\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^6} \, dx\) [62]

Optimal result

Integrand size = 24, antiderivative size = 219 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=-\frac {3 (b-2 c x) \sqrt {a x^2+b x^3+c x^4}}{4 x^2}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{2 x^5}-\frac {3 \left (b^2+4 a c\right ) x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}+\frac {3 b \sqrt {c} x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {a x^2+b x^3+c x^4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.73 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\frac {\sqrt {x^2 (a+x (b+c x))} \left (3 \left (b^2+4 a c\right ) x^2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\sqrt {a} \left ((2 a+x (5 b-4 c x)) \sqrt {a+x (b+c x)}+6 b \sqrt {c} x^2 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{4 \sqrt {a} x^3 \sqrt {a+x (b+c x)}} \] Input:

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6,x]
 

Output:

(Sqrt[x^2*(a + x*(b + c*x))]*(3*(b^2 + 4*a*c)*x^2*ArcTanh[(Sqrt[c]*x - Sqr 
t[a + x*(b + c*x)])/Sqrt[a]] - Sqrt[a]*((2*a + x*(5*b - 4*c*x))*Sqrt[a + x 
*(b + c*x)] + 6*b*Sqrt[c]*x^2*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c* 
x)]])))/(4*Sqrt[a]*x^3*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.86, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1967, 1988, 25, 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.

\(\displaystyle \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx\)

\(\Big \downarrow \) 1967

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

\(\Big \downarrow \) 1988

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1980

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

\(\Big \downarrow \) 1269

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6,x]
 

Output:

-1/2*(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^5 + (3*(-(((b - 2*c*x)*Sqrt[a*x^2 + b 
*x^3 + c*x^4])/x^2) + (x*Sqrt[a + b*x + c*x^2]*(-(((b^2 + 4*a*c)*ArcTanh[( 
2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/Sqrt[a]) + 4*b*Sqrt[c]*ArcT 
anh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])]))/(2*Sqrt[a*x^2 + b*x^3 
 + c*x^4])))/4
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 1967

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*q 
+ 1)), x] - Simp[(n - q)*(p/(m + p*q + 1))   Int[x^(m + n)*(b + 2*c*x^(n - 
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] && LeQ[m + p*q + 1, -(n - q) + 
 1] && NeQ[m + p*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 1988

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_ 
.)*((A_) + (B_.)*(x_)^(r_.)), x_Symbol] :> Simp[x^(m + 1)*(A*(m + p*q + (n 
- q)*(2*p + 1) + 1) + B*(m + p*q + 1)*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n 
 - q))^p/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1))), x] + Simp[(n - 
 q)*(p/((m + p*q + 1)*(m + p*q + (n - q)*(2*p + 1) + 1)))   Int[x^(n + m)*S 
imp[2*a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(2*p + 1) + 1) + (b*B*(m + 
 p*q + 1) - 2*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 - 1), 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] && IG 
tQ[n, 0] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q, -(n - q)] && NeQ[m 
 + p*q + 1, 0] && NeQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]
 

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.64

method	result	size
pseudoelliptic	\(-\frac {3 \left (x^{2} \left (a c +\frac {b^{2}}{4}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )-\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b \,x^{2} \sqrt {a}\, \sqrt {c}+\frac {\left (a^{\frac {3}{2}}+\left (-2 c \,x^{2}+\frac {5}{2} b x \right ) \sqrt {a}\right ) \sqrt {c \,x^{2}+b x +a}}{3}-\ln \left (2\right ) x^{2} \left (a c +\frac {b^{2}}{4}\right )\right )}{2 \sqrt {a}\, x^{2}}\)	\(140\)
risch	\(-\frac {\left (5 b x +2 a \right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{4 x^{3}}+\frac {\left (-\frac {3 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) c}{2}-\frac {3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{2}}{8 \sqrt {a}}+\frac {3 b \sqrt {c}\, \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2}+c \sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{x \sqrt {c \,x^{2}+b x +a}}\)	\(181\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (12 c^{\frac {5}{2}} a^{\frac {5}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) x^{2}-2 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b \,x^{3}-4 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,x^{2}-6 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a b \,x^{3}+3 c^{\frac {3}{2}} a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{2} x^{2}-12 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a^{2} x^{2}+2 c^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b x -2 c^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} x^{2}+4 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,c^{\frac {3}{2}}-6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} x^{2}-12 c^{2} \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{2} b \,x^{2}\right )}{8 x^{5} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} c^{\frac {3}{2}}}\)	\(338\)

Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x,method=_RETURNVERBOSE)
 

Output:

-3/2/a^(1/2)*(x^2*(a*c+1/4*b^2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2)) 
/x/a^(1/2))-ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*b*x^2*a^(1/2)*c^(1/2 
)+1/3*(a^(3/2)+(-2*c*x^2+5/2*b*x)*a^(1/2))*(c*x^2+b*x+a)^(1/2)-ln(2)*x^2*( 
a*c+1/4*b^2))/x^2
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 757, normalized size of antiderivative = 3.46 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx =\text {Too large to display} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="fricas")
 

Output:

[1/16*(12*a*b*sqrt(c)*x^3*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) + 3*(b^2 + 4*a*c)*sq 
rt(a)*x^3*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(4 
*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^3), -1/16*(24*a*b*sqrt(-c)*x^3*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)) - 3*(b^2 + 4*a*c)*sqrt(a)*x^3*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*sqrt 
(c*x^4 + b*x^3 + a*x^2)*(4*a*c*x^2 - 5*a*b*x - 2*a^2))/(a*x^3), 1/8*(6*a*b 
*sqrt(c)*x^3*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) + 3*(b^2 + 4*a*c)*sqrt(-a)*x^3*ar 
ctan(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*sqrt(c*x^4 + b*x^3 + a*x^2)*(4*a*c*x^2 - 5*a*b*x - 2*a^2) 
)/(a*x^3), -1/8*(12*a*b*sqrt(-c)*x^3*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)) - 3*(b^2 + 4*a*c)*sqrt 
(-a)*x^3*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*sqrt(c*x^4 + b*x^3 + a*x^2)*(4*a*c*x^2 - 5*a*b 
*x - 2*a^2))/(a*x^3)]
 

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{6}}\, dx \] Input:

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**6,x)
 

Output:

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**6, x)
 

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\int { \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{6}} \,d x } \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^6, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\text {Timed out} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^6} \,d x \] Input:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6,x)
                                                                                    
                                                                                    
 

Output:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^6} \, dx=\frac {-4 \sqrt {c \,x^{2}+b x +a}\, a^{2}-10 \sqrt {c \,x^{2}+b x +a}\, a b x +8 \sqrt {c \,x^{2}+b x +a}\, a c \,x^{2}+12 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a c \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{2} x^{2}-12 \sqrt {a}\, \mathrm {log}\left (x \right ) a c \,x^{2}-3 \sqrt {a}\, \mathrm {log}\left (x \right ) b^{2} x^{2}+12 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a b \,x^{2}}{8 a \,x^{2}} \] Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^6,x)
 

Output:

( - 4*sqrt(a + b*x + c*x**2)*a**2 - 10*sqrt(a + b*x + c*x**2)*a*b*x + 8*sq 
rt(a + b*x + c*x**2)*a*c*x**2 + 12*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c* 
x**2) - 2*a - b*x)*a*c*x**2 + 3*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x** 
2) - 2*a - b*x)*b**2*x**2 - 12*sqrt(a)*log(x)*a*c*x**2 - 3*sqrt(a)*log(x)* 
b**2*x**2 + 12*sqrt(c)*log( - 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x 
)*a*b*x**2)/(8*a*x**2)