### 3.326 $$\int \frac{\sqrt{3-x+2 x^2} (2+x+3 x^2-x^3+5 x^4)}{5+2 x} \, dx$$

Optimal. Leaf size=149 $\frac{1}{16} \left (2 x^2-x+3\right )^{3/2} (2 x+5)^2-\frac{127}{128} \left (2 x^2-x+3\right )^{3/2} (2 x+5)+\frac{4535}{768} \left (2 x^2-x+3\right )^{3/2}+\frac{(489587-80844 x) \sqrt{2 x^2-x+3}}{4096}-\frac{11001 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{16 \sqrt{2}}+\frac{5627989 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8192 \sqrt{2}}$

[Out]

((489587 - 80844*x)*Sqrt[3 - x + 2*x^2])/4096 + (4535*(3 - x + 2*x^2)^(3/2))/768 - (127*(5 + 2*x)*(3 - x + 2*x
^2)^(3/2))/128 + ((5 + 2*x)^2*(3 - x + 2*x^2)^(3/2))/16 + (5627989*ArcSinh[(1 - 4*x)/Sqrt[23]])/(8192*Sqrt[2])
- (11001*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(16*Sqrt[2])

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Rubi [A]  time = 0.240436, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 40, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.175, Rules used = {1653, 814, 843, 619, 215, 724, 206} $\frac{1}{16} \left (2 x^2-x+3\right )^{3/2} (2 x+5)^2-\frac{127}{128} \left (2 x^2-x+3\right )^{3/2} (2 x+5)+\frac{4535}{768} \left (2 x^2-x+3\right )^{3/2}+\frac{(489587-80844 x) \sqrt{2 x^2-x+3}}{4096}-\frac{11001 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{2 x^2-x+3}}\right )}{16 \sqrt{2}}+\frac{5627989 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8192 \sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((489587 - 80844*x)*Sqrt[3 - x + 2*x^2])/4096 + (4535*(3 - x + 2*x^2)^(3/2))/768 - (127*(5 + 2*x)*(3 - x + 2*x
^2)^(3/2))/128 + ((5 + 2*x)^2*(3 - x + 2*x^2)^(3/2))/16 + (5627989*ArcSinh[(1 - 4*x)/Sqrt[23]])/(8192*Sqrt[2])
- (11001*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])/(16*Sqrt[2])

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 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{\sqrt{3-x+2 x^2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx &=\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}+\frac{1}{160} \int \frac{\sqrt{3-x+2 x^2} \left (-805-6490 x-9300 x^2-5080 x^3\right )}{5+2 x} \, dx\\ &=-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}+\frac{\int \frac{\sqrt{3-x+2 x^2} \left (-127720+824160 x+725600 x^2\right )}{5+2 x} \, dx}{10240}\\ &=\frac{4535}{768} \left (3-x+2 x^2\right )^{3/2}-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}+\frac{\int \frac{(7818720-19402560 x) \sqrt{3-x+2 x^2}}{5+2 x} \, dx}{245760}\\ &=\frac{(489587-80844 x) \sqrt{3-x+2 x^2}}{4096}+\frac{4535}{768} \left (3-x+2 x^2\right )^{3/2}-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}-\frac{\int \frac{-5428921920+10805738880 x}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx}{7864320}\\ &=\frac{(489587-80844 x) \sqrt{3-x+2 x^2}}{4096}+\frac{4535}{768} \left (3-x+2 x^2\right )^{3/2}-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}-\frac{5627989 \int \frac{1}{\sqrt{3-x+2 x^2}} \, dx}{8192}+\frac{33003}{8} \int \frac{1}{(5+2 x) \sqrt{3-x+2 x^2}} \, dx\\ &=\frac{(489587-80844 x) \sqrt{3-x+2 x^2}}{4096}+\frac{4535}{768} \left (3-x+2 x^2\right )^{3/2}-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}-\frac{33003}{4} \operatorname{Subst}\left (\int \frac{1}{288-x^2} \, dx,x,\frac{17-22 x}{\sqrt{3-x+2 x^2}}\right )-\frac{5627989 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{23}}} \, dx,x,-1+4 x\right )}{8192 \sqrt{46}}\\ &=\frac{(489587-80844 x) \sqrt{3-x+2 x^2}}{4096}+\frac{4535}{768} \left (3-x+2 x^2\right )^{3/2}-\frac{127}{128} (5+2 x) \left (3-x+2 x^2\right )^{3/2}+\frac{1}{16} (5+2 x)^2 \left (3-x+2 x^2\right )^{3/2}+\frac{5627989 \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{8192 \sqrt{2}}-\frac{11001 \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{2} \sqrt{3-x+2 x^2}}\right )}{16 \sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.148461, size = 91, normalized size = 0.61 $\frac{4 \sqrt{2 x^2-x+3} \left (6144 x^4-21120 x^3+79840 x^2-300404 x+1561161\right )-16897536 \sqrt{2} \tanh ^{-1}\left (\frac{17-22 x}{12 \sqrt{4 x^2-2 x+6}}\right )+16883967 \sqrt{2} \sinh ^{-1}\left (\frac{1-4 x}{\sqrt{23}}\right )}{49152}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(1561161 - 300404*x + 79840*x^2 - 21120*x^3 + 6144*x^4) + 16883967*Sqrt[2]*ArcSinh[(1 -
4*x)/Sqrt[23]] - 16897536*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/49152

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Maple [A]  time = 0.056, size = 127, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{4} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}-{\frac{47\,x}{64} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}+{\frac{1925}{768} \left ( 2\,{x}^{2}-x+3 \right ) ^{{\frac{3}{2}}}}-{\frac{-20211+80844\,x}{4096}\sqrt{2\,{x}^{2}-x+3}}-{\frac{5627989\,\sqrt{2}}{16384}{\it Arcsinh} \left ({\frac{4\,\sqrt{23}}{23} \left ( x-{\frac{1}{4}} \right ) } \right ) }+{\frac{3667}{32}\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}-{\frac{11001\,\sqrt{2}}{32}{\it Artanh} \left ({\frac{\sqrt{2}}{12} \left ({\frac{17}{2}}-11\,x \right ){\frac{1}{\sqrt{2\, \left ( x+5/2 \right ) ^{2}-11\,x-{\frac{19}{2}}}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^4-x^3+3*x^2+x+2)*(2*x^2-x+3)^(1/2)/(5+2*x),x)

[Out]

1/4*x^2*(2*x^2-x+3)^(3/2)-47/64*x*(2*x^2-x+3)^(3/2)+1925/768*(2*x^2-x+3)^(3/2)-20211/4096*(-1+4*x)*(2*x^2-x+3)
^(1/2)-5627989/16384*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+3667/32*(2*(x+5/2)^2-11*x-19/2)^(1/2)-11001/32*2^(
1/2)*arctanh(1/12*(17/2-11*x)*2^(1/2)/(2*(x+5/2)^2-11*x-19/2)^(1/2))

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Maxima [A]  time = 1.59072, size = 173, normalized size = 1.16 \begin{align*} \frac{1}{4} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}} x^{2} - \frac{47}{64} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}} x + \frac{1925}{768} \,{\left (2 \, x^{2} - x + 3\right )}^{\frac{3}{2}} - \frac{20211}{1024} \, \sqrt{2 \, x^{2} - x + 3} x - \frac{5627989}{16384} \, \sqrt{2} \operatorname{arsinh}\left (\frac{4}{23} \, \sqrt{23} x - \frac{1}{23} \, \sqrt{23}\right ) + \frac{11001}{32} \, \sqrt{2} \operatorname{arsinh}\left (\frac{22 \, \sqrt{23} x}{23 \,{\left | 2 \, x + 5 \right |}} - \frac{17 \, \sqrt{23}}{23 \,{\left | 2 \, x + 5 \right |}}\right ) + \frac{489587}{4096} \, \sqrt{2 \, x^{2} - x + 3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*x^2 - x + 3)^(3/2)*x^2 - 47/64*(2*x^2 - x + 3)^(3/2)*x + 1925/768*(2*x^2 - x + 3)^(3/2) - 20211/1024*sq
rt(2*x^2 - x + 3)*x - 5627989/16384*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 11001/32*sqrt(2)*arcsin
h(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 489587/4096*sqrt(2*x^2 - x + 3)

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Fricas [A]  time = 1.38627, size = 394, normalized size = 2.64 \begin{align*} \frac{1}{12288} \,{\left (6144 \, x^{4} - 21120 \, x^{3} + 79840 \, x^{2} - 300404 \, x + 1561161\right )} \sqrt{2 \, x^{2} - x + 3} + \frac{5627989}{32768} \, \sqrt{2} \log \left (4 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + \frac{11001}{64} \, \sqrt{2} \log \left (-\frac{24 \, \sqrt{2} \sqrt{2 \, x^{2} - x + 3}{\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12288*(6144*x^4 - 21120*x^3 + 79840*x^2 - 300404*x + 1561161)*sqrt(2*x^2 - x + 3) + 5627989/32768*sqrt(2)*lo
g(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 25) + 11001/64*sqrt(2)*log(-(24*sqrt(2)*sqrt(2*x^2
- x + 3)*(22*x - 17) + 1060*x^2 - 1036*x + 1153)/(4*x^2 + 20*x + 25))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{2 x^{2} - x + 3} \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{2 x + 5}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**4-x**3+3*x**2+x+2)*(2*x**2-x+3)**(1/2)/(5+2*x),x)

[Out]

Integral(sqrt(2*x**2 - x + 3)*(5*x**4 - x**3 + 3*x**2 + x + 2)/(2*x + 5), x)

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Giac [A]  time = 1.16418, size = 174, normalized size = 1.17 \begin{align*} \frac{1}{12288} \,{\left (4 \,{\left (8 \,{\left (12 \,{\left (16 \, x - 55\right )} x + 2495\right )} x - 75101\right )} x + 1561161\right )} \sqrt{2 \, x^{2} - x + 3} + \frac{5627989}{16384} \, \sqrt{2} \log \left (-4 \, \sqrt{2} x + \sqrt{2} + 4 \, \sqrt{2 \, x^{2} - x + 3}\right ) - \frac{11001}{32} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x + \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) + \frac{11001}{32} \, \sqrt{2} \log \left ({\left | -2 \, \sqrt{2} x - 11 \, \sqrt{2} + 2 \, \sqrt{2 \, x^{2} - x + 3} \right |}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12288*(4*(8*(12*(16*x - 55)*x + 2495)*x - 75101)*x + 1561161)*sqrt(2*x^2 - x + 3) + 5627989/16384*sqrt(2)*lo
g(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x^2 - x + 3)) - 11001/32*sqrt(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2
*x^2 - x + 3))) + 11001/32*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3)))