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

Optimal. Leaf size=144 \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{a^2 x^2 \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{2 a^{5/2}}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}} \]

[Out]

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

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Rubi [A]  time = 0.163171, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {1924, 1951, 12, 1904, 206} \[ -\frac{\left (3 b^2-8 a c\right ) \sqrt{a x^2+b x^3+c x^4}}{a^2 x^2 \left (b^2-4 a c\right )}+\frac{3 b \tanh ^{-1}\left (\frac{x (2 a+b x)}{2 \sqrt{a} \sqrt{a x^2+b x^3+c x^4}}\right )}{2 a^{5/2}}+\frac{2 \left (-2 a c+b^2+b c x\right )}{a \left (b^2-4 a c\right ) \sqrt{a x^2+b x^3+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1924

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[1/(a*(n - q)*(p + 1)*(b^2 - 4*a*c)), Int[x^(m - q)*(b^2*(m + p*q + (n - q)*(p + 1) + 1) - 2*a*c*(m + p
*q + 2*(n - q)*(p + 1) + 1) + b*c*(m + p*q + (n - q)*(2*p + 3) + 1)*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] && LtQ[p, -1] && RationalQ[m, q] && LtQ[m + p*q + 1, n - q]

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

Mathematica [A]  time = 0.105634, size = 138, normalized size = 0.96 \[ \frac{2 \sqrt{a} \left (-4 a^2 c+a \left (b^2-10 b c x-8 c^2 x^2\right )+3 b^2 x (b+c x)\right )-3 b x \left (b^2-4 a c\right ) \sqrt{a+x (b+c x)} \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{2 a^{5/2} \left (4 a c-b^2\right ) \sqrt{x^2 (a+x (b+c x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a]*(-4*a^2*c + 3*b^2*x*(b + c*x) + a*(b^2 - 10*b*c*x - 8*c^2*x^2)) - 3*b*(b^2 - 4*a*c)*x*Sqrt[a + x*(b
+ c*x)]*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/(2*a^(5/2)*(-b^2 + 4*a*c)*Sqrt[x^2*(a + x*(b
+ c*x))])

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Maple [A]  time = 0.006, size = 201, normalized size = 1.4 \begin{align*} -{\frac{{x}^{2} \left ( c{x}^{2}+bx+a \right ) }{8\,ac-2\,{b}^{2}} \left ( 16\,{a}^{5/2}{x}^{2}{c}^{2}-6\,{a}^{3/2}{x}^{2}{b}^{2}c+20\,{a}^{5/2}xbc-6\,{a}^{3/2}x{b}^{3}-12\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}x{a}^{2}bc+3\,\ln \left ({\frac{2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a}}{x}} \right ) \sqrt{c{x}^{2}+bx+a}xa{b}^{3}+8\,{a}^{7/2}c-2\,{a}^{5/2}{b}^{2} \right ) \left ( c{x}^{4}+b{x}^{3}+a{x}^{2} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*(c*x^2+b*x+a)*(16*a^(5/2)*x^2*c^2-6*a^(3/2)*x^2*b^2*c+20*a^(5/2)*x*b*c-6*a^(3/2)*x*b^3-12*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)*x*a^2*b*c+3*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)*x*a*b^3+8*a^(7/2)*c-2*a^(5/2)*b^2)/(c*x^4+b*x^3+a*x^2)^(3/2)/a^(7/2)/(4*a*c-b^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\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/(c*x^4+b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.87612, size = 1041, normalized size = 7.23 \begin{align*} \left [\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{3} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\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^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x\right )}}{4 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{4} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{3} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}, -\frac{3 \,{\left ({\left (b^{3} c - 4 \, a b c^{2}\right )} x^{4} +{\left (b^{4} - 4 \, a b^{2} c\right )} x^{3} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2}\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^{2} b^{2} - 4 \, a^{3} c +{\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x^{2} +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x\right )}}{2 \,{\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{4} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{3} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*((b^3*c - 4*a*b*c^2)*x^4 + (b^4 - 4*a*b^2*c)*x^3 + (a*b^3 - 4*a^2*b*c)*x^2)*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^2*b^2 - 4*a^3*c + (3*a*b^2*c - 8*a^2*c^2)*x^2 + (3*a*b^3 - 10*a^2*b*c)*x))/((a^3*b^2*c - 4*a^4*c^2)
*x^4 + (a^3*b^3 - 4*a^4*b*c)*x^3 + (a^4*b^2 - 4*a^5*c)*x^2), -1/2*(3*((b^3*c - 4*a*b*c^2)*x^4 + (b^4 - 4*a*b^2
*c)*x^3 + (a*b^3 - 4*a^2*b*c)*x^2)*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^2*b^2 - 4*a^3*c + (3*a*b^2*c - 8*a^2*c^2)*x^2 + (3*a
*b^3 - 10*a^2*b*c)*x))/((a^3*b^2*c - 4*a^4*c^2)*x^4 + (a^3*b^3 - 4*a^4*b*c)*x^3 + (a^4*b^2 - 4*a^5*c)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/(c*x**4+b*x**3+a*x**2)**(3/2),x)

[Out]

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

[Out]

Exception raised: NotImplementedError