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

Optimal. Leaf size=131 \[ -\frac{\left (3 x^2-x+2\right )^{5/2}}{13 (2 x+1)}-\frac{1}{104} (23-38 x) \left (3 x^2-x+2\right )^{3/2}-\frac{1}{192} (349-294 x) \sqrt{3 x^2-x+2}+\frac{25}{32} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-\frac{2327 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{384 \sqrt{3}} \]

[Out]

-((349 - 294*x)*Sqrt[2 - x + 3*x^2])/192 - ((23 - 38*x)*(2 - x + 3*x^2)^(3/2))/104 - (2 - x + 3*x^2)^(5/2)/(13
*(1 + 2*x)) - (2327*ArcSinh[(1 - 6*x)/Sqrt[23]])/(384*Sqrt[3]) + (25*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sq
rt[2 - x + 3*x^2])])/32

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Rubi [A]  time = 0.140166, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{\left (3 x^2-x+2\right )^{5/2}}{13 (2 x+1)}-\frac{1}{104} (23-38 x) \left (3 x^2-x+2\right )^{3/2}-\frac{1}{192} (349-294 x) \sqrt{3 x^2-x+2}+\frac{25}{32} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-\frac{2327 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{384 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((349 - 294*x)*Sqrt[2 - x + 3*x^2])/192 - ((23 - 38*x)*(2 - x + 3*x^2)^(3/2))/104 - (2 - x + 3*x^2)^(5/2)/(13
*(1 + 2*x)) - (2327*ArcSinh[(1 - 6*x)/Sqrt[23]])/(384*Sqrt[3]) + (25*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sq
rt[2 - x + 3*x^2])])/32

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{\left (2-x+3 x^2\right )^{3/2} \left (1+3 x+4 x^2\right )}{(1+2 x)^2} \, dx &=-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}-\frac{1}{13} \int \frac{\left (-\frac{13}{2}-38 x\right ) \left (2-x+3 x^2\right )^{3/2}}{1+2 x} \, dx\\ &=-\frac{1}{104} (23-38 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}+\frac{\int \frac{(-78+7644 x) \sqrt{2-x+3 x^2}}{1+2 x} \, dx}{1248}\\ &=-\frac{1}{192} (349-294 x) \sqrt{2-x+3 x^2}-\frac{1}{104} (23-38 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}-\frac{\int \frac{245388-726024 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx}{59904}\\ &=-\frac{1}{192} (349-294 x) \sqrt{2-x+3 x^2}-\frac{1}{104} (23-38 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}+\frac{2327}{384} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx-\frac{325}{32} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=-\frac{1}{192} (349-294 x) \sqrt{2-x+3 x^2}-\frac{1}{104} (23-38 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}+\frac{325}{16} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )+\frac{2327 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{384 \sqrt{69}}\\ &=-\frac{1}{192} (349-294 x) \sqrt{2-x+3 x^2}-\frac{1}{104} (23-38 x) \left (2-x+3 x^2\right )^{3/2}-\frac{\left (2-x+3 x^2\right )^{5/2}}{13 (1+2 x)}-\frac{2327 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{384 \sqrt{3}}+\frac{25}{32} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0899109, size = 103, normalized size = 0.79 \[ \frac{\frac{6 \sqrt{3 x^2-x+2} \left (288 x^4-96 x^3+564 x^2-332 x-493\right )}{2 x+1}+900 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )+2327 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{1152} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*Sqrt[2 - x + 3*x^2]*(-493 - 332*x + 564*x^2 - 96*x^3 + 288*x^4))/(1 + 2*x) + 2327*Sqrt[3]*ArcSinh[(-1 + 6*
x)/Sqrt[23]] + 900*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/1152

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Maple [A]  time = 0.055, size = 179, normalized size = 1.4 \begin{align*}{\frac{-1+6\,x}{24} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{-23+138\,x}{192}\sqrt{3\,{x}^{2}-x+2}}+{\frac{2327\,\sqrt{3}}{1152}{\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{5}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}-{\frac{25}{156} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}}}+{\frac{-13+78\,x}{96}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}-{\frac{25}{32}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}+{\frac{25\,\sqrt{13}}{32}{\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} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(-1+6*x)*(3*x^2-x+2)^(3/2)+23/192*(-1+6*x)*(3*x^2-x+2)^(1/2)+2327/1152*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)^(5/2)-25/156*(3*(x+1/2)^2-4*x+5/4)^(3/2)+13/96*(-1+6*x)*(3*(x+1/2)^2-4
*x+5/4)^(1/2)-25/32*(12*(x+1/2)^2-16*x+5)^(1/2)+25/32*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-1
6*x+5)^(1/2))+1/52*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(3/2)

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Maxima [A]  time = 1.49578, size = 178, normalized size = 1.36 \begin{align*} \frac{1}{4} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x - \frac{1}{8} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} + \frac{49}{32} \, \sqrt{3 \, x^{2} - x + 2} x + \frac{2327}{1152} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) - \frac{25}{32} \, \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{349}{192} \, \sqrt{3 \, x^{2} - x + 2} - \frac{{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}}{4 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*x^2 - x + 2)^(3/2)*x - 1/8*(3*x^2 - x + 2)^(3/2) + 49/32*sqrt(3*x^2 - x + 2)*x + 2327/1152*sqrt(3)*arcs
inh(6/23*sqrt(23)*x - 1/23*sqrt(23)) - 25/32*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x + 1) - 9/23*sqrt(23)/abs
(2*x + 1)) - 349/192*sqrt(3*x^2 - x + 2) - 1/4*(3*x^2 - x + 2)^(3/2)/(2*x + 1)

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Fricas [A]  time = 1.71441, size = 396, normalized size = 3.02 \begin{align*} \frac{2327 \, \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 ) + 900 \, \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 ) + 12 \,{\left (288 \, x^{4} - 96 \, x^{3} + 564 \, x^{2} - 332 \, x - 493\right )} \sqrt{3 \, x^{2} - x + 2}}{2304 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2304*(2327*sqrt(3)*(2*x + 1)*log(-4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x - 25) + 900*sqrt(1
3)*(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)) + 12*(2
88*x^4 - 96*x^3 + 564*x^2 - 332*x - 493)*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{\left (3 x^{2} - x + 2\right )^{\frac{3}{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((3*x**2-x+2)**(3/2)*(4*x**2+3*x+1)/(1+2*x)**2,x)

[Out]

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

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Giac [B]  time = 1.84953, size = 770, normalized size = 5.88 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/32*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
)) - 2327/1152*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)) - 13/32*sqrt(-8
/(2*x + 1) + 13/(2*x + 1)^2 + 3)*sgn(1/(2*x + 1)) + 1/192*(5929*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqr
t(13)/(2*x + 1))^7*sgn(1/(2*x + 1)) - 7272*sqrt(13)*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x +
1))^6*sgn(1/(2*x + 1)) + 25101*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^5*sgn(1/(2*x +
1)) - 48*sqrt(13)*(sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^4*sgn(1/(2*x + 1)) + 112359*(
sqrt(-8/(2*x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^3*sgn(1/(2*x + 1)) - 69336*sqrt(13)*(sqrt(-8/(2*
x + 1) + 13/(2*x + 1)^2 + 3) + sqrt(13)/(2*x + 1))^2*sgn(1/(2*x + 1)) + 71955*(sqrt(-8/(2*x + 1) + 13/(2*x + 1
)^2 + 3) + sqrt(13)/(2*x + 1))*sgn(1/(2*x + 1)) + 24624*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)^4