\(\int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx\) [32]

Optimal result

Integrand size = 24, antiderivative size = 119 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

-1/8*(-4*a*c+b^2)*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)+1/4*(2*c*x+b)*(c*x^4+b*x^3+a*x^2)^(1/2)/c/x

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 119, 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 = {1932, 1928, 635, 212} \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \]

[In]

Int[Sqrt[a*x^2 + b*x^3 + c*x^4]/x,x]

[Out]

((b + 2*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(4*c*x) - ((b^2 - 4*a*c)*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(b + 2*c*x)
/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*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 1932

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

Rubi steps \begin{align*} \text {integral}& = \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{8 c} \\ & = \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (\left (b^2-4 a c\right ) x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (\left (b^2-4 a c\right ) 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 )}{4 c \sqrt {a x^2+b x^3+c x^4}} \\ & = \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) 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 )}{8 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {\sqrt {x^2 (a+x (b+c x))} \left (\sqrt {c} (b+2 c x)+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a}-\sqrt {a+x (b+c x)}}\right )}{\sqrt {a+x (b+c x)}}\right )}{4 c^{3/2} x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {4 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {3}{2}} x +4 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) a c -\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b^{2}+2 b \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{8 c^{\frac {3}{2}}}\)	\(100\)
risch	\(\frac {\left (2 c x +b \right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{4 c x}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{8 c^{\frac {3}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(103\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (4 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, x +2 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b +4 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,c^{2}-\ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{2} c \right )}{8 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, x}\)	\(146\)

[In]

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

[Out]

1/8/c^(3/2)*(4*(c*x^2+b*x+a)^(1/2)*c^(3/2)*x+4*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*a*c-ln(2*(c*x^2+b*x+a
)^(1/2)*c^(1/2)+2*c*x+b)*b^2+2*b*(c*x^2+b*x+a)^(1/2)*c^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x + b c\right )}}{16 \, c^{2} x}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x + b c\right )}}{8 \, c^{2} x}\right ] \]

[In]

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

[Out]

[-1/16*((b^2 - 4*a*c)*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)*(2*c^2*x + b*c))/(c^2*x), 1/8*((b^2 - 4*a*c)*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)*(2*c^2*x + b*c))/(c^2*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {1}{8} \, {\left (2 \, \sqrt {c x^{2} + b x + a} {\left (2 \, x + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} \mathrm {sgn}\left (x\right ) - \frac {{\left (b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{8 \, c^{\frac {3}{2}}} \]

[In]

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

[Out]

1/8*(2*sqrt(c*x^2 + b*x + a)*(2*x + b/c) + (b^2 - 4*a*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c)
+ b))/c^(3/2))*sgn(x) - 1/8*(b^2*log(abs(b - 2*sqrt(a)*sqrt(c))) - 4*a*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 2*
sqrt(a)*b*sqrt(c))*sgn(x)/c^(3/2)

Mupad [F(-1)]

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

[In]

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

[Out]

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