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

Optimal. Leaf size=153 \[ -\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{384 c^{7/2}}+\frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{192 c^3}-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c} \]

[Out]

-5/72*b*(c*x^6+b*x^3+a)^(3/2)/c^2+1/12*x^3*(c*x^6+b*x^3+a)^(3/2)/c-1/384*(-4*a*c+b^2)*(-4*a*c+5*b^2)*arctanh(1
/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(7/2)+1/192*(-4*a*c+5*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c
^3

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Rubi [A]  time = 0.14, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1357, 742, 640, 612, 621, 206} \[ \frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{192 c^3}-\frac {\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{384 c^{7/2}}-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c} \]

Antiderivative was successfully verified.

[In]

Int[x^8*Sqrt[a + b*x^3 + c*x^6],x]

[Out]

((5*b^2 - 4*a*c)*(b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(192*c^3) - (5*b*(a + b*x^3 + c*x^6)^(3/2))/(72*c^2) +
(x^3*(a + b*x^3 + c*x^6)^(3/2))/(12*c) - ((b^2 - 4*a*c)*(5*b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt
[a + b*x^3 + c*x^6])])/(384*c^(7/2))

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 612

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

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 x^8 \sqrt {a+b x^3+c x^6} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int x^2 \sqrt {a+b x+c x^2} \, dx,x,x^3\right )\\ &=\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c}+\frac {\operatorname {Subst}\left (\int \left (-a-\frac {5 b x}{2}\right ) \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{12 c}\\ &=-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c}+\frac {\left (5 b^2-4 a c\right ) \operatorname {Subst}\left (\int \sqrt {a+b x+c x^2} \, dx,x,x^3\right )}{48 c^2}\\ &=\frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{192 c^3}-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x+c x^2}} \, dx,x,x^3\right )}{384 c^3}\\ &=\frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{192 c^3}-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c}-\frac {\left (\left (b^2-4 a c\right ) \left (5 b^2-4 a c\right )\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 )}{192 c^3}\\ &=\frac {\left (5 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{192 c^3}-\frac {5 b \left (a+b x^3+c x^6\right )^{3/2}}{72 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{3/2}}{12 c}-\frac {\left (b^2-4 a c\right ) \left (5 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 )}{384 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 136, normalized size = 0.89 \[ \frac {2 \sqrt {c} \sqrt {a+b x^3+c x^6} \left (b \left (8 c^2 x^6-52 a c\right )+24 c^2 x^3 \left (a+2 c x^6\right )+15 b^3-10 b^2 c x^3\right )-3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{1152 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]*(15*b^3 - 10*b^2*c*x^3 + 24*c^2*x^3*(a + 2*c*x^6) + b*(-52*a*c + 8*c^2*x^6)
) - 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(1152*c^(7
/2))

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fricas [A]  time = 1.04, size = 303, normalized size = 1.98 \[ \left [\frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} x^{9} + 8 \, b c^{3} x^{6} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{2304 \, c^{4}}, \frac {3 \, {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} x^{9} + 8 \, b c^{3} x^{6} + 15 \, b^{3} c - 52 \, a b c^{2} - 2 \, {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{1152 \, c^{4}}\right ] \]

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]

[1/2304*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*x^9 + 8*b*c^3*x^6 + 15*b^3*c - 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a
*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^4, 1/1152*(3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*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*(48*c^4*x^9 + 8*b*c^3*x^6 + 15*b^3*c
- 52*a*b*c^2 - 2*(5*b^2*c^2 - 12*a*c^3)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^4]

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

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(sqrt(c*x^6 + b*x^3 + a)*x^8, x)

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

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 positive, negative or zero?

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mupad [B]  time = 1.59, size = 193, normalized size = 1.26 \[ \frac {x^3\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}}{12\,c}-\frac {a\,\left (\left (\frac {b}{4\,c}+\frac {x^3}{2}\right )\,\sqrt {c\,x^6+b\,x^3+a}+\frac {\ln \left (\sqrt {c\,x^6+b\,x^3+a}+\frac {c\,x^3+\frac {b}{2}}{\sqrt {c}}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{12\,c}-\frac {5\,b\,\left (\frac {\left (8\,c\,\left (c\,x^6+a\right )-3\,b^2+2\,b\,c\,x^3\right )\,\sqrt {c\,x^6+b\,x^3+a}}{24\,c^2}+\frac {\ln \left (2\,\sqrt {c\,x^6+b\,x^3+a}+\frac {2\,c\,x^3+b}{\sqrt {c}}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}\right )}{24\,c} \]

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]

(x^3*(a + b*x^3 + c*x^6)^(3/2))/(12*c) - (a*((b/(4*c) + x^3/2)*(a + b*x^3 + c*x^6)^(1/2) + (log((a + b*x^3 + c
*x^6)^(1/2) + (b/2 + c*x^3)/c^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(12*c) - (5*b*(((8*c*(a + c*x^6) - 3*b^2 + 2
*b*c*x^3)*(a + b*x^3 + c*x^6)^(1/2))/(24*c^2) + (log(2*(a + b*x^3 + c*x^6)^(1/2) + (b + 2*c*x^3)/c^(1/2))*(b^3
- 4*a*b*c))/(16*c^(5/2))))/(24*c)

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

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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