3.221 \(\int \frac {x^8}{\sqrt {a+b x^3+c x^6}} \, dx\)

Optimal. Leaf size=104 \[ \frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{24 c^{5/2}}-\frac {b \sqrt {a+b x^3+c x^6}}{4 c^2}+\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c} \]

[Out]

1/24*(-4*a*c+3*b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(5/2)-1/4*b*(c*x^6+b*x^3+a)^(1/2)
/c^2+1/6*x^3*(c*x^6+b*x^3+a)^(1/2)/c

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Rubi [A]  time = 0.09, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1357, 742, 640, 621, 206} \[ \frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{24 c^{5/2}}-\frac {b \sqrt {a+b x^3+c x^6}}{4 c^2}+\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x^8/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

-(b*Sqrt[a + b*x^3 + c*x^6])/(4*c^2) + (x^3*Sqrt[a + b*x^3 + c*x^6])/(6*c) + ((3*b^2 - 4*a*c)*ArcTanh[(b + 2*c
*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(24*c^(5/2))

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 742

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 1)), x] + Dist[1/(c*(m + 2*p + 1)), Int[(d + e*x)^(m - 2)*Simp[c*d^2
*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]*(a + b*x + c*x^2)^p, x], x] /;
FreeQ[{a, b, c, d, e, m, p}, 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]
&& If[RationalQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadraticQ[a, b, c, d, e, m,
p, x]

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}{\sqrt {a+b x^3+c x^6}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )\\ &=\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c}+\frac {\operatorname {Subst}\left (\int \frac {-a-\frac {3 b x}{2}}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{6 c}\\ &=-\frac {b \sqrt {a+b x^3+c x^6}}{4 c^2}+\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c}+\frac {\left (3 b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{24 c^2}\\ &=-\frac {b \sqrt {a+b x^3+c x^6}}{4 c^2}+\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c}+\frac {\left (3 b^2-4 a c\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x^3}{\sqrt {a+b x^3+c x^6}}\right )}{12 c^2}\\ &=-\frac {b \sqrt {a+b x^3+c x^6}}{4 c^2}+\frac {x^3 \sqrt {a+b x^3+c x^6}}{6 c}+\frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{24 c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 88, normalized size = 0.85 \[ \frac {\left (3 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )+2 \sqrt {c} \left (2 c x^3-3 b\right ) \sqrt {a+b x^3+c x^6}}{24 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/Sqrt[a + b*x^3 + c*x^6],x]

[Out]

(2*Sqrt[c]*(-3*b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6] + (3*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a
+ b*x^3 + c*x^6])])/(24*c^(5/2))

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fricas [A]  time = 0.64, size = 203, normalized size = 1.95 \[ \left [-\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c^{2} x^{3} - 3 \, b c\right )}}{48 \, c^{3}}, -\frac {{\left (3 \, b^{2} - 4 \, a c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c^{2} x^{3} - 3 \, b c\right )}}{24 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^6+b*x^3+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*((3*b^2 - 4*a*c)*sqrt(c)*log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 + 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqr
t(c) - 4*a*c) - 4*sqrt(c*x^6 + b*x^3 + a)*(2*c^2*x^3 - 3*b*c))/c^3, -1/24*((3*b^2 - 4*a*c)*sqrt(-c)*arctan(1/2
*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 2*sqrt(c*x^6 + b*x^3 + a)*(2*c^2*
x^3 - 3*b*c))/c^3]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\sqrt {c x^{6} + b x^{3} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(c*x^6+b*x^3+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^8/sqrt(c*x^6 + b*x^3 + a), x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\sqrt {c \,x^{6}+b \,x^{3}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(c*x^6+b*x^3+a)^(1/2),x)

[Out]

int(x^8/(c*x^6+b*x^3+a)^(1/2),x)

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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)^(1/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^8}{\sqrt {c\,x^6+b\,x^3+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^3 + c*x^6)^(1/2),x)

[Out]

int(x^8/(a + b*x^3 + c*x^6)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\sqrt {a + b x^{3} + c x^{6}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(c*x**6+b*x**3+a)**(1/2),x)

[Out]

Integral(x**8/sqrt(a + b*x**3 + c*x**6), x)

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