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

Optimal. Leaf size=64 \[ \frac {b \tanh ^{-1}\left (\frac {b-2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {\log \left (a-b x^2+c x^4\right )}{4 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1114, 634, 618, 206, 628} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {b-2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {\log \left (a-b x^2+c x^4\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(a - b*x^2 + c*x^4),x]

[Out]

(b*ArcTanh[(b - 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(2*c*Sqrt[b^2 - 4*a*c]) + Log[a - b*x^2 + c*x^4]/(4*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 1114

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 65, normalized size = 1.02 \begin {gather*} \frac {\frac {2 b \tan ^{-1}\left (\frac {2 c x^2-b}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}+\log \left (a-b x^2+c x^4\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a - b*x^2 + c*x^4),x]

[Out]

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

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.69, size = 206, normalized size = 3.22 \begin {gather*} \left [\frac {\sqrt {b^{2} - 4 \, a c} b \log \left (\frac {2 \, c^{2} x^{4} - 2 \, b c x^{2} + b^{2} - 2 \, a c - {\left (2 \, c x^{2} - b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{4} - b x^{2} + a}\right ) + {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} - b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}, -\frac {2 \, \sqrt {-b^{2} + 4 \, a c} b \arctan \left (-\frac {{\left (2 \, c x^{2} - b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{2} - 4 \, a c\right )} \log \left (c x^{4} - b x^{2} + a\right )}{4 \, {\left (b^{2} c - 4 \, a c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.57, size = 62, normalized size = 0.97 \begin {gather*} \frac {b \arctan \left (\frac {2 \, c x^{2} - b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c} + \frac {\log \left (c x^{4} - b x^{2} + a\right )}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 63, normalized size = 0.98 \begin {gather*} \frac {b \arctan \left (\frac {2 c \,x^{2}-b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c}+\frac {\ln \left (c \,x^{4}-b \,x^{2}+a \right )}{4 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3/(c*x^4-b*x^2+a),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 = 4.40, size = 120, normalized size = 1.88 \begin {gather*} \frac {4\,a\,c\,\ln \left (c\,x^4-b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b^2\,\ln \left (c\,x^4-b\,x^2+a\right )}{16\,a\,c^2-4\,b^2\,c}-\frac {b\,\mathrm {atan}\left (\frac {b}{\sqrt {4\,a\,c-b^2}}-\frac {2\,c\,x^2}{\sqrt {4\,a\,c-b^2}}\right )}{2\,c\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a - b*x^2 + c*x^4),x)

[Out]

(4*a*c*log(a - b*x^2 + c*x^4))/(16*a*c^2 - 4*b^2*c) - (b^2*log(a - b*x^2 + c*x^4))/(16*a*c^2 - 4*b^2*c) - (b*a
tan(b/(4*a*c - b^2)^(1/2) - (2*c*x^2)/(4*a*c - b^2)^(1/2)))/(2*c*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 1.45, size = 223, normalized size = 3.48 \begin {gather*} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {8 a c \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) - 2 a - 2 b^{2} \left (- \frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} + \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) \log {\left (x^{2} + \frac {8 a c \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right ) - 2 a - 2 b^{2} \left (\frac {b \sqrt {- 4 a c + b^{2}}}{4 c \left (4 a c - b^{2}\right )} + \frac {1}{4 c}\right )}{b} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c))*log(x**2 + (8*a*c*(-b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c
- b**2)) + 1/(4*c)) - 2*a - 2*b**2*(-b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c)))/b) + (b*sqrt(-4*a*
c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c))*log(x**2 + (8*a*c*(b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4
*c)) - 2*a - 2*b**2*(b*sqrt(-4*a*c + b**2)/(4*c*(4*a*c - b**2)) + 1/(4*c)))/b)

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