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

Optimal. Leaf size=69 \[ -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {\log \left (a x^4+2 a x^2+a-b\right )}{2 a}+\frac {x^2}{2 a} \]

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Rubi [A]  time = 0.08, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1114, 703, 634, 618, 206, 628} \begin {gather*} -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}}-\frac {\log \left (a x^4+2 a x^2+a-b\right )}{2 a}+\frac {x^2}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/(a - b + 2*a*x^2 + a*x^4),x]

[Out]

x^2/(2*a) - ((a + b)*ArcTanh[(Sqrt[a]*(1 + x^2))/Sqrt[b]])/(2*a^(3/2)*Sqrt[b]) - Log[a - b + 2*a*x^2 + a*x^4]/
(2*a)

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

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Mathematica [A]  time = 0.04, size = 62, normalized size = 0.90 \begin {gather*} \frac {x^2-\log \left (a \left (x^2+1\right )^2-b\right )}{2 a}-\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \left (x^2+1\right )}{\sqrt {b}}\right )}{2 a^{3/2} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a - b + 2*a*x^2 + a*x^4),x]

[Out]

-1/2*((a + b)*ArcTanh[(Sqrt[a]*(1 + x^2))/Sqrt[b]])/(a^(3/2)*Sqrt[b]) + (x^2 - Log[-b + a*(1 + x^2)^2])/(2*a)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.05, size = 156, normalized size = 2.26 \begin {gather*} \left [\frac {2 \, a b x^{2} - 2 \, a b \log \left (a x^{4} + 2 \, a x^{2} + a - b\right ) + \sqrt {a b} {\left (a + b\right )} \log \left (\frac {a x^{4} + 2 \, a x^{2} - 2 \, \sqrt {a b} {\left (x^{2} + 1\right )} + a + b}{a x^{4} + 2 \, a x^{2} + a - b}\right )}{4 \, a^{2} b}, \frac {a b x^{2} - a b \log \left (a x^{4} + 2 \, a x^{2} + a - b\right ) + \sqrt {-a b} {\left (a + b\right )} \arctan \left (\frac {\sqrt {-a b}}{a x^{2} + a}\right )}{2 \, a^{2} b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.26, size = 60, normalized size = 0.87 \begin {gather*} \frac {x^{2}}{2 \, a} + \frac {{\left (a + b\right )} \arctan \left (\frac {a x^{2} + a}{\sqrt {-a b}}\right )}{2 \, \sqrt {-a b} a} - \frac {\log \left (a x^{4} + 2 \, a x^{2} + a - b\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 86, normalized size = 1.25 \begin {gather*} -\frac {b \arctanh \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{2 \sqrt {a b}\, a}+\frac {x^{2}}{2 a}-\frac {\arctanh \left (\frac {2 a \,x^{2}+2 a}{2 \sqrt {a b}}\right )}{2 \sqrt {a b}}-\frac {\ln \left (a \,x^{4}+2 a \,x^{2}+a -b \right )}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 3.00, size = 74, normalized size = 1.07 \begin {gather*} \frac {x^{2}}{2 \, a} + \frac {{\left (a + b\right )} \log \left (\frac {a x^{2} + a - \sqrt {a b}}{a x^{2} + a + \sqrt {a b}}\right )}{4 \, \sqrt {a b} a} - \frac {\log \left (a x^{4} + 2 \, a x^{2} + a - b\right )}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 166, normalized size = 2.41 \begin {gather*} \frac {x^2}{2\,a}-\ln \left (a\,\sqrt {a^3\,b}-b\,\sqrt {a^3\,b}-a^2\,b\,x^2+a\,x^2\,\sqrt {a^3\,b}\right )\,\left (\frac {\frac {a^2}{2}+\frac {\sqrt {a^3\,b}}{4}}{a^3}+\frac {\sqrt {a^3\,b}}{4\,a^2\,b}\right )-\ln \left (a\,\sqrt {a^3\,b}-b\,\sqrt {a^3\,b}+a^2\,b\,x^2+a\,x^2\,\sqrt {a^3\,b}\right )\,\left (\frac {\frac {a^2}{2}-\frac {\sqrt {a^3\,b}}{4}}{a^3}-\frac {\sqrt {a^3\,b}}{4\,a^2\,b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a - b + 2*a*x^2 + a*x^4),x)

[Out]

x^2/(2*a) - log(a*(a^3*b)^(1/2) - b*(a^3*b)^(1/2) - a^2*b*x^2 + a*x^2*(a^3*b)^(1/2))*((a^2/2 + (a^3*b)^(1/2)/4
)/a^3 + (a^3*b)^(1/2)/(4*a^2*b)) - log(a*(a^3*b)^(1/2) - b*(a^3*b)^(1/2) + a^2*b*x^2 + a*x^2*(a^3*b)^(1/2))*((
a^2/2 - (a^3*b)^(1/2)/4)/a^3 - (a^3*b)^(1/2)/(4*a^2*b))

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sympy [B]  time = 1.79, size = 138, normalized size = 2.00 \begin {gather*} \left (- \frac {1}{2 a} - \frac {\sqrt {a^{3} b} \left (a + b\right )}{4 a^{3} b}\right ) \log {\left (x^{2} + \frac {- 4 a b \left (- \frac {1}{2 a} - \frac {\sqrt {a^{3} b} \left (a + b\right )}{4 a^{3} b}\right ) + a - b}{a + b} \right )} + \left (- \frac {1}{2 a} + \frac {\sqrt {a^{3} b} \left (a + b\right )}{4 a^{3} b}\right ) \log {\left (x^{2} + \frac {- 4 a b \left (- \frac {1}{2 a} + \frac {\sqrt {a^{3} b} \left (a + b\right )}{4 a^{3} b}\right ) + a - b}{a + b} \right )} + \frac {x^{2}}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/(2*a) - sqrt(a**3*b)*(a + b)/(4*a**3*b))*log(x**2 + (-4*a*b*(-1/(2*a) - sqrt(a**3*b)*(a + b)/(4*a**3*b)) +
a - b)/(a + b)) + (-1/(2*a) + sqrt(a**3*b)*(a + b)/(4*a**3*b))*log(x**2 + (-4*a*b*(-1/(2*a) + sqrt(a**3*b)*(a
+ b)/(4*a**3*b)) + a - b)/(a + b)) + x**2/(2*a)

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