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

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

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Rubi [A]  time = 0.16, antiderivative size = 144, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 12

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

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 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 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]

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*}

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Mathematica [A]  time = 0.10, size = 138, normalized size = 0.96 \begin {gather*} \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))}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 2.34, size = 157, normalized size = 1.09 \begin {gather*} \frac {3 b \log (x)}{a^{5/2}}+\frac {\sqrt {a x^2+b x^3+c x^4} \left (-4 a^2 c+a b^2-10 a b c x-8 a c^2 x^2+3 b^3 x+3 b^2 c x^2\right )}{a^2 x^2 \left (4 a c-b^2\right ) \left (a+b x+c x^2\right )}-\frac {3 b \log \left (-2 a^{5/2} \sqrt {a x^2+b x^3+c x^4}+2 a^3 x+a^2 b x^2\right )}{2 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 496, normalized size = 3.44 \begin {gather*} \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 {gather*}

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

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 >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0,0]%%%}+%%%{-2,[0,1,0,1]%%%}+%%%{-2,[0,0,1,2]%%%},0,%%%{1,[0,2,0,2]%%%}+%%%{2,[0,1,1,3]%%%}+%%%{1,[0,0,2,4
]%%%}] at parameters values [54.7579903365,-49,-33,-70]2*((-(8*a^2*c-2*a*b^2)/(16*a^3*c-4*a^2*b^2)/x-(20*a*b*c
-6*b^3)/(16*a^3*c-4*a^2*b^2))/x-(16*a*c^2-6*b^2*c)/(16*a^3*c-4*a^2*b^2))*sqrt(a*(1/x)^2+b/x+c)/(a*(1/x)^2+b/x+
c)-6*b/4/a^2/sqrt(a)*ln(abs(2*sqrt(a)*(sqrt(a*(1/x)^2+b/x+c)-sqrt(a)/x)-b))

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maple [A]  time = 0.01, size = 201, normalized size = 1.40 \begin {gather*} \frac {\left (c \,x^{2}+b x +a \right ) \left (-16 a^{\frac {5}{2}} c^{2} x^{2}+6 a^{\frac {3}{2}} b^{2} c \,x^{2}+12 \sqrt {c \,x^{2}+b x +a}\, a^{2} b c x \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-3 \sqrt {c \,x^{2}+b x +a}\, a \,b^{3} x \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )-20 a^{\frac {5}{2}} b c x +6 a^{\frac {3}{2}} b^{3} x -8 a^{\frac {7}{2}} c +2 a^{\frac {5}{2}} b^{2}\right ) x^{2}}{2 \left (c \,x^{4}+b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} \left (4 a c -b^{2}\right ) a^{\frac {7}{2}}} \end {gather*}

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+12*(c*x^2+b*x+a)^(1/2)*ln((b*x+2*a+2*(c*x^2+b*x
+a)^(1/2)*a^(1/2))/x)*x*a^2*b*c-3*(c*x^2+b*x+a)^(1/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)*x*a*b^3-20
*a^(5/2)*x*b*c+6*a^(3/2)*x*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (c x^{4} + b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\left (c\,x^4+b\,x^3+a\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (x^{2} \left (a + b x + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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