3.54 \(\int \frac{1}{x^2 \sqrt{a x^2+b x^3+c x^4}} \, dx\)

Optimal. Leaf size=119 \[ -\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{8 a^{5/2}}+\frac{3 b \sqrt{a x^2+b x^3+c x^4}}{4 a^2 x^2}-\frac{\sqrt{a x^2+b x^3+c x^4}}{2 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.148656, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {1929, 1951, 12, 1904, 206} \[ -\frac{\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{8 a^{5/2}}+\frac{3 b \sqrt{a x^2+b x^3+c x^4}}{4 a^2 x^2}-\frac{\sqrt{a x^2+b x^3+c x^4}}{2 a x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1929

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

Rule 1951

Int[(x_)^(m_.)*((c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.) + (a_.)*(x_)^(q_.))^(p_.)*((A_) + (B_.)*(x_)^(r_.)), x_Sym
bol] :> Simp[(A*x^(m - q + 1)*(a*x^q + b*x^n + c*x^(2*n - q))^(p + 1))/(a*(m + p*q + 1)), x] + Dist[1/(a*(m +
p*q + 1)), Int[x^(m + n - q)*Simp[a*B*(m + p*q + 1) - A*b*(m + p*q + (n - q)*(p + 1) + 1) - A*c*(m + p*q + 2*(
n - q)*(p + 1) + 1)*x^(n - q), x]*(a*x^q + b*x^n + c*x^(2*n - q))^p, x], x] /; FreeQ[{a, b, c, A, B}, x] && Eq
Q[r, n - q] && EqQ[j, 2*n - q] &&  !IntegerQ[p] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && RationalQ[m, p, q] &&
((GeQ[p, -1] && LtQ[p, 0]) || EqQ[m + p*q + (n - q)*(2*p + 1) + 1, 0]) && LeQ[m + p*q, -(n - q)] && NeQ[m + p*
q + 1, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0955229, size = 112, normalized size = 0.94 \[ \frac{x^2 \left (-\left (3 b^2-4 a c\right )\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )-2 \sqrt{a} (2 a-3 b x) (a+x (b+c x))}{8 a^{5/2} x \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 152, normalized size = 1.3 \begin{align*} -{\frac{1}{8\,x}\sqrt{c{x}^{2}+bx+a} \left ( 4\,\sqrt{c{x}^{2}+bx+a}{a}^{5/2}-6\,{a}^{3/2}\sqrt{c{x}^{2}+bx+a}xb-4\,c\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){a}^{2}{x}^{2}+3\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ){x}^{2}a{b}^{2} \right ){\frac{1}{\sqrt{c{x}^{4}+b{x}^{3}+a{x}^{2}}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*x^2+b*x+a)^(1/2)*(4*(c*x^2+b*x+a)^(1/2)*a^(5/2)-6*a^(3/2)*(c*x^2+b*x+a)^(1/2)*x*b-4*c*ln((2*a+b*x+2*a^
(1/2)*(c*x^2+b*x+a)^(1/2))/x)*a^2*x^2+3*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)*x^2*a*b^2)/x/(c*x^4+b*x^
3+a*x^2)^(1/2)/a^(7/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c x^{4} + b x^{3} + a x^{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.7908, size = 527, normalized size = 4.43 \begin{align*} \left [-\frac{{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt{a} x^{3} \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 (3 \, a b x - 2 \, a^{2}\right )}}{16 \, a^{3} x^{3}}, \frac{{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt{-a} x^{3} \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 (3 \, a b x - 2 \, a^{2}\right )}}{8 \, a^{3} x^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{x^{2} \left (a + b x + c x^{2}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: NotImplementedError