\(\int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx\) [36]

Optimal result

Integrand size = 24, antiderivative size = 155 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{5/2}} \]

[Out]

-1/16*b*(-4*a*c+b^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)-1/3*(c*x^4+b*x^3+a*x^2
)^(1/2)/x^4-1/12*b*(c*x^4+b*x^3+a*x^2)^(1/2)/a/x^3+1/24*(-8*a*c+3*b^2)*(c*x^4+b*x^3+a*x^2)^(1/2)/a^2/x^2

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 155, 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 = {1934, 1965, 12, 1918, 212} \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=-\frac {b \left (b^2-4 a c\right ) \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{5/2}}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}-\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4} \]

[In]

Int[Sqrt[a*x^2 + b*x^3 + c*x^4]/x^5,x]

[Out]

-1/3*Sqrt[a*x^2 + b*x^3 + c*x^4]/x^4 - (b*Sqrt[a*x^2 + b*x^3 + c*x^4])/(12*a*x^3) + ((3*b^2 - 8*a*c)*Sqrt[a*x^
2 + b*x^3 + c*x^4])/(24*a^2*x^2) - (b*(b^2 - 4*a*c)*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*
x^4])])/(16*a^(5/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 1918

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-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 1934

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] - Dist[(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 1965

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> 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] + Dist[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] && Eq
Q[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}+\frac {1}{6} \int \frac {b+2 c x}{x^2 \sqrt {a x^2+b x^3+c x^4}} \, dx \\ & = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}-\frac {\int \frac {\frac {1}{2} \left (3 b^2-8 a c\right )+b c x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{12 a} \\ & = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}+\frac {\int \frac {3 b \left (b^2-4 a c\right )}{4 \sqrt {a x^2+b x^3+c x^4}} \, dx}{12 a^2} \\ & = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}+\frac {\left (b \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{16 a^2} \\ & = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {\left (b \left (b^2-4 a c\right )\right ) \text {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {x (2 a+b x)}{\sqrt {a x^2+b x^3+c x^4}}\right )}{8 a^2} \\ & = -\frac {\sqrt {a x^2+b x^3+c x^4}}{3 x^4}-\frac {b \sqrt {a x^2+b x^3+c x^4}}{12 a x^3}+\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{24 a^2 x^2}-\frac {b \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{16 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\frac {\sqrt {x^2 (a+x (b+c x))} \left (\sqrt {a} \sqrt {a+x (b+c x)} \left (-8 a^2+3 b^2 x^2-2 a x (b+4 c x)\right )+3 b \left (b^2-4 a c\right ) x^3 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )}{24 a^{5/2} x^4 \sqrt {a+x (b+c x)}} \]

[In]

Integrate[Sqrt[a*x^2 + b*x^3 + c*x^4]/x^5,x]

[Out]

(Sqrt[x^2*(a + x*(b + c*x))]*(Sqrt[a]*Sqrt[a + x*(b + c*x)]*(-8*a^2 + 3*b^2*x^2 - 2*a*x*(b + 4*c*x)) + 3*b*(b^
2 - 4*a*c)*x^3*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]]))/(24*a^(5/2)*x^4*Sqrt[a + x*(b + c*x)])

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {b \,x^{3} \left (a c -\frac {b^{2}}{4}\right ) \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )+\left (-\frac {x \left (4 c x +b \right ) a^{\frac {3}{2}}}{3}+\frac {\sqrt {a}\, b^{2} x^{2}}{2}-\frac {4 a^{\frac {5}{2}}}{3}\right ) \sqrt {c \,x^{2}+b x +a}-\ln \left (2\right ) x^{3} b \left (a c -\frac {b^{2}}{4}\right )}{4 a^{\frac {5}{2}} x^{3}}\)	\(115\)
risch	\(-\frac {\left (8 a c \,x^{2}-3 b^{2} x^{2}+2 a b x +8 a^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{24 x^{4} a^{2}}+\frac {\left (4 a c -b^{2}\right ) b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{16 a^{\frac {5}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(128\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (12 c \,a^{\frac {3}{2}} \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b \,x^{3}+6 c \sqrt {c \,x^{2}+b x +a}\, b^{2} x^{4}-12 c \sqrt {c \,x^{2}+b x +a}\, a b \,x^{3}-3 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) b^{3} x^{3}-6 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b^{2} x^{2}+6 \sqrt {c \,x^{2}+b x +a}\, b^{3} x^{3}+12 a \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b x -16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} a^{2}\right )}{48 x^{4} \sqrt {c \,x^{2}+b x +a}\, a^{3}}\)	\(234\)

[In]

int((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 272, normalized size of antiderivative = 1.75 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {a} x^{4} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, a^{2} b x + 8 \, a^{3} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{2}\right )}}{96 \, a^{3} x^{4}}, \frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-a} x^{4} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, a^{2} b x + 8 \, a^{3} - {\left (3 \, a b^{2} - 8 \, a^{2} c\right )} x^{2}\right )}}{48 \, a^{3} x^{4}}\right ] \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\int \frac {\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}{x^{5}}\, dx \]

[In]

integrate((c*x**4+b*x**3+a*x**2)**(1/2)/x**5,x)

[Out]

Integral(sqrt(x**2*(a + b*x + c*x**2))/x**5, x)

Maxima [F]

\[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\int { \frac {\sqrt {c x^{4} + b x^{3} + a x^{2}}}{x^{5}} \,d x } \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^3 + a*x^2)/x^5, x)

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((c*x^4+b*x^3+a*x^2)^(1/2)/x^5,x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Not invertible Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^5} \, dx=\int \frac {\sqrt {c\,x^4+b\,x^3+a\,x^2}}{x^5} \,d x \]

[In]

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^5,x)

[Out]

int((a*x^2 + b*x^3 + c*x^4)^(1/2)/x^5, x)