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

Optimal. Leaf size=83 \[ -\frac{\sqrt{3 x^2-x+2}}{13 (2 x+1)}+\frac{9 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{26 \sqrt{13}}-\frac{\sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{\sqrt{3}} \]

[Out]

-Sqrt[2 - x + 3*x^2]/(13*(1 + 2*x)) - ArcSinh[(1 - 6*x)/Sqrt[23]]/Sqrt[3] + (9*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*S
qrt[2 - x + 3*x^2])])/(26*Sqrt[13])

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Rubi [A]  time = 0.0950249, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1650, 843, 619, 215, 724, 206} \[ -\frac{\sqrt{3 x^2-x+2}}{13 (2 x+1)}+\frac{9 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{26 \sqrt{13}}-\frac{\sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[2 - x + 3*x^2]/(13*(1 + 2*x)) - ArcSinh[(1 - 6*x)/Sqrt[23]]/Sqrt[3] + (9*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*S
qrt[2 - x + 3*x^2])])/(26*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 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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)^2 \sqrt{2-x+3 x^2}} \, dx &=-\frac{\sqrt{2-x+3 x^2}}{13 (1+2 x)}-\frac{1}{13} \int \frac{-\frac{17}{2}-26 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{\sqrt{2-x+3 x^2}}{13 (1+2 x)}-\frac{9}{26} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx+\int \frac{1}{\sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{\sqrt{2-x+3 x^2}}{13 (1+2 x)}+\frac{9}{13} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{\sqrt{69}}\\ &=-\frac{\sqrt{2-x+3 x^2}}{13 (1+2 x)}-\frac{\sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{\sqrt{3}}+\frac{9 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{26 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0518867, size = 82, normalized size = 0.99 \[ -\frac{\sqrt{3 x^2-x+2}}{13 (2 x+1)}+\frac{9 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{26 \sqrt{13}}+\frac{\sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[2 - x + 3*x^2]/(13*(1 + 2*x)) + ArcSinh[(-1 + 6*x)/Sqrt[23]]/Sqrt[3] + (9*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*
Sqrt[2 - x + 3*x^2])])/(26*Sqrt[13])

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Maple [A]  time = 0.056, size = 67, normalized size = 0.8 \begin{align*}{\frac{\sqrt{3}}{3}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }+{\frac{9\,\sqrt{13}}{338}{\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 ) }-{\frac{1}{26}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))+9/338*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5
)^(1/2))-1/26/(x+1/2)*(3*(x+1/2)^2-4*x+5/4)^(1/2)

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Maxima [A]  time = 1.54286, size = 100, normalized size = 1.2 \begin{align*} \frac{1}{3} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{9}{338} \, \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}}{13 \,{\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)^2/(3*x^2-x+2)^(1/2),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) - 9/338*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9
/23*sqrt(23)/abs(2*x + 1)) - 1/13*sqrt(3*x^2 - x + 2)/(2*x + 1)

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Fricas [A]  time = 1.70077, size = 336, normalized size = 4.05 \begin{align*} \frac{338 \, \sqrt{3}{\left (2 \, x + 1\right )} \log \left (-4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) + 27 \, \sqrt{13}{\left (2 \, 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 ) - 156 \, \sqrt{3 \, x^{2} - x + 2}}{2028 \,{\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)^2/(3*x^2-x+2)^(1/2),x, algorithm="fricas")

[Out]

1/2028*(338*sqrt(3)*(2*x + 1)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 27*sqrt(13)
*(2*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)) - 156*sqr
t(3*x^2 - x + 2))/(2*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 )^{2} \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)**2/(3*x**2-x+2)**(1/2),x)

[Out]

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

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

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)^(1/2),x, algorithm="giac")

[Out]

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