3.1.41 \(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^2} \, dx\) [41]

3.1.41.1 Optimal result

Integrand size = 24, antiderivative size = 198 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^2} \, dx=\frac {3 b \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{128 c^3 x}-\frac {b (b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{16 c^2 x^3}+\frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {3 b \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2} \sqrt {a x^2+b x^3+c x^4}} \]

output

-1/16*b*(2*c*x+b)*(c*x^4+b*x^3+a*x^2)^(3/2)/c^2/x^3+1/5*(c*x^4+b*x^3+a*x^2 
)^(5/2)/c/x^5-3/256*b*(-4*a*c+b^2)^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^(7/2)/(c*x^4+b*x^3+a*x^2)^(1/2)+3/12 
8*b*(-4*a*c+b^2)*(2*c*x+b)*(c*x^4+b*x^3+a*x^2)^(1/2)/c^3/x
 

3.1.41.2 Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^2} \, dx=\frac {x \sqrt {a+x (b+c x)} \left (2 \sqrt {c} \sqrt {a+x (b+c x)} \left (15 b^4-10 b^3 c x+128 c^2 \left (a+c x^2\right )^2+4 b^2 c \left (-25 a+2 c x^2\right )+8 b c^2 x \left (7 a+22 c x^2\right )\right )+15 b \left (b^2-4 a c\right )^2 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{1280 c^{7/2} \sqrt {x^2 (a+x (b+c x))}} \]

input

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

output

(x*Sqrt[a + x*(b + c*x)]*(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^4 - 10*b^3 
*c*x + 128*c^2*(a + c*x^2)^2 + 4*b^2*c*(-25*a + 2*c*x^2) + 8*b*c^2*x*(7*a 
+ 22*c*x^2)) + 15*b*(b^2 - 4*a*c)^2*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*( 
b + c*x)]]))/(1280*c^(7/2)*Sqrt[x^2*(a + x*(b + c*x))])
 

3.1.41.3 Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1964, 1965, 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.

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

\(\Big \downarrow \) 1964

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

\(\Big \downarrow \) 1965

\(\displaystyle \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \left (\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {3 \left (b^2-4 a c\right ) \int \frac {\sqrt {c x^4+b x^3+a x^2}}{x}dx}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 1965

\(\displaystyle \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \left (\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {\left (b^2-4 a c\right ) \int \frac {x}{\sqrt {c x^4+b x^3+a x^2}}dx}{8 c}\right )}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 1961

\(\displaystyle \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \left (\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{8 c \sqrt {a x^2+b x^3+c x^4}}\right )}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \left (\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{4 c \sqrt {a x^2+b x^3+c x^4}}\right )}{16 c}\right )}{2 c}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \left (\frac {(b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{8 c x^3}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 c x}-\frac {x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 c}\right )}{2 c}\)

input

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

output

(a*x^2 + b*x^3 + c*x^4)^(5/2)/(5*c*x^5) - (b*(((b + 2*c*x)*(a*x^2 + b*x^3 
+ c*x^4)^(3/2))/(8*c*x^3) - (3*(b^2 - 4*a*c)*(((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])))/(16*c)))/(2*c)
 

3.1.41.3.1 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 1964

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

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]
 

3.1.41.4 Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(\frac {-\frac {15 b \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right )}{16}+\left (\left (\frac {1}{16} b^{2} x^{2}+\frac {7}{16} a b x +a^{2}\right ) c^{\frac {5}{2}}-\frac {25 b^{2} \left (\frac {b x}{10}+a \right ) c^{\frac {3}{2}}}{32}+\left (\frac {11}{8} b \,x^{3}+2 a \,x^{2}\right ) c^{\frac {7}{2}}+c^{\frac {9}{2}} x^{4}+\frac {15 \sqrt {c}\, b^{4}}{128}\right ) \sqrt {c \,x^{2}+b x +a}}{5 c^{\frac {7}{2}}}\)	\(126\)
risch	\(\frac {\left (128 c^{4} x^{4}+176 b \,c^{3} x^{3}+256 a \,c^{3} x^{2}+8 b^{2} c^{2} x^{2}+56 a b \,c^{2} x -10 b^{3} c x +128 a^{2} c^{2}-100 a \,b^{2} c +15 b^{4}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{640 c^{3} x}-\frac {3 b \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{256 c^{\frac {7}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(180\)
default	\(\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (256 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} c^{\frac {7}{2}}-160 c^{\frac {7}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b x -80 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2}-240 c^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, a b x +60 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{3} x -120 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a \,b^{2}+30 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{4}-240 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{2} b \,c^{3}+120 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,b^{3} c^{2}-15 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{5} c \right )}{1280 x^{3} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(289\)

input

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

output

1/5/c^(7/2)*(-15/16*b*(a*c-1/4*b^2)^2*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c 
*x+b)+((1/16*b^2*x^2+7/16*a*b*x+a^2)*c^(5/2)-25/32*b^2*(1/10*b*x+a)*c^(3/2 
)+(11/8*b*x^3+2*a*x^2)*c^(7/2)+c^(9/2)*x^4+15/128*c^(1/2)*b^4)*(c*x^2+b*x+ 
a)^(1/2))
 

3.1.41.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^2} \, dx=\left [\frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{2560 \, c^{4} x}, \frac {15 \, {\left (b^{5} - 8 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, {\left (128 \, c^{5} x^{4} + 176 \, b c^{4} x^{3} + 15 \, b^{4} c - 100 \, a b^{2} c^{2} + 128 \, a^{2} c^{3} + 8 \, {\left (b^{2} c^{3} + 32 \, a c^{4}\right )} x^{2} - 2 \, {\left (5 \, b^{3} c^{2} - 28 \, a b c^{3}\right )} x\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{1280 \, c^{4} x}\right ] \]

input

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

output

[1/2560*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c - 100*a*b^2*c^2 + 128 
*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 - 28*a*b*c^3)*x)*sqrt 
(c*x^4 + b*x^3 + a*x^2))/(c^4*x), 1/1280*(15*(b^5 - 8*a*b^3*c + 16*a^2*b*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*(128*c^5*x^4 + 176*b*c^4*x^3 + 15*b^4*c 
- 100*a*b^2*c^2 + 128*a^2*c^3 + 8*(b^2*c^3 + 32*a*c^4)*x^2 - 2*(5*b^3*c^2 
- 28*a*b*c^3)*x)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^4*x)]
 

3.1.41.6 Sympy [F]

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

input

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

output

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

3.1.41.7 Maxima [F]

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

input

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

output

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

3.1.41.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^2} \, dx=\frac {1}{640} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, c x \mathrm {sgn}\left (x\right ) + 11 \, b \mathrm {sgn}\left (x\right )\right )} x + \frac {b^{2} c^{3} \mathrm {sgn}\left (x\right ) + 32 \, a c^{4} \mathrm {sgn}\left (x\right )}{c^{4}}\right )} x - \frac {5 \, b^{3} c^{2} \mathrm {sgn}\left (x\right ) - 28 \, a b c^{3} \mathrm {sgn}\left (x\right )}{c^{4}}\right )} x + \frac {15 \, b^{4} c \mathrm {sgn}\left (x\right ) - 100 \, a b^{2} c^{2} \mathrm {sgn}\left (x\right ) + 128 \, a^{2} c^{3} \mathrm {sgn}\left (x\right )}{c^{4}}\right )} + \frac {3 \, {\left (b^{5} \mathrm {sgn}\left (x\right ) - 8 \, a b^{3} c \mathrm {sgn}\left (x\right ) + 16 \, a^{2} b c^{2} \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{256 \, c^{\frac {7}{2}}} - \frac {{\left (15 \, b^{5} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 120 \, a b^{3} c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 240 \, a^{2} b c^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 30 \, \sqrt {a} b^{4} \sqrt {c} - 200 \, a^{\frac {3}{2}} b^{2} c^{\frac {3}{2}} + 256 \, a^{\frac {5}{2}} c^{\frac {5}{2}}\right )} \mathrm {sgn}\left (x\right )}{1280 \, c^{\frac {7}{2}}} \]

input

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

output

1/640*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*c*x*sgn(x) + 11*b*sgn(x))*x + (b^2 
*c^3*sgn(x) + 32*a*c^4*sgn(x))/c^4)*x - (5*b^3*c^2*sgn(x) - 28*a*b*c^3*sgn 
(x))/c^4)*x + (15*b^4*c*sgn(x) - 100*a*b^2*c^2*sgn(x) + 128*a^2*c^3*sgn(x) 
)/c^4) + 3/256*(b^5*sgn(x) - 8*a*b^3*c*sgn(x) + 16*a^2*b*c^2*sgn(x))*log(a 
bs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2) - 1/1280*(1 
5*b^5*log(abs(b - 2*sqrt(a)*sqrt(c))) - 120*a*b^3*c*log(abs(b - 2*sqrt(a)* 
sqrt(c))) + 240*a^2*b*c^2*log(abs(b - 2*sqrt(a)*sqrt(c))) + 30*sqrt(a)*b^4 
*sqrt(c) - 200*a^(3/2)*b^2*c^(3/2) + 256*a^(5/2)*c^(5/2))*sgn(x)/c^(7/2)
 

3.1.41.9 Mupad [F(-1)]

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

input

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

output

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