\(\int x \sqrt {a x^2+b x^3+c x^4} \, dx\) [47]

Optimal result

Integrand size = 22, antiderivative size = 177 \[ \int x \sqrt {a x^2+b x^3+c x^4} \, dx=-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{96 c^2}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{192 c^3 x}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {x (b+2 c x)}{2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}}\right )}{128 c^{7/2}} \] Output:

-1/96*(-12*a*c+5*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/c^2+1/192*b*(-52*a*c+15*b^ 
2)*(c*x^4+b*x^3+a*x^2)^(1/2)/c^3/x+1/24*x*(6*c*x+b)*(c*x^4+b*x^3+a*x^2)^(1 
/2)/c-1/128*(-4*a*c+b^2)*(-4*a*c+5*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^(7/2)
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.85 \[ \int x \sqrt {a x^2+b x^3+c x^4} \, dx=\frac {2 \sqrt {c} x (a+x (b+c x)) \left (15 b^3-10 b^2 c x+24 c^2 x \left (a+2 c x^2\right )+b \left (-52 a c+8 c^2 x^2\right )\right )+3 \left (5 b^4-24 a b^2 c+16 a^2 c^2\right ) x \sqrt {a+x (b+c x)} \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )}{384 c^{7/2} \sqrt {x^2 (a+x (b+c x))}} \] Input:

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

Output:

(2*Sqrt[c]*x*(a + x*(b + c*x))*(15*b^3 - 10*b^2*c*x + 24*c^2*x*(a + 2*c*x^ 
2) + b*(-52*a*c + 8*c^2*x^2)) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*x*Sqrt 
[a + x*(b + c*x)]*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(384*c 
^(7/2)*Sqrt[x^2*(a + x*(b + c*x))])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.23, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1966, 27, 1996, 27, 1996, 27, 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.

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

\(\Big \downarrow \) 1966

\(\displaystyle \frac {\int -\frac {x^2 \left (4 a b+\left (5 b^2-12 a c\right ) x\right )}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{24 c}+\frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\int \frac {x^2 \left (4 a b+\left (5 b^2-12 a c\right ) x\right )}{\sqrt {c x^4+b x^3+a x^2}}dx}{48 c}\)

\(\Big \downarrow \) 1996

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\int \frac {x \left (2 a \left (5 b^2-12 a c\right )+b \left (15 b^2-52 a c\right ) x\right )}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{2 c}}{48 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\int \frac {x \left (2 a \left (5 b^2-12 a c\right )+b \left (15 b^2-52 a c\right ) x\right )}{\sqrt {c x^4+b x^3+a x^2}}dx}{4 c}}{48 c}\)

\(\Big \downarrow \) 1996

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {\int \frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) x}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{c}}{4 c}}{48 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {3 \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \int \frac {x}{\sqrt {c x^4+b x^3+a x^2}}dx}{2 c}}{4 c}}{48 c}\)

\(\Big \downarrow \) 1961

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {3 x \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{2 c \sqrt {a x^2+b x^3+c x^4}}}{4 c}}{48 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {3 x \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c \sqrt {a x^2+b x^3+c x^4}}}{4 c}}{48 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {x (b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{24 c}-\frac {\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {3 x \left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}}{4 c}}{48 c}\)

Input:

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

Output:

(x*(b + 6*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(24*c) - (((5*b^2 - 12*a*c)*Sq 
rt[a*x^2 + b*x^3 + c*x^4])/(2*c) - ((b*(15*b^2 - 52*a*c)*Sqrt[a*x^2 + b*x^ 
3 + c*x^4])/(c*x) - (3*(b^2 - 4*a*c)*(5*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])])/(2*c^(3/2)*Sqrt 
[a*x^2 + b*x^3 + c*x^4]))/(4*c))/(48*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 1966

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ 
), x_Symbol] :> Simp[x^(m - n + q + 1)*(b*(n - q)*p + c*(m + p*q + (n - q)* 
(2*p - 1) + 1)*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^p/(c*(m + p*(2*n 
 - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1))), x] + Simp[(n - q)*(p/(c*(m 
+ p*(2*n - q) + 1)*(m + p*q + (n - q)*(2*p - 1) + 1)))   Int[x^(m - (n - 2* 
q))*Simp[(-a)*b*(m + p*q - n + q + 1) + (2*a*c*(m + p*q + (n - q)*(2*p - 1) 
 + 1) - b^2*(m + p*q + (n - q)*(p - 1) + 1))*x^(n - q), x]*(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] && GtQ[m + p*q + 1, n - q] && NeQ[m + p*(2*n - q) 
+ 1, 0] && NeQ[m + p*q + (n - q)*(2*p - 1) + 1, 0]
 

rule 1996

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[B*x^(m - n + 1)*((a*x^q + b 
*x^n + c*x^(2*n - q))^(p + 1)/(c*(m + p*q + (n - q)*(2*p + 1) + 1))), x] - 
Simp[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1))   Int[x^(m - n + q)*Simp[a*B*( 
m + p*q - n + q + 1) + (b*B*(m + p*q + (n - q)*p + 1) - A*c*(m + p*q + (n - 
 q)*(2*p + 1) + 1))*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] 
 /; FreeQ[{a, b, c, A, B}, x] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !Inte 
gerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && 
RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (n - q)*(2*p + 
1) + 1, 0]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.60

method	result	size
pseudoelliptic	\(-\frac {\left (a^{2} c^{2}-\frac {3}{2} a \,b^{2} c +\frac {5}{16} b^{4}\right ) \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right )+\frac {13 \left (b \left (\frac {5 b x}{26}+a \right ) c^{\frac {3}{2}}-\frac {6 x \left (\frac {b x}{3}+a \right ) c^{\frac {5}{2}}}{13}-\frac {15 \sqrt {c}\, b^{3}}{52}-\frac {12 c^{\frac {7}{2}} x^{3}}{13}\right ) \sqrt {c \,x^{2}+b x +a}}{6}}{8 c^{\frac {7}{2}}}\)	\(106\)
risch	\(-\frac {\left (-48 c^{3} x^{3}-8 b \,c^{2} x^{2}-24 a \,c^{2} x +10 b^{2} c x +52 a b c -15 b^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{192 c^{3} x}-\frac {\left (16 a^{2} c^{2}-24 a \,b^{2} c +5 b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{128 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(150\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (96 x \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {7}{2}}-48 c^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, a x -80 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b +60 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{2} x -24 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a b +30 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{3}-48 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{2} c^{3}+72 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,b^{2} c^{2}-15 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{4} c \right )}{384 x \sqrt {c \,x^{2}+b x +a}\, c^{\frac {9}{2}}}\)	\(265\)

Input:

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

Output:

-1/8/c^(7/2)*((a^2*c^2-3/2*a*b^2*c+5/16*b^4)*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1 
/2)+2*c*x+b)+13/6*(b*(5/26*b*x+a)*c^(3/2)-6/13*x*(1/3*b*x+a)*c^(5/2)-15/52 
*c^(1/2)*b^3-12/13*c^(7/2)*x^3)*(c*x^2+b*x+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.84 \[ \int x \sqrt {a x^2+b x^3+c x^4} \, dx=\left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{768 \, c^{4} x}, \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, {\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{384 \, c^{4} x}\right ] \] Input:

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

Output:

[1/768*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*x^3 + 8*b*c^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c 
^2 - 12*a*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^4*x), 1/384*(3*(5*b^4 - 
24*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*x^3 + 8*b*c 
^3*x^2 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x)*sqrt(c*x^4 + 
b*x^3 + a*x^2))/(c^4*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(x*(c*x^4+b*x^3+a*x^2)^(1/2),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 = 222, normalized size of antiderivative = 1.25 \[ \int x \sqrt {a x^2+b x^3+c x^4} \, dx=\frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, x \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{c}\right )} x - \frac {5 \, b^{2} c \mathrm {sgn}\left (x\right ) - 12 \, a c^{2} \mathrm {sgn}\left (x\right )}{c^{3}}\right )} x + \frac {15 \, b^{3} \mathrm {sgn}\left (x\right ) - 52 \, a b c \mathrm {sgn}\left (x\right )}{c^{3}}\right )} + \frac {{\left (5 \, b^{4} \mathrm {sgn}\left (x\right ) - 24 \, a b^{2} c \mathrm {sgn}\left (x\right ) + 16 \, a^{2} c^{2} \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{128 \, c^{\frac {7}{2}}} - \frac {{\left (15 \, b^{4} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 72 \, a b^{2} c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 48 \, a^{2} c^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 30 \, \sqrt {a} b^{3} \sqrt {c} - 104 \, a^{\frac {3}{2}} b c^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{384 \, c^{\frac {7}{2}}} \] Input:

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

Output:

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*x*sgn(x) + b*sgn(x)/c)*x - (5*b^2*c*s 
gn(x) - 12*a*c^2*sgn(x))/c^3)*x + (15*b^3*sgn(x) - 52*a*b*c*sgn(x))/c^3) + 
 1/128*(5*b^4*sgn(x) - 24*a*b^2*c*sgn(x) + 16*a^2*c^2*sgn(x))*log(abs(2*(s 
qrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2) - 1/384*(15*b^4*lo 
g(abs(b - 2*sqrt(a)*sqrt(c))) - 72*a*b^2*c*log(abs(b - 2*sqrt(a)*sqrt(c))) 
 + 48*a^2*c^2*log(abs(b - 2*sqrt(a)*sqrt(c))) + 30*sqrt(a)*b^3*sqrt(c) - 1 
04*a^(3/2)*b*c^(3/2))*sgn(x)/c^(7/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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