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

Optimal. Leaf size=138 \[ -\frac{\left (3 x^2-x+2\right )^{5/2}}{26 (2 x+1)^2}+\frac{(122 x+151) \left (3 x^2-x+2\right )^{3/2}}{312 (2 x+1)}+\frac{1}{624} (1858-771 x) \sqrt{3 x^2-x+2}-\frac{1153 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{64 \sqrt{13}}+\frac{1519 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{192 \sqrt{3}} \]

[Out]

((1858 - 771*x)*Sqrt[2 - x + 3*x^2])/624 + ((151 + 122*x)*(2 - x + 3*x^2)^(3/2))/(312*(1 + 2*x)) - (2 - x + 3*
x^2)^(5/2)/(26*(1 + 2*x)^2) + (1519*ArcSinh[(1 - 6*x)/Sqrt[23]])/(192*Sqrt[3]) - (1153*ArcTanh[(9 - 8*x)/(2*Sq
rt[13]*Sqrt[2 - x + 3*x^2])])/(64*Sqrt[13])

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Rubi [A]  time = 0.139158, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1650, 812, 814, 843, 619, 215, 724, 206} \[ -\frac{\left (3 x^2-x+2\right )^{5/2}}{26 (2 x+1)^2}+\frac{(122 x+151) \left (3 x^2-x+2\right )^{3/2}}{312 (2 x+1)}+\frac{1}{624} (1858-771 x) \sqrt{3 x^2-x+2}-\frac{1153 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )}{64 \sqrt{13}}+\frac{1519 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{192 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1858 - 771*x)*Sqrt[2 - x + 3*x^2])/624 + ((151 + 122*x)*(2 - x + 3*x^2)^(3/2))/(312*(1 + 2*x)) - (2 - x + 3*
x^2)^(5/2)/(26*(1 + 2*x)^2) + (1519*ArcSinh[(1 - 6*x)/Sqrt[23]])/(192*Sqrt[3]) - (1153*ArcTanh[(9 - 8*x)/(2*Sq
rt[13]*Sqrt[2 - x + 3*x^2])])/(64*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 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)^3} \, dx &=-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}-\frac{1}{26} \int \frac{\left (-\frac{31}{2}-61 x\right ) \left (2-x+3 x^2\right )^{3/2}}{(1+2 x)^2} \, dx\\ &=\frac{(151+122 x) \left (2-x+3 x^2\right )^{3/2}}{312 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}+\frac{1}{208} \int \frac{(639-1028 x) \sqrt{2-x+3 x^2}}{1+2 x} \, dx\\ &=\frac{1}{624} (1858-771 x) \sqrt{2-x+3 x^2}+\frac{(151+122 x) \left (2-x+3 x^2\right )^{3/2}}{312 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}-\frac{\int \frac{-100880+157976 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx}{9984}\\ &=\frac{1}{624} (1858-771 x) \sqrt{2-x+3 x^2}+\frac{(151+122 x) \left (2-x+3 x^2\right )^{3/2}}{312 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}-\frac{1519}{192} \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx+\frac{1153}{64} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{1}{624} (1858-771 x) \sqrt{2-x+3 x^2}+\frac{(151+122 x) \left (2-x+3 x^2\right )^{3/2}}{312 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}-\frac{1153}{32} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )-\frac{1519 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{192 \sqrt{69}}\\ &=\frac{1}{624} (1858-771 x) \sqrt{2-x+3 x^2}+\frac{(151+122 x) \left (2-x+3 x^2\right )^{3/2}}{312 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{5/2}}{26 (1+2 x)^2}+\frac{1519 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{192 \sqrt{3}}-\frac{1153 \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )}{64 \sqrt{13}}\\ \end{align*}

Mathematica [A]  time = 0.103753, size = 103, normalized size = 0.75 \[ \frac{\frac{156 \sqrt{3 x^2-x+2} \left (96 x^4-68 x^3+390 x^2+627 x+182\right )}{(2 x+1)^2}-10377 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-19747 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{7488} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((156*Sqrt[2 - x + 3*x^2]*(182 + 627*x + 390*x^2 - 68*x^3 + 96*x^4))/(1 + 2*x)^2 - 19747*Sqrt[3]*ArcSinh[(-1 +
6*x)/Sqrt[23]] - 10377*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/7488

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Maple [A]  time = 0.063, size = 162, normalized size = 1.2 \begin{align*}{\frac{15}{338} \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{1153}{4056} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}}}-{\frac{-257+1542\,x}{1248}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}-{\frac{1519\,\sqrt{3}}{576}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }+{\frac{1153}{832}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}-{\frac{1153\,\sqrt{13}}{832}{\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{-15+90\,x}{676} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}}}-{\frac{1}{104} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{5}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-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)^3,x)

[Out]

15/338/(x+1/2)*(3*(x+1/2)^2-4*x+5/4)^(5/2)+1153/4056*(3*(x+1/2)^2-4*x+5/4)^(3/2)-257/1248*(-1+6*x)*(3*(x+1/2)^
2-4*x+5/4)^(1/2)-1519/576*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))+1153/832*(12*(x+1/2)^2-16*x+5)^(1/2)-1153/832
*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x+5)^(1/2))-15/676*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(
3/2)-1/104/(x+1/2)^2*(3*(x+1/2)^2-4*x+5/4)^(5/2)

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Maxima [A]  time = 1.49185, size = 193, normalized size = 1.4 \begin{align*} \frac{61}{312} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} - \frac{{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}}}{26 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} - \frac{257}{208} \, \sqrt{3 \, x^{2} - x + 2} x - \frac{1519}{576} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{1153}{832} \, \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{929}{312} \, \sqrt{3 \, x^{2} - x + 2} + \frac{15 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}}}{52 \,{\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)^3,x, algorithm="maxima")

[Out]

61/312*(3*x^2 - x + 2)^(3/2) - 1/26*(3*x^2 - x + 2)^(5/2)/(4*x^2 + 4*x + 1) - 257/208*sqrt(3*x^2 - x + 2)*x -
1519/576*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) + 1153/832*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x
+ 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 929/312*sqrt(3*x^2 - x + 2) + 15/52*(3*x^2 - x + 2)^(3/2)/(2*x + 1)

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Fricas [A]  time = 1.68986, size = 433, normalized size = 3.14 \begin{align*} \frac{19747 \, \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 ) + 10377 \, \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 ) + 312 \,{\left (96 \, x^{4} - 68 \, x^{3} + 390 \, x^{2} + 627 \, x + 182\right )} \sqrt{3 \, x^{2} - x + 2}}{14976 \,{\left (4 \, x^{2} + 4 \, 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)^3,x, algorithm="fricas")

[Out]

1/14976*(19747*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) + 1
0377*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)) + 312*(96*x^4 - 68*x^3 + 390*x^2 + 627*x + 182)*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{\left (3 x^{2} - x + 2\right )^{\frac{3}{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((3*x**2-x+2)**(3/2)*(4*x**2+3*x+1)/(1+2*x)**3,x)

[Out]

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

[Out]

Exception raised: TypeError