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

Optimal. Leaf size=128 \[ \frac {92239 \sqrt {2 x^2-x+3}}{27648 (2 x+5)}-\frac {3667 \sqrt {2 x^2-x+3}}{1152 (2 x+5)^2}+\frac {5}{16} \sqrt {2 x^2-x+3}-\frac {1546507 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{331776 \sqrt {2}}+\frac {149 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]

[Out]

149/64*arcsinh(1/23*(1-4*x)*23^(1/2))*2^(1/2)-1546507/663552*arctanh(1/24*(17-22*x)*2^(1/2)/(2*x^2-x+3)^(1/2))
*2^(1/2)+5/16*(2*x^2-x+3)^(1/2)-3667/1152*(2*x^2-x+3)^(1/2)/(5+2*x)^2+92239/27648*(2*x^2-x+3)^(1/2)/(5+2*x)

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Rubi [A]  time = 0.21, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {1650, 1653, 843, 619, 215, 724, 206} \[ \frac {92239 \sqrt {2 x^2-x+3}}{27648 (2 x+5)}-\frac {3667 \sqrt {2 x^2-x+3}}{1152 (2 x+5)^2}+\frac {5}{16} \sqrt {2 x^2-x+3}-\frac {1546507 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {2 x^2-x+3}}\right )}{331776 \sqrt {2}}+\frac {149 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*Sqrt[3 - x + 2*x^2])/16 - (3667*Sqrt[3 - x + 2*x^2])/(1152*(5 + 2*x)^2) + (92239*Sqrt[3 - x + 2*x^2])/(2764
8*(5 + 2*x)) + (149*ArcSinh[(1 - 4*x)/Sqrt[23]])/(32*Sqrt[2]) - (1546507*ArcTanh[(17 - 22*x)/(12*Sqrt[2]*Sqrt[
3 - x + 2*x^2])])/(331776*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 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]

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 {2+x+3 x^2-x^3+5 x^4}{(5+2 x)^3 \sqrt {3-x+2 x^2}} \, dx &=-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}-\frac {1}{144} \int \frac {\frac {20347}{16}-\frac {6917 x}{4}+972 x^2-360 x^3}{(5+2 x)^2 \sqrt {3-x+2 x^2}} \, dx\\ &=-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}+\frac {92239 \sqrt {3-x+2 x^2}}{27648 (5+2 x)}+\frac {\int \frac {\frac {647841}{16}-67392 x+12960 x^2}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{10368}\\ &=\frac {5}{16} \sqrt {3-x+2 x^2}-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}+\frac {92239 \sqrt {3-x+2 x^2}}{27648 (5+2 x)}+\frac {\int \frac {\frac {777441}{2}-772416 x}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{82944}\\ &=\frac {5}{16} \sqrt {3-x+2 x^2}-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}+\frac {92239 \sqrt {3-x+2 x^2}}{27648 (5+2 x)}-\frac {149}{32} \int \frac {1}{\sqrt {3-x+2 x^2}} \, dx+\frac {1546507 \int \frac {1}{(5+2 x) \sqrt {3-x+2 x^2}} \, dx}{55296}\\ &=\frac {5}{16} \sqrt {3-x+2 x^2}-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}+\frac {92239 \sqrt {3-x+2 x^2}}{27648 (5+2 x)}-\frac {1546507 \operatorname {Subst}\left (\int \frac {1}{288-x^2} \, dx,x,\frac {17-22 x}{\sqrt {3-x+2 x^2}}\right )}{27648}-\frac {149 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{23}}} \, dx,x,-1+4 x\right )}{32 \sqrt {46}}\\ &=\frac {5}{16} \sqrt {3-x+2 x^2}-\frac {3667 \sqrt {3-x+2 x^2}}{1152 (5+2 x)^2}+\frac {92239 \sqrt {3-x+2 x^2}}{27648 (5+2 x)}+\frac {149 \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{32 \sqrt {2}}-\frac {1546507 \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {2} \sqrt {3-x+2 x^2}}\right )}{331776 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 88, normalized size = 0.69 \[ \frac {\frac {24 \sqrt {2 x^2-x+3} \left (34560 x^2+357278 x+589187\right )}{(2 x+5)^2}-1546507 \sqrt {2} \tanh ^{-1}\left (\frac {17-22 x}{12 \sqrt {4 x^2-2 x+6}}\right )+1544832 \sqrt {2} \sinh ^{-1}\left (\frac {1-4 x}{\sqrt {23}}\right )}{663552} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((24*Sqrt[3 - x + 2*x^2]*(589187 + 357278*x + 34560*x^2))/(5 + 2*x)^2 + 1544832*Sqrt[2]*ArcSinh[(1 - 4*x)/Sqrt
[23]] - 1546507*Sqrt[2]*ArcTanh[(17 - 22*x)/(12*Sqrt[6 - 2*x + 4*x^2])])/663552

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fricas [A]  time = 0.96, size = 149, normalized size = 1.16 \[ \frac {1544832 \, \sqrt {2} {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (4 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (4 \, x - 1\right )} - 32 \, x^{2} + 16 \, x - 25\right ) + 1546507 \, \sqrt {2} {\left (4 \, x^{2} + 20 \, x + 25\right )} \log \left (-\frac {24 \, \sqrt {2} \sqrt {2 \, x^{2} - x + 3} {\left (22 \, x - 17\right )} + 1060 \, x^{2} - 1036 \, x + 1153}{4 \, x^{2} + 20 \, x + 25}\right ) + 48 \, {\left (34560 \, x^{2} + 357278 \, x + 589187\right )} \sqrt {2 \, x^{2} - x + 3}}{1327104 \, {\left (4 \, x^{2} + 20 \, x + 25\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1327104*(1544832*sqrt(2)*(4*x^2 + 20*x + 25)*log(4*sqrt(2)*sqrt(2*x^2 - x + 3)*(4*x - 1) - 32*x^2 + 16*x - 2
5) + 1546507*sqrt(2)*(4*x^2 + 20*x + 25)*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*(34560*x^2 + 357278*x + 589187)*sqrt(2*x^2 - x + 3))/(4*x^2 + 20*x + 25)

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giac [B]  time = 0.26, size = 248, normalized size = 1.94 \[ \frac {149}{64} \, \sqrt {2} \log \left (-2 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 1\right ) - \frac {1546507}{663552} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x + \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {1546507}{663552} \, \sqrt {2} \log \left ({\left | -2 \, \sqrt {2} x - 11 \, \sqrt {2} + 2 \, \sqrt {2 \, x^{2} - x + 3} \right |}\right ) + \frac {5}{16} \, \sqrt {2 \, x^{2} - x + 3} + \frac {\sqrt {2} {\left (2381290 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{3} + 16628406 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} - 25697445 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} + 16720645\right )}}{55296 \, {\left (2 \, {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )}^{2} + 10 \, \sqrt {2} {\left (\sqrt {2} x - \sqrt {2 \, x^{2} - x + 3}\right )} - 11\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

149/64*sqrt(2)*log(-2*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1) - 1546507/663552*sqrt(2)*log(abs(-2*sqrt(
2)*x + sqrt(2) + 2*sqrt(2*x^2 - x + 3))) + 1546507/663552*sqrt(2)*log(abs(-2*sqrt(2)*x - 11*sqrt(2) + 2*sqrt(2
*x^2 - x + 3))) + 5/16*sqrt(2*x^2 - x + 3) + 1/55296*sqrt(2)*(2381290*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)
)^3 + 16628406*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 - 25697445*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) + 1672
0645)/(2*(sqrt(2)*x - sqrt(2*x^2 - x + 3))^2 + 10*sqrt(2)*(sqrt(2)*x - sqrt(2*x^2 - x + 3)) - 11)^2

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maple [A]  time = 0.01, size = 102, normalized size = 0.80 \[ -\frac {149 \sqrt {2}\, \arcsinh \left (\frac {4 \sqrt {23}\, \left (x -\frac {1}{4}\right )}{23}\right )}{64}-\frac {1546507 \sqrt {2}\, \arctanh \left (\frac {\left (-11 x +\frac {17}{2}\right ) \sqrt {2}}{12 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}\right )}{663552}+\frac {5 \sqrt {2 x^{2}-x +3}}{16}+\frac {92239 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{55296 \left (x +\frac {5}{2}\right )}-\frac {3667 \sqrt {-11 x +2 \left (x +\frac {5}{2}\right )^{2}-\frac {19}{2}}}{4608 \left (x +\frac {5}{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*(2*x^2-x+3)^(1/2)-149/64*2^(1/2)*arcsinh(4/23*23^(1/2)*(x-1/4))+92239/55296/(x+5/2)*(-11*x+2*(x+5/2)^2-19
/2)^(1/2)-1546507/663552*2^(1/2)*arctanh(1/12*(-11*x+17/2)*2^(1/2)/(-11*x+2*(x+5/2)^2-19/2)^(1/2))-3667/4608/(
x+5/2)^2*(-11*x+2*(x+5/2)^2-19/2)^(1/2)

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maxima [A]  time = 0.98, size = 114, normalized size = 0.89 \[ -\frac {149}{64} \, \sqrt {2} \operatorname {arsinh}\left (\frac {4}{23} \, \sqrt {23} x - \frac {1}{23} \, \sqrt {23}\right ) + \frac {1546507}{663552} \, \sqrt {2} \operatorname {arsinh}\left (\frac {22 \, \sqrt {23} x}{23 \, {\left | 2 \, x + 5 \right |}} - \frac {17 \, \sqrt {23}}{23 \, {\left | 2 \, x + 5 \right |}}\right ) + \frac {5}{16} \, \sqrt {2 \, x^{2} - x + 3} - \frac {3667 \, \sqrt {2 \, x^{2} - x + 3}}{1152 \, {\left (4 \, x^{2} + 20 \, x + 25\right )}} + \frac {92239 \, \sqrt {2 \, x^{2} - x + 3}}{27648 \, {\left (2 \, x + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-149/64*sqrt(2)*arcsinh(4/23*sqrt(23)*x - 1/23*sqrt(23)) + 1546507/663552*sqrt(2)*arcsinh(22/23*sqrt(23)*x/abs
(2*x + 5) - 17/23*sqrt(23)/abs(2*x + 5)) + 5/16*sqrt(2*x^2 - x + 3) - 3667/1152*sqrt(2*x^2 - x + 3)/(4*x^2 + 2
0*x + 25) + 92239/27648*sqrt(2*x^2 - x + 3)/(2*x + 5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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