3.3.19 \(\int \frac {(2-x+3 x^2)^{3/2} (1+3 x+4 x^2)}{(1+2 x)^3} \, dx\) [219]

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

[Out]

1/312*(151+122*x)*(3*x^2-x+2)^(3/2)/(1+2*x)-1/26*(3*x^2-x+2)^(5/2)/(1+2*x)^2+1519/576*arcsinh(1/23*(1-6*x)*23^
(1/2))*3^(1/2)-1153/832*arctanh(1/26*(9-8*x)*13^(1/2)/(3*x^2-x+2)^(1/2))*13^(1/2)+1/624*(1858-771*x)*(3*x^2-x+
2)^(1/2)

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Rubi [A]
time = 0.08, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1664, 826, 828, 857, 633, 221, 738, 212} \begin {gather*} -\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}} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 221

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 633

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 738

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 826

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 828

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 857

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 1664

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]

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} \text {Subst}\left (\int \frac {1}{52-x^2} \, dx,x,\frac {9-8 x}{\sqrt {2-x+3 x^2}}\right )-\frac {1519 \text {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*}

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Mathematica [A]
time = 0.58, size = 121, normalized size = 0.88 \begin {gather*} \frac {\frac {156 \sqrt {2-x+3 x^2} \left (182+627 x+390 x^2-68 x^3+96 x^4\right )}{(1+2 x)^2}+20754 \sqrt {13} \tanh ^{-1}\left (\frac {\sqrt {3}+2 \sqrt {3} x-2 \sqrt {2-x+3 x^2}}{\sqrt {13}}\right )+19747 \sqrt {3} \log \left (1-6 x+2 \sqrt {6-3 x+9 x^2}\right )}{7488} \end {gather*}

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 + 20754*Sqrt[13]*ArcTanh[(Sqr
t[3] + 2*Sqrt[3]*x - 2*Sqrt[2 - x + 3*x^2])/Sqrt[13]] + 19747*Sqrt[3]*Log[1 - 6*x + 2*Sqrt[6 - 3*x + 9*x^2]])/
7488

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Maple [A]
time = 0.16, size = 162, normalized size = 1.17

method	result	size
risch	\(\frac {288 x^{6}-300 x^{5}+1430 x^{4}+1355 x^{3}+699 x^{2}+1072 x +364}{48 \left (2 x +1\right )^{2} \sqrt {3 x^{2}-x +2}}-\frac {1519 \sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{576}-\frac {1153 \sqrt {13}\, \arctanh \left (\frac {2 \left (\frac {9}{2}-4 x \right ) \sqrt {13}}{13 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-16 x +5}}\right )}{832}\)	\(97\)
trager	\(\frac {\left (96 x^{4}-68 x^{3}+390 x^{2}+627 x +182\right ) \sqrt {3 x^{2}-x +2}}{48 \left (2 x +1\right )^{2}}+\frac {1153 \RootOf \left (\textit {\_Z}^{2}-13\right ) \ln \left (\frac {8 \RootOf \left (\textit {\_Z}^{2}-13\right ) x +26 \sqrt {3 x^{2}-x +2}-9 \RootOf \left (\textit {\_Z}^{2}-13\right )}{2 x +1}\right )}{832}-\frac {1519 \RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (6 \RootOf \left (\textit {\_Z}^{2}-3\right ) x +6 \sqrt {3 x^{2}-x +2}-\RootOf \left (\textit {\_Z}^{2}-3\right )\right )}{576}\)	\(133\)
default	\(\frac {1153 \left (3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}\right )^{\frac {3}{2}}}{4056}-\frac {257 \left (6 x -1\right ) \sqrt {3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}}}{1248}-\frac {1519 \sqrt {3}\, \arcsinh \left (\frac {6 \sqrt {23}\, \left (x -\frac {1}{6}\right )}{23}\right )}{576}+\frac {1153 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-16 x +5}}{832}-\frac {1153 \sqrt {13}\, \arctanh \left (\frac {2 \left (\frac {9}{2}-4 x \right ) \sqrt {13}}{13 \sqrt {12 \left (x +\frac {1}{2}\right )^{2}-16 x +5}}\right )}{832}+\frac {15 \left (3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}\right )^{\frac {5}{2}}}{338 \left (x +\frac {1}{2}\right )}-\frac {15 \left (6 x -1\right ) \left (3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}\right )^{\frac {3}{2}}}{676}-\frac {\left (3 \left (x +\frac {1}{2}\right )^{2}-4 x +\frac {5}{4}\right )^{\frac {5}{2}}}{104 \left (x +\frac {1}{2}\right )^{2}}\)	\(162\)

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)/(2*x+1)^3,x,method=_RETURNVERBOSE)

[Out]

1153/4056*(3*(x+1/2)^2-4*x+5/4)^(3/2)-257/1248*(6*x-1)*(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)/(1
2*(x+1/2)^2-16*x+5)^(1/2))+15/338/(x+1/2)*(3*(x+1/2)^2-4*x+5/4)^(5/2)-15/676*(6*x-1)*(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 = 0.52, size = 143, normalized size = 1.04 \begin {gather*} \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 {gather*}

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 = 0.44, size = 159, normalized size = 1.15 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3 x^{2} - x + 2\right )^{\frac {3}{2}} \cdot \left (4 x^{2} + 3 x + 1\right )}{\left (2 x + 1\right )^{3}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Unable to divide, perhaps due to rounding error%%%{-18688512,[6]%%%}+%%%{%%{[56065536,0]:[1,0,-3]%%},[5]%%%
}+%%%{-2803

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (3\,x^2-x+2\right )}^{3/2}\,\left (4\,x^2+3\,x+1\right )}{{\left (2\,x+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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