3.138 \(\int \frac {x^8}{a+b x^3+c x^6} \, dx\)
Optimal. Leaf size=81 \[ -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^3+c x^6\right )}{6 c^2}+\frac {x^3}{3 c} \]
[Out]
1/3*x^3/c-1/6*b*ln(c*x^6+b*x^3+a)/c^2-1/3*(-2*a*c+b^2)*arctanh((2*c*x^3+b)/(-4*a*c+b^2)^(1/2))/c^2/(-4*a*c+b^2
)^(1/2)
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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 =
{1357, 703, 634, 618, 206, 628} \[ -\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^3+c x^6\right )}{6 c^2}+\frac {x^3}{3 c} \]
Antiderivative was successfully verified.
[In]
Int[x^8/(a + b*x^3 + c*x^6),x]
[Out]
x^3/(3*c) - ((b^2 - 2*a*c)*ArcTanh[(b + 2*c*x^3)/Sqrt[b^2 - 4*a*c]])/(3*c^2*Sqrt[b^2 - 4*a*c]) - (b*Log[a + b*
x^3 + c*x^6])/(6*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 1357
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[
b^2 - 4*a*c, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rubi steps
\begin {align*} \int \frac {x^8}{a+b x^3+c x^6} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{a+b x+c x^2} \, dx,x,x^3\right )\\ &=\frac {x^3}{3 c}+\frac {\operatorname {Subst}\left (\int \frac {-a-b x}{a+b x+c x^2} \, dx,x,x^3\right )}{3 c}\\ &=\frac {x^3}{3 c}-\frac {b \operatorname {Subst}\left (\int \frac {b+2 c x}{a+b x+c x^2} \, dx,x,x^3\right )}{6 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^3\right )}{6 c^2}\\ &=\frac {x^3}{3 c}-\frac {b \log \left (a+b x^3+c x^6\right )}{6 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^3\right )}{3 c^2}\\ &=\frac {x^3}{3 c}-\frac {\left (b^2-2 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{\sqrt {b^2-4 a c}}\right )}{3 c^2 \sqrt {b^2-4 a c}}-\frac {b \log \left (a+b x^3+c x^6\right )}{6 c^2}\\ \end {align*}
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Mathematica [A] time = 0.05, size = 78, normalized size = 0.96 \[ \frac {\frac {2 \left (b^2-2 a c\right ) \tan ^{-1}\left (\frac {b+2 c x^3}{\sqrt {4 a c-b^2}}\right )}{\sqrt {4 a c-b^2}}-b \log \left (a+b x^3+c x^6\right )+2 c x^3}{6 c^2} \]
Antiderivative was successfully verified.
[In]
Integrate[x^8/(a + b*x^3 + c*x^6),x]
[Out]
(2*c*x^3 + (2*(b^2 - 2*a*c)*ArcTan[(b + 2*c*x^3)/Sqrt[-b^2 + 4*a*c]])/Sqrt[-b^2 + 4*a*c] - b*Log[a + b*x^3 + c
*x^6])/(6*c^2)
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fricas [A] time = 0.93, size = 254, normalized size = 3.14 \[ \left [\frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} - {\left (b^{2} - 2 \, a c\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{6} + 2 \, b c x^{3} + b^{2} - 2 \, a c + {\left (2 \, c x^{3} + b\right )} \sqrt {b^{2} - 4 \, a c}}{c x^{6} + b x^{3} + a}\right ) - {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}, \frac {2 \, {\left (b^{2} c - 4 \, a c^{2}\right )} x^{3} - 2 \, {\left (b^{2} - 2 \, a c\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {{\left (2 \, c x^{3} + b\right )} \sqrt {-b^{2} + 4 \, a c}}{b^{2} - 4 \, a c}\right ) - {\left (b^{3} - 4 \, a b c\right )} \log \left (c x^{6} + b x^{3} + a\right )}{6 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8/(c*x^6+b*x^3+a),x, algorithm="fricas")
[Out]
[1/6*(2*(b^2*c - 4*a*c^2)*x^3 - (b^2 - 2*a*c)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^6 + 2*b*c*x^3 + b^2 - 2*a*c + (2*
c*x^3 + b)*sqrt(b^2 - 4*a*c))/(c*x^6 + b*x^3 + a)) - (b^3 - 4*a*b*c)*log(c*x^6 + b*x^3 + a))/(b^2*c^2 - 4*a*c^
3), 1/6*(2*(b^2*c - 4*a*c^2)*x^3 - 2*(b^2 - 2*a*c)*sqrt(-b^2 + 4*a*c)*arctan(-(2*c*x^3 + b)*sqrt(-b^2 + 4*a*c)
/(b^2 - 4*a*c)) - (b^3 - 4*a*b*c)*log(c*x^6 + b*x^3 + a))/(b^2*c^2 - 4*a*c^3)]
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giac [A] time = 1.14, size = 75, normalized size = 0.93 \[ \frac {x^{3}}{3 \, c} - \frac {b \log \left (c x^{6} + b x^{3} + a\right )}{6 \, c^{2}} + \frac {{\left (b^{2} - 2 \, a c\right )} \arctan \left (\frac {2 \, c x^{3} + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{3 \, \sqrt {-b^{2} + 4 \, a c} c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8/(c*x^6+b*x^3+a),x, algorithm="giac")
[Out]
1/3*x^3/c - 1/6*b*log(c*x^6 + b*x^3 + a)/c^2 + 1/3*(b^2 - 2*a*c)*arctan((2*c*x^3 + b)/sqrt(-b^2 + 4*a*c))/(sqr
t(-b^2 + 4*a*c)*c^2)
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maple [A] time = 0.01, size = 111, normalized size = 1.37 \[ \frac {x^{3}}{3 c}-\frac {2 a \arctan \left (\frac {2 c \,x^{3}+b}{\sqrt {4 a c -b^{2}}}\right )}{3 \sqrt {4 a c -b^{2}}\, c}+\frac {b^{2} \arctan \left (\frac {2 c \,x^{3}+b}{\sqrt {4 a c -b^{2}}}\right )}{3 \sqrt {4 a c -b^{2}}\, c^{2}}-\frac {b \ln \left (c \,x^{6}+b \,x^{3}+a \right )}{6 c^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^8/(c*x^6+b*x^3+a),x)
[Out]
1/3*x^3/c-1/6*b*ln(c*x^6+b*x^3+a)/c^2-2/3/c/(4*a*c-b^2)^(1/2)*arctan((2*c*x^3+b)/(4*a*c-b^2)^(1/2))*a+1/3/c^2/
(4*a*c-b^2)^(1/2)*arctan((2*c*x^3+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 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8/(c*x^6+b*x^3+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 = 1.98, size = 1758, normalized size = 21.70 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^8/(a + b*x^3 + c*x^6),x)
[Out]
x^3/(3*c) + (log(a + b*x^3 + c*x^6)*(3*b^3 - 12*a*b*c))/(2*(36*a*c^3 - 9*b^2*c^2)) + (atan((4*c^3*x^3*(4*a*c -
b^2)^(3/2)*((b*((b^5 + a^2*b*c^2 - 2*a*b^3*c)/c^3 + ((3*b^3 - 12*a*b*c)*((6*a^2*c^4 + 12*b^4*c^2 - 18*a*b^2*c
^3)/c^3 + ((3*b^3 - 12*a*b*c)*((45*b^3*c^4 - 36*a*b*c^5)/c^3 + (27*b^2*c^3*(3*b^3 - 12*a*b*c))/(36*a*c^3 - 9*b
^2*c^2)))/(2*(36*a*c^3 - 9*b^2*c^2))))/(2*(36*a*c^3 - 9*b^2*c^2)) - ((((2*a*c - b^2)*((45*b^3*c^4 - 36*a*b*c^5
)/c^3 + (27*b^2*c^3*(3*b^3 - 12*a*b*c))/(36*a*c^3 - 9*b^2*c^2)))/(6*c^2*(4*a*c - b^2)^(1/2)) + (9*b^2*c*(3*b^3
- 12*a*b*c)*(2*a*c - b^2))/(2*(4*a*c - b^2)^(1/2)*(36*a*c^3 - 9*b^2*c^2)))*(2*a*c - b^2))/(6*c^2*(4*a*c - b^2
)^(1/2)) - (3*b^2*(3*b^3 - 12*a*b*c)*(2*a*c - b^2)^2)/(4*c*(4*a*c - b^2)*(36*a*c^3 - 9*b^2*c^2))))/(4*a^2*c) +
((2*a*c - b^2)*(((3*b^3 - 12*a*b*c)*(((2*a*c - b^2)*((45*b^3*c^4 - 36*a*b*c^5)/c^3 + (27*b^2*c^3*(3*b^3 - 12*
a*b*c))/(36*a*c^3 - 9*b^2*c^2)))/(6*c^2*(4*a*c - b^2)^(1/2)) + (9*b^2*c*(3*b^3 - 12*a*b*c)*(2*a*c - b^2))/(2*(
4*a*c - b^2)^(1/2)*(36*a*c^3 - 9*b^2*c^2))))/(2*(36*a*c^3 - 9*b^2*c^2)) - (b^2*(2*a*c - b^2)^3)/(4*c^3*(4*a*c
- b^2)^(3/2)) + (((6*a^2*c^4 + 12*b^4*c^2 - 18*a*b^2*c^3)/c^3 + ((3*b^3 - 12*a*b*c)*((45*b^3*c^4 - 36*a*b*c^5)
/c^3 + (27*b^2*c^3*(3*b^3 - 12*a*b*c))/(36*a*c^3 - 9*b^2*c^2)))/(2*(36*a*c^3 - 9*b^2*c^2)))*(2*a*c - b^2))/(6*
c^2*(4*a*c - b^2)^(1/2))))/(4*a^2*c*(4*a*c - b^2)^(1/2))))/(b^6 - 8*a^3*c^3 + 12*a^2*b^2*c^2 - 6*a*b^4*c) - (c
^2*(2*a*c - b^2)*(4*a*c - b^2)*(((3*b^3 - 12*a*b*c)*((((36*a^2*c^5 - 72*a*b^2*c^4)/c^3 - (54*a*b*c^3*(3*b^3 -
12*a*b*c))/(36*a*c^3 - 9*b^2*c^2))*(2*a*c - b^2))/(6*c^2*(4*a*c - b^2)^(1/2)) - (9*a*b*c*(3*b^3 - 12*a*b*c)*(2
*a*c - b^2))/((4*a*c - b^2)^(1/2)*(36*a*c^3 - 9*b^2*c^2))))/(2*(36*a*c^3 - 9*b^2*c^2)) - (((15*a*b^3*c^2 - 12*
a^2*b*c^3)/c^3 - ((3*b^3 - 12*a*b*c)*((36*a^2*c^5 - 72*a*b^2*c^4)/c^3 - (54*a*b*c^3*(3*b^3 - 12*a*b*c))/(36*a*
c^3 - 9*b^2*c^2)))/(2*(36*a*c^3 - 9*b^2*c^2)))*(2*a*c - b^2))/(6*c^2*(4*a*c - b^2)^(1/2)) + (a*b*(2*a*c - b^2)
^3)/(2*c^3*(4*a*c - b^2)^(3/2))))/(a^2*(b^6 - 8*a^3*c^3 + 12*a^2*b^2*c^2 - 6*a*b^4*c)) + (b*c^2*(4*a*c - b^2)^
(3/2)*((a*b^4 - a^2*b^2*c)/c^3 + ((3*b^3 - 12*a*b*c)*((15*a*b^3*c^2 - 12*a^2*b*c^3)/c^3 - ((3*b^3 - 12*a*b*c)*
((36*a^2*c^5 - 72*a*b^2*c^4)/c^3 - (54*a*b*c^3*(3*b^3 - 12*a*b*c))/(36*a*c^3 - 9*b^2*c^2)))/(2*(36*a*c^3 - 9*b
^2*c^2))))/(2*(36*a*c^3 - 9*b^2*c^2)) + (((((36*a^2*c^5 - 72*a*b^2*c^4)/c^3 - (54*a*b*c^3*(3*b^3 - 12*a*b*c))/
(36*a*c^3 - 9*b^2*c^2))*(2*a*c - b^2))/(6*c^2*(4*a*c - b^2)^(1/2)) - (9*a*b*c*(3*b^3 - 12*a*b*c)*(2*a*c - b^2)
)/((4*a*c - b^2)^(1/2)*(36*a*c^3 - 9*b^2*c^2)))*(2*a*c - b^2))/(6*c^2*(4*a*c - b^2)^(1/2)) - (3*a*b*(3*b^3 - 1
2*a*b*c)*(2*a*c - b^2)^2)/(2*c*(4*a*c - b^2)*(36*a*c^3 - 9*b^2*c^2))))/(a^2*(b^6 - 8*a^3*c^3 + 12*a^2*b^2*c^2
- 6*a*b^4*c)))*(2*a*c - b^2))/(3*c^2*(4*a*c - b^2)^(1/2))
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sympy [B] time = 2.70, size = 316, normalized size = 3.90 \[ \left (- \frac {b}{6 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{3} + \frac {- a b - 12 a c^{2} \left (- \frac {b}{6 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right ) + 3 b^{2} c \left (- \frac {b}{6 c^{2}} - \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \left (- \frac {b}{6 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right ) \log {\left (x^{3} + \frac {- a b - 12 a c^{2} \left (- \frac {b}{6 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right ) + 3 b^{2} c \left (- \frac {b}{6 c^{2}} + \frac {\sqrt {- 4 a c + b^{2}} \left (2 a c - b^{2}\right )}{6 c^{2} \left (4 a c - b^{2}\right )}\right )}{2 a c - b^{2}} \right )} + \frac {x^{3}}{3 c} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**8/(c*x**6+b*x**3+a),x)
[Out]
(-b/(6*c**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(6*c**2*(4*a*c - b**2)))*log(x**3 + (-a*b - 12*a*c**2*(-b/(6
*c**2) - sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(6*c**2*(4*a*c - b**2))) + 3*b**2*c*(-b/(6*c**2) - sqrt(-4*a*c + b
**2)*(2*a*c - b**2)/(6*c**2*(4*a*c - b**2))))/(2*a*c - b**2)) + (-b/(6*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b*
*2)/(6*c**2*(4*a*c - b**2)))*log(x**3 + (-a*b - 12*a*c**2*(-b/(6*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(6
*c**2*(4*a*c - b**2))) + 3*b**2*c*(-b/(6*c**2) + sqrt(-4*a*c + b**2)*(2*a*c - b**2)/(6*c**2*(4*a*c - b**2))))/
(2*a*c - b**2)) + x**3/(3*c)
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