\(\int \frac {(a x^2+b x^3+c x^4)^{3/2}}{x^7} \, dx\) [63]

Optimal result

Integrand size = 24, antiderivative size = 232 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=-\frac {\left (2 a b+\left (b^2+8 a c\right ) x\right ) \sqrt {a x^2+b x^3+c x^4}}{8 a x^3}-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}+\frac {b \left (b^2-12 a c\right ) \left (a x^2+b x^3+c x^4\right )^{3/2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{3/2} x^3 \left (a+b x+c x^2\right )^{3/2}}+\frac {c^{3/2} \left (a x^2+b x^3+c x^4\right )^{3/2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{x^3 \left (a+b x+c x^2\right )^{3/2}} \] Output:

-1/8*(2*a*b+(8*a*c+b^2)*x)*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^3-1/3*(c*x^4+b*x^ 
3+a*x^2)^(3/2)/x^6+1/16*b*(-12*a*c+b^2)*(c*x^4+b*x^3+a*x^2)^(3/2)*arctanh( 
1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(3/2)/x^3/(c*x^2+b*x+a)^(3/2) 
+c^(3/2)*(c*x^4+b*x^3+a*x^2)^(3/2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b* 
x+a)^(1/2))/x^3/(c*x^2+b*x+a)^(3/2)
 

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=-\frac {\sqrt {x^2 (a+x (b+c x))} \left (3 b \left (b^2-12 a c\right ) x^3 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+\sqrt {a} \left (\sqrt {a+x (b+c x)} \left (8 a^2+3 b^2 x^2+2 a x (7 b+16 c x)\right )+24 a c^{3/2} x^3 \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{24 a^{3/2} x^4 \sqrt {a+x (b+c x)}} \] Input:

Integrate[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7,x]
 

Output:

-1/24*(Sqrt[x^2*(a + x*(b + c*x))]*(3*b*(b^2 - 12*a*c)*x^3*ArcTanh[(Sqrt[c 
]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + Sqrt[a]*(Sqrt[a + x*(b + c*x)]*(8* 
a^2 + 3*b^2*x^2 + 2*a*x*(7*b + 16*c*x)) + 24*a*c^(3/2)*x^3*Log[b + 2*c*x - 
 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])))/(a^(3/2)*x^4*Sqrt[a + x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.01, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {1967, 1998, 27, 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.

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

\(\Big \downarrow \) 1967

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

\(\Big \downarrow \) 1998

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\left (b^2-2 c x b-8 a c\right ) \sqrt {c x^4+b x^3+a x^2}}{2 x^3}dx}{2 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\left (b^2-2 c x b-8 a c\right ) \sqrt {c x^4+b x^3+a x^2}}{x^3}dx}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 1988

\(\displaystyle \frac {1}{2} \left (-\frac {-\frac {1}{2} \int -\frac {b \left (b^2-12 a c\right )-16 a c^2 x}{\sqrt {c x^4+b x^3+a x^2}}dx-\frac {\sqrt {a x^2+b x^3+c x^4} \left (-8 a c+b^2+2 b c x\right )}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {1}{2} \int \frac {b \left (b^2-12 a c\right )-16 a c^2 x}{\sqrt {c x^4+b x^3+a x^2}}dx-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 1980

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \int \frac {b \left (b^2-12 a c\right )-16 a c^2 x}{x \sqrt {c x^2+b x+a}}dx}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \left (b \left (b^2-12 a c\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-16 a c^2 \int \frac {1}{\sqrt {c x^2+b x+a}}dx\right )}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 1092

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \left (b \left (b^2-12 a c\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-32 a c^2 \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}\right )}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \left (b \left (b^2-12 a c\right ) \int \frac {1}{x \sqrt {c x^2+b x+a}}dx-16 a c^{3/2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\right )}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 1154

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \left (-2 b \left (b^2-12 a c\right ) \int \frac {1}{4 a-\frac {(2 a+b x)^2}{c x^2+b x+a}}d\frac {2 a+b x}{\sqrt {c x^2+b x+a}}-16 a c^{3/2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\right )}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{2} \left (-\frac {\frac {x \sqrt {a+b x+c x^2} \left (-\frac {b \left (b^2-12 a c\right ) \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a}}-16 a c^{3/2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )\right )}{2 \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (-8 a c+b^2+2 b c x\right ) \sqrt {a x^2+b x^3+c x^4}}{x^2}}{4 a}-\frac {b \left (a x^2+b x^3+c x^4\right )^{3/2}}{2 a x^5}\right )-\frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{3 x^6}\)

Input:

Int[(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7,x]
 

Output:

-1/3*(a*x^2 + b*x^3 + c*x^4)^(3/2)/x^6 + (-1/2*(b*(a*x^2 + b*x^3 + c*x^4)^ 
(3/2))/(a*x^5) - (-(((b^2 - 8*a*c + 2*b*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/ 
x^2) + (x*Sqrt[a + b*x + c*x^2]*(-((b*(b^2 - 12*a*c)*ArcTanh[(2*a + b*x)/( 
2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/Sqrt[a]) - 16*a*c^(3/2)*ArcTanh[(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*a))/2
 

Defintions of rubi rules used

rule 25

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

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 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]
 

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 = 150, normalized size of antiderivative = 0.65

method	result	size
pseudoelliptic	\(-\frac {3 \left (b \,x^{3} \left (a c -\frac {b^{2}}{12}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )-\frac {4 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) a^{\frac {3}{2}} c^{\frac {3}{2}} x^{3}}{3}+\left (\frac {7 \left (\frac {16 c x}{7}+b \right ) x \,a^{\frac {3}{2}}}{9}+\frac {\sqrt {a}\, b^{2} x^{2}}{6}+\frac {4 a^{\frac {5}{2}}}{9}\right ) \sqrt {c \,x^{2}+b x +a}-\ln \left (2\right ) \left (a c -\frac {b^{2}}{12}\right ) b \,x^{3}\right )}{4 a^{\frac {3}{2}} x^{3}}\)	\(150\)
risch	\(-\frac {\left (32 a c \,x^{2}+3 b^{2} x^{2}+14 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{24 x^{4} a}+\frac {\left (-\frac {b \left (12 a c -b^{2}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{\sqrt {a}}+16 a \,c^{\frac {3}{2}} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{16 a x \sqrt {c \,x^{2}+b x +a}}\)	\(165\)
default	\(-\frac {\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (36 a^{\frac {5}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) c^{\frac {5}{2}} b \,x^{3}-32 c^{\frac {7}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a \,x^{4}-48 c^{\frac {7}{2}} \sqrt {c \,x^{2}+b x +a}\, a^{2} x^{4}+2 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} x^{4}-3 a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) c^{\frac {3}{2}} b^{3} x^{3}+32 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a \,x^{2}-28 c^{\frac {5}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a b \,x^{3}+6 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} x^{4}-60 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, a^{2} b \,x^{3}-2 c^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} b^{2} x^{2}+2 c^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{3} x^{3}-4 c^{\frac {3}{2}} \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a b x +6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} x^{3}+16 \left (c \,x^{2}+b x +a \right )^{\frac {5}{2}} a^{2} c^{\frac {3}{2}}-48 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a^{3} c^{3} x^{3}\right )}{48 x^{6} \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{3} c^{\frac {3}{2}}}\)	\(435\)

Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x,method=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-3/4/a^(3/2)*(b*x^3*(a*c-1/12*b^2)*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/ 
2))/x/a^(1/2))-4/3*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*a^(3/2)*c^(3/ 
2)*x^3+(7/9*(16/7*c*x+b)*x*a^(3/2)+1/6*a^(1/2)*b^2*x^2+4/9*a^(5/2))*(c*x^2 
+b*x+a)^(1/2)-ln(2)*(a*c-1/12*b^2)*b*x^3)/x^3
 

Fricas [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 815, normalized size of antiderivative = 3.51 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx =\text {Too large to display} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}{x^{7}}\, dx \] Input:

integrate((c*x**4+b*x**3+a*x**2)**(3/2)/x**7,x)
 

Output:

Integral((x**2*(a + b*x + c*x**2))**(3/2)/x**7, x)
 

Maxima [F]

\[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=\int { \frac {{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}{x^{7}} \,d x } \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^3 + a*x^2)^(3/2)/x^7, x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=\int \frac {{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}}{x^7} \,d x \] Input:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7,x)
 

Output:

int((a*x^2 + b*x^3 + c*x^4)^(3/2)/x^7, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a x^2+b x^3+c x^4\right )^{3/2}}{x^7} \, dx=\frac {-16 \sqrt {c \,x^{2}+b x +a}\, a^{3}-28 \sqrt {c \,x^{2}+b x +a}\, a^{2} b x -64 \sqrt {c \,x^{2}+b x +a}\, a^{2} c \,x^{2}-6 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2} x^{2}+36 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) a b c \,x^{3}-3 \sqrt {a}\, \mathrm {log}\left (2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}-2 a -b x \right ) b^{3} x^{3}-36 \sqrt {a}\, \mathrm {log}\left (x \right ) a b c \,x^{3}+3 \sqrt {a}\, \mathrm {log}\left (x \right ) b^{3} x^{3}+48 \sqrt {c}\, \mathrm {log}\left (-2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}-b -2 c x \right ) a^{2} c \,x^{3}}{48 a^{2} x^{3}} \] Input:

int((c*x^4+b*x^3+a*x^2)^(3/2)/x^7,x)
 

Output:

( - 16*sqrt(a + b*x + c*x**2)*a**3 - 28*sqrt(a + b*x + c*x**2)*a**2*b*x - 
64*sqrt(a + b*x + c*x**2)*a**2*c*x**2 - 6*sqrt(a + b*x + c*x**2)*a*b**2*x* 
*2 + 36*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*a*b*c*x* 
*3 - 3*sqrt(a)*log(2*sqrt(a)*sqrt(a + b*x + c*x**2) - 2*a - b*x)*b**3*x**3 
 - 36*sqrt(a)*log(x)*a*b*c*x**3 + 3*sqrt(a)*log(x)*b**3*x**3 + 48*sqrt(c)* 
log( - 2*sqrt(c)*sqrt(a + b*x + c*x**2) - b - 2*c*x)*a**2*c*x**3)/(48*a**2 
*x**3)