\(\int \frac {1}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [79]

Optimal result

Integrand size = 20, antiderivative size = 209 \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) x \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a^2 \left (b^2-4 a c\right ) x^3}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{4 a^3 \left (b^2-4 a c\right ) x^2}-\frac {3 \left (5 b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^{7/2}} \] Output:

2*(b*c*x-2*a*c+b^2)/a/(-4*a*c+b^2)/x/(c*x^4+b*x^3+a*x^2)^(1/2)-1/2*(-12*a* 
c+5*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^2/(-4*a*c+b^2)/x^3+1/4*b*(-52*a*c+15* 
b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^3/(-4*a*c+b^2)/x^2-3/8*(-4*a*c+5*b^2)*arc 
tanh(1/2*x*(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 1.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {\sqrt {a} \left (-8 a^3 c-15 b^3 x^2 (b+c x)+2 a^2 \left (b^2+10 b c x-12 c^2 x^2\right )+a b x \left (-5 b^2+62 b c x+52 c^2 x^2\right )\right )-3 \left (5 b^4-24 a b^2 c+16 a^2 c^2\right ) x^2 \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{4 a^{7/2} \left (b^2-4 a c\right ) x \sqrt {x^2 (a+x (b+c x))}} \] Input:

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

Output:

-1/4*(Sqrt[a]*(-8*a^3*c - 15*b^3*x^2*(b + c*x) + 2*a^2*(b^2 + 10*b*c*x - 1 
2*c^2*x^2) + a*b*x*(-5*b^2 + 62*b*c*x + 52*c^2*x^2)) - 3*(5*b^4 - 24*a*b^2 
*c + 16*a^2*c^2)*x^2*Sqrt[a + x*(b + c*x)]*ArcTanh[(Sqrt[c]*x - Sqrt[a + x 
*(b + c*x)])/Sqrt[a]])/(a^(7/2)*(b^2 - 4*a*c)*x*Sqrt[x^2*(a + x*(b + c*x)) 
])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1954, 27, 1998, 27, 1998, 27, 1951, 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.

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

\(\Big \downarrow \) 1954

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1998

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1998

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1951

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(2*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*x*Sqrt[a*x^2 + b*x^3 + c*x^4]) 
+ (-1/2*((5*b^2 - 12*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(a*x^3) - (-((b*(15 
*b^2 - 52*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(a*x^2)) + (3*(b^2 - 4*a*c)*(5 
*b^2 - 4*a*c)*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^ 
4])])/(2*a^(3/2)))/(4*a))/(a*(b^2 - 4*a*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 1951

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] : 
> Simp[-2/(n - 2)   Subst[Int[1/(4*a - x^2), x], x, x*((2*a + b*x^(n - 2))/ 
Sqrt[a*x^2 + b*x^n + c*x^r])], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r, 2* 
n - 2] && PosQ[n - 2] && NeQ[b^2 - 4*a*c, 0]
 

rule 1954

Int[((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol 
] :> Simp[(-x^(-q + 1))*(b^2 - 2*a*c + b*c*x^(n - q))*((a*x^q + b*x^n + c*x 
^(2*n - q))^(p + 1)/(a*(n - q)*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(a*(n - 
 q)*(p + 1)*(b^2 - 4*a*c))   Int[(((p*q + 1)*(b^2 - 2*a*c) + (n - q)*(p + 1 
)*(b^2 - 4*a*c) + b*c*(p*q + (n - q)*(2*p + 3) + 1)*x^(n - q))*(a*x^q + b*x 
^n + c*x^(2*n - q))^(p + 1))/x^q, x], x] /; FreeQ[{a, b, c, n, q}, x] && Eq 
Q[r, 2*n - q] && PosQ[n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && LtQ 
[p, -1]
 

rule 1998

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-\frac {5 \left (-\frac {52}{5} c^{2} x^{2}-\frac {62}{5} b c x +b^{2}\right ) b x \,a^{\frac {3}{2}}}{4}+6 \left (-c^{2} x^{2}+\frac {5}{6} b c x +\frac {1}{12} b^{2}\right ) a^{\frac {5}{2}}-2 a^{\frac {7}{2}} c +6 \left (-\frac {5 b^{3} \left (c x +b \right ) \sqrt {a}}{8}+\left (-\ln \left (2\right )+\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )\right ) \left (a c -\frac {5 b^{2}}{4}\right ) \sqrt {c \,x^{2}+b x +a}\, \left (a c -\frac {b^{2}}{4}\right )\right ) x^{2}}{a^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, \left (4 a c -b^{2}\right ) x^{2}}\)	\(173\)
default	\(-\frac {x \left (c \,x^{2}+b x +a \right ) \left (-104 a^{\frac {5}{2}} b \,c^{2} x^{3}+30 a^{\frac {3}{2}} b^{3} c \,x^{3}+48 a^{\frac {7}{2}} c^{2} x^{2}-124 a^{\frac {5}{2}} b^{2} c \,x^{2}+30 a^{\frac {3}{2}} b^{4} x^{2}-48 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{3} c^{2} x^{2}+72 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} c \,x^{2}-15 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a \,b^{4} x^{2}-40 a^{\frac {7}{2}} b c x +10 a^{\frac {5}{2}} b^{3} x +16 a^{\frac {9}{2}} c -4 a^{\frac {7}{2}} b^{2}\right )}{8 \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {9}{2}} \left (4 a c -b^{2}\right )}\)	\(292\)
risch	\(-\frac {\left (c \,x^{2}+b x +a \right ) \left (-7 b x +2 a \right )}{4 a^{3} x \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}+\frac {\left (-\frac {2 c \,b^{3} x}{a^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {b^{4}}{a^{3} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {c}{a^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {6 c^{2} b x}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {3 c \,b^{2}}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {b^{2}}{a^{3} \sqrt {c \,x^{2}+b x +a}}+\frac {3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) c}{2 a^{\frac {5}{2}}}-\frac {15 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{2}}{8 a^{\frac {7}{2}}}\right ) x \sqrt {c \,x^{2}+b x +a}}{\sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(317\)

Input:

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

Output:

6/(c*x^2+b*x+a)^(1/2)/a^(7/2)*(-5/24*(-52/5*c^2*x^2-62/5*b*c*x+b^2)*b*x*a^ 
(3/2)+(-c^2*x^2+5/6*b*c*x+1/12*b^2)*a^(5/2)-1/3*a^(7/2)*c+(-5/8*b^3*(c*x+b 
)*a^(1/2)+(-ln(2)+ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x/a^(1/2)))*( 
a*c-5/4*b^2)*(c*x^2+b*x+a)^(1/2)*(a*c-1/4*b^2))*x^2)/(4*a*c-b^2)/x^2
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 630, normalized size of antiderivative = 3.01 \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{5} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{4} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{3}\right )} \sqrt {a} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, {\left (2 \, a^{3} b^{2} - 8 \, a^{4} c - {\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{3} - {\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{2} - 5 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{16 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{5} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{4} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{3}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{5} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{4} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 2 \, {\left (2 \, a^{3} b^{2} - 8 \, a^{4} c - {\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{3} - {\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{2} - 5 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{8 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{5} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{4} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{3}\right )}}\right ] \] Input:

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

Output:

[-1/16*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^5 + (5*b^5 - 24*a*b^3*c 
 + 16*a^2*b*c^2)*x^4 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*x^3)*sqrt(a)* 
log(-(8*a*b*x^2 + (b^2 + 4*a*c)*x^3 + 8*a^2*x + 4*sqrt(c*x^4 + b*x^3 + a*x 
^2)*(b*x + 2*a)*sqrt(a))/x^3) + 4*(2*a^3*b^2 - 8*a^4*c - (15*a*b^3*c - 52* 
a^2*b*c^2)*x^3 - (15*a*b^4 - 62*a^2*b^2*c + 24*a^3*c^2)*x^2 - 5*(a^2*b^3 - 
 4*a^3*b*c)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/((a^4*b^2*c - 4*a^5*c^2)*x^5 + 
 (a^4*b^3 - 4*a^5*b*c)*x^4 + (a^5*b^2 - 4*a^6*c)*x^3), 1/8*(3*((5*b^4*c - 
24*a*b^2*c^2 + 16*a^2*c^3)*x^5 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^4 + 
 (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*x^3)*sqrt(-a)*arctan(1/2*sqrt(c*x^4 
 + b*x^3 + a*x^2)*(b*x + 2*a)*sqrt(-a)/(a*c*x^3 + a*b*x^2 + a^2*x)) - 2*(2 
*a^3*b^2 - 8*a^4*c - (15*a*b^3*c - 52*a^2*b*c^2)*x^3 - (15*a*b^4 - 62*a^2* 
b^2*c + 24*a^3*c^2)*x^2 - 5*(a^2*b^3 - 4*a^3*b*c)*x)*sqrt(c*x^4 + b*x^3 + 
a*x^2))/((a^4*b^2*c - 4*a^5*c^2)*x^5 + (a^4*b^3 - 4*a^5*b*c)*x^4 + (a^5*b^ 
2 - 4*a^6*c)*x^3)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3} + c x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:

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

Output:

Integral((a*x**2 + b*x**3 + c*x**4)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 696, normalized size of antiderivative = 3.33 \[ \int \frac {1}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 16*sqrt(a + b*x + c*x**2)*a**4*c + 4*sqrt(a + b*x + c*x**2)*a**3*b**2 
+ 40*sqrt(a + b*x + c*x**2)*a**3*b*c*x - 48*sqrt(a + b*x + c*x**2)*a**3*c* 
*2*x**2 - 10*sqrt(a + b*x + c*x**2)*a**2*b**3*x + 124*sqrt(a + b*x + c*x** 
2)*a**2*b**2*c*x**2 + 104*sqrt(a + b*x + c*x**2)*a**2*b*c**2*x**3 - 30*sqr 
t(a + b*x + c*x**2)*a*b**4*x**2 - 30*sqrt(a + b*x + c*x**2)*a*b**3*c*x**3 
+ 48*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a**3*c** 
2*x**2 - 72*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a 
**2*b**2*c*x**2 + 48*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a 
 - b*x)*a**2*b*c**2*x**3 + 48*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x + c*x* 
*2) - 2*a - b*x)*a**2*c**3*x**4 + 15*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + b*x 
 + c*x**2) - 2*a - b*x)*a*b**4*x**2 - 72*sqrt(a)*log( - 2*sqrt(a)*sqrt(a + 
 b*x + c*x**2) - 2*a - b*x)*a*b**3*c*x**3 - 72*sqrt(a)*log( - 2*sqrt(a)*sq 
rt(a + b*x + c*x**2) - 2*a - b*x)*a*b**2*c**2*x**4 + 15*sqrt(a)*log( - 2*s 
qrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**5*x**3 + 15*sqrt(a)*log( - 2 
*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**4*c*x**4 - 48*sqrt(a)*log( 
x)*a**3*c**2*x**2 + 72*sqrt(a)*log(x)*a**2*b**2*c*x**2 - 48*sqrt(a)*log(x) 
*a**2*b*c**2*x**3 - 48*sqrt(a)*log(x)*a**2*c**3*x**4 - 15*sqrt(a)*log(x)*a 
*b**4*x**2 + 72*sqrt(a)*log(x)*a*b**3*c*x**3 + 72*sqrt(a)*log(x)*a*b**2*c* 
*2*x**4 - 15*sqrt(a)*log(x)*b**5*x**3 - 15*sqrt(a)*log(x)*b**4*c*x**4)/(8* 
a**4*x**2*(4*a**2*c - a*b**2 + 4*a*b*c*x + 4*a*c**2*x**2 - b**3*x - b**...