\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^8} \, dx\) [64]

Optimal result

Integrand size = 24, antiderivative size = 197 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{32 a x^3}+\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{64 a^2 x^2}-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{8 x^4}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}-\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{128 a^{5/2}} \] Output:

-1/32*(-12*a*c+b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^3+1/64*b*(-20*a*c+3*b^2) 
*(c*x^4+b*x^3+a*x^2)^(1/2)/a^2/x^2-1/8*(6*c*x+b)*(c*x^4+b*x^3+a*x^2)^(1/2) 
/x^4-1/4*(c*x^4+b*x^3+a*x^2)^(3/2)/x^7-3/128*(-4*a*c+b^2)^2*arctanh(1/2*x* 
(b*x+2*a)/a^(1/2)/(c*x^4+b*x^3+a*x^2)^(1/2))/a^(5/2)
 

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.72 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\frac {\sqrt {x^2 (a+x (b+c x))} \left (-\sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)} \left (8 a^2-3 b^2 x^2+4 a x (2 b+5 c x)\right )+3 \left (b^2-4 a c\right )^2 x^4 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )}{64 a^{5/2} x^5 \sqrt {a+x (b+c x)}} \] Input:

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x]
 

Output:

(Sqrt[x^2*(a + x*(b + c*x))]*(-(Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)]* 
(8*a^2 - 3*b^2*x^2 + 4*a*x*(2*b + 5*c*x))) + 3*(b^2 - 4*a*c)^2*x^4*ArcTanh 
[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]]))/(64*a^(5/2)*x^5*Sqrt[a + x 
*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1967, 1988, 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.

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

\(\Big \downarrow \) 1967

\(\displaystyle \frac {3}{8} \int \frac {(b+2 c x) \sqrt {c x^4+b x^3+a x^2}}{x^5}dx-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 1988

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \int \frac {b^2-4 c x b-12 a c}{x^2 \sqrt {c x^4+b x^3+a x^2}}dx-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 1998

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {\int \frac {b \left (3 b^2-20 a c\right )+2 c \left (b^2-12 a c\right ) x}{2 x \sqrt {c x^4+b x^3+a x^2}}dx}{2 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {\int \frac {b \left (3 b^2-20 a c\right )+2 c \left (b^2-12 a c\right ) x}{x \sqrt {c x^4+b x^3+a x^2}}dx}{4 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 1998

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {-\frac {\int \frac {3 \left (b^2-4 a c\right )^2}{2 \sqrt {c x^4+b x^3+a x^2}}dx}{a}-\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a x^2}}{4 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {-\frac {3 \left (b^2-4 a c\right )^2 \int \frac {1}{\sqrt {c x^4+b x^3+a x^2}}dx}{2 a}-\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a x^2}}{4 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 1951

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {\frac {3 \left (b^2-4 a c\right )^2 \int \frac {1}{4 a-\frac {x^2 (2 a+b x)^2}{c x^4+b x^3+a x^2}}d\frac {x (2 a+b x)}{\sqrt {c x^4+b x^3+a x^2}}}{a}-\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a x^2}}{4 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {3}{8} \left (\frac {1}{6} \left (-\frac {\frac {3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{3/2}}-\frac {b \left (3 b^2-20 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a x^2}}{4 a}-\frac {\left (b^2-12 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{2 a x^3}\right )-\frac {(b+6 c x) \sqrt {a x^2+b x^3+c x^4}}{3 x^4}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{4 x^7}\)

Input:

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x]
 

Output:

-1/4*(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7 + (3*(-1/3*((b + 6*c*x)*Sqrt[a*x^2 
+ b*x^3 + c*x^4])/x^4 + (-1/2*((b^2 - 12*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]) 
/(a*x^3) - (-((b*(3*b^2 - 20*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(a*x^2)) + 
(3*(b^2 - 4*a*c)^2*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))/6))/8
 

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 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 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]
 

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 = 0.22 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(-\frac {3 \left (x^{4} \left (a c -\frac {b^{2}}{4}\right )^{2} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )+\left (\frac {b \,x^{2} \left (10 c x +b \right ) a^{\frac {3}{2}}}{12}+x \left (\frac {5 c x}{3}+b \right ) a^{\frac {5}{2}}-\frac {\sqrt {a}\, b^{3} x^{3}}{8}+\frac {2 a^{\frac {7}{2}}}{3}\right ) \sqrt {c \,x^{2}+b x +a}-\ln \left (2\right ) x^{4} \left (a c -\frac {b^{2}}{4}\right )^{2}\right )}{8 a^{\frac {5}{2}} x^{4}}\)	\(131\)
risch	\(-\frac {\left (20 a b c \,x^{3}-3 b^{3} x^{3}+40 a^{2} c \,x^{2}+2 a \,b^{2} x^{2}+24 b \,a^{2} x +16 a^{3}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{64 x^{5} a^{2}}-\frac {3 \left (16 a^{2} c^{2}-8 a \,b^{2} c +b^{4}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{128 a^{\frac {5}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(157\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (48 c^{2} a^{\frac {7}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) x^{4}+24 c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b \,x^{5}-24 c \,a^{\frac {5}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{2} x^{4}-16 c^{2} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2} x^{4}+24 c^{2} \sqrt {c \,x^{2}+b x +a}\, a^{2} b \,x^{5}-2 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} x^{5}-48 c^{2} \sqrt {c \,x^{2}+b x +a}\, a^{3} x^{4}-24 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a b \,x^{3}+20 c \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,b^{2} x^{4}-6 c \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} x^{5}+3 a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{4} x^{4}+16 c \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} x^{2}+36 c \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} x^{4}+2 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{3} x^{3}-2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{4} x^{4}+4 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,b^{2} x^{2}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{4} x^{4}-16 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} b x +32 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{3}\right )}{128 x^{7} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{4}}\)	\(501\)

Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x,method=_RETURNVERBOSE)
 

Output:

-3/8/a^(5/2)*(x^4*(a*c-1/4*b^2)^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2 
))/x/a^(1/2))+(1/12*b*x^2*(10*c*x+b)*a^(3/2)+x*(5/3*c*x+b)*a^(5/2)-1/8*a^( 
1/2)*b^3*x^3+2/3*a^(7/2))*(c*x^2+b*x+a)^(1/2)-ln(2)*x^4*(a*c-1/4*b^2)^2)/x 
^4
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.69 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\left [\frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {a} x^{5} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, {\left (24 \, a^{3} b x + 16 \, a^{4} - {\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{3} + 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{256 \, a^{3} x^{5}}, \frac {3 \, {\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 2 \, {\left (24 \, a^{3} b x + 16 \, a^{4} - {\left (3 \, a b^{3} - 20 \, a^{2} b c\right )} x^{3} + 2 \, {\left (a^{2} b^{2} + 20 \, a^{3} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{128 \, a^{3} x^{5}}\right ] \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="fricas")
 

Output:

[1/256*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*sqrt(a)*x^5*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*(24*a^3*b*x + 16*a^4 - (3*a*b^3 - 20*a^2*b*c)*x^3 + 2*(a^2*b^ 
2 + 20*a^3*c)*x^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(a^3*x^5), 1/128*(3*(b^4 - 
 8*a*b^2*c + 16*a^2*c^2)*sqrt(-a)*x^5*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*(24*a^3*b*x + 16* 
a^4 - (3*a*b^3 - 20*a^2*b*c)*x^3 + 2*(a^2*b^2 + 20*a^3*c)*x^2)*sqrt(c*x^4 
+ b*x^3 + a*x^2))/(a^3*x^5)]
 

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{8}}\, dx \] Input:

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**8,x)
 

Output:

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**8, x)
 

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\int { \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{8}} \,d x } \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^8, x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\text {Timed out} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^8} \,d x \] Input:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8,x)
 

Output:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^8, x)
 

Reduce [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.39 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^8} \, dx=\frac {-32 \sqrt {c \,x^{2}+b x +a}\, a^{4}-48 \sqrt {c \,x^{2}+b x +a}\, a^{3} b x -80 \sqrt {c \,x^{2}+b x +a}\, a^{3} c \,x^{2}-4 \sqrt {c \,x^{2}+b x +a}\, a^{2} b^{2} x^{2}-40 \sqrt {c \,x^{2}+b x +a}\, a^{2} b c \,x^{3}+6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} x^{3}+48 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a^{2} c^{2} x^{4}-24 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a \,b^{2} c \,x^{4}+3 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{4} x^{4}-48 \sqrt {a}\, \mathrm {log}\left (x \right ) a^{2} c^{2} x^{4}+24 \sqrt {a}\, \mathrm {log}\left (x \right ) a \,b^{2} c \,x^{4}-3 \sqrt {a}\, \mathrm {log}\left (x \right ) b^{4} x^{4}}{128 a^{3} x^{4}} \] Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^8,x)
 

Output:

( - 32*sqrt(a + b*x + c*x**2)*a**4 - 48*sqrt(a + b*x + c*x**2)*a**3*b*x - 
80*sqrt(a + b*x + c*x**2)*a**3*c*x**2 - 4*sqrt(a + b*x + c*x**2)*a**2*b**2 
*x**2 - 40*sqrt(a + b*x + c*x**2)*a**2*b*c*x**3 + 6*sqrt(a + b*x + c*x**2) 
*a*b**3*x**3 + 48*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x 
)*a**2*c**2*x**4 - 24*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - 
 b*x)*a*b**2*c*x**4 + 3*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a 
 - b*x)*b**4*x**4 - 48*sqrt(a)*log(x)*a**2*c**2*x**4 + 24*sqrt(a)*log(x)*a 
*b**2*c*x**4 - 3*sqrt(a)*log(x)*b**4*x**4)/(128*a**3*x**4)