\(\int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx\) [33]

Optimal result

Integrand size = 24, antiderivative size = 173 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx=\frac {\sqrt {a x^2+b x^3+c x^4}}{x}-\frac {\sqrt {a} x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}+\frac {b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1935, 1947, 857, 635, 212, 738} \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx=-\frac {\sqrt {a} x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{\sqrt {a x^2+b x^3+c x^4}}+\frac {b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}}+\frac {\sqrt {a x^2+b x^3+c x^4}}{x} \]

[In]

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

[Out]

Sqrt[a*x^2 + b*x^3 + c*x^4]/x - (Sqrt[a]*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x +
c*x^2])])/Sqrt[a*x^2 + b*x^3 + c*x^4] + (b*x*Sqrt[a + b*x + c*x^2]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*
x + c*x^2])])/(2*Sqrt[c]*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 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-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] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1935

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*(2*n - q) + 1)), x] + Dist[(n - q)*(p/(m + p*(2*n - q) + 1)), Int[x^(m +
q)*(2*a + b*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] && Gt
Q[m + p*q + 1, -(n - q)] && NeQ[m + p*(2*n - q) + 1, 0]

Rule 1947

Int[((A_) + (B_.)*(x_)^(j_.))/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[(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]

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx=\frac {x \sqrt {a+x (b+c x)} \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+4 \sqrt {a} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-b \log \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{2 \sqrt {c} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

(x*Sqrt[a + x*(b + c*x)]*(2*Sqrt[c]*Sqrt[a + x*(b + c*x)] + 4*Sqrt[a]*Sqrt[c]*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*
(b + c*x)])/Sqrt[a]] - b*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]]))/(2*Sqrt[c]*Sqrt[x^2*(a + x*(b + c*
x))])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.55

method	result	size
pseudoelliptic	\(\frac {-2 \left (-\ln \left (2\right )+\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )\right ) \sqrt {c}\, \sqrt {a}+\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\)	\(95\)
default	\(-\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (2 \sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c}-b \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right )-2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}\right )}{2 x \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}\)	\(126\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 638, normalized size of antiderivative = 3.69 \[ \int \frac {\sqrt {a x^2+b x^3+c x^4}}{x^2} \, dx=\left [\frac {b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 2 \, \sqrt {a} c x \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{4 \, c x}, -\frac {b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - \sqrt {a} c x \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{2 \, c x}, \frac {4 \, \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{4 \, c x}, \frac {2 \, \sqrt {-a} c x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) - b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{2 \, c x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Error: Bad Argument Type

Mupad [F(-1)]

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

[In]

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

[Out]

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