\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^4} \, dx\) [60]

Optimal result

Integrand size = 24, antiderivative size = 227 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^4} \, dx=\frac {\left (b^2+8 a c+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 c x}+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}-\frac {a^{3/2} x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}-\frac {b \left (b^2-12 a c\right ) x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{3/2} \sqrt {a x^2+b x^3+c x^4}} \] Output:

1/8*(2*b*c*x+8*a*c+b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/c/x+1/3*(c*x^4+b*x^3+a*x 
^2)^(3/2)/x^3-a^(3/2)*x*(c*x^2+b*x+a)^(1/2)*arctanh(1/2*(b*x+2*a)/a^(1/2)/ 
(c*x^2+b*x+a)^(1/2))/(c*x^4+b*x^3+a*x^2)^(1/2)-1/16*b*(-12*a*c+b^2)*x*(c*x 
^2+b*x+a)^(1/2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(3/2) 
/(c*x^4+b*x^3+a*x^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^4} \, dx=\frac {x \sqrt {a+x (b+c x)} \left (-3 b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )+2 \sqrt {c} \left (\sqrt {a+x (b+c x)} \left (3 b^2+14 b c x+8 c \left (4 a+c x^2\right )\right )+48 a^{3/2} c \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )\right )}{48 c^{3/2} \sqrt {x^2 (a+x (b+c x))}} \] Input:

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

Output:

(x*Sqrt[a + x*(b + c*x)]*(-3*b*(b^2 - 12*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[ 
c]*Sqrt[a + x*(b + c*x)])] + 2*Sqrt[c]*(Sqrt[a + x*(b + c*x)]*(3*b^2 + 14* 
b*c*x + 8*c*(4*a + c*x^2)) + 48*a^(3/2)*c*ArcTanh[(Sqrt[c]*x - Sqrt[a + x* 
(b + c*x)])/Sqrt[a]])))/(48*c^(3/2)*Sqrt[x^2*(a + x*(b + c*x))])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1968, 1992, 27, 1980, 1269, 1092, 219, 1154, 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.

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

\(\Big \downarrow \) 1968

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

\(\Big \downarrow \) 1992

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {16 a^2 c-b \left (b^2-12 a c\right ) x}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{4 c}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (\frac {\int \frac {16 a^2 c-b \left (b^2-12 a c\right ) x}{\sqrt {c x^4+b x^3+a x^2}}dx}{8 c}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 1980

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \int \frac {16 a^2 c-b \left (b^2-12 a c\right ) x}{x \sqrt {c x^2+b x+a}}dx}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \left (16 a^2 c \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-b \left (b^2-12 a c\right ) \int \frac {1}{\sqrt {c x^2+b x+a}}dx\right )}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \left (16 a^2 c \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-2 b \left (b^2-12 a c\right ) \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}\right )}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \left (16 a^2 c \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \left (-32 a^2 c \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (\frac {x \sqrt {a+b x+c x^2} \left (-16 a^{3/2} c \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{\sqrt {c}}\right )}{8 c \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (8 a c+b^2+2 b c x\right )}{4 c x}\right )+\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^3}\)

Input:

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

Output:

(a*x^2 + b*x^3 + c*x^4)^(3/2)/(3*x^3) + (((b^2 + 8*a*c + 2*b*c*x)*Sqrt[a*x 
^2 + b*x^3 + c*x^4])/(4*c*x) + (x*Sqrt[a + b*x + c*x^2]*(-16*a^(3/2)*c*Arc 
Tanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])] - (b*(b^2 - 12*a*c)*Ar 
cTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/Sqrt[c]))/(8*c*Sqrt[ 
a*x^2 + b*x^3 + c*x^4]))/2
 

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 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

rule 1968

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

rule 1980

Int[((A_) + (B_.)*(x_)^(j_.))/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[(A + B*x^(n - q))/(x^(q 
/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; FreeQ[{a, b, c, A, 
B, n, q}, x] && EqQ[j, n - q] && EqQ[r, 2*n - q] && PosQ[n - q] && EqQ[n, 3 
] && EqQ[q, 2]
 

rule 1992

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.85

method	result	size
pseudoelliptic	\(\frac {16 x^{2} \sqrt {c \,x^{2}+b x +a}\, c^{\frac {5}{2}}-48 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right ) a^{\frac {3}{2}} c^{\frac {3}{2}}+48 \ln \left (2\right ) a^{\frac {3}{2}} c^{\frac {3}{2}}+28 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b x +64 a \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}+6 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}\, b^{2}+36 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) a b c -3 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b^{3}}{48 c^{\frac {3}{2}}}\)	\(192\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 c^{\frac {5}{2}} a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )-16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}-12 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b x -48 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a -6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{2}-36 c^{2} \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a b +3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{3} c \right )}{48 x^{3} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}}\)	\(222\)

Input:

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

Output:

1/48*(16*x^2*(c*x^2+b*x+a)^(1/2)*c^(5/2)-48*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b 
*x+a)^(1/2))/x/a^(1/2))*a^(3/2)*c^(3/2)+48*ln(2)*a^(3/2)*c^(3/2)+28*c^(3/2 
)*(c*x^2+b*x+a)^(1/2)*b*x+64*a*c^(3/2)*(c*x^2+b*x+a)^(1/2)+6*c^(1/2)*(c*x^ 
2+b*x+a)^(1/2)*b^2+36*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*a*b*c-3*ln 
(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*b^3)/c^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 791, normalized size of antiderivative = 3.48 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/96*(48*a^(3/2)*c^2*x*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) - 3*(b^3 - 12*a*b*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*(8*c^3*x^2 + 14*b*c^2*x + 3*b^2 
*c + 32*a*c^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^2*x), 1/48*(24*a^(3/2)*c^2* 
x*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) + 3*(b^3 - 12*a*b*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*(8*c^3*x^2 + 14*b*c^2*x + 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^3 
+ a*x^2))/(c^2*x), 1/96*(96*sqrt(-a)*a*c^2*x*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)) - 3*(b^3 - 12* 
a*b*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*(8*c^3*x^2 + 14*b*c^2*x + 
 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^2*x), 1/48*(48*sqrt(- 
a)*a*c^2*x*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)) + 3*(b^3 - 12*a*b*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*(8*c^3*x^2 + 14*b*c^2*x + 3*b^2*c + 32*a*c^2)*sqrt(c*x^4 + b*x^3 + a*x^ 
2))/(c^2*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio 
n over extensionNot implemented, e.g. for multivariate mod/approx polynomi 
alsError:
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 1018, normalized size of antiderivative = 4.48 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^4} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 48*sqrt(a)*sqrt(4*sqrt(c)*sqrt(a)*b - 4*a*c - b**2)*atan((2*sqrt(c)*sq 
rt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*sqrt(c)*sqrt(a)*b - 4*a*c - b**2) 
)*a*b*c**2 - 96*sqrt(c)*sqrt(4*sqrt(c)*sqrt(a)*b - 4*a*c - b**2)*atan((2*s 
qrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)/sqrt(4*sqrt(c)*sqrt(a)*b - 4*a* 
c - b**2))*a**2*c**2 - 24*sqrt(a)*sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b**2) 
*log( - sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b**2) + 2*sqrt(c)*sqrt(a + b*x 
+ c*x**2) + b + 2*c*x)*a*b*c**2 + 24*sqrt(a)*sqrt(4*sqrt(c)*sqrt(a)*b + 4* 
a*c + b**2)*log(sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b**2) + 2*sqrt(c)*sqrt( 
a + b*x + c*x**2) + b + 2*c*x)*a*b*c**2 + 48*sqrt(c)*sqrt(4*sqrt(c)*sqrt(a 
)*b + 4*a*c + b**2)*log( - sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b**2) + 2*sq 
rt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)*a**2*c**2 - 48*sqrt(c)*sqrt(4*sq 
rt(c)*sqrt(a)*b + 4*a*c + b**2)*log(sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b** 
2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)*a**2*c**2 + 256*sqrt(a 
+ b*x + c*x**2)*a**2*c**3 - 40*sqrt(a + b*x + c*x**2)*a*b**2*c**2 + 112*sq 
rt(a + b*x + c*x**2)*a*b*c**3*x + 64*sqrt(a + b*x + c*x**2)*a*c**4*x**2 - 
6*sqrt(a + b*x + c*x**2)*b**4*c - 28*sqrt(a + b*x + c*x**2)*b**3*c**2*x - 
16*sqrt(a + b*x + c*x**2)*b**2*c**3*x**2 + 96*sqrt(a)*log( - sqrt(4*sqrt(c 
)*sqrt(a)*b + 4*a*c + b**2) + 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x 
)*a**2*c**3 - 24*sqrt(a)*log( - sqrt(4*sqrt(c)*sqrt(a)*b + 4*a*c + b**2) + 
 2*sqrt(c)*sqrt(a + b*x + c*x**2) + b + 2*c*x)*a*b**2*c**2 + 96*sqrt(a)...