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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1931, 1932, 1928, 635, 212} \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^2} \, dx=-\frac {3 b x \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{7/2} \sqrt {a x^2+b x^3+c x^4}}+\frac {3 b \left (b^2-4 a c\right ) (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{128 c^3 x}-\frac {b (b+2 c x) \left (a x^2+b x^3+c x^4\right )^{3/2}}{16 c^2 x^3}+\frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5} \]

[In]

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

[Out]

(3*b*(b^2 - 4*a*c)*(b + 2*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(128*c^3*x) - (b*(b + 2*c*x)*(a*x^2 + b*x^3 + c*x^
4)^(3/2))/(16*c^2*x^3) + (a*x^2 + b*x^3 + c*x^4)^(5/2)/(5*c*x^5) - (3*b*(b^2 - 4*a*c)^2*x*Sqrt[a + b*x + c*x^2
]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(256*c^(7/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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

Rule 1928

Int[(x_)^(m_.)/Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[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 1931

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] - Dist[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 1932

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] - Dist[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] && Eq
Q[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] && EqQ[m + p*q + 1, n - q]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a x^2+b x^3+c x^4\right )^{5/2}}{5 c x^5}-\frac {b \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^3} \, dx}{2 c} \\ & = -\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 {\left (3 b \left (b^2-4 a c\right )\right ) \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x} \, dx}{32 c^2} \\ & = \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 {\left (3 b \left (b^2-4 a c\right )^2\right ) \int \frac {x}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{256 c^3} \\ & = \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 {\left (3 b \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^3 \sqrt {a x^2+b x^3+c x^4}} \\ & = \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 {\left (3 b \left (b^2-4 a c\right )^2 x \sqrt {a+b x+c x^2}\right ) \text {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{128 c^3 \sqrt {a x^2+b x^3+c x^4}} \\ & = \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} \tanh ^{-1}\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}} \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

(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))])

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	\(\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (256 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} c^{\frac {7}{2}}-160 c^{\frac {7}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b x -80 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2}-240 c^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, a b x +60 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{3} x -120 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a \,b^{2}+30 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{4}-240 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{2} b \,c^{3}+120 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,b^{3} c^{2}-15 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{5} c \right )}{1280 x^{3} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {9}{2}}}\)	\(289\)

[In]

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

[Out]

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))

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

[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 + 1
28*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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

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(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*sqrt(c) + b))/c^(7/2) - 1/1280*(15*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)

Mupad [F(-1)]

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

[In]

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

[Out]

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