3.336 \(\int \frac {(3-x+2 x^2)^{3/2} (2+x+3 x^2-x^3+5 x^4)}{5+2 x} \, dx\)

Optimal. Leaf size=172 \[ \frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}-\frac {311}{448} (2 x+5) \left (2 x^2-x+3\right )^{5/2}+\frac {3505}{896} \left (2 x^2-x+3\right )^{5/2}+\frac {(500141-123060 x) \left (2 x^2-x+3\right )^{3/2}}{12288}+\frac {(141051019-23482924 x) \sqrt {2 x^2-x+3}}{65536}-\frac {99009 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{8 \sqrt {2}}+\frac {1622009981 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{131072 \sqrt {2}} \]

[Out]

1/12288*(500141-123060*x)*(2*x^2-x+3)^(3/2)+3505/896*(2*x^2-x+3)^(5/2)-311/448*(5+2*x)*(2*x^2-x+3)^(5/2)+5/112
*(5+2*x)^2*(2*x^2-x+3)^(5/2)[phone]/262144*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-99009/16*arctanh(1/24*(1
7-22*x)*2^(1/2)/(2*x^2-x+3)^(1/2))*2^(1/2)+1/65536*(141051019-23482924*x)*(2*x^2-x+3)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {5}{112} (2 x+5)^2 \left (2 x^2-x+3\right )^{5/2}-\frac {311}{448} (2 x+5) \left (2 x^2-x+3\right )^{5/2}+\frac {3505}{896} \left (2 x^2-x+3\right )^{5/2}+\frac {(500141-123060 x) \left (2 x^2-x+3\right )^{3/2}}{12288}+\frac {(141051019-23482924 x) \sqrt {2 x^2-x+3}}{65536}-\frac {99009 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{8 \sqrt {2}}+\frac {1622009981 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{131072 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((141051019 - 23482924*x)*Sqrt[3 - x + 2*x^2])/65536 + ((500141 - 123060*x)*(3 - x + 2*x^2)^(3/2))/12288 + (35
05*(3 - x + 2*x^2)^(5/2))/896 - (311*(5 + 2*x)*(3 - x + 2*x^2)^(5/2))/448 + (5*(5 + 2*x)^2*(3 - x + 2*x^2)^(5/
2))/112 + (1622009981*ArcSinh[(1 - 4*x)/Sqrt[23]])/(131072*Sqrt[2]) - (99009*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*S
qrt[3 - x + 2*x^2])])/(8*Sqrt[2])

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])

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 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 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 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 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]))

Rubi steps

\begin {align*} \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (2+x+3 x^2-x^3+5 x^4\right )}{5+2 x} \, dx &=\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {1}{224} \int \frac {\left (3-x+2 x^2\right )^{3/2} \left (573-9926 x-14508 x^2-7464 x^3\right )}{5+2 x} \, dx\\ &=-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {\int \frac {\left (3-x+2 x^2\right )^{3/2} \left (-430152+2062560 x+1682400 x^2\right )}{5+2 x} \, dx}{21504}\\ &=\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {\int \frac {(24853920-68913600 x) \left (3-x+2 x^2\right )^{3/2}}{5+2 x} \, dx}{860160}\\ &=\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}-\frac {\int \frac {(-29846322240+78902624640 x) \sqrt {3-x+2 x^2}}{5+2 x} \, dx}{55050240}\\ &=\frac {(141051019-23482924 x) \sqrt {3-x+2 x^2}}{65536}+\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {\int \frac {21812190368640-43599628289280 x}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{1761607680}\\ &=\frac {(141051019-23482924 x) \sqrt {3-x+2 x^2}}{65536}+\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}-\frac {1622009981 \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx}{131072}+\frac {297027}{4} \int \frac {1}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx\\ &=\frac {(141051019-23482924 x) \sqrt {3-x+2 x^2}}{65536}+\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}-\frac {297027}{2} \operatorname {Subst}\left (\int \frac {1}{288-x^2} \, dx,x,\frac {17-22 x}{\sqrt {3-x+2 x^2}}\right )-\frac {1622009981 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{131072 \sqrt {46}}\\ &=\frac {(141051019-23482924 x) \sqrt {3-x+2 x^2}}{65536}+\frac {(500141-123060 x) \left (3-x+2 x^2\right )^{3/2}}{12288}+\frac {3505}{896} \left (3-x+2 x^2\right )^{5/2}-\frac {311}{448} (5+2 x) \left (3-x+2 x^2\right )^{5/2}+\frac {5}{112} (5+2 x)^2 \left (3-x+2 x^2\right )^{5/2}+\frac {1622009981 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{131072 \sqrt {2}}-\frac {99009 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{8 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 101, normalized size = 0.59 \[ \frac {-34065432576 \sqrt {2} \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {4 x^2-2 x+6}}\right )+4 \sqrt {2 x^2-x+3} \left (983040 x^6-3710976 x^5+14493696 x^4-46476672 x^3+159973408 x^2-609499532 x+3149403255\right )[phone] \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{5505024} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[3 - x + 2*x^2]*(3149403255 - 609499532*x + 159973408*x^2 - 46476672*x^3 + 14493696*x^4 - 3710976*x^5 +
983040*x^6) + 34062209601*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]] - 34065432576*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*S
qrt[6 - 2*x + 4*x^2])])/5505024

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fricas [A]  time = 0.62, size = 135, normalized size = 0.78 \[ \frac {1}{1376256} \, {\left (983040 \, x^{6} - 3710976 \, x^{5} + 14493696 \, x^{4} - 46476672 \, x^{3} + 159973408 \, x^{2} - 609499532 \, x + 3149403255\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1622009981}{524288} \, \sqrt {2} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + \frac {99009}{32} \, \sqrt {2} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1376256*(983040*x^6 - 3710976*x^5 + 14493696*x^4 - 46476672*x^3 + 159973408*x^2 - 609499532*x + 3149403255)*
sqrt(2*x^2 - x + 3) + 1622009981/524288*sqrt(2)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x -
25) + 99009/32*sqrt(2)*log(-(24*sqrt(2)*sqrt(2*x^2 - x + 3)*(22*x - 17) + 1060*x^2 - 1036*x + 1153)/(4*x^2 + 2
0*x + 25))

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giac [A]  time = 0.23, size = 139, normalized size = 0.81 \[ \frac {1}{1376256} \, {\left (4 \, {\left (8 \, {\left (12 \, {\left (16 \, {\left (4 \, {\left (40 \, x - 151\right )} x + 2359\right )} x - 121033\right )} x + 4999169\right )} x - 152374883\right )} x + 3149403255\right )} \sqrt {2 \, x^{2} - x + 3} + \frac {1622009981}{262144} \, \sqrt {2} \log \left (-4 \, \sqrt {2} x + \sqrt {2} + 4 \, \sqrt {2 \, x^{2} - x + 3}\right ) - \frac {99009}{16} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {99009}{16} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1376256*(4*(8*(12*(16*(4*(40*x - 151)*x + 2359)*x - 121033)*x + 4999169)*x - 152374883)*x + 3149403255)*sqrt
(2*x^2 - x + 3) + 1622009981/262144*sqrt(2)*log(-4*sqrt(2)*x + sqrt(2) + 4*sqrt(2*x^2 - x + 3)) - 99009/16*sqr
t(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 99009/16*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqr
t(2) + 2*sqrt(2*x^2 - x + 3)))

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maple [A]  time = 0.01, size = 183, normalized size = 1.06 \[ \frac {5 \left (2 x^{2}-x +3\right )^{\frac {5}{2}} x^{2}}{28}-\frac {111 \left (2 x^{2}-x +3\right )^{\frac {5}{2}} x}{224}-\frac {1622009981 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{262144}-\frac {99009 \sqrt {2}\, \arctanh \left (\frac {\left (-11 x +\frac {17}{2}\right ) \sqrt {2}}{12 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}\right )}{16}+\frac {1395 \left (2 x^{2}-x +3\right )^{\frac {5}{2}}}{896}-\frac {10255 \left (4 x -1\right ) \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}{4096}-\frac {707595 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{65536}+\frac {3667 \left (-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}\right )^{\frac {3}{2}}}{96}-\frac {40337 \left (4 x -1\right ) \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{512}+\frac {33003 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/28*(2*x^2-x+3)^(5/2)*x^2-111/224*(2*x^2-x+3)^(5/2)*x+1395/896*(2*x^2-x+3)^(5/2)-10255/4096*(4*x-1)*(2*x^2-x+
3)^(3/2)-707595/65536*(4*x-1)*(2*x^2-x+3)^(1/2)-1622009981/262144*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+3667/
96*(-11*x+2*(x+5/2)^2-19/2)^(3/2)-40337/512*(4*x-1)*(-11*x+2*(x+5/2)^2-19/2)^(1/2)+33003/16*(-11*x+2*(x+5/2)^2
-19/2)^(1/2)-99009/16*2^(1/2)*arctanh(1/12*(-11*x+17/2)*2^(1/2)/(-11*x+2*(x+5/2)^2-19/2)^(1/2))

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maxima [A]  time = 1.00, size = 157, normalized size = 0.91 \[ \frac {5}{28} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} x^{2} - \frac {111}{224} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} x + \frac {1395}{896} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {5}{2}} - \frac {10255}{1024} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} x + \frac {500141}{12288} \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {5870731}{16384} \, \sqrt {2 \, x^{2} - x + 3} x - \frac {1622009981}{262144} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) + \frac {99009}{16} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) + \frac {141051019}{65536} \, \sqrt {2 \, x^{2} - x + 3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/28*(2*x^2 - x + 3)^(5/2)*x^2 - 111/224*(2*x^2 - x + 3)^(5/2)*x + 1395/896*(2*x^2 - x + 3)^(5/2) - 10255/1024
*(2*x^2 - x + 3)^(3/2)*x + 500141/12288*(2*x^2 - x + 3)^(3/2) - 5870731/16384*sqrt(2*x^2 - x + 3)*x - 16220099
81/262144*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 99009/16*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x
+ 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 141051019/65536*sqrt(2*x^2 - x + 3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (2\,x^2-x+3\right )}^{3/2}\,\left (5\,x^4-x^3+3\,x^2+x+2\right )}{2\,x+5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (2 x^{2} - x + 3\right )^{\frac {3}{2}} \left (5 x^{4} - x^{3} + 3 x^{2} + x + 2\right )}{2 x + 5}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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