\(\int \frac {x}{(a x^2+b x^3+c x^4)^{3/2}} \, dx\) [61]

Optimal result

Integrand size = 22, antiderivative size = 144 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {2 \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 \left (b^2-4 a c\right ) x^2}+\frac {3 b \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1938, 1965, 12, 1918, 212} \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {3 b \text {arctanh}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{5/2}}-\frac {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 x^2 \left (b^2-4 a c\right )}+\frac {2 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}} \]

[In]

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

[Out]

(2*(b^2 - 2*a*c + b*c*x))/(a*(b^2 - 4*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]) - ((3*b^2 - 8*a*c)*Sqrt[a*x^2 + b*x^3
+ c*x^4])/(a^2*(b^2 - 4*a*c)*x^2) + (3*b*ArcTanh[(x*(2*a + b*x))/(2*Sqrt[a]*Sqrt[a*x^2 + b*x^3 + c*x^4])])/(2*
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 1938

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[1/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - q)*(b^2*(m + p*q + (n - q)*(p + 1) + 1) - 2*a*c*(m +
p*q + 2*(n - q)*(p + 1) + 1) + b*c*(m + p*q + (n - q)*(2*p + 3) + 1)*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] && LtQ[p, -1] && RationalQ[m, q] && LtQ[m + p*q + 1, n - q]

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 {2 \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 \int \frac {-\frac {3 b^2}{2}+4 a c-b c x}{x \sqrt {a x^2+b x^3+c x^4}} \, dx}{a \left (b^2-4 a c\right )} \\ & = \frac {2 \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 {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 \left (b^2-4 a c\right ) x^2}+\frac {2 \int -\frac {3 b \left (b^2-4 a c\right )}{4 \sqrt {a x^2+b x^3+c x^4}} \, dx}{a^2 \left (b^2-4 a c\right )} \\ & = \frac {2 \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 {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 \left (b^2-4 a c\right ) x^2}-\frac {(3 b) \int \frac {1}{\sqrt {a x^2+b x^3+c x^4}} \, dx}{2 a^2} \\ & = \frac {2 \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 {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 \left (b^2-4 a c\right ) x^2}+\frac {(3 b) \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^2} \\ & = \frac {2 \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 {\left (3 b^2-8 a c\right ) \sqrt {a x^2+b x^3+c x^4}}{a^2 \left (b^2-4 a c\right ) x^2}+\frac {3 b \tanh ^{-1}\left (\frac {x (2 a+b x)}{2 \sqrt {a} \sqrt {a x^2+b x^3+c x^4}}\right )}{2 a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-4 a^2 c+3 b^2 x (b+c x)+a \left (b^2-10 b c x-8 c^2 x^2\right )\right )+3 b \left (b^2-4 a c\right ) x \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{a^{5/2} \left (-b^2+4 a c\right ) \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

(Sqrt[a]*(-4*a^2*c + 3*b^2*x*(b + c*x) + a*(b^2 - 10*b*c*x - 8*c^2*x^2)) + 3*b*(b^2 - 4*a*c)*x*Sqrt[a + x*(b +
c*x)]*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]])/(a^(5/2)*(-b^2 + 4*a*c)*Sqrt[x^2*(a + x*(b + c*x)
)])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {\frac {\left (-4 c^{2} x^{2}-5 b c x +\frac {1}{2} b^{2}\right ) a^{\frac {3}{2}}}{2}-a^{\frac {5}{2}} c +\frac {3 x \left (\frac {b \sqrt {a}\, \left (c x +b \right )}{2}+\left (-\ln \left (2\right )+\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x \sqrt {a}}\right )\right ) \sqrt {c \,x^{2}+b x +a}\, \left (a c -\frac {b^{2}}{4}\right )\right ) b}{2}}{\sqrt {c \,x^{2}+b x +a}\, a^{\frac {5}{2}} x \left (a c -\frac {b^{2}}{4}\right )}\)	\(137\)
default	\(-\frac {x^{2} \left (c \,x^{2}+b x +a \right ) \left (16 a^{\frac {5}{2}} c^{2} x^{2}-6 a^{\frac {3}{2}} b^{2} c \,x^{2}+20 a^{\frac {5}{2}} b c x -6 a^{\frac {3}{2}} b^{3} x -12 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2} b c x +3 \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} x +8 a^{\frac {7}{2}} c -2 a^{\frac {5}{2}} b^{2}\right )}{2 \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {7}{2}} \left (4 a c -b^{2}\right )}\)	\(201\)
risch	\(-\frac {c \,x^{2}+b x +a}{a^{2} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}+\frac {\left (\frac {2 b^{2} c x}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {b^{3}}{a^{2} \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {4 c^{2} x}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {2 c b}{a \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {b}{a^{2} \sqrt {c \,x^{2}+b x +a}}+\frac {3 b \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )}{2 a^{\frac {5}{2}}}\right ) x \sqrt {c \,x^{2}+b x +a}}{\sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(246\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.44 \[ \int \frac {x}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{3} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {a} \log \left (-\frac {8 \, a b x^{2} + {\left (b^{2} + 4 \, a c\right )} x^{3} + 8 \, a^{2} x + 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {a}}{x^{3}}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x\right )}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{4} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{3} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}, -\frac {3 \, {\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} + {\left (b^{4} - 4 \, a b^{2} c\right )} x^{3} + {\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{3} + a b x^{2} + a^{2} x\right )}}\right ) + 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (a^{2} b^{2} - 4 \, a^{3} c + {\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} + {\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x\right )}}{2 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{4} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{3} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/4*(3*((b^3*c - 4*a*b*c^2)*x^4 + (b^4 - 4*a*b^2*c)*x^3 + (a*b^3 - 4*a^2*b*c)*x^2)*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^2*b^2 - 4*a^3*c + (3*a*b^2*c - 8*a^2*c^2)*x^2 + (3*a*b^3 - 10*a^2*b*c)*x))/((a^3*b^2*c - 4*a^4*c^2)
*x^4 + (a^3*b^3 - 4*a^4*b*c)*x^3 + (a^4*b^2 - 4*a^5*c)*x^2), -1/2*(3*((b^3*c - 4*a*b*c^2)*x^4 + (b^4 - 4*a*b^2
*c)*x^3 + (a*b^3 - 4*a^2*b*c)*x^2)*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^2*b^2 - 4*a^3*c + (3*a*b^2*c - 8*a^2*c^2)*x^2 + (3*a
*b^3 - 10*a^2*b*c)*x))/((a^3*b^2*c - 4*a^4*c^2)*x^4 + (a^3*b^3 - 4*a^4*b*c)*x^3 + (a^4*b^2 - 4*a^5*c)*x^2)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [F(-1)]

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

[In]

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

[Out]

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