\(\int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx\) [49]

Optimal result

Integrand size = 24, antiderivative size = 143 \[ \int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {\sqrt {a x^2+b x^3+c x^4}}{2 c}-\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 c^2 x}+\frac {\left (3 b^2-4 a c\right ) x \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} \sqrt {a x^2+b x^3+c x^4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1942, 1963, 12, 1928, 635, 212} \[ \int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \left (3 b^2-4 a c\right ) \sqrt {a+b x+c x^2} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{5/2} \sqrt {a x^2+b x^3+c x^4}}-\frac {3 b \sqrt {a x^2+b x^3+c x^4}}{4 c^2 x}+\frac {\sqrt {a x^2+b x^3+c x^4}}{2 c} \]

[In]

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

[Out]

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

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 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 1928

Int[(x_)^(m_.)/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[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]))

Rule 1942

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[x^(m - 2*n + q
+ 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/(c*(m + p*q + 2*(n - q)*p + 1))), x] - Dist[1/(c*(m + p*q + 2*(
n - q)*p + 1)), Int[x^(m - 2*(n - q))*(a*(m + p*q - 2*(n - q) + 1) + b*(m + p*q + (n - q)*(p - 1) + 1)*x^(n -
q))*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c}, x] && EqQ[r, 2*n - q] && PosQ[n - q] &&  !In
tegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GtQ[m + p*q +
1, 2*(n - q)]

Rule 1963

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[B*x^(m - n + 1)*((a*x^q + b*x^n + c*x^(2*n - q))^(p + 1)/(c*(m + p*q + (n - q)*(2*p + 1) + 1))),
x] - Dist[1/(c*(m + p*q + (n - q)*(2*p + 1) + 1)), Int[x^(m - n + q)*Simp[a*B*(m + p*q - n + q + 1) + (b*B*(m
+ p*q + (n - q)*p + 1) - A*c*(m + p*q + (n - q)*(2*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] && EqQ[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c
, 0] && IGtQ[n, 0] && GeQ[p, -1] && LtQ[p, 0] && RationalQ[m, q] && GeQ[m + p*q, n - q - 1] && NeQ[m + p*q + (
n - q)*(2*p + 1) + 1, 0]

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {x \left (2 \sqrt {c} (-3 b+2 c x) (a+x (b+c x))+\left (-3 b^2+4 a c\right ) \sqrt {a+x (b+c x)} \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )\right )}{8 c^{5/2} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {4 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {3}{2}} x -6 b \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}-4 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) a c +3 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b^{2}}{8 c^{\frac {5}{2}}}\)	\(100\)
risch	\(-\frac {\left (-2 c x +3 b \right ) \left (c \,x^{2}+b x +a \right ) x}{4 c^{2} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}-\frac {\left (4 a c -3 b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) x \sqrt {c \,x^{2}+b x +a}}{8 c^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}\)	\(111\)
default	\(\frac {x \sqrt {c \,x^{2}+b x +a}\, \left (4 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, x -6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b -4 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a \,c^{2}+3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{2} c \right )}{8 \sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, c^{\frac {7}{2}}}\)	\(144\)

[In]

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

[Out]

1/8*(4*(c*x^2+b*x+a)^(1/2)*c^(3/2)*x-6*b*(c*x^2+b*x+a)^(1/2)*c^(1/2)-4*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+
b)*a*c+3*ln(2*(c*x^2+b*x+a)^(1/2)*c^(1/2)+2*c*x+b)*b^2)/c^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.58 \[ \int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\left [-\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x - 3 \, b c\right )}}{16 \, c^{3} x}, -\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - 2 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c^{2} x - 3 \, b c\right )}}{8 \, c^{3} x}\right ] \]

[In]

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

[Out]

[-1/16*((3*b^2 - 4*a*c)*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)*(2*c^2*x - 3*b*c))/(c^3*x), -1/8*((3*b^2 - 4*a*c)*sq
rt(-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)*(2*c^2*x - 3*b*c))/(c^3*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int \frac {x^3}{\sqrt {a x^2+b x^3+c x^4}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x + a} {\left (\frac {2 \, x}{c \mathrm {sgn}\left (x\right )} - \frac {3 \, b}{c^{2} \mathrm {sgn}\left (x\right )}\right )} + \frac {{\left (3 \, b^{2} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 4 \, a c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b \sqrt {c}\right )} \mathrm {sgn}\left (x\right )}{8 \, c^{\frac {5}{2}}} - \frac {{\left (3 \, b^{2} - 4 \, a c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/4*sqrt(c*x^2 + b*x + a)*(2*x/(c*sgn(x)) - 3*b/(c^2*sgn(x))) + 1/8*(3*b^2*log(abs(b - 2*sqrt(a)*sqrt(c))) - 4
*a*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 6*sqrt(a)*b*sqrt(c))*sgn(x)/c^(5/2) - 1/8*(3*b^2 - 4*a*c)*log(abs(2*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/(c^(5/2)*sgn(x))

Mupad [F(-1)]

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

[In]

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

[Out]

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