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

Optimal. Leaf size=148 \[ \frac {b \log \left (a+b x+c x^2\right )}{a^3}-\frac {2 b \log (x)}{a^3}-\frac {2 \left (b^2-3 a c\right )}{a^2 x \left (b^2-4 a c\right )}-\frac {2 \left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}+\frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \]

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Rubi [A]  time = 0.20, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1585, 740, 800, 634, 618, 206, 628} \begin {gather*} -\frac {2 \left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac {2 \left (b^2-3 a c\right )}{a^2 x \left (b^2-4 a c\right )}+\frac {b \log \left (a+b x+c x^2\right )}{a^3}-\frac {2 b \log (x)}{a^3}+\frac {-2 a c+b^2+b c x}{a x \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a*x^2 + b*x^3 + c*x^4)^2,x]

[Out]

(-2*(b^2 - 3*a*c))/(a^2*(b^2 - 4*a*c)*x) + (b^2 - 2*a*c + b*c*x)/(a*(b^2 - 4*a*c)*x*(a + b*x + c*x^2)) - (2*(b
^4 - 6*a*b^2*c + 6*a^2*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(a^3*(b^2 - 4*a*c)^(3/2)) - (2*b*Log[x])/a
^3 + (b*Log[a + b*x + c*x^2])/a^3

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 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

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

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 800

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[((d + e*x)^m*(f + g*x))/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^2} \, dx &=\int \frac {1}{x^2 \left (a+b x+c x^2\right )^2} \, dx\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {\int \frac {-2 \left (b^2-3 a c\right )-2 b c x}{x^2 \left (a+b x+c x^2\right )} \, dx}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {2 \left (-b^2+3 a c\right )}{a x^2}-\frac {2 b \left (-b^2+4 a c\right )}{a^2 x}+\frac {2 \left (-b^4+5 a b^2 c-3 a^2 c^2-b c \left (b^2-4 a c\right ) x\right )}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx}{a \left (b^2-4 a c\right )}\\ &=-\frac {2 \left (b^2-3 a c\right )}{a^2 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {2 b \log (x)}{a^3}-\frac {2 \int \frac {-b^4+5 a b^2 c-3 a^2 c^2-b c \left (b^2-4 a c\right ) x}{a+b x+c x^2} \, dx}{a^3 \left (b^2-4 a c\right )}\\ &=-\frac {2 \left (b^2-3 a c\right )}{a^2 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {2 b \log (x)}{a^3}+\frac {b \int \frac {b+2 c x}{a+b x+c x^2} \, dx}{a^3}+\frac {\left (b^4-6 a b^2 c+6 a^2 c^2\right ) \int \frac {1}{a+b x+c x^2} \, dx}{a^3 \left (b^2-4 a c\right )}\\ &=-\frac {2 \left (b^2-3 a c\right )}{a^2 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {2 b \log (x)}{a^3}+\frac {b \log \left (a+b x+c x^2\right )}{a^3}-\frac {\left (2 \left (b^4-6 a b^2 c+6 a^2 c^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b^2-4 a c-x^2} \, dx,x,b+2 c x\right )}{a^3 \left (b^2-4 a c\right )}\\ &=-\frac {2 \left (b^2-3 a c\right )}{a^2 \left (b^2-4 a c\right ) x}+\frac {b^2-2 a c+b c x}{a \left (b^2-4 a c\right ) x \left (a+b x+c x^2\right )}-\frac {2 \left (b^4-6 a b^2 c+6 a^2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a^3 \left (b^2-4 a c\right )^{3/2}}-\frac {2 b \log (x)}{a^3}+\frac {b \log \left (a+b x+c x^2\right )}{a^3}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 131, normalized size = 0.89 \begin {gather*} -\frac {\frac {2 \left (6 a^2 c^2-6 a b^2 c+b^4\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\left (4 a c-b^2\right )^{3/2}}+\frac {a \left (-3 a b c-2 a c^2 x+b^3+b^2 c x\right )}{\left (b^2-4 a c\right ) (a+x (b+c x))}-b \log (a+x (b+c x))+\frac {a}{x}+2 b \log (x)}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a/x + (a*(b^3 - 3*a*b*c + b^2*c*x - 2*a*c^2*x))/((b^2 - 4*a*c)*(a + x*(b + c*x))) + (2*(b^4 - 6*a*b^2*c + 6
*a^2*c^2)*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]])/(-b^2 + 4*a*c)^(3/2) + 2*b*Log[x] - b*Log[a + x*(b + c*x)])/
a^3)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2}{\left (a x^2+b x^3+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.36, size = 975, normalized size = 6.59 \begin {gather*} \left [-\frac {a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + 2 \, {\left (a b^{4} c - 7 \, a^{2} b^{2} c^{2} + 12 \, a^{3} c^{3}\right )} x^{2} + {\left ({\left (b^{4} c - 6 \, a b^{2} c^{2} + 6 \, a^{2} c^{3}\right )} x^{3} + {\left (b^{5} - 6 \, a b^{3} c + 6 \, a^{2} b c^{2}\right )} x^{2} + {\left (a b^{4} - 6 \, a^{2} b^{2} c + 6 \, a^{3} c^{2}\right )} x\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (2 \, a b^{5} - 15 \, a^{2} b^{3} c + 28 \, a^{3} b c^{2}\right )} x - {\left ({\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} x^{2} + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left ({\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} x^{2} + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )} \log \relax (x)}{{\left (a^{3} b^{4} c - 8 \, a^{4} b^{2} c^{2} + 16 \, a^{5} c^{3}\right )} x^{3} + {\left (a^{3} b^{5} - 8 \, a^{4} b^{3} c + 16 \, a^{5} b c^{2}\right )} x^{2} + {\left (a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2}\right )} x}, -\frac {a^{2} b^{4} - 8 \, a^{3} b^{2} c + 16 \, a^{4} c^{2} + 2 \, {\left (a b^{4} c - 7 \, a^{2} b^{2} c^{2} + 12 \, a^{3} c^{3}\right )} x^{2} + 2 \, {\left ({\left (b^{4} c - 6 \, a b^{2} c^{2} + 6 \, a^{2} c^{3}\right )} x^{3} + {\left (b^{5} - 6 \, a b^{3} c + 6 \, a^{2} b c^{2}\right )} x^{2} + {\left (a b^{4} - 6 \, a^{2} b^{2} c + 6 \, a^{3} c^{2}\right )} x\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (2 \, a b^{5} - 15 \, a^{2} b^{3} c + 28 \, a^{3} b c^{2}\right )} x - {\left ({\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} x^{2} + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )} \log \left (c x^{2} + b x + a\right ) + 2 \, {\left ({\left (b^{5} c - 8 \, a b^{3} c^{2} + 16 \, a^{2} b c^{3}\right )} x^{3} + {\left (b^{6} - 8 \, a b^{4} c + 16 \, a^{2} b^{2} c^{2}\right )} x^{2} + {\left (a b^{5} - 8 \, a^{2} b^{3} c + 16 \, a^{3} b c^{2}\right )} x\right )} \log \relax (x)}{{\left (a^{3} b^{4} c - 8 \, a^{4} b^{2} c^{2} + 16 \, a^{5} c^{3}\right )} x^{3} + {\left (a^{3} b^{5} - 8 \, a^{4} b^{3} c + 16 \, a^{5} b c^{2}\right )} x^{2} + {\left (a^{4} b^{4} - 8 \, a^{5} b^{2} c + 16 \, a^{6} c^{2}\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*x^2 + ((b^4*c - 6*a*b^2*c^2 +
6*a^2*c^3)*x^3 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*x^2 + (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*x)*sqrt(b^2 - 4*a*c)
*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + (2*a*b^5 - 15*a^
2*b^3*c + 28*a^3*b*c^2)*x - ((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2
+ (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(c*x^2 + b*x + a) + 2*((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3
+ (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(x))/((a^3*b^4*c - 8*a^4
*b^2*c^2 + 16*a^5*c^3)*x^3 + (a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2 + (a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2)
*x), -(a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2 + 2*(a*b^4*c - 7*a^2*b^2*c^2 + 12*a^3*c^3)*x^2 + 2*((b^4*c - 6*a*b^2
*c^2 + 6*a^2*c^3)*x^3 + (b^5 - 6*a*b^3*c + 6*a^2*b*c^2)*x^2 + (a*b^4 - 6*a^2*b^2*c + 6*a^3*c^2)*x)*sqrt(-b^2 +
4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + (2*a*b^5 - 15*a^2*b^3*c + 28*a^3*b*c^2)*x - ((
b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^2*c^2)*x^2 + (a*b^5 - 8*a^2*b^3*c + 16*a
^3*b*c^2)*x)*log(c*x^2 + b*x + a) + 2*((b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*x^3 + (b^6 - 8*a*b^4*c + 16*a^2*b^
2*c^2)*x^2 + (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*x)*log(x))/((a^3*b^4*c - 8*a^4*b^2*c^2 + 16*a^5*c^3)*x^3 + (
a^3*b^5 - 8*a^4*b^3*c + 16*a^5*b*c^2)*x^2 + (a^4*b^4 - 8*a^5*b^2*c + 16*a^6*c^2)*x)]

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giac [A]  time = 0.45, size = 171, normalized size = 1.16 \begin {gather*} \frac {2 \, {\left (b^{4} - 6 \, a b^{2} c + 6 \, a^{2} c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (a^{3} b^{2} - 4 \, a^{4} c\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {2 \, b^{2} c x^{2} - 6 \, a c^{2} x^{2} + 2 \, b^{3} x - 7 \, a b c x + a b^{2} - 4 \, a^{2} c}{{\left (a^{2} b^{2} - 4 \, a^{3} c\right )} {\left (c x^{3} + b x^{2} + a x\right )}} + \frac {b \log \left (c x^{2} + b x + a\right )}{a^{3}} - \frac {2 \, b \log \left ({\left | x \right |}\right )}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^4 - 6*a*b^2*c + 6*a^2*c^2)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((a^3*b^2 - 4*a^4*c)*sqrt(-b^2 + 4*a*c)
) - (2*b^2*c*x^2 - 6*a*c^2*x^2 + 2*b^3*x - 7*a*b*c*x + a*b^2 - 4*a^2*c)/((a^2*b^2 - 4*a^3*c)*(c*x^3 + b*x^2 +
a*x)) + b*log(c*x^2 + b*x + a)/a^3 - 2*b*log(abs(x))/a^3

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maple [B]  time = 0.02, size = 328, normalized size = 2.22 \begin {gather*} -\frac {2 c^{2} x}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a}-\frac {12 c^{2} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} a}+\frac {b^{2} c x}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a^{2}}+\frac {12 b^{2} c \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} a^{2}}-\frac {2 b^{4} \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\left (4 a c -b^{2}\right )^{\frac {3}{2}} a^{3}}-\frac {3 b c}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a}+\frac {b^{3}}{\left (c \,x^{2}+b x +a \right ) \left (4 a c -b^{2}\right ) a^{2}}+\frac {4 b c \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right ) a^{2}}-\frac {b^{3} \ln \left (c \,x^{2}+b x +a \right )}{\left (4 a c -b^{2}\right ) a^{3}}-\frac {2 b \ln \relax (x )}{a^{3}}-\frac {1}{a^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a^2/x-2/a^3*b*ln(x)-2/a/(c*x^2+b*x+a)*c^2/(4*a*c-b^2)*x+1/a^2/(c*x^2+b*x+a)*c/(4*a*c-b^2)*x*b^2-3/a/(c*x^2+
b*x+a)*b/(4*a*c-b^2)*c+1/a^2/(c*x^2+b*x+a)*b^3/(4*a*c-b^2)+4/a^2/(4*a*c-b^2)*c*ln(c*x^2+b*x+a)*b-1/a^3/(4*a*c-
b^2)*ln(c*x^2+b*x+a)*b^3-12/a/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*c^2+12/a^2/(4*a*c-b^2)^(3/
2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^2*c-2/a^3/(4*a*c-b^2)^(3/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2))*b^4

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [B]  time = 2.83, size = 775, normalized size = 5.24 \begin {gather*} \ln \left (2\,a\,b^7+2\,b^8\,x+2\,a\,b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-23\,a^2\,b^5\,c-108\,a^4\,b\,c^3+24\,a^4\,c^4\,x+2\,b^5\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+87\,a^3\,b^3\,c^2+3\,a^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-9\,a^2\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+97\,a^2\,b^4\,c^2\,x-138\,a^3\,b^2\,c^3\,x-24\,a\,b^6\,c\,x-12\,a\,b^3\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+15\,a^2\,b\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+6\,a^2\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-6\,a\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^6\,c^3+48\,a^5\,b^2\,c^2-12\,a^4\,b^4\,c+a^3\,b^6}+\frac {b}{a^3}\right )-\frac {\frac {1}{a}-\frac {x\,\left (2\,b^3-7\,a\,b\,c\right )}{a^2\,\left (4\,a\,c-b^2\right )}+\frac {2\,c\,x^2\,\left (3\,a\,c-b^2\right )}{a^2\,\left (4\,a\,c-b^2\right )}}{c\,x^3+b\,x^2+a\,x}-\ln \left (2\,a\,b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-2\,b^8\,x-2\,a\,b^7+23\,a^2\,b^5\,c+108\,a^4\,b\,c^3-24\,a^4\,c^4\,x+2\,b^5\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-87\,a^3\,b^3\,c^2+3\,a^3\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-9\,a^2\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-97\,a^2\,b^4\,c^2\,x+138\,a^3\,b^2\,c^3\,x+24\,a\,b^6\,c\,x-12\,a\,b^3\,c\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+15\,a^2\,b\,c^2\,x\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}\right )\,\left (\frac {b^4\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}+6\,a^2\,c^2\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}-6\,a\,b^2\,c\,\sqrt {-{\left (4\,a\,c-b^2\right )}^3}}{-64\,a^6\,c^3+48\,a^5\,b^2\,c^2-12\,a^4\,b^4\,c+a^3\,b^6}-\frac {b}{a^3}\right )-\frac {2\,b\,\ln \relax (x)}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^2 + b*x^3 + c*x^4)^2,x)

[Out]

log(2*a*b^7 + 2*b^8*x + 2*a*b^4*(-(4*a*c - b^2)^3)^(1/2) - 23*a^2*b^5*c - 108*a^4*b*c^3 + 24*a^4*c^4*x + 2*b^5
*x*(-(4*a*c - b^2)^3)^(1/2) + 87*a^3*b^3*c^2 + 3*a^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 9*a^2*b^2*c*(-(4*a*c - b^2
)^3)^(1/2) + 97*a^2*b^4*c^2*x - 138*a^3*b^2*c^3*x - 24*a*b^6*c*x - 12*a*b^3*c*x*(-(4*a*c - b^2)^3)^(1/2) + 15*
a^2*b*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*((b^4*(-(4*a*c - b^2)^3)^(1/2) + 6*a^2*c^2*(-(4*a*c - b^2)^3)^(1/2) - 6*
a*b^2*c*(-(4*a*c - b^2)^3)^(1/2))/(a^3*b^6 - 64*a^6*c^3 - 12*a^4*b^4*c + 48*a^5*b^2*c^2) + b/a^3) - (1/a - (x*
(2*b^3 - 7*a*b*c))/(a^2*(4*a*c - b^2)) + (2*c*x^2*(3*a*c - b^2))/(a^2*(4*a*c - b^2)))/(a*x + b*x^2 + c*x^3) -
log(2*a*b^4*(-(4*a*c - b^2)^3)^(1/2) - 2*b^8*x - 2*a*b^7 + 23*a^2*b^5*c + 108*a^4*b*c^3 - 24*a^4*c^4*x + 2*b^5
*x*(-(4*a*c - b^2)^3)^(1/2) - 87*a^3*b^3*c^2 + 3*a^3*c^2*(-(4*a*c - b^2)^3)^(1/2) - 9*a^2*b^2*c*(-(4*a*c - b^2
)^3)^(1/2) - 97*a^2*b^4*c^2*x + 138*a^3*b^2*c^3*x + 24*a*b^6*c*x - 12*a*b^3*c*x*(-(4*a*c - b^2)^3)^(1/2) + 15*
a^2*b*c^2*x*(-(4*a*c - b^2)^3)^(1/2))*((b^4*(-(4*a*c - b^2)^3)^(1/2) + 6*a^2*c^2*(-(4*a*c - b^2)^3)^(1/2) - 6*
a*b^2*c*(-(4*a*c - b^2)^3)^(1/2))/(a^3*b^6 - 64*a^6*c^3 - 12*a^4*b^4*c + 48*a^5*b^2*c^2) - b/a^3) - (2*b*log(x
))/a^3

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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