3*x^2 - x + 2)*(28*x + 1))/(4*x^2 + 4*x + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{4 x^{2} + 3 x + 1}{\left (2 x + 1\right )^{3} \sqrt{3 x^{2} - x + 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.22116, size = 275, normalized size = 3.09 \begin{align*} \frac{581}{8788} \, \sqrt{13} \log \left (-\frac{{\left | -4 \, \sqrt{3} x - 2 \, \sqrt{13} - 2 \, \sqrt{3} + 4 \, \sqrt{3 \, x^{2} - x + 2} \right |}}{2 \,{\left (2 \, \sqrt{3} x - \sqrt{13} + \sqrt{3} - 2 \, \sqrt{3 \, x^{2} - x + 2}\right )}}\right ) + \frac{190 \,{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )}^{3} - 53 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )}^{2} - 489 \, \sqrt{3} x + 289 \, \sqrt{3} + 489 \, \sqrt{3 \, x^{2} - x + 2}}{338 \,{\left (2 \,{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )}^{2} + 2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )} - 5\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

581/8788*sqrt(13)*log(-1/2*abs(-4*sqrt(3)*x - 2*sqrt(13) - 2*sqrt(3) + 4*sqrt(3*x^2 - x + 2))/(2*sqrt(3)*x - s
qrt(13) + sqrt(3) - 2*sqrt(3*x^2 - x + 2))) + 1/338*(190*(sqrt(3)*x - sqrt(3*x^2 - x + 2))^3 - 53*sqrt(3)*(sqr
t(3)*x - sqrt(3*x^2 - x + 2))^2 - 489*sqrt(3)*x + 289*sqrt(3) + 489*sqrt(3*x^2 - x + 2))/(2*(sqrt(3)*x - sqrt(
3*x^2 - x + 2))^2 + 2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) - 5)^2