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

Optimal. Leaf size=173 \[ -\frac{\sqrt{a x^2+b x^3+c x^4}}{x^2}-\frac{b x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}+\frac{\sqrt{c} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.124074, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1920, 1933, 843, 621, 206, 724} \[ -\frac{\sqrt{a x^2+b x^3+c x^4}}{x^2}-\frac{b x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}+\frac{\sqrt{c} x \sqrt{a+b x+c x^2} \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{\sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1920

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a*
x^q + b*x^n + c*x^(2*n - q))^p)/(m + p*q + 1), x] - Dist[((n - q)*p)/(m + p*q + 1), Int[x^(m + n)*(b + 2*c*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] && GtQ[p, 0] && RationalQ[m, q] && LeQ[m + p*q + 1, -
(n - q) + 1] && NeQ[m + p*q + 1, 0]

Rule 1933

Int[((A_) + (B_.)*(x_)^(j_.))/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[(A + B*x^(n - q))/(
x^(q/2)*Sqrt[a + b*x^(n - q) + c*x^(2*(n - q))]), x], x] /; FreeQ[{a, b, c, A, B, n, q}, x] && EqQ[j, n - q] &
& EqQ[r, 2*n - q] && PosQ[n - q] && EqQ[n, 3] && EqQ[q, 2]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

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])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [A]  time = 0.127059, size = 131, normalized size = 0.76 \[ -\frac{\sqrt{a+x (b+c x)} \left (b x \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )+2 \sqrt{a} \left (\sqrt{a+x (b+c x)}-\sqrt{c} x \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )\right )\right )}{2 \sqrt{a} \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 174, normalized size = 1. \begin{align*}{\frac{1}{2\,a{x}^{2}}\sqrt{c{x}^{4}+b{x}^{3}+a{x}^{2}} \left ( 2\,{c}^{5/2}\sqrt{c{x}^{2}+bx+a}{x}^{2}-{c}^{{\frac{3}{2}}}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) xb-2\, \left ( c{x}^{2}+bx+a \right ) ^{3/2}{c}^{3/2}+2\,{c}^{3/2}\sqrt{c{x}^{2}+bx+a}xb+2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{c{x}^{2}+bx+a}\sqrt{c}+2\,cx+b}{\sqrt{c}}} \right ) xa{c}^{2} \right ){\frac{1}{\sqrt{c{x}^{2}+bx+a}}}{c}^{-{\frac{3}{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^3,x)

[Out]

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

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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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.96342, size = 1489, normalized size = 8.61 \begin{align*} \left [\frac{2 \, a \sqrt{c} x^{2} \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 ) + \sqrt{a} b x^{2} \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}} a}{4 \, a x^{2}}, -\frac{4 \, a \sqrt{-c} x^{2} \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 ) - \sqrt{a} b x^{2} \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}} a}{4 \, a x^{2}}, \frac{\sqrt{-a} b x^{2} \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 ) + a \sqrt{c} x^{2} \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 ) - 2 \, \sqrt{c x^{4} + b x^{3} + a x^{2}} a}{2 \, a x^{2}}, \frac{\sqrt{-a} b x^{2} \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 \, a \sqrt{-c} x^{2} \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}} a}{2 \, a x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError