\(\int \sqrt {a x^2+b x^3+c x^4} \, dx\) [31]

Optimal result

Integrand size = 20, antiderivative size = 163 \[ \int \sqrt {a x^2+b x^3+c x^4} \, dx=-\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x}+\frac {b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} x \sqrt {a+b x+c x^2}} \]

[Out]

-1/8*b*(2*c*x+b)*(c*x^4+b*x^3+a*x^2)^(1/2)/c^2/x+1/3*(c*x^2+b*x+a)*(c*x^4+b*x^3+a*x^2)^(1/2)/c/x+1/16*b*(-4*a*
c+b^2)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))*(c*x^4+b*x^3+a*x^2)^(1/2)/c^(5/2)/x/(c*x^2+b*x+a)^(1
/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1917, 654, 626, 635, 212} \[ \int \sqrt {a x^2+b x^3+c x^4} \, dx=\frac {b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4} \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} x \sqrt {a+b x+c x^2}}-\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x} \]

[In]

Int[Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

-1/8*(b*(b + 2*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(c^2*x) + ((a + b*x + c*x^2)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(3*
c*x) + (b*(b^2 - 4*a*c)*Sqrt[a*x^2 + b*x^3 + c*x^4]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(1
6*c^(5/2)*x*Sqrt[a + b*x + c*x^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 626

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x)*((a + b*x + c*x^2)^p/(2*c*(2*p + 1
))), x] - Dist[p*((b^2 - 4*a*c)/(2*c*(2*p + 1))), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 654

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*((a + b*x + c*x^2)^(p +
1)/(2*c*(p + 1))), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1917

Int[Sqrt[(b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[Sqrt[a*x^q + b*x^n + c*x^(
2*n - q)]/(x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), Int[x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q)
)], x], x] /; FreeQ[{a, b, c, n, q}, x] && EqQ[r, 2*n - q] && PosQ[n - q]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a x^2+b x^3+c x^4} \int x \sqrt {a+b x+c x^2} \, dx}{x \sqrt {a+b x+c x^2}} \\ & = \frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x}-\frac {\left (b \sqrt {a x^2+b x^3+c x^4}\right ) \int \sqrt {a+b x+c x^2} \, dx}{2 c x \sqrt {a+b x+c x^2}} \\ & = -\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x}+\frac {\left (b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{16 c^2 x \sqrt {a+b x+c x^2}} \\ & = -\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x}+\frac {\left (b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4}\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 )}{8 c^2 x \sqrt {a+b x+c x^2}} \\ & = -\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x}+\frac {b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} x \sqrt {a+b x+c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74 \[ \int \sqrt {a x^2+b x^3+c x^4} \, dx=\frac {2 \sqrt {c} x (a+x (b+c x)) \left (-3 b^2+2 b c x+8 c \left (a+c x^2\right )\right )-3 \left (b^3-4 a b c\right ) x \sqrt {a+x (b+c x)} \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{48 c^{5/2} \sqrt {x^2 (a+x (b+c x))}} \]

[In]

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

[Out]

(2*Sqrt[c]*x*(a + x*(b + c*x))*(-3*b^2 + 2*b*c*x + 8*c*(a + c*x^2)) - 3*(b^3 - 4*a*b*c)*x*Sqrt[a + x*(b + c*x)
]*Log[c^2*(b + 2*c*x - 2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(48*c^(5/2)*Sqrt[x^2*(a + x*(b + c*x))])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {\left (8 c^{2} x^{2}+2 b c x +8 a c -3 b^{2}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{24 c^{2} x}-\frac {b \left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right ) \sqrt {x^{2} \left (c \,x^{2}+b x +a \right )}}{16 c^{\frac {5}{2}} x \sqrt {c \,x^{2}+b x +a}}\)	\(121\)
pseudoelliptic	\(\frac {16 x^{2} \sqrt {c \,x^{2}+b x +a}\, c^{\frac {5}{2}}+4 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b x +16 a \,c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}-6 \sqrt {c}\, \sqrt {c \,x^{2}+b x +a}\, b^{2}-12 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) a b c +3 \ln \left (2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b \right ) b^{3}}{48 c^{\frac {5}{2}}}\)	\(142\)
default	\(\frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}-12 c^{\frac {5}{2}} \sqrt {c \,x^{2}+b x +a}\, b x -6 c^{\frac {3}{2}} \sqrt {c \,x^{2}+b x +a}\, b^{2}-12 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) a b \,c^{2}+3 \ln \left (\frac {2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}+2 c x +b}{2 \sqrt {c}}\right ) b^{3} c \right )}{48 x \sqrt {c \,x^{2}+b x +a}\, c^{\frac {7}{2}}}\)	\(167\)

[In]

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

[Out]

1/24*(8*c^2*x^2+2*b*c*x+8*a*c-3*b^2)/c^2*(x^2*(c*x^2+b*x+a))^(1/2)/x-1/16*b*(4*a*c-b^2)/c^(5/2)*ln((1/2*b+c*x)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))*(x^2*(c*x^2+b*x+a))^(1/2)/x/(c*x^2+b*x+a)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.60 \[ \int \sqrt {a x^2+b x^3+c x^4} \, dx=\left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{3} x}, -\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - 2 \, {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{3} x}\right ] \]

[In]

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

[Out]

[-1/96*(3*(b^3 - 4*a*b*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)*sq
rt(c) + (b^2 + 4*a*c)*x)/x) - 4*(8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c + 8*a*c^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^3*
x), -1/48*(3*(b^3 - 4*a*b*c)*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*(8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c + 8*a*c^2)*sqrt(c*x^4 + b*x^3 + a*x^2))/(c^3*x)]

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(a*x**2 + b*x**3 + c*x**4), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.98 \[ \int \sqrt {a x^2+b x^3+c x^4} \, dx=\frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, x \mathrm {sgn}\left (x\right ) + \frac {b \mathrm {sgn}\left (x\right )}{c}\right )} x - \frac {3 \, b^{2} \mathrm {sgn}\left (x\right ) - 8 \, a c \mathrm {sgn}\left (x\right )}{c^{2}}\right )} - \frac {{\left (b^{3} \mathrm {sgn}\left (x\right ) - 4 \, a b c \mathrm {sgn}\left (x\right )\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{16 \, c^{\frac {5}{2}}} + \frac {{\left (3 \, b^{3} \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 12 \, a b c \log \left ({\left | b - 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b^{2} \sqrt {c} - 16 \, a^{\frac {3}{2}} c^{\frac {3}{2}}\right )} \mathrm {sgn}\left (x\right )}{48 \, c^{\frac {5}{2}}} \]

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*x*sgn(x) + b*sgn(x)/c)*x - (3*b^2*sgn(x) - 8*a*c*sgn(x))/c^2) - 1/16*(b^3*sgn
(x) - 4*a*b*c*sgn(x))*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2) + 1/48*(3*b^3*log(ab
s(b - 2*sqrt(a)*sqrt(c))) - 12*a*b*c*log(abs(b - 2*sqrt(a)*sqrt(c))) + 6*sqrt(a)*b^2*sqrt(c) - 16*a^(3/2)*c^(3
/2))*sgn(x)/c^(5/2)

Mupad [F(-1)]

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

[In]

int((a*x^2 + b*x^3 + c*x^4)^(1/2),x)

[Out]

int((a*x^2 + b*x^3 + c*x^4)^(1/2), x)