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

Optimal. Leaf size=108 \[ -\frac{\left (3 x^2-x+2\right )^{3/2}}{13 (2 x+1)}-\frac{1}{156} (67-96 x) \sqrt{3 x^2-x+2}+\frac{17 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{8 \sqrt{13}}-\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

[Out]

-((67 - 96*x)*Sqrt[2 - x + 3*x^2])/156 - (2 - x + 3*x^2)^(3/2)/(13*(1 + 2*x)) - (11*ArcSinh[(1 - 6*x)/Sqrt[23]
])/(6*Sqrt[3]) + (17*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(8*Sqrt[13])

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Rubi [A]  time = 0.116414, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {1650, 814, 843, 619, 215, 724, 206} \[ -\frac{\left (3 x^2-x+2\right )^{3/2}}{13 (2 x+1)}-\frac{1}{156} (67-96 x) \sqrt{3 x^2-x+2}+\frac{17 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{8 \sqrt{13}}-\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((67 - 96*x)*Sqrt[2 - x + 3*x^2])/156 - (2 - x + 3*x^2)^(3/2)/(13*(1 + 2*x)) - (11*ArcSinh[(1 - 6*x)/Sqrt[23]
])/(6*Sqrt[3]) + (17*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(8*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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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{\sqrt{2-x+3 x^2} \left (1+3 x+4 x^2\right )}{(1+2 x)^2} \, dx &=-\frac{\left (2-x+3 x^2\right )^{3/2}}{13 (1+2 x)}-\frac{1}{13} \int \frac{\left (-\frac{15}{2}-32 x\right ) \sqrt{2-x+3 x^2}}{1+2 x} \, dx\\ &=-\frac{1}{156} (67-96 x) \sqrt{2-x+3 x^2}-\frac{\left (2-x+3 x^2\right )^{3/2}}{13 (1+2 x)}+\frac{1}{624} \int \frac{-182+2288 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{1}{156} (67-96 x) \sqrt{2-x+3 x^2}-\frac{\left (2-x+3 x^2\right )^{3/2}}{13 (1+2 x)}+\frac{11}{6} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx-\frac{17}{8} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{1}{156} (67-96 x) \sqrt{2-x+3 x^2}-\frac{\left (2-x+3 x^2\right )^{3/2}}{13 (1+2 x)}+\frac{17}{4} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )+\frac{11 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{6 \sqrt{69}}\\ &=-\frac{1}{156} (67-96 x) \sqrt{2-x+3 x^2}-\frac{\left (2-x+3 x^2\right )^{3/2}}{13 (1+2 x)}-\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{6 \sqrt{3}}+\frac{17 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{8 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0875472, size = 92, normalized size = 0.85 \[ \frac{\sqrt{3 x^2-x+2} \left (12 x^2-2 x-7\right )}{24 x+12}+\frac{17 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{8 \sqrt{13}}+\frac{11 \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{6 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 - x + 3*x^2]*(-7 - 2*x + 12*x^2))/(12 + 24*x) + (11*ArcSinh[(-1 + 6*x)/Sqrt[23]])/(6*Sqrt[3]) + (17*Ar
cTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(8*Sqrt[13])

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Maple [A]  time = 0.055, size = 123, normalized size = 1.1 \begin{align*}{\frac{-1+6\,x}{12}\sqrt{3\,{x}^{2}-x+2}}+{\frac{11\,\sqrt{3}}{18}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }-{\frac{1}{26} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}-{\frac{17}{104}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}+{\frac{17\,\sqrt{13}}{104}{\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+6\,x}{52}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(-1+6*x)*(3*x^2-x+2)^(1/2)+11/18*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))-1/26/(x+1/2)*(3*(x+1/2)^2-4*x+5/4
)^(3/2)-17/104*(12*(x+1/2)^2-16*x+5)^(1/2)+17/104*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+
5)^(1/2))+1/52*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(1/2)

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Maxima [A]  time = 1.63886, size = 139, normalized size = 1.29 \begin{align*} \frac{1}{2} \, \sqrt{3 \, x^{2} - x + 2} x + \frac{11}{18} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{17}{104} \, \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{1}{3} \, \sqrt{3 \, x^{2} - x + 2} - \frac{\sqrt{3 \, x^{2} - x + 2}}{4 \,{\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)*(3*x^2-x+2)^(1/2)/(1+2*x)^2,x, algorithm="maxima")

[Out]

1/2*sqrt(3*x^2 - x + 2)*x + 11/18*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) - 17/104*sqrt(13)*arcsinh(8
/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs(2*x + 1)) - 1/3*sqrt(3*x^2 - x + 2) - 1/4*sqrt(3*x^2 - x + 2)/
(2*x + 1)

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Fricas [A]  time = 1.62976, size = 363, normalized size = 3.36 \begin{align*} \frac{572 \, \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 ) + 153 \, \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 \,{\left (12 \, x^{2} - 2 \, x - 7\right )} \sqrt{3 \, x^{2} - x + 2}}{1872 \,{\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)*(3*x^2-x+2)^(1/2)/(1+2*x)^2,x, algorithm="fricas")

[Out]

1/1872*(572*sqrt(3)*(2*x + 1)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 153*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*(1
2*x^2 - 2*x - 7)*sqrt(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{\sqrt{3 x^{2} - x + 2} \left (4 x^{2} + 3 x + 1\right )}{\left (2 x + 1\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.6724, size = 513, normalized size = 4.75 \begin{align*} \frac{17}{104} \, \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 ) \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) - \frac{11}{18} \, \sqrt{3} \log \left (\frac{{\left | -2 \, \sqrt{3} + 2 \, \sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac{2 \, \sqrt{13}}{2 \, x + 1} \right |}}{2 \,{\left (\sqrt{3} + \sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac{\sqrt{13}}{2 \, x + 1}\right )}}\right ) \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) - \frac{1}{8} \, \sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) + \frac{67 \,{\left (\sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac{\sqrt{13}}{2 \, x + 1}\right )}^{3} \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) - 57 \, \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 )}^{2} \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) + 129 \,{\left (\sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac{\sqrt{13}}{2 \, x + 1}\right )} \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right ) + 27 \, \sqrt{13} \mathrm{sgn}\left (\frac{1}{2 \, x + 1}\right )}{12 \,{\left ({\left (\sqrt{-\frac{8}{2 \, x + 1} + \frac{13}{{\left (2 \, x + 1\right )}^{2}} + 3} + \frac{\sqrt{13}}{2 \, x + 1}\right )}^{2} - 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/104*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)) - 11/18*sqrt(3)*log(1/2*abs(-2*sqrt(3) + 2*sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + 2*sqrt(13)/(2*x + 1))
/(sqrt(3) + sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1)))*sgn(1/(2*x + 1)) - 1/8*sqrt(-8/(2*x
+ 1) + 13/(2*x + 1)^2 + 3)*sgn(1/(2*x + 1)) + 1/12*(67*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2
*x + 1))^3*sgn(1/(2*x + 1)) - 57*sqrt(13)*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^2*sgn
(1/(2*x + 1)) + 129*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))*sgn(1/(2*x + 1)) + 27*sqrt(
13)*sgn(1/(2*x + 1)))/((sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^2 - 3)^2