\(\int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx\) [50]

Optimal result

Integrand size = 24, antiderivative size = 103 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[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] (verified)

Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1931, 1928, 635, 212} \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 1928

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 1931

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*} \text {integral}& = \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 ) \text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \left (2 \sqrt {c} (a+x (b+c x))-b \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{2 c^{3/2} \sqrt {x^2 (a+x (b+c x))}} \]

[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))])

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.49

method	result	size
pseudoelliptic	\(-\frac {\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b -2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 c^{\frac {3}{2}}}\)	\(50\)
default	\(\frac {x \sqrt {c \,x^{2}+b x +a}\, \left (2 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {3}{2}}-b \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) c \right )}{2 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, c^{\frac {5}{2}}}\)	\(88\)
risch	\(\frac {\left (c \,x^{2}+b x +a \right ) x}{c \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) x \sqrt {c \,x^{2}+b x +a}}{2 c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(93\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.83 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\left [\frac {b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{4 \, c^{2} x}, \frac {b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{2 \, c^{2} x}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int \frac {x^{2}}{\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {c x^{4} + b x^{3} + a x^{2}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {{\left (b \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{2 \, c^{\frac {3}{2}}} + \frac {b \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} + \frac {\sqrt {c x^{2} + b x + a}}{c \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/2*(b*log(abs(b - 2*sqrt(a)*sqrt(c))) + 2*sqrt(a)*sqrt(c))*sgn(x)/c^(3/2) + 1/2*b*log(abs(2*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(3/2)*sgn(x)) + sqrt(c*x^2 + b*x + a)/(c*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int \frac {x^2}{\sqrt {c\,x^4+b\,x^3+a\,x^2}} \,d x \]

[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)