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

Optimal. Leaf size=141 \[ \frac{1}{12} \left (3 x^2-x+2\right )^{5/2} (2 x+1)^3+\frac{8}{63} \left (3 x^2-x+2\right )^{5/2} (2 x+1)^2+\frac{13 (50 x+29) \left (3 x^2-x+2\right )^{5/2}}{2520}+\frac{91 (1-6 x) \left (3 x^2-x+2\right )^{3/2}}{3456}+\frac{2093 (1-6 x) \sqrt{3 x^2-x+2}}{27648}+\frac{48139 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{55296 \sqrt{3}} \]

[Out]

(2093*(1 - 6*x)*Sqrt[2 - x + 3*x^2])/27648 + (91*(1 - 6*x)*(2 - x + 3*x^2)^(3/2))/3456 + (8*(1 + 2*x)^2*(2 - x
+ 3*x^2)^(5/2))/63 + ((1 + 2*x)^3*(2 - x + 3*x^2)^(5/2))/12 + (13*(29 + 50*x)*(2 - x + 3*x^2)^(5/2))/2520 + (
48139*ArcSinh[(1 - 6*x)/Sqrt[23]])/(55296*Sqrt[3])

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Rubi [A]  time = 0.12095, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1653, 832, 779, 612, 619, 215} \[ \frac{1}{12} \left (3 x^2-x+2\right )^{5/2} (2 x+1)^3+\frac{8}{63} \left (3 x^2-x+2\right )^{5/2} (2 x+1)^2+\frac{13 (50 x+29) \left (3 x^2-x+2\right )^{5/2}}{2520}+\frac{91 (1-6 x) \left (3 x^2-x+2\right )^{3/2}}{3456}+\frac{2093 (1-6 x) \sqrt{3 x^2-x+2}}{27648}+\frac{48139 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{55296 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2093*(1 - 6*x)*Sqrt[2 - x + 3*x^2])/27648 + (91*(1 - 6*x)*(2 - x + 3*x^2)^(3/2))/3456 + (8*(1 + 2*x)^2*(2 - x
+ 3*x^2)^(5/2))/63 + ((1 + 2*x)^3*(2 - x + 3*x^2)^(5/2))/12 + (13*(29 + 50*x)*(2 - x + 3*x^2)^(5/2))/2520 + (
48139*ArcSinh[(1 - 6*x)/Sqrt[23]])/(55296*Sqrt[3])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 832

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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]

Rubi steps

\begin{align*} \int (1+2 x)^2 \left (2-x+3 x^2\right )^{3/2} \left (1+3 x+4 x^2\right ) \, dx &=\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{96} \int (1+2 x)^2 (20+256 x) \left (2-x+3 x^2\right )^{3/2} \, dx\\ &=\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{\int (1+2 x) (-988+4680 x) \left (2-x+3 x^2\right )^{3/2} \, dx}{2016}\\ &=\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{13 (29+50 x) \left (2-x+3 x^2\right )^{5/2}}{2520}-\frac{91}{144} \int \left (2-x+3 x^2\right )^{3/2} \, dx\\ &=\frac{91 (1-6 x) \left (2-x+3 x^2\right )^{3/2}}{3456}+\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{13 (29+50 x) \left (2-x+3 x^2\right )^{5/2}}{2520}-\frac{2093 \int \sqrt{2-x+3 x^2} \, dx}{2304}\\ &=\frac{2093 (1-6 x) \sqrt{2-x+3 x^2}}{27648}+\frac{91 (1-6 x) \left (2-x+3 x^2\right )^{3/2}}{3456}+\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{13 (29+50 x) \left (2-x+3 x^2\right )^{5/2}}{2520}-\frac{48139 \int \frac{1}{\sqrt{2-x+3 x^2}} \, dx}{55296}\\ &=\frac{2093 (1-6 x) \sqrt{2-x+3 x^2}}{27648}+\frac{91 (1-6 x) \left (2-x+3 x^2\right )^{3/2}}{3456}+\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{13 (29+50 x) \left (2-x+3 x^2\right )^{5/2}}{2520}-\frac{\left (2093 \sqrt{\frac{23}{3}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+6 x\right )}{55296}\\ &=\frac{2093 (1-6 x) \sqrt{2-x+3 x^2}}{27648}+\frac{91 (1-6 x) \left (2-x+3 x^2\right )^{3/2}}{3456}+\frac{8}{63} (1+2 x)^2 \left (2-x+3 x^2\right )^{5/2}+\frac{1}{12} (1+2 x)^3 \left (2-x+3 x^2\right )^{5/2}+\frac{13 (29+50 x) \left (2-x+3 x^2\right )^{5/2}}{2520}+\frac{48139 \sinh ^{-1}\left (\frac{1-6 x}{\sqrt{23}}\right )}{55296 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.0433892, size = 75, normalized size = 0.53 \[ \frac{6 \sqrt{3 x^2-x+2} \left (5806080 x^7+9262080 x^6+10656000 x^5+12173952 x^4+10119792 x^3+5694024 x^2+2735918 x+1517367\right )-1684865 \sqrt{3} \sinh ^{-1}\left (\frac{6 x-1}{\sqrt{23}}\right )}{5806080} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[2 - x + 3*x^2]*(1517367 + 2735918*x + 5694024*x^2 + 10119792*x^3 + 12173952*x^4 + 10656000*x^5 + 92620
80*x^6 + 5806080*x^7) - 1684865*Sqrt[3]*ArcSinh[(-1 + 6*x)/Sqrt[23]])/5806080

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Maple [A]  time = 0.055, size = 117, normalized size = 0.8 \begin{align*}{\frac{2\,{x}^{3}}{3} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{5}{2}}}}+{\frac{95\,{x}^{2}}{63} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{5}{2}}}}+{\frac{319\,x}{252} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{5}{2}}}}-{\frac{-2093+12558\,x}{27648}\sqrt{3\,{x}^{2}-x+2}}-{\frac{48139\,\sqrt{3}}{165888}{\it Arcsinh} \left ({\frac{6\,\sqrt{23}}{23} \left ( x-{\frac{1}{6}} \right ) } \right ) }-{\frac{-91+546\,x}{3456} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{3}{2}}}}+{\frac{907}{2520} \left ( 3\,{x}^{2}-x+2 \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*x^3*(3*x^2-x+2)^(5/2)+95/63*x^2*(3*x^2-x+2)^(5/2)+319/252*x*(3*x^2-x+2)^(5/2)-2093/27648*(-1+6*x)*(3*x^2-x
+2)^(1/2)-48139/165888*3^(1/2)*arcsinh(6/23*23^(1/2)*(x-1/6))-91/3456*(-1+6*x)*(3*x^2-x+2)^(3/2)+907/2520*(3*x
^2-x+2)^(5/2)

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Maxima [A]  time = 1.50437, size = 186, normalized size = 1.32 \begin{align*} \frac{2}{3} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}} x^{3} + \frac{95}{63} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}} x^{2} + \frac{319}{252} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}} x + \frac{907}{2520} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{5}{2}} - \frac{91}{576} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} x + \frac{91}{3456} \,{\left (3 \, x^{2} - x + 2\right )}^{\frac{3}{2}} - \frac{2093}{4608} \, \sqrt{3 \, x^{2} - x + 2} x - \frac{48139}{165888} \, \sqrt{3} \operatorname{arsinh}\left (\frac{1}{23} \, \sqrt{23}{\left (6 \, x - 1\right )}\right ) + \frac{2093}{27648} \, \sqrt{3 \, x^{2} - x + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*x^2 - x + 2)^(5/2)*x^3 + 95/63*(3*x^2 - x + 2)^(5/2)*x^2 + 319/252*(3*x^2 - x + 2)^(5/2)*x + 907/2520*(
3*x^2 - x + 2)^(5/2) - 91/576*(3*x^2 - x + 2)^(3/2)*x + 91/3456*(3*x^2 - x + 2)^(3/2) - 2093/4608*sqrt(3*x^2 -
x + 2)*x - 48139/165888*sqrt(3)*arcsinh(1/23*sqrt(23)*(6*x - 1)) + 2093/27648*sqrt(3*x^2 - x + 2)

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Fricas [A]  time = 1.62997, size = 308, normalized size = 2.18 \begin{align*} \frac{1}{967680} \,{\left (5806080 \, x^{7} + 9262080 \, x^{6} + 10656000 \, x^{5} + 12173952 \, x^{4} + 10119792 \, x^{3} + 5694024 \, x^{2} + 2735918 \, x + 1517367\right )} \sqrt{3 \, x^{2} - x + 2} + \frac{48139}{331776} \, \sqrt{3} \log \left (4 \, \sqrt{3} \sqrt{3 \, x^{2} - x + 2}{\left (6 \, x - 1\right )} - 72 \, x^{2} + 24 \, x - 25\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/967680*(5806080*x^7 + 9262080*x^6 + 10656000*x^5 + 12173952*x^4 + 10119792*x^3 + 5694024*x^2 + 2735918*x + 1
517367)*sqrt(3*x^2 - x + 2) + 48139/331776*sqrt(3)*log(4*sqrt(3)*sqrt(3*x^2 - x + 2)*(6*x - 1) - 72*x^2 + 24*x
- 25)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (2 x + 1\right )^{2} \left (3 x^{2} - x + 2\right )^{\frac{3}{2}} \left (4 x^{2} + 3 x + 1\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15427, size = 112, normalized size = 0.79 \begin{align*} \frac{1}{967680} \,{\left (2 \,{\left (12 \,{\left (2 \,{\left (8 \,{\left (30 \,{\left (12 \,{\left (42 \, x + 67\right )} x + 925\right )} x + 31703\right )} x + 210829\right )} x + 237251\right )} x + 1367959\right )} x + 1517367\right )} \sqrt{3 \, x^{2} - x + 2} + \frac{48139}{165888} \, \sqrt{3} \log \left (-2 \, \sqrt{3}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} - x + 2}\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/967680*(2*(12*(2*(8*(30*(12*(42*x + 67)*x + 925)*x + 31703)*x + 210829)*x + 237251)*x + 1367959)*x + 1517367
)*sqrt(3*x^2 - x + 2) + 48139/165888*sqrt(3)*log(-2*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 - x + 2)) + 1)