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

Optimal. Leaf size=163 \[ \frac {b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} x \sqrt {a+b x+c x^2}}-\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x} \]

[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)

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Rubi [A]  time = 0.06, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1903, 640, 612, 621, 206} \[ \frac {b \left (b^2-4 a c\right ) \sqrt {a x^2+b x^3+c x^4} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{16 c^{5/2} x \sqrt {a+b x+c x^2}}-\frac {b (b+2 c x) \sqrt {a x^2+b x^3+c x^4}}{8 c^2 x}+\frac {\left (a+b x+c x^2\right ) \sqrt {a x^2+b x^3+c x^4}}{3 c x} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3 + c*x^4],x]

[Out]

-(b*(b + 2*c*x)*Sqrt[a*x^2 + b*x^3 + c*x^4])/(8*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])])/(16*
c^(5/2)*x*Sqrt[a + b*x + c*x^2])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

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 621

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 640

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 1903

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*} \int \sqrt {a x^2+b x^3+c x^4} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.21, size = 119, normalized size = 0.73 \[ \frac {2 \sqrt {c} x (a+x (b+c x)) \left (8 c \left (a+c x^2\right )-3 b^2+2 b c x\right )+3 b x \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{48 c^{5/2} \sqrt {x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

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

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fricas [A]  time = 0.71, size = 260, normalized size = 1.60 \[ \left [-\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {c} x \log \left (-\frac {8 \, c^{2} x^{3} + 8 \, b c x^{2} - 4 \, \sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {c} + {\left (b^{2} + 4 \, a c\right )} x}{x}\right ) - 4 \, {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{96 \, c^{3} x}, -\frac {3 \, {\left (b^{3} - 4 \, a b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{4} + b x^{3} + a x^{2}} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{3} + b c x^{2} + a c x\right )}}\right ) - 2 \, {\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c + 8 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{3} + a x^{2}}}{48 \, c^{3} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.88, size = 166, normalized size = 1.02 \[ \frac {1}{24} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, x \mathrm {sgn}\relax (x) + \frac {b \mathrm {sgn}\relax (x)}{c}\right )} x - \frac {3 \, b^{2} \mathrm {sgn}\relax (x) - 8 \, a c \mathrm {sgn}\relax (x)}{c^{2}}\right )} - \frac {{\left (b^{3} \mathrm {sgn}\relax (x) - 4 \, a b c \mathrm {sgn}\relax (x)\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{16 \, c^{\frac {5}{2}}} + \frac {{\left (3 \, b^{3} \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) - 12 \, a b c \log \left ({\left | -b + 2 \, \sqrt {a} \sqrt {c} \right |}\right ) + 6 \, \sqrt {a} b^{2} \sqrt {c} - 16 \, a^{\frac {3}{2}} c^{\frac {3}{2}}\right )} \mathrm {sgn}\relax (x)}{48 \, c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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(a
bs(-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)

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maple [A]  time = 0.01, size = 167, normalized size = 1.02 \[ \frac {\sqrt {c \,x^{4}+b \,x^{3}+a \,x^{2}}\, \left (-12 a b \,c^{2} \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )+3 b^{3} c \ln \left (\frac {2 c x +b +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {c}}{2 \sqrt {c}}\right )-12 \sqrt {c \,x^{2}+b x +a}\, b \,c^{\frac {5}{2}} x -6 \sqrt {c \,x^{2}+b x +a}\, b^{2} c^{\frac {3}{2}}+16 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} c^{\frac {5}{2}}\right )}{48 \sqrt {c \,x^{2}+b x +a}\, c^{\frac {7}{2}} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(c*x^4+b*x^3+a*x^2)^(1/2)*(16*(c*x^2+b*x+a)^(3/2)*c^(5/2)-12*c^(5/2)*(c*x^2+b*x+a)^(1/2)*x*b-6*c^(3/2)*(c
*x^2+b*x+a)^(1/2)*b^2-12*ln(1/2*(2*c*x+b+2*(c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*a*b*c^2+3*ln(1/2*(2*c*x+b+2*(
c*x^2+b*x+a)^(1/2)*c^(1/2))/c^(1/2))*b^3*c)/x/(c*x^2+b*x+a)^(1/2)/c^(7/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {c x^{4} + b x^{3} + a x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {c\,x^4+b\,x^3+a\,x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a x^{2} + b x^{3} + c x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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