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

Optimal. Leaf size=56 \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x+c x^2\right )}{2 c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 56, 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 = {1585, 634, 618, 206, 628} \[ \frac {b \tanh ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}+\frac {\log \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a*x^2 + b*x^3 + c*x^4),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 57, normalized size = 1.02 \[ \frac {\log (a+x (b+c x))-\frac {2 b \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.93, size = 185, normalized size = 3.30 \[ \left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, \frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.42, size = 55, normalized size = 0.98 \[ -\frac {b \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} c} + \frac {\log \left (c x^{2} + b x + a\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4*a*c)*c) + 1/2*log(c*x^2 + b*x + a)/c

________________________________________________________________________________________

maple [A]  time = 0.00, size = 56, normalized size = 1.00 \[ -\frac {b \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {\ln \left (c \,x^{2}+b x +a \right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(c*x^4+b*x^3+a*x^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?

________________________________________________________________________________________

mupad [B]  time = 0.13, size = 112, normalized size = 2.00 \[ \frac {2\,a\,c\,\ln \left (c\,x^2+b\,x+a\right )}{4\,a\,c^2-b^2\,c}-\frac {b\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}+\frac {2\,c\,x}{\sqrt {4\,a\,c-b^2}}\right )}{c\,\sqrt {4\,a\,c-b^2}}-\frac {b^2\,\ln \left (c\,x^2+b\,x+a\right )}{2\,\left (4\,a\,c^2-b^2\,c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a*x^2 + b*x^3 + c*x^4),x)

[Out]

(2*a*c*log(a + b*x + c*x^2))/(4*a*c^2 - b^2*c) - (b*atan(b/(4*a*c - b^2)^(1/2) + (2*c*x)/(4*a*c - b^2)^(1/2)))
/(c*(4*a*c - b^2)^(1/2)) - (b^2*log(a + b*x + c*x^2))/(2*(4*a*c^2 - b^2*c))

________________________________________________________________________________________

sympy [B]  time = 0.33, size = 216, normalized size = 3.86 \[ \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right ) \log {\left (x + \frac {- 4 a c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right ) + 2 a + b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right )}{b} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right ) \log {\left (x + \frac {- 4 a c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right ) + 2 a + b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{2 c \left (4 a c - b^{2}\right )} + \frac {1}{2 c}\right )}{b} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)) + 1/(2*c))*log(x + (-4*a*c*(-b*sqrt(-4*a*c + b**2)/(2*c*(4*a*c -
b**2)) + 1/(2*c)) + 2*a + b**2*(-b*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)) + 1/(2*c)))/b) + (b*sqrt(-4*a*c +
b**2)/(2*c*(4*a*c - b**2)) + 1/(2*c))*log(x + (-4*a*c*(b*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)) + 1/(2*c)) +
2*a + b**2*(b*sqrt(-4*a*c + b**2)/(2*c*(4*a*c - b**2)) + 1/(2*c)))/b)

________________________________________________________________________________________