Integrand size = 24, antiderivative size = 75 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b \text {arctanh}\left (\frac {x (b+2 c x)}{2 \sqrt {c} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 c^{3/2}} \] Output:
(c*x^4+b*x^3+a*x^2)^(1/2)/c/x-1/2*b*arctanh(1/2*x*(2*c*x+b)/c^(1/2)/(c*x^4 +b*x^3+a*x^2)^(1/2))/c^(3/2)
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \left (2 \sqrt {c} (a+x (b+c x))-b \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )\right )}{2 c^{3/2} \sqrt {x^2 (a+x (b+c x))}} \] Input:
Integrate[x^2/Sqrt[a*x^2 + b*x^3 + c*x^4],x]
Output:
(x*(2*Sqrt[c]*(a + x*(b + c*x)) - b*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c *x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(2*c^(3/2)*Sqrt[x^2*(a + x*(b + c *x))])
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.37, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1964, 1961, 1092, 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.
\(\displaystyle \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx\) |
\(\Big \downarrow \) 1964 |
\(\displaystyle \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b \int \frac {x}{\sqrt {c x^4+b x^3+a x^2}}dx}{2 c}\) |
\(\Big \downarrow \) 1961 |
\(\displaystyle \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{2 c \sqrt {a x^2+b x^3+c x^4}}\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c \sqrt {a x^2+b x^3+c x^4}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a x^2+b x^3+c x^4}}{c x}-\frac {b x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2} \sqrt {a x^2+b x^3+c x^4}}\) |
Input:
Int[x^2/Sqrt[a*x^2 + b*x^3 + c*x^4],x]
Output:
Sqrt[a*x^2 + b*x^3 + c*x^4]/(c*x) - (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*c^(3/2)*Sqrt[a*x^2 + b*x^3 + c*x^4])
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])
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]
Int[(x_)^(m_.)/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[x^(m - q/2)/Sqrt[a + b*x^(n - q) + c*x ^(2*(n - q))], x], x] /; FreeQ[{a, b, c, m, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q] && ((EqQ[m, 1] && EqQ[n, 3] && EqQ[q, 2]) || ((EqQ[m + 1/2] || EqQ[m, 3/2] || EqQ[m, 1/2] || EqQ[m, 5/2]) && EqQ[n, 3] && EqQ[q, 1]))
Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_ ), x_Symbol] :> Simp[x^(m - n)*((a*x^(n - 1) + b*x^n + c*x^(n + 1))^(p + 1) /(2*c*(p + 1))), x] - Simp[b/(2*c) Int[x^(m - 1)*(a*x^(n - 1) + b*x^n + c *x^(n + 1))^p, 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] && RationalQ[m, p , q] && EqQ[m + p*(n - 1) - 1, 0]
Time = 0.13 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.67
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b -2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 c^{\frac {3}{2}}}\) | \(50\) |
default | \(\frac {x \sqrt {c \,x^{2}+b x +a}\, \left (2 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {3}{2}}-b \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) c \right )}{2 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, c^{\frac {5}{2}}}\) | \(88\) |
risch | \(\frac {\left (c \,x^{2}+b x +a \right ) x}{c \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) x \sqrt {c \,x^{2}+b x +a}}{2 c^{\frac {3}{2}} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\) | \(93\) |
Input:
int(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-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))/c^(3/2)
Time = 0.08 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.51 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\left [\frac {b \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{4 \, c^{2} x}, \frac {b \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} c}{2 \, c^{2} x}\right ] \] Input:
integrate(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="fricas")
Output:
[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) + 4*sqrt(c*x^4 + b*x^3 + a*x ^2)*c)/(c^2*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)) + 2*sqrt(c*x^4 + b*x^3 + a*x^2)*c)/(c^2*x)]
\[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int \frac {x^{2}}{\sqrt {x^{2} \left (a + b x + c x^{2}\right )}}\, dx \] Input:
integrate(x**2/(c*x**4+b*x**3+a*x**2)**(1/2),x)
Output:
Integral(x**2/sqrt(x**2*(a + b*x + c*x**2)), x)
\[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {c x^{4} + b x^{3} + a x^{2}}} \,d x } \] Input:
integrate(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="maxima")
Output:
integrate(x^2/sqrt(c*x^4 + b*x^3 + a*x^2), x)
Time = 0.15 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=-\frac {{\left (b \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 2 \, \sqrt {a} \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{2 \, c^{\frac {3}{2}}} + \frac {b \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{2 \, c^{\frac {3}{2}} \mathrm {sgn}\left (x\right )} + \frac {\sqrt {c x^{2} + b x + a}}{c \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x, algorithm="giac")
Output:
-1/2*(b*log(abs(b - 2*sqrt(a)*sqrt(c))) + 2*sqrt(a)*sqrt(c))*sgn(x)/c^(3/2 ) + 1/2*b*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^( 3/2)*sgn(x)) + sqrt(c*x^2 + b*x + a)/(c*sgn(x))
Timed out. \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\int \frac {x^2}{\sqrt {c\,x^4+b\,x^3+a\,x^2}} \,d x \] Input:
int(x^2/(a*x^2 + b*x^3 + c*x^4)^(1/2),x)
Output:
int(x^2/(a*x^2 + b*x^3 + c*x^4)^(1/2), x)
Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.81 \[ \int \frac {x^2}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {2 \sqrt {c \,x^{2}+b x +a}\, c -\sqrt {c}\, \mathrm {log}\left (\frac {2 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}+b +2 c x}{\sqrt {4 a c -b^{2}}}\right ) b}{2 c^{2}} \] Input:
int(x^2/(c*x^4+b*x^3+a*x^2)^(1/2),x)
Output:
(2*sqrt(a + b*x + c*x**2)*c - sqrt(c)*log((2*sqrt(c)*sqrt(a + b*x + c*x**2 ) + b + 2*c*x)/sqrt(4*a*c - b**2))*b)/(2*c**2)