\(\int \frac {x^2}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [60]

Optimal result

Integrand size = 24, antiderivative size = 94 \[ \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {\text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{a^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1936, 1918, 212} \[ \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 x \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {\text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{a^{3/2}} \]

[In]

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

[Out]

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

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 1936

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(-x^(m - q + 1
))*(b^2 - 2*a*c + b*c*x^(n - q))*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/(a*(n - q)*(p + 1)*(b^2 - 4*a*c))),
x] + Dist[(2*a*c - b^2*(p + 2))/(a*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - 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] && LtQ[p, -1] && RationalQ[m, p, q] && EqQ[m + p*q + 1, (-(n - q))*(2*p + 3)]

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.15 \[ \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 x \left (\sqrt {a} \left (b^2-2 a c+b c x\right )+\left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )\right )}{a^{3/2} \left (-b^2+4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.14

method	result	size
pseudoelliptic	\(-\frac {4 \left (-a^{\frac {3}{2}} c +\frac {b \sqrt {a}\, \left (c x +b \right )}{2}+\left (-\ln \left (2\right )+\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )\right ) \sqrt {c \,x^{2}+b x +a}\, \left (a c -\frac {b^{2}}{4}\right )\right )}{\sqrt {c \,x^{2}+b x +a}\, a^{\frac {3}{2}} \left (4 a c -b^{2}\right )}\)	\(107\)
default	\(\frac {x^{3} \left (c \,x^{2}+b x +a \right ) \left (-2 a^{\frac {3}{2}} b c x +4 a^{\frac {5}{2}} c -2 a^{\frac {3}{2}} b^{2}-4 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2} c +\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a \,b^{2}\right )}{\left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {5}{2}} \left (4 a c -b^{2}\right )}\)	\(164\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 201 vs. \(2 (84) = 168\).

Time = 0.31 (sec) , antiderivative size = 411, normalized size of antiderivative = 4.37 \[ \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {a} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c x + a b^{2} - 2 \, a^{2} c\right )}}{2 \, {\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{3} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}, \frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} + {\left (b^{3} - 4 \, a b c\right )} x^{2} + {\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a b c x + a b^{2} - 2 \, a^{2} c\right )}}{{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{3} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2} + {\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x}\right ] \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (84) = 168\).

Time = 0.34 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.12 \[ \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=-\frac {2 \, {\left (a b^{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) - 4 \, a^{2} c \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + \sqrt {-a} \sqrt {a} b^{2} - 2 \, \sqrt {-a} a^{\frac {3}{2}} c\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{2} b^{2} - 4 \, \sqrt {-a} a^{3} c} + \frac {2 \, {\left (\frac {a b c x \mathrm {sgn}\left (x\right )}{a^{2} b^{2} - 4 \, a^{3} c} + \frac {a b^{2} \mathrm {sgn}\left (x\right ) - 2 \, a^{2} c \mathrm {sgn}\left (x\right )}{a^{2} b^{2} - 4 \, a^{3} c}\right )}}{\sqrt {c x^{2} + b x + a}} + \frac {2 \, \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-2*(a*b^2*arctan(sqrt(a)/sqrt(-a)) - 4*a^2*c*arctan(sqrt(a)/sqrt(-a)) + sqrt(-a)*sqrt(a)*b^2 - 2*sqrt(-a)*a^(3
/2)*c)*sgn(x)/(sqrt(-a)*a^2*b^2 - 4*sqrt(-a)*a^3*c) + 2*(a*b*c*x*sgn(x)/(a^2*b^2 - 4*a^3*c) + (a*b^2*sgn(x) -
2*a^2*c*sgn(x))/(a^2*b^2 - 4*a^3*c))/sqrt(c*x^2 + b*x + a) + 2*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqr
t(-a))/(sqrt(-a)*a*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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