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

Optimal. Leaf size=158 \[ -\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{5971968 (2 x+5)}+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{82944 (2 x+5)^2}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}-\frac {(44378877-7400779 x) \sqrt {2 x^2-x+3}}{5971968}+\frac {170114729 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{3981312 \sqrt {2}}-\frac {10939 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{256 \sqrt {2}} \]

[Out]

-3667/1728*(2*x^2-x+3)^(3/2)/(5+2*x)^3+158527/82944*(2*x^2-x+3)^(3/2)/(5+2*x)^2-6467659/5971968*(2*x^2-x+3)^(3
/2)/(5+2*x)-10939/512*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)[phone]/7962624*arctanh(1/24*(17-22*x)*2^(1/2)/
(2*x^2-x+3)^(1/2))*2^(1/2)-1/5971968*(44378877-7400779*x)*(2*x^2-x+3)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 158, normalized size of antiderivative = 1.00, 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 = {1650, 814, 843, 619, 215, 724, 206} \[ -\frac {6467659 \left (2 x^2-x+3\right )^{3/2}}{5971968 (2 x+5)}+\frac {158527 \left (2 x^2-x+3\right )^{3/2}}{82944 (2 x+5)^2}-\frac {3667 \left (2 x^2-x+3\right )^{3/2}}{1728 (2 x+5)^3}-\frac {(44378877-7400779 x) \sqrt {2 x^2-x+3}}{5971968}+\frac {170114729 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{3981312 \sqrt {2}}-\frac {10939 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{256 \sqrt {2}} \]

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)^4,x]

[Out]

-((44378877 - 7400779*x)*Sqrt[3 - x + 2*x^2])/5971968 - (3667*(3 - x + 2*x^2)^(3/2))/(1728*(5 + 2*x)^3) + (158
527*(3 - x + 2*x^2)^(3/2))/(82944*(5 + 2*x)^2) - (6467659*(3 - x + 2*x^2)^(3/2))/(5971968*(5 + 2*x)) - (10939*
ArcSinh[(1 - 4*x)/Sqrt[23]])/(256*Sqrt[2]) + (170114729*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[3 - x + 2*x^2])])
/(3981312*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 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]

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)^4} \, dx &=-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}-\frac {1}{216} \int \frac {\sqrt {3-x+2 x^2} \left (\frac {36021}{16}-3969 x+1458 x^2-540 x^3\right )}{(5+2 x)^3} \, dx\\ &=-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}+\frac {\int \frac {\sqrt {3-x+2 x^2} \left (\frac {2672127}{16}-\frac {1284285 x}{4}+38880 x^2\right )}{(5+2 x)^2} \, dx}{31104}\\ &=-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}-\frac {\int \frac {\left (\frac {66297447}{16}-\frac {22202337 x}{2}\right ) \sqrt {3-x+2 x^2}}{5+2 x} \, dx}{2239488}\\ &=-\frac {(44378877-7400779 x) \sqrt {3-x+2 x^2}}{5971968}-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}+\frac {\int \frac {-3061291212+6124439808 x}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{71663616}\\ &=-\frac {(44378877-7400779 x) \sqrt {3-x+2 x^2}}{5971968}-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}+\frac {10939}{256} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx-\frac {170114729 \int \frac {1}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{663552}\\ &=-\frac {(44378877-7400779 x) \sqrt {3-x+2 x^2}}{5971968}-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}+\frac {170114729 \operatorname {Subst}\left (\int \frac {1}{288-x^2} \, dx,x,\frac {17-22 x}{\sqrt {3-x+2 x^2}}\right )}{331776}+\frac {10939 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{256 \sqrt {46}}\\ &=-\frac {(44378877-7400779 x) \sqrt {3-x+2 x^2}}{5971968}-\frac {3667 \left (3-x+2 x^2\right )^{3/2}}{1728 (5+2 x)^3}+\frac {158527 \left (3-x+2 x^2\right )^{3/2}}{82944 (5+2 x)^2}-\frac {6467659 \left (3-x+2 x^2\right )^{3/2}}{5971968 (5+2 x)}-\frac {10939 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{256 \sqrt {2}}+\frac {170114729 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{3981312 \sqrt {2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.16, size = 98, normalized size = 0.62 \[ \frac {170114729 \sqrt {2} \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {4 x^2-2 x+6}}\right )+\frac {24 \sqrt {2 x^2-x+3} \left (414720 x^4-5453568 x^3-97682900 x^2-329667508 x-327735797\right )}{(2 x+5)^3}-170123328 \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{7962624} \]

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)^4,x]

[Out]

((24*Sqrt[3 - x + 2*x^2]*(-327735797 - 329667508*x - 97682900*x^2 - 5453568*x^3 + 414720*x^4))/(5 + 2*x)^3 - 1
70123328*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt[23]] + 170114729*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2]
)])/7962624

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 173, normalized size = 1.09 \[ \frac {170123328 \, \sqrt {2} {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (-4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 170114729 \, \sqrt {2} {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )} \log \left (\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} - 1060 \, x^{2} + 1036 \, x - 1153}{4 \, x^{2} + 20 \, x + 25}\right ) + 48 \, {\left (414720 \, x^{4} - 5453568 \, x^{3} - 97682900 \, x^{2} - 329667508 \, x - 327735797\right )} \sqrt {2 \, x^{2} - x + 3}}{15925248 \, {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} \]

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)^4,x, algorithm="fricas")

[Out]

1/15925248*(170123328*sqrt(2)*(8*x^3 + 60*x^2 + 150*x + 125)*log(-4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32
*x^2 + 16*x - 25) + 170114729*sqrt(2)*(8*x^3 + 60*x^2 + 150*x + 125)*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)) + 48*(414720*x^4 - 5453568*x^3 - 97682900*x^2 - 329667
508*x - 327735797)*sqrt(2*x^2 - x + 3))/(8*x^3 + 60*x^2 + 150*x + 125)

________________________________________________________________________________________

giac [B]  time = 0.26, size = 304, normalized size = 1.92 \[ \frac {1}{128} \, \sqrt {2 \, x^{2} - x + 3} {\left (20 \, x - 413\right )} - \frac {10939}{512} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) + \frac {170114729}{7962624} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) - \frac {170114729}{7962624} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) - \frac {\sqrt {2} {\left (575810908 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{5} + 9206213116 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{4} + 9688786604 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{3} - 73157325092 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 49481952947 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 20269228621\right )}}{663552 \, {\left (2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 10 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 11\right )}^{3}} \]

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)^4,x, algorithm="giac")

[Out]

1/128*sqrt(2*x^2 - x + 3)*(20*x - 413) - 10939/512*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) +
1) + 170114729/7962624*sqrt(2)*log(abs(-2*sqrt(2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) - 170114729/7962624*sq
rt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2*x^2 - x + 3))) - 1/663552*sqrt(2)*(575810908*sqrt(2)*(sqrt(
2)*x - sqrt(2*x^2 - x + 3))^5 + 9206213116*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^4 + 9688786604*sqrt(2)*(sqrt(2)*x
- sqrt(2*x^2 - x + 3))^3 - 73157325092*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 49481952947*sqrt(2)*(sqrt(2)*x -
sqrt(2*x^2 - x + 3)) - 20269228621)/(2*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10*sqrt(2)*(sqrt(2)*x - sqrt(2*x
^2 - x + 3)) - 11)^3

________________________________________________________________________________________

maple [A]  time = 0.01, size = 165, normalized size = 1.04 \[ \frac {10939 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{512}+\frac {170114729 \sqrt {2}\, \arctanh \left (\frac {\left (-11 x +\frac {17}{2}\right ) \sqrt {2}}{12 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}\right )}{7962624}+\frac {5 \left (4 x -1\right ) \sqrt {2 x^{2}-x +3}}{128}-\frac {6467659 \left (-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}\right )^{\frac {3}{2}}}{11943936 \left (x +\frac {5}{2}\right )}-\frac {170114729 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{23887872}+\frac {6467659 \left (4 x -1\right ) \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{23887872}+\frac {158527 \left (-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}\right )^{\frac {3}{2}}}{331776 \left (x +\frac {5}{2}\right )^{2}}-\frac {3667 \left (-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}\right )^{\frac {3}{2}}}{13824 \left (x +\frac {5}{2}\right )^{3}} \]

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)^4,x)

[Out]

5/128*(4*x-1)*(2*x^2-x+3)^(1/2)+10939/512*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))-6467659/11943936/(x+5/2)*(-11
*x+2*(x+5/2)^2-19/2)^(3/2)-170114729/23887872*(-11*x+2*(x+5/2)^2-19/2)^(1/2)[phone]/7962624*2^(1/2)*arctanh
(1/12*(-11*x+17/2)*2^(1/2)/(-11*x+2*(x+5/2)^2-19/2)^(1/2))+6467659/23887872*(4*x-1)*(-11*x+2*(x+5/2)^2-19/2)^(
1/2)+158527/331776/(x+5/2)^2*(-11*x+2*(x+5/2)^2-19/2)^(3/2)-3667/13824/(x+5/2)^3*(-11*x+2*(x+5/2)^2-19/2)^(3/2
)

________________________________________________________________________________________

maxima [A]  time = 0.98, size = 160, normalized size = 1.01 \[ \frac {5}{32} \, \sqrt {2 \, x^{2} - x + 3} x + \frac {10939}{512} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) - \frac {170114729}{7962624} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) - \frac {693775}{165888} \, \sqrt {2 \, x^{2} - x + 3} - \frac {3667 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{1728 \, {\left (8 \, x^{3} + 60 \, x^{2} + 150 \, x + 125\right )}} + \frac {158527 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}}{82944 \, {\left (4 \, x^{2} + 20 \, x + 25\right )}} - \frac {6467659 \, \sqrt {2 \, x^{2} - x + 3}}{331776 \, {\left (2 \, x + 5\right )}} \]

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)^4,x, algorithm="maxima")

[Out]

5/32*sqrt(2*x^2 - x + 3)*x + 10939/512*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) - 170114729/7962624*sq
rt(2)*arcsinh(22/23*sqrt(23)*x/abs(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) - 693775/165888*sqrt(2*x^2 - x + 3)
- 3667/1728*(2*x^2 - x + 3)^(3/2)/(8*x^3 + 60*x^2 + 150*x + 125) + 158527/82944*(2*x^2 - x + 3)^(3/2)/(4*x^2
+ 20*x + 25) - 6467659/331776*sqrt(2*x^2 - x + 3)/(2*x + 5)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)**4,x)

[Out]

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

________________________________________________________________________________________