3.4.26 \(\int \frac {x^8}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\) [326]

Optimal. Leaf size=71 \[ -\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {128 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^3} \]

[Out]

128/81*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))*c^(1/2)/d^3-2/27*c/d^3/(d*x^3+c)^(1/2)-2/3*(d*x^3+c)^(1/2)/d^3

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Rubi [A]
time = 0.05, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {457, 89, 65, 212} \begin {gather*} -\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {128 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(-2*c)/(27*d^3*Sqrt[c + d*x^3]) - (2*Sqrt[c + d*x^3])/(3*d^3) + (128*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c
])])/(81*d^3)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 89

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], (c + d*x)^n*((e + f*x)^IntegerPart[p]/(a + b*x)), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (\frac {c}{9 d^2 (c+d x)^{3/2}}-\frac {1}{d^2 \sqrt {c+d x}}+\frac {64 c}{9 d^2 (8 c-d x) \sqrt {c+d x}}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {(64 c) \text {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27 d^2}\\ &=-\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {(128 c) \text {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{27 d^3}\\ &=-\frac {2 c}{27 d^3 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{3 d^3}+\frac {128 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^3}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 59, normalized size = 0.83 \begin {gather*} \frac {2 \left (-\frac {3 \left (10 c+9 d x^3\right )}{\sqrt {c+d x^3}}+64 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{81 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*((-3*(10*c + 9*d*x^3))/Sqrt[c + d*x^3] + 64*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/(81*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.42, size = 498, normalized size = 7.01 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/d*(2/3/d^2*c/((x^3+c/d)*d)^(1/2)+2/3*(d*x^3+c)^(1/2)/d^2)+16/3*c/d^3/(d*x^3+c)^(1/2)-64*c^2/d^2*(2/27/d/c/(
(x^3+c/d)*d)^(1/2)+1/243*I/d^3/c^2*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c
*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1
/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c
*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*E
llipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/
2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/
3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),
_alpha=RootOf(_Z^3*d-8*c)))

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Maxima [A]
time = 0.49, size = 68, normalized size = 0.96 \begin {gather*} -\frac {2 \, {\left (32 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 27 \, \sqrt {d x^{3} + c} + \frac {3 \, c}{\sqrt {d x^{3} + c}}\right )}}{81 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

-2/81*(32*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 27*sqrt(d*x^3 + c) + 3*c/
sqrt(d*x^3 + c))/d^3

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Fricas [A]
time = 2.48, size = 161, normalized size = 2.27 \begin {gather*} \left [\frac {2 \, {\left (32 \, {\left (d x^{3} + c\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - 3 \, {\left (9 \, d x^{3} + 10 \, c\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{4} x^{3} + c d^{3}\right )}}, -\frac {2 \, {\left (64 \, {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (9 \, d x^{3} + 10 \, c\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{4} x^{3} + c d^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[2/81*(32*(d*x^3 + c)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - 3*(9*d*x^3 + 10*
c)*sqrt(d*x^3 + c))/(d^4*x^3 + c*d^3), -2/81*(64*(d*x^3 + c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) +
3*(9*d*x^3 + 10*c)*sqrt(d*x^3 + c))/(d^4*x^3 + c*d^3)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)

[Out]

Timed out

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Giac [A]
time = 0.75, size = 58, normalized size = 0.82 \begin {gather*} -\frac {128 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{3}} - \frac {2 \, \sqrt {d x^{3} + c}}{3 \, d^{3}} - \frac {2 \, c}{27 \, \sqrt {d x^{3} + c} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-128/81*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/3*sqrt(d*x^3 + c)/d^3 - 2/27*c/(sqrt(d*x^3 +
c)*d^3)

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Mupad [B]
time = 3.71, size = 75, normalized size = 1.06 \begin {gather*} \frac {64\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,d^3}-\frac {2\,c}{27\,d^3\,\sqrt {d\,x^3+c}}-\frac {2\,\sqrt {d\,x^3+c}}{3\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((c + d*x^3)^(3/2)*(8*c - d*x^3)),x)

[Out]

(64*c^(1/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(81*d^3) - (2*c)/(27*d^3*(c + d*x
^3)^(1/2)) - (2*(c + d*x^3)^(1/2))/(3*d^3)

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