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

Optimal. Leaf size=94 \[ \frac{2 x \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{\tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0676333, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1922, 1904, 206} \[ \frac{2 x \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{\tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1922

Int[(x_)^(m_.)*((b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(p_), x_Symbol] :> -Simp[(x^(m - q + 1
)*(b^2 - 2*a*c + b*c*x^(n - q))*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), x]
+ Dist[(2*a*c - b^2*(p + 2))/(a*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - 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] && I
GtQ[n, 0] && LtQ[p, -1] && RationalQ[m, p, q] && EqQ[m + p*q + 1, -((n - q)*(2*p + 3))]

Rule 1904

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(r_.)], x_Symbol] :> Dist[-2/(n - 2), Subst[Int[1/(4*a
- x^2), x], x, (x*(2*a + b*x^(n - 2)))/Sqrt[a*x^2 + b*x^n + c*x^r]], x] /; FreeQ[{a, b, c, n, r}, x] && EqQ[r
, 2*n - 2] && PosQ[n - 2] && 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{x^2}{\left (a x^2+b x^3+c x^4\right )^{3/2}} \, dx &=\frac{2 x \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}+\frac{\int \frac{1}{\sqrt{a x^2+b x^3+c x^4}} \, dx}{a}\\ &=\frac{2 x \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{2 \operatorname{Subst}\left (\int \frac{1}{4 a-x^2} \, dx,x,\frac{x (2 a+b x)}{\sqrt{a x^2+b x^3+c x^4}}\right )}{a}\\ &=\frac{2 x \left (b^2-2 a c+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}}-\frac{\tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.123031, size = 109, normalized size = 1.16 \[ \frac{x \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{a} x \left (-2 a c+b^2+b c x\right )}{a^{3/2} \left (4 a c-b^2\right ) \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 164, normalized size = 1.7 \begin{align*}{\frac{{x}^{3} \left ( c{x}^{2}+bx+a \right ) }{4\,ac-{b}^{2}} \left ( 4\,{a}^{5/2}c-2\,{a}^{3/2}xbc-2\,{a}^{3/2}{b}^{2}-4\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}{a}^{2}c+\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) \sqrt{c{x}^{2}+bx+a}a{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.06535, size = 869, normalized size = 9.24 \begin{align*} \left [\frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} +{\left (b^{3} - 4 \, a b c\right )} x^{2} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt{a} \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}}{\left (a b c x + a b^{2} - 2 \, a^{2} c\right )}}{2 \,{\left ({\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{3} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x\right )}}, \frac{{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} +{\left (b^{3} - 4 \, a b c\right )} x^{2} +{\left (a b^{2} - 4 \, a^{2} c\right )} x\right )} \sqrt{-a} \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 \, \sqrt{c x^{4} + b x^{3} + a x^{2}}{\left (a b c x + a b^{2} - 2 \, a^{2} c\right )}}{{\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{3} +{\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2} +{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError