### 3.225 $$\int \frac{(2-x+3 x^2)^{5/2} (1+3 x+4 x^2)}{(1+2 x)^3} \, dx$$

Optimal. Leaf size=161 $-\frac{\left (3 x^2-x+2\right )^{7/2}}{26 (2 x+1)^2}+\frac{(134 x+257) \left (3 x^2-x+2\right )^{5/2}}{520 (2 x+1)}+\frac{1}{832} (1227-838 x) \left (3 x^2-x+2\right )^{3/2}+\frac{(21317-10470 x) \sqrt{3 x^2-x+2}}{1536}-\frac{1631}{256} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )+\frac{118423 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{3072 \sqrt{3}}$

[Out]

((21317 - 10470*x)*Sqrt[2 - x + 3*x^2])/1536 + ((1227 - 838*x)*(2 - x + 3*x^2)^(3/2))/832 + ((257 + 134*x)*(2
- x + 3*x^2)^(5/2))/(520*(1 + 2*x)) - (2 - x + 3*x^2)^(7/2)/(26*(1 + 2*x)^2) + (118423*ArcSinh[(1 - 6*x)/Sqrt[
23]])/(3072*Sqrt[3]) - (1631*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/256

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Rubi [A]  time = 0.162784, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 9, 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} $-\frac{\left (3 x^2-x+2\right )^{7/2}}{26 (2 x+1)^2}+\frac{(134 x+257) \left (3 x^2-x+2\right )^{5/2}}{520 (2 x+1)}+\frac{1}{832} (1227-838 x) \left (3 x^2-x+2\right )^{3/2}+\frac{(21317-10470 x) \sqrt{3 x^2-x+2}}{1536}-\frac{1631}{256} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )+\frac{118423 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{3072 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[((2 - x + 3*x^2)^(5/2)*(1 + 3*x + 4*x^2))/(1 + 2*x)^3,x]

[Out]

((21317 - 10470*x)*Sqrt[2 - x + 3*x^2])/1536 + ((1227 - 838*x)*(2 - x + 3*x^2)^(3/2))/832 + ((257 + 134*x)*(2
- x + 3*x^2)^(5/2))/(520*(1 + 2*x)) - (2 - x + 3*x^2)^(7/2)/(26*(1 + 2*x)^2) + (118423*ArcSinh[(1 - 6*x)/Sqrt[
23]])/(3072*Sqrt[3]) - (1631*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 - x + 3*x^2])])/256

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 )^{5/2} \left (1+3 x+4 x^2\right )}{(1+2 x)^3} \, dx &=-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}-\frac{1}{26} \int \frac{\left (-\frac{29}{2}-67 x\right ) \left (2-x+3 x^2\right )^{5/2}}{(1+2 x)^2} \, dx\\ &=\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}+\frac{1}{208} \int \frac{(793-1676 x) \left (2-x+3 x^2\right )^{3/2}}{1+2 x} \, dx\\ &=\frac{1}{832} (1227-838 x) \left (2-x+3 x^2\right )^{3/2}+\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}-\frac{\int \frac{(-236652+544440 x) \sqrt{2-x+3 x^2}}{1+2 x} \, dx}{19968}\\ &=\frac{(21317-10470 x) \sqrt{2-x+3 x^2}}{1536}+\frac{1}{832} (1227-838 x) \left (2-x+3 x^2\right )^{3/2}+\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}+\frac{\int \frac{42436056-73895952 x}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx}{958464}\\ &=\frac{(21317-10470 x) \sqrt{2-x+3 x^2}}{1536}+\frac{1}{832} (1227-838 x) \left (2-x+3 x^2\right )^{3/2}+\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}-\frac{118423 \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx}{3072}+\frac{21203}{256} \int \frac{1}{(1+2 x) \sqrt{2-x+3 x^2}} \, dx\\ &=\frac{(21317-10470 x) \sqrt{2-x+3 x^2}}{1536}+\frac{1}{832} (1227-838 x) \left (2-x+3 x^2\right )^{3/2}+\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}-\frac{21203}{128} \operatorname{Subst}\left (\int \frac{1}{52-x^2} \, dx,x,\frac{9-8 x}{\sqrt{2-x+3 x^2}}\right )-\frac{118423 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{3072 \sqrt{69}}\\ &=\frac{(21317-10470 x) \sqrt{2-x+3 x^2}}{1536}+\frac{1}{832} (1227-838 x) \left (2-x+3 x^2\right )^{3/2}+\frac{(257+134 x) \left (2-x+3 x^2\right )^{5/2}}{520 (1+2 x)}-\frac{\left (2-x+3 x^2\right )^{7/2}}{26 (1+2 x)^2}+\frac{118423 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{3072 \sqrt{3}}-\frac{1631}{256} \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{2-x+3 x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.108018, size = 113, normalized size = 0.7 $\frac{\frac{6 \sqrt{3 x^2-x+2} \left (27648 x^6-22464 x^5+83616 x^4-76200 x^3+256564 x^2+464446 x+142057\right )}{(2 x+1)^2}-293580 \sqrt{13} \tanh ^{-1}\left (\frac{9-8 x}{2 \sqrt{13} \sqrt{3 x^2-x+2}}\right )-592115 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{46080}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[((2 - x + 3*x^2)^(5/2)*(1 + 3*x + 4*x^2))/(1 + 2*x)^3,x]

[Out]

((6*Sqrt[2 - x + 3*x^2]*(142057 + 464446*x + 256564*x^2 - 76200*x^3 + 83616*x^4 - 22464*x^5 + 27648*x^6))/(1 +
2*x)^2 - 592115*Sqrt[3]*ArcSinh[(-1 + 6*x)/Sqrt[23]] - 293580*Sqrt[13]*ArcTanh[(9 - 8*x)/(2*Sqrt[13]*Sqrt[2 -
x + 3*x^2])])/46080

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Maple [A]  time = 0.059, size = 199, normalized size = 1.2 \begin{align*}{\frac{1631}{6760} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{5}{2}}}}-{\frac{-19+114\,x}{676} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{5}{2}}}}+{\frac{19}{338} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{7}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-1}}-{\frac{1}{104} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{7}{2}}} \left ( x+{\frac{1}{2}} \right ) ^{-2}}-{\frac{-1745+10470\,x}{1536}\sqrt{3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}}}}-{\frac{-419+2514\,x}{2496} \left ( 3\, \left ( x+1/2 \right ) ^{2}-4\,x+{\frac{5}{4}} \right ) ^{{\frac{3}{2}}}}-{\frac{1631\,\sqrt{13}}{256}{\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{118423\,\sqrt{3}}{9216}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }+{\frac{1631}{256}\sqrt{12\, \left ( x+1/2 \right ) ^{2}-16\,x+5}}+{\frac{1631}{1248} \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)^(5/2)*(4*x^2+3*x+1)/(1+2*x)^3,x)

[Out]

1631/6760*(3*(x+1/2)^2-4*x+5/4)^(5/2)-19/676*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(5/2)+19/338/(x+1/2)*(3*(x+1/2)^2-
4*x+5/4)^(7/2)-1/104/(x+1/2)^2*(3*(x+1/2)^2-4*x+5/4)^(7/2)-1745/1536*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(1/2)-419/
2496*(-1+6*x)*(3*(x+1/2)^2-4*x+5/4)^(3/2)-1631/256*13^(1/2)*arctanh(2/13*(9/2-4*x)*13^(1/2)/(12*(x+1/2)^2-16*x
+5)^(1/2))-118423/9216*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))+1631/256*(12*(x+1/2)^2-16*x+5)^(1/2)+1631/1248*(
3*(x+1/2)^2-4*x+5/4)^(3/2)

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Maxima [A]  time = 1.59254, size = 232, normalized size = 1.44 \begin{align*} \frac{67}{520} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}} - \frac{{\left (3 \, x^{2} - x + 2\right )}^{\frac{7}{2}}}{26 \,{\left (4 \, x^{2} + 4 \, x + 1\right )}} - \frac{419}{416} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x + \frac{1227}{832} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} + \frac{19 \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}}}{52 \,{\left (2 \, x + 1\right )}} - \frac{1745}{256} \, \sqrt{3 \, x^{2} - x + 2} x - \frac{118423}{9216} \, \sqrt{3} \operatorname{arsinh}\left (\frac{6}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{1631}{256} \, \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{21317}{1536} \, \sqrt{3 \, x^{2} - x + 2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2-x+2)^(5/2)*(4*x^2+3*x+1)/(1+2*x)^3,x, algorithm="maxima")

[Out]

67/520*(3*x^2 - x + 2)^(5/2) - 1/26*(3*x^2 - x + 2)^(7/2)/(4*x^2 + 4*x + 1) - 419/416*(3*x^2 - x + 2)^(3/2)*x
+ 1227/832*(3*x^2 - x + 2)^(3/2) + 19/52*(3*x^2 - x + 2)^(5/2)/(2*x + 1) - 1745/256*sqrt(3*x^2 - x + 2)*x - 11
8423/9216*sqrt(3)*arcsinh(6/23*sqrt(23)*x - 1/23*sqrt(23)) + 1631/256*sqrt(13)*arcsinh(8/23*sqrt(23)*x/abs(2*x
+ 1) - 9/23*sqrt(23)/abs(2*x + 1)) + 21317/1536*sqrt(3*x^2 - x + 2)

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Fricas [A]  time = 1.49135, size = 487, normalized size = 3.02 \begin{align*} \frac{592115 \, \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 ) + 293580 \, \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 ) + 12 \,{\left (27648 \, x^{6} - 22464 \, x^{5} + 83616 \, x^{4} - 76200 \, x^{3} + 256564 \, x^{2} + 464446 \, x + 142057\right )} \sqrt{3 \, x^{2} - x + 2}}{92160 \,{\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)^(5/2)*(4*x^2+3*x+1)/(1+2*x)^3,x, algorithm="fricas")

[Out]

1/92160*(592115*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) +
293580*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)) + 12*(27648*x^6 - 22464*x^5 + 83616*x^4 - 76200*x^3 + 256564*x^2 + 464446*x + 142057)*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{5}{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)**(5/2)*(4*x**2+3*x+1)/(1+2*x)**3,x)

[Out]

Integral((3*x**2 - x + 2)**(5/2)*(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)^(5/2)*(4*x^2+3*x+1)/(1+2*x)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError