\(\int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx\) [51]

Optimal result

Integrand size = 22, antiderivative size = 71 \[ \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1928, 635, 212} \[ \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c} \sqrt {a x^2+b x^3+c x^4}} \]

[In]

Int[x/Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{\sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {\left (2 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 )}{\sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {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 )}{\sqrt {c} \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {x \sqrt {a+b x+c x^2} \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+b x+c x^2}\right )}{\sqrt {c} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.41

method	result	size
pseudoelliptic	\(\frac {\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right )}{\sqrt {c}}\)	\(29\)
default	\(\frac {x \sqrt {c \,x^{2}+b x +a}\, \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right )}{\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \sqrt {c}}\)	\(65\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.82 \[ \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\left [\frac {\log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right )}{2 \, \sqrt {c}}, -\frac {\sqrt {-c} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right )}{c}\right ] \]

[In]

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

[Out]

[1/2*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)/sqr
t(c), -sqrt(-c)*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))/c]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {\log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {c}} - \frac {\log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{\sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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