3.50 \(\int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx\)

Optimal. Leaf size=103 \[ \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

-1/2*b*x*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*(c*x^2+b*x+a)^(1/2)/c^(3/2)/(c*x^4+b*x^3+a*x^2)^(1
/2)+(c*x^4+b*x^3+a*x^2)^(1/2)/c/x

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1917, 1914, 621, 206} \[ \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

Sqrt[a*x^2 + b*x^3 + c*x^4]/(c*x) - (b*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c
*x^2])])/(2*c^(3/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])

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 621

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

Rule 1914

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[(x^(q/2)*Sqrt[a
+ b*x^(n - q) + c*x^(2*(n - q))])/Sqrt[a*x^q + b*x^n + c*x^(2*n - q)], Int[x^(m - q/2)/Sqrt[a + b*x^(n - q) +
c*x^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] &&
EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q,
1]))

Rule 1917

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m - n)*(a*
x^(n - 1) + b*x^n + c*x^(n + 1))^(p + 1))/(2*c*(p + 1)), x] - Dist[b/(2*c), Int[x^(m - 1)*(a*x^(n - 1) + b*x^n
+ c*x^(n + 1))^p, x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2
- 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] && EqQ[m + p*(n - 1) - 1, 0]

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx &=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{2 c}\\ &=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {\left (b x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{2 c \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {\left (b x \sqrt {a+b x+c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{c \sqrt {a x^2+b x^3+c x^4}}\\ &=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.05, size = 89, normalized size = 0.86 \[ \frac {x \left (2 \sqrt {c} (a+x (b+c x))-b \sqrt {a+x (b+c x)} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{2 c^{3/2} \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

(x*(2*Sqrt[c]*(a + x*(b + c*x)) - b*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]
)]))/(2*c^(3/2)*Sqrt[x^2*(a + x*(b + c*x))])

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 188, normalized size = 1.83 \[ \left [\frac {b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{4 \, c^{2} x}, \frac {b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{2 \, c^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(b*sqrt(c)*x*log(-(8*c^2*x^3 + 8*b*c*x^2 - 4*sqrt(c*x^4 + b*x^3 + a*x^2)*(2*c*x + b)*sqrt(c) + (b^2 + 4*a
*c)*x)/x) + 4*sqrt(c*x^4 + b*x^3 + a*x^2)*c)/(c^2*x), 1/2*(b*sqrt(-c)*x*arctan(1/2*sqrt(c*x^4 + b*x^3 + a*x^2)
*(2*c*x + b)*sqrt(-c)/(c^2*x^3 + b*c*x^2 + a*c*x)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*c)/(c^2*x)]

________________________________________________________________________________________

giac [A]  time = 0.91, size = 108, normalized size = 1.05 \[ \frac {b \arctan \left (\frac {\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}} - \frac {\sqrt {a}}{x}}{\sqrt {-c}}\right )}{\sqrt {-c} c} + \frac {b {\left (\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}} - \frac {\sqrt {a}}{x}\right )} - 2 \, \sqrt {a} c}{{\left ({\left (\sqrt {c + \frac {b}{x} + \frac {a}{x^{2}}} - \frac {\sqrt {a}}{x}\right )}^{2} - c\right )} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*arctan((sqrt(c + b/x + a/x^2) - sqrt(a)/x)/sqrt(-c))/(sqrt(-c)*c) + (b*(sqrt(c + b/x + a/x^2) - sqrt(a)/x) -
2*sqrt(a)*c)/(((sqrt(c + b/x + a/x^2) - sqrt(a)/x)^2 - c)*c)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 88, normalized size = 0.85 \[ \frac {\sqrt {c \,x^{2}+b x +a}\, \left (-b c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+2 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {3}{2}}\right ) x}{2 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x)

[Out]

1/2*x*(c*x^2+b*x+a)^(1/2)*(2*(c*x^2+b*x+a)^(1/2)*c^(3/2)-b*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1
/2))*c)/(c*x^4+b*x^3+a*x^2)^(1/2)/c^(5/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {c x^{4} + b x^{3} + a x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/sqrt(c*x^4 + b*x^3 + a*x^2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^2}{\sqrt {c\,x^4+b\,x^3+a\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^2 + b*x^3 + c*x^4)^(1/2),x)

[Out]

int(x^2/(a*x^2 + b*x^3 + c*x^4)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(c*x**4+b*x**3+a*x**2)**(1/2),x)

[Out]

Integral(x**2/sqrt(x**2*(a + b*x + c*x**2)), x)

________________________________________________________________________________________