3.236 \(\int \frac {x^8}{(a+b x^3+c x^6)^{3/2}} \, dx\)

Optimal. Leaf size=120 \[ \frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 120, 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, 738, 640, 621, 206} \[ \frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^8/(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

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

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 738

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(
d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), 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] && LtQ[p, -1] && GtQ[m, 1] && 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}{\left (a+b x^3+c x^6\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{\left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 \operatorname {Subst}\left (\int \frac {2 a+b x}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{3 \left (b^2-4 a c\right )}\\ &=\frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{3 c}\\ &=\frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {2 \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 )}{3 c}\\ &=\frac {2 x^3 \left (2 a+b x^3\right )}{3 \left (b^2-4 a c\right ) \sqrt {a+b x^3+c x^6}}-\frac {2 b \sqrt {a+b x^3+c x^6}}{3 c \left (b^2-4 a c\right )}+\frac {\tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 c^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 107, normalized size = 0.89 \[ \frac {\frac {2 \sqrt {c} \left (a \left (b-2 c x^3\right )+b^2 x^3\right )}{\sqrt {a+b x^3+c x^6}}-\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3 c^{3/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/(a + b*x^3 + c*x^6)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.05, size = 387, normalized size = 3.22 \[ \left [\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{6} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + a b^{2} - 4 \, a^{2} c\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left ({\left (b^{2} c - 2 \, a c^{2}\right )} x^{3} + a b c\right )}}{6 \, {\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} + a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{3}\right )}}, -\frac {{\left ({\left (b^{2} c - 4 \, a c^{2}\right )} x^{6} + {\left (b^{3} - 4 \, a b c\right )} x^{3} + a b^{2} - 4 \, a^{2} c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) + 2 \, \sqrt {c x^{6} + b x^{3} + a} {\left ({\left (b^{2} c - 2 \, a c^{2}\right )} x^{3} + a b c\right )}}{3 \, {\left ({\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{6} + a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(((b^2*c - 4*a*c^2)*x^6 + (b^3 - 4*a*b*c)*x^3 + a*b^2 - 4*a^2*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)*sqrt(c) - 4*a*c) - 4*sqrt(c*x^6 + b*x^3 + a)*((b^2*c - 2*a*c^2)*x^3
+ a*b*c))/((b^2*c^3 - 4*a*c^4)*x^6 + a*b^2*c^2 - 4*a^2*c^3 + (b^3*c^2 - 4*a*b*c^3)*x^3), -1/3*(((b^2*c - 4*a*
c^2)*x^6 + (b^3 - 4*a*b*c)*x^3 + a*b^2 - 4*a^2*c)*sqrt(-c)*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b)*sq
rt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) + 2*sqrt(c*x^6 + b*x^3 + a)*((b^2*c - 2*a*c^2)*x^3 + a*b*c))/((b^2*c^3 - 4*a
*c^4)*x^6 + a*b^2*c^2 - 4*a^2*c^3 + (b^3*c^2 - 4*a*b*c^3)*x^3)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^8/(c*x^6+b*x^3+a)^(3/2),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)^(3/2),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 [B]  time = 1.66, size = 84, normalized size = 0.70 \[ \frac {\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )}{3\,c^{3/2}}+\frac {\frac {a\,b}{2}-x^3\,\left (a\,c-\frac {b^2}{2}\right )}{3\,c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^6+b\,x^3+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(a + b*x^3 + c*x^6)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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