3.7.52 \(\int \frac {x^5}{a+b x^2+c x^4} \, dx\)

Optimal. Leaf size=81 \[ -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac {x^2}{2 c} \]

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Rubi [A]  time = 0.08, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1114, 703, 634, 618, 206, 628} \begin {gather*} -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^2+c x^4\right )}{4 c^2}+\frac {x^2}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^2 + c*x^4),x]

[Out]

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

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 703

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

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

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Mathematica [A]  time = 0.04, size = 78, normalized size = 0.96 \begin {gather*} \frac {\frac {2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x^2}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-b \log \left (a+b x^2+c x^4\right )+2 c x^2}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^2 + c*x^4),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^5}{a+b x^2+c x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x^5/(a + b*x^2 + c*x^4),x]

[Out]

IntegrateAlgebraic[x^5/(a + b*x^2 + c*x^4), x]

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fricas [A]  time = 1.01, size = 254, normalized size = 3.14 \begin {gather*} \left [\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} - {\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} \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^{3} - 4 \, a b c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{2} - 2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{2} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{4} + b x^{2} + a\right )}{4 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a),x, algorithm="fricas")

[Out]

[1/4*(2*(b^2*c - 4*a*c^2)*x^2 - (b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*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^3 - 4*a*b*c)*log(c*x^4 + b*x^2 + a))/(b^2*c^2 - 4*a*c^
3), 1/4*(2*(b^2*c - 4*a*c^2)*x^2 - 2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^2 + b)*sqrt(-b^2 + 4*a*c)
/(b^2 - 4*a*c)) - (b^3 - 4*a*b*c)*log(c*x^4 + b*x^2 + a))/(b^2*c^2 - 4*a*c^3)]

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giac [A]  time = 0.62, size = 75, normalized size = 0.93 \begin {gather*} \frac {x^{2}}{2 \, c} - \frac {b \log \left (c x^{4} + b x^{2} + a\right )}{4 \, c^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{2} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{2 \, \sqrt {-b^{2} + 4 \, a c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^4+b*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 111, normalized size = 1.37 \begin {gather*} -\frac {a \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}\, c}+\frac {b^{2} \arctan \left (\frac {2 c \,x^{2}+b}{\sqrt {4 a c -b^{2}}}\right )}{2 \sqrt {4 a c -b^{2}}\, c^{2}}+\frac {x^{2}}{2 c}-\frac {b \ln \left (c \,x^{4}+b \,x^{2}+a \right )}{4 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^4+b*x^2+a),x)

[Out]

1/2/c*x^2-1/4*b*ln(c*x^4+b*x^2+a)/c^2-1/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*a+1/2/c^2/(4
*a*c-b^2)^(1/2)*arctan((2*c*x^2+b)/(4*a*c-b^2)^(1/2))*b^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^5/(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 = 4.75, size = 655, normalized size = 8.09 \begin {gather*} \frac {x^2}{2\,c}+\frac {\ln \left (c\,x^4+b\,x^2+a\right )\,\left (2\,b^3-8\,a\,b\,c\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}+\frac {\mathrm {atan}\left (\frac {2\,c^2\,\left (4\,a\,c-b^2\right )\,\left (\frac {\frac {\left (8\,a\,b+\frac {8\,a\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )\,\left (2\,a\,c-b^2\right )}{8\,c^2\,\sqrt {4\,a\,c-b^2}}+\frac {a\,\left (2\,b^3-8\,a\,b\,c\right )\,\left (2\,a\,c-b^2\right )}{\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}}{a}-x^2\,\left (\frac {\frac {\left (2\,a\,c-b^2\right )\,\left (\frac {4\,a\,c^3-6\,b^2\,c^2}{c^2}-\frac {4\,b\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )}{8\,c^2\,\sqrt {4\,a\,c-b^2}}-\frac {b\,\left (2\,b^3-8\,a\,b\,c\right )\,\left (2\,a\,c-b^2\right )}{2\,\sqrt {4\,a\,c-b^2}\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}}{a}+\frac {b\,\left (\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (\frac {4\,a\,c^3-6\,b^2\,c^2}{c^2}-\frac {4\,b\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}-\frac {b^3-a\,b\,c}{c^2}+\frac {b\,{\left (2\,a\,c-b^2\right )}^2}{2\,c^2\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )+\frac {b\,\left (\frac {a\,b^2}{c^2}+\frac {\left (2\,b^3-8\,a\,b\,c\right )\,\left (8\,a\,b+\frac {8\,a\,c^2\,\left (2\,b^3-8\,a\,b\,c\right )}{16\,a\,c^3-4\,b^2\,c^2}\right )}{2\,\left (16\,a\,c^3-4\,b^2\,c^2\right )}-\frac {a\,{\left (2\,a\,c-b^2\right )}^2}{c^2\,\left (4\,a\,c-b^2\right )}\right )}{2\,a\,\sqrt {4\,a\,c-b^2}}\right )}{4\,a^2\,c^2-4\,a\,b^2\,c+b^4}\right )\,\left (2\,a\,c-b^2\right )}{2\,c^2\,\sqrt {4\,a\,c-b^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x^2 + c*x^4),x)

[Out]

x^2/(2*c) + (log(a + b*x^2 + c*x^4)*(2*b^3 - 8*a*b*c))/(2*(16*a*c^3 - 4*b^2*c^2)) + (atan((2*c^2*(4*a*c - b^2)
*((((8*a*b + (8*a*c^2*(2*b^3 - 8*a*b*c))/(16*a*c^3 - 4*b^2*c^2))*(2*a*c - b^2))/(8*c^2*(4*a*c - b^2)^(1/2)) +
(a*(2*b^3 - 8*a*b*c)*(2*a*c - b^2))/((4*a*c - b^2)^(1/2)*(16*a*c^3 - 4*b^2*c^2)))/a - x^2*((((2*a*c - b^2)*((4
*a*c^3 - 6*b^2*c^2)/c^2 - (4*b*c^2*(2*b^3 - 8*a*b*c))/(16*a*c^3 - 4*b^2*c^2)))/(8*c^2*(4*a*c - b^2)^(1/2)) - (
b*(2*b^3 - 8*a*b*c)*(2*a*c - b^2))/(2*(4*a*c - b^2)^(1/2)*(16*a*c^3 - 4*b^2*c^2)))/a + (b*(((2*b^3 - 8*a*b*c)*
((4*a*c^3 - 6*b^2*c^2)/c^2 - (4*b*c^2*(2*b^3 - 8*a*b*c))/(16*a*c^3 - 4*b^2*c^2)))/(2*(16*a*c^3 - 4*b^2*c^2)) -
(b^3 - a*b*c)/c^2 + (b*(2*a*c - b^2)^2)/(2*c^2*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))) + (b*((a*b^2)/c^2
+ ((2*b^3 - 8*a*b*c)*(8*a*b + (8*a*c^2*(2*b^3 - 8*a*b*c))/(16*a*c^3 - 4*b^2*c^2)))/(2*(16*a*c^3 - 4*b^2*c^2))
- (a*(2*a*c - b^2)^2)/(c^2*(4*a*c - b^2))))/(2*a*(4*a*c - b^2)^(1/2))))/(b^4 + 4*a^2*c^2 - 4*a*b^2*c))*(2*a*c
- b^2))/(2*c^2*(4*a*c - b^2)^(1/2))

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sympy [B]  time = 2.14, size = 316, normalized size = 3.90 \begin {gather*} \left (- \frac {b}{4 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- a b - 8 a c^{2} \left (- \frac {b}{4 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} c \left (- \frac {b}{4 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \left (- \frac {b}{4 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{2} + \frac {- a b - 8 a c^{2} \left (- \frac {b}{4 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right ) + 2 b^{2} c \left (- \frac {b}{4 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{4 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \frac {x^{2}}{2 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**4+b*x**2+a),x)

[Out]

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

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