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

Optimal. Leaf size=119 \[ \frac{(b+2 c x) \sqrt{a x^2+b x^3+c x^4}}{4 c x}-\frac{x \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0777735, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1918, 1914, 621, 206} \[ \frac{(b+2 c x) \sqrt{a x^2+b x^3+c x^4}}{4 c x}-\frac{x \left (b^2-4 a c\right ) \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{3/2} \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a*x^2 + b*x^3 + c*x^4]/x,x]

[Out]

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

Rule 1918

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

Rule 1914

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.14577, size = 100, normalized size = 0.84 \[ \frac{x \left (2 \sqrt{c} (b+2 c x) (a+x (b+c x))-\left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )\right )}{8 c^{3/2} \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 146, normalized size = 1.2 \begin{align*}{\frac{1}{8\,x}\sqrt{c{x}^{4}+b{x}^{3}+a{x}^{2}} \left ( 4\,\sqrt{c{x}^{2}+bx+a}{c}^{5/2}x+2\,\sqrt{c{x}^{2}+bx+a}{c}^{3/2}b+4\,\ln \left ( 1/2\,{\frac{2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b}{\sqrt{c}}} \right ) a{c}^{2}-\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b \right ){\frac{1}{\sqrt{c}}}} \right ){b}^{2}c \right ){\frac{1}{\sqrt{c{x}^{2}+bx+a}}}{c}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c x^{4} + b x^{3} + a x^{2}}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.66874, size = 502, normalized size = 4.22 \begin{align*} \left [-\frac{{\left (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 + b c\right )}}{16 \, c^{2} x}, \frac{{\left (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 + b c\right )}}{8 \, c^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{x^{2} \left (a + b x + c x^{2}\right )}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16641, size = 169, normalized size = 1.42 \begin{align*} \frac{1}{8} \,{\left (2 \, \sqrt{c x^{2} + b x + a}{\left (2 \, x + \frac{b}{c}\right )} + \frac{{\left (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 )}{c^{\frac{3}{2}}}\right )} \mathrm{sgn}\left (x\right ) - \frac{{\left (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 ) + 2 \, \sqrt{a} b \sqrt{c}\right )} \mathrm{sgn}\left (x\right )}{8 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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