\(\int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx\) [49]

Optimal result

Integrand size = 24, antiderivative size = 91 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {x (b+2 c x)}{2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {\sqrt {x^2 (a+x (b+c x))} \left (\sqrt {c} (b+2 c x)+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {a}-\sqrt {a+x (b+c x)}}\right )}{\sqrt {a+x (b+c x)}}\right )}{4 c^{3/2} x} \] Input:

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

Output:

(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)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1965, 1961, 1092, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 1965

\(\displaystyle \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) \int \frac {x}{\sqrt {c x^4+b x^3+a x^2}}dx}{8 c}\)

\(\Big \downarrow \) 1961

\(\displaystyle \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c \sqrt {a x^2+b x^3+c x^4}}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c \sqrt {a x^2+b x^3+c x^4}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}\)

Input:

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

Output:

((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])
 

Defintions of rubi rules used

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1092

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

rule 1961

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)] 
, x_Symbol] :> Simp[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 1965

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] - Simp[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] && EqQ[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] && E 
qQ[m + p*q + 1, n - q]
 

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10

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	\(\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (4 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, x +2 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b +4 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,c^{2}-\ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{2} c \right )}{8 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, x}\)	\(146\)

Input:

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

Output:

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)

Time = 0.08 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.42 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\left [-\frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x + b c\right )}}{16 \, c^{2} x}, \frac {{\left (b^{2} - 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x + b c\right )}}{8 \, c^{2} x}\right ] \] Input:

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

Output:

[-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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {1}{8} \, {\left (2 \, \sqrt {c x^{2} + b x + a} {\left (2 \, x + \frac {b}{c}\right )} + \frac {{\left (b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {3}{2}}}\right )} \mathrm {sgn}\left (x\right ) - \frac {{\left (b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} b \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{8 \, c^{\frac {3}{2}}} \] Input:

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

Output:

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*sq 
rt(a)*b*sqrt(c))*sgn(x)/c^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.35 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx=\frac {2 \sqrt {c \,x^{2}+b x +a}\, b c +4 \sqrt {c \,x^{2}+b x +a}\, c^{2} x +4 \sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) a c -\sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b^{2}}{8 c^{2}} \] Input:

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

Output:

(2*sqrt(a + b*x + c*x**2)*b*c + 4*sqrt(a + b*x + c*x**2)*c**2*x + 4*sqrt(c 
)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c - b**2))*a 
*c - sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*a*c 
 - b**2))*b**2)/(8*c**2)