### 3.213 $$\int \frac{\sqrt{2-x+3 x^2} (1+3 x+4 x^2)}{(1+2 x)^3} \, dx$$

Optimal. Leaf size=115 $-\frac{\left (3 x^2-x+2\right )^{3/2}}{26 (2 x+1)^2}+\frac{11 (10 x+7) \sqrt{3 x^2-x+2}}{104 (2 x+1)}-\frac{803 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{208 \sqrt{13}}+\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{8 \sqrt{3}}$

[Out]

(11*(7 + 10*x)*Sqrt[2 - x + 3*x^2])/(104*(1 + 2*x)) - (2 - x + 3*x^2)^(3/2)/(26*(1 + 2*x)^2) + (11*ArcSinh[(1
- 6*x)/Sqrt[23]])/(8*Sqrt[3]) - (803*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(208*Sqrt[13])

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Rubi [A]  time = 0.116874, antiderivative size = 115, 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, 812, 843, 619, 215, 724, 206} $-\frac{\left (3 x^2-x+2\right )^{3/2}}{26 (2 x+1)^2}+\frac{11 (10 x+7) \sqrt{3 x^2-x+2}}{104 (2 x+1)}-\frac{803 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{208 \sqrt{13}}+\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{8 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(11*(7 + 10*x)*Sqrt[2 - x + 3*x^2])/(104*(1 + 2*x)) - (2 - x + 3*x^2)^(3/2)/(26*(1 + 2*x)^2) + (11*ArcSinh[(1
- 6*x)/Sqrt[23]])/(8*Sqrt[3]) - (803*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/(208*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 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*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] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 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)^3} \, dx &=-\frac{\left (2-x+3 x^2\right )^{3/2}}{26 (1+2 x)^2}-\frac{1}{26} \int \frac{\left (-\frac{33}{2}-55 x\right ) \sqrt{2-x+3 x^2}}{(1+2 x)^2} \, dx\\ &=\frac{11 (7+10 x) \sqrt{2-x+3 x^2}}{104 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{3/2}}{26 (1+2 x)^2}+\frac{1}{208} \int \frac{517-572 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{11 (7+10 x) \sqrt{2-x+3 x^2}}{104 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{3/2}}{26 (1+2 x)^2}-\frac{11}{8} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx+\frac{803}{208} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{11 (7+10 x) \sqrt{2-x+3 x^2}}{104 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{3/2}}{26 (1+2 x)^2}-\frac{803}{104} \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 )}{8 \sqrt{69}}\\ &=\frac{11 (7+10 x) \sqrt{2-x+3 x^2}}{104 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{3/2}}{26 (1+2 x)^2}+\frac{11 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{8 \sqrt{3}}-\frac{803 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{208 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.0815547, size = 93, normalized size = 0.81 $\frac{\frac{78 \sqrt{3 x^2-x+2} \left (208 x^2+268 x+69\right )}{(2 x+1)^2}-2409 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-3718 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{8112}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((78*Sqrt[2 - x + 3*x^2]*(69 + 268*x + 208*x^2))/(1 + 2*x)^2 - 3718*Sqrt[3]*ArcSinh[(-1 + 6*x)/Sqrt[23]] - 240
9*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/8112

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Maple [A]  time = 0.057, size = 125, normalized size = 1.1 \begin{align*}{\frac{11}{338} \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{803}{2704}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}-{\frac{11\,\sqrt{3}}{24}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }-{\frac{803\,\sqrt{13}}{2704}{\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{-11+66\,x}{676}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}-{\frac{1}{104} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-2}} \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)^3,x)

[Out]

11/338/(x+1/2)*(3*(x+1/2)^2-4*x+5/4)^(3/2)+803/2704*(12*(x+1/2)^2-16*x+5)^(1/2)-11/24*3^(1/2)*arcsinh(6/23*23^
(1/2)*(x-1/6))-803/2704*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))-11/676*(-1+6*x)*
(3*(x+1/2)^2-4*x+5/4)^(1/2)-1/104/(x+1/2)^2*(3*(x+1/2)^2-4*x+5/4)^(3/2)

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Maxima [A]  time = 1.49728, size = 154, normalized size = 1.34 \begin{align*} -\frac{11}{24} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{803}{2704} \, \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{55}{104} \, \sqrt{3 \, x^{2} - x + 2} - \frac{{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}}{26 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} + \frac{11 \, \sqrt{3 \, x^{2} - x + 2}}{52 \,{\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)^3,x, algorithm="maxima")

[Out]

-11/24*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) + 803/2704*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x +
1) - 9/23*sqrt(23)/abs(2*x + 1)) + 55/104*sqrt(3*x^2 - x + 2) - 1/26*(3*x^2 - x + 2)^(3/2)/(4*x^2 + 4*x + 1) +
11/52*sqrt(3*x^2 - x + 2)/(2*x + 1)

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Fricas [A]  time = 1.65216, size = 405, normalized size = 3.52 \begin{align*} \frac{3718 \, \sqrt{3}{\left (4 \, x^{2} + 4 \, 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 ) + 2409 \, \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 ) + 156 \,{\left (208 \, x^{2} + 268 \, x + 69\right )} \sqrt{3 \, x^{2} - x + 2}}{16224 \,{\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)*(3*x^2-x+2)^(1/2)/(1+2*x)^3,x, algorithm="fricas")

[Out]

1/16224*(3718*sqrt(3)*(4*x^2 + 4*x + 1)*log(4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 24
09*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)) + 156*(208*x^2 + 268*x + 69)*sqrt(3*x^2 - x + 2))/(4*x^2 + 4*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 )^{3}}\, 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)**3,x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError