3.309 \(\int \frac {x^8}{(8 c-d x^3) \sqrt {c+d x^3}} \, dx\)

Optimal. Leaf size=71 \[ \frac {128 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3}-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]

[Out]

-2/9*(d*x^3+c)^(3/2)/d^3+128/9*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^3-14/3*c*(d*x^3+c)^(1/2)/d^3

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Rubi [A]  time = 0.07, 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 = {446, 88, 63, 206} \[ \frac {128 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3}-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*c*Sqrt[c + d*x^3])/(3*d^3) - (2*(c + d*x^3)^(3/2))/(9*d^3) + (128*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt
[c])])/(9*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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -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 ) \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {7 c}{d^2 \sqrt {c+d x}}+\frac {64 c^2}{d^2 (8 c-d x) \sqrt {c+d x}}-\frac {\sqrt {c+d x}}{d^2}\right ) \, dx,x,x^3\right )\\ &=-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (64 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (128 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d^3}\\ &=-\frac {14 c \sqrt {c+d x^3}}{3 d^3}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {128 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 58, normalized size = 0.82 \[ \frac {128 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-2 \sqrt {c+d x^3} \left (22 c+d x^3\right )}{9 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3]*(22*c + d*x^3) + 128*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^3)

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fricas [A]  time = 0.98, size = 121, normalized size = 1.70 \[ \left [\frac {2 \, {\left (32 \, c^{\frac {3}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (d x^{3} + 22 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d^{3}}, -\frac {2 \, {\left (64 \, \sqrt {-c} c \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (d x^{3} + 22 \, c\right )} \sqrt {d x^{3} + c}\right )}}{9 \, d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[2/9*(32*c^(3/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (d*x^3 + 22*c)*sqrt(d*x^3 + c
))/d^3, -2/9*(64*sqrt(-c)*c*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (d*x^3 + 22*c)*sqrt(d*x^3 + c))/d^3]

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giac [A]  time = 0.19, size = 65, normalized size = 0.92 \[ -\frac {128 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d^{3}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{6} + 21 \, \sqrt {d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

-128/9*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/9*((d*x^3 + c)^(3/2)*d^6 + 21*sqrt(d*x^3 +
c)*c*d^6)/d^9

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maple [C]  time = 0.16, size = 468, normalized size = 6.59 \[ -\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, x^{3}}{9 d}-\frac {4 \sqrt {d \,x^{3}+c}\, c}{9 d^{2}}\right ) d +\frac {16 \sqrt {d \,x^{3}+c}\, c}{3 d}}{d^{2}}-\frac {64 i c \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{27 d^{5} \sqrt {d \,x^{3}+c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)

[Out]

-1/d^2*((2/9*(d*x^3+c)^(1/2)/d*x^3-4/9*(d*x^3+c)^(1/2)*c/d^2)*d+16/3*(d*x^3+c)^(1/2)*c/d)-64/27*I*c/d^5*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 = 1.31, size = 66, normalized size = 0.93 \[ -\frac {2 \, {\left (32 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 21 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

-2/9*(32*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + (d*x^3 + c)^(3/2) + 21*sqr
t(d*x^3 + c)*c)/d^3

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mupad [B]  time = 3.39, size = 78, normalized size = 1.10 \[ \frac {64\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{9\,d^3}-\frac {44\,c\,\sqrt {d\,x^3+c}}{9\,d^3}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((c + d*x^3)^(1/2)*(8*c - d*x^3)),x)

[Out]

(64*c^(3/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(9*d^3) - (44*c*(c + d*x^3)^(1/2)
)/(9*d^3) - (2*x^3*(c + d*x^3)^(1/2))/(9*d^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)

[Out]

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

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