3.3.95 \(\int \frac {x^8}{(8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=83 \[ \frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {22}{81 d^3 \sqrt {c+d x^3}}-\frac {32 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 \sqrt {c} d^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {446, 89, 78, 63, 206} \begin {gather*} \frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {22}{81 d^3 \sqrt {c+d x^3}}-\frac {32 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 \sqrt {c} d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-22/(81*d^3*Sqrt[c + d*x^3]) + (64*c)/(27*d^3*(8*c - d*x^3)*Sqrt[c + d*x^3]) - (32*ArcTanh[Sqrt[c + d*x^3]/(3*
Sqrt[c])])/(243*Sqrt[c]*d^3)

Rule 63

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 446

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 )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\operatorname {Subst}\left (\int \frac {-24 c^2 d+9 c d^2 x}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{27 c d^3}\\ &=-\frac {22}{81 d^3 \sqrt {c+d x^3}}+\frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {16 \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{81 d^2}\\ &=-\frac {22}{81 d^3 \sqrt {c+d x^3}}+\frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {32 \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{81 d^3}\\ &=-\frac {22}{81 d^3 \sqrt {c+d x^3}}+\frac {64 c}{27 d^3 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {32 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 \sqrt {c} d^3}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 71, normalized size = 0.86 \begin {gather*} \frac {2 \left (\frac {3 \left (8 c+11 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {16 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )}{243 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((3*(8*c + 11*d*x^3))/((8*c - d*x^3)*Sqrt[c + d*x^3]) - (16*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/Sqrt[c]))
/(243*d^3)

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IntegrateAlgebraic [A]  time = 0.08, size = 115, normalized size = 1.39 \begin {gather*} \frac {\frac {16 c}{81 d^3}+\left (\frac {32 x^3}{243 \sqrt {c} d^2}-\frac {256 \sqrt {c}}{243 d^3}\right ) \sqrt {c+d x^3} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+\frac {22 x^3}{81 d^2}}{8 c \sqrt {c+d x^3}-d x^3 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*c)/(81*d^3) + (22*x^3)/(81*d^2) + ((-256*Sqrt[c])/(243*d^3) + (32*x^3)/(243*Sqrt[c]*d^2))*Sqrt[c + d*x^3]
*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(8*c*Sqrt[c + d*x^3] - d*x^3*Sqrt[c + d*x^3])

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fricas [A]  time = 0.71, size = 223, normalized size = 2.69 \begin {gather*} \left [\frac {2 \, {\left (8 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\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 (11 \, c d x^{3} + 8 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{243 \, {\left (c d^{5} x^{6} - 7 \, c^{2} d^{4} x^{3} - 8 \, c^{3} d^{3}\right )}}, \frac {2 \, {\left (16 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 3 \, {\left (11 \, c d x^{3} + 8 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{243 \, {\left (c d^{5} x^{6} - 7 \, c^{2} d^{4} x^{3} - 8 \, c^{3} 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)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[2/243*(8*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c))
- 3*(11*c*d*x^3 + 8*c^2)*sqrt(d*x^3 + c))/(c*d^5*x^6 - 7*c^2*d^4*x^3 - 8*c^3*d^3), 2/243*(16*(d^2*x^6 - 7*c*d*
x^3 - 8*c^2)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 3*(11*c*d*x^3 + 8*c^2)*sqrt(d*x^3 + c))/(c*d^5*
x^6 - 7*c^2*d^4*x^3 - 8*c^3*d^3)]

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giac [A]  time = 0.17, size = 67, normalized size = 0.81 \begin {gather*} \frac {32 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{243 \, \sqrt {-c} d^{3}} - \frac {2 \, {\left (11 \, d x^{3} + 8 \, c\right )}}{81 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c\right )} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/243*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/81*(11*d*x^3 + 8*c)/(((d*x^3 + c)^(3/2) - 9*sqr
t(d*x^3 + c)*c)*d^3)

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maple [C]  time = 0.18, size = 926, normalized size = 11.16

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d^3/(d*x^3+c)^(1/2)+64*c^2/d^2*(-1/243*(d*x^3+c)^(1/2)/(d*x^3-8*c)/c^2/d-2/243/((x^3+c/d)*d)^(1/2)/c^2/d-
1/1458*I/d^3/c^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)
^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^
(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/
3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I
*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c*d^2)^(
1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I*3^
(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d-8*c
)))+16*c/d^2*(2/27/((x^3+c/d)*d)^(1/2)/c/d+1/243*I/c^2/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*
(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(
-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x
^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)
-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c
*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)
*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(
1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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maxima [A]  time = 1.41, size = 81, normalized size = 0.98 \begin {gather*} \frac {2 \, {\left (\frac {8 \, \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {3 \, {\left (11 \, d x^{3} + 8 \, c\right )}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c}\right )}}{243 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/243*(8*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c)))/sqrt(c) - 3*(11*d*x^3 + 8*c)/((d*x^3
+ c)^(3/2) - 9*sqrt(d*x^3 + c)*c))/d^3

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mupad [B]  time = 4.30, size = 94, normalized size = 1.13 \begin {gather*} \frac {16\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{243\,\sqrt {c}\,d^3}+\frac {\sqrt {d\,x^3+c}\,\left (\frac {16\,c}{81\,d^3}+\frac {22\,x^3}{81\,d^2}\right )}{8\,c^2+7\,c\,d\,x^3-d^2\,x^6} \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)^2),x)

[Out]

(16*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(243*c^(1/2)*d^3) + ((c + d*x^3)^(1/2)*((
16*c)/(81*d^3) + (22*x^3)/(81*d^2)))/(8*c^2 - d^2*x^6 + 7*c*d*x^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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