3.399 \(\int \frac{x^8 \sqrt{c+d x^3}}{(8 c-d x^3)^2} \, dx\)
Optimal. Leaf size=102 \[ -\frac{352 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}+\frac{352 c \sqrt{c+d x^3}}{27 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]
[Out]
(352*c*Sqrt[c + d*x^3])/(27*d^3) + (2*(c + d*x^3)^(3/2))/(9*d^3) + (64*c*(c + d*x^3)^(3/2))/(27*d^3*(8*c - d*x
^3)) - (352*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^3)
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Rubi [A] time = 0.0758963, antiderivative size = 102, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used =
{446, 89, 80, 50, 63, 206} \[ -\frac{352 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}+\frac{352 c \sqrt{c+d x^3}}{27 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]
Antiderivative was successfully verified.
[In]
Int[(x^8*Sqrt[c + d*x^3])/(8*c - d*x^3)^2,x]
[Out]
(352*c*Sqrt[c + d*x^3])/(27*d^3) + (2*(c + d*x^3)^(3/2))/(9*d^3) + (64*c*(c + d*x^3)^(3/2))/(27*d^3*(8*c - d*x
^3)) - (352*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*d^3)
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]]
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 80
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]
Rule 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]
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 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])
Rubi steps
\begin{align*} \int \frac{x^8 \sqrt{c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{c+d x}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{c+d x} \left (104 c^2 d+9 c d^2 x\right )}{8 c-d x} \, dx,x,x^3\right )}{27 c d^3}\\ &=\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac{(176 c) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{8 c-d x} \, dx,x,x^3\right )}{27 d^2}\\ &=\frac{352 c \sqrt{c+d x^3}}{27 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac{\left (176 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{352 c \sqrt{c+d x^3}}{27 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac{\left (352 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{352 c \sqrt{c+d x^3}}{27 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{64 c \left (c+d x^3\right )^{3/2}}{27 d^3 \left (8 c-d x^3\right )}-\frac{352 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^3}\\ \end{align*}
Mathematica [A] time = 0.0496656, size = 90, normalized size = 0.88 \[ \frac{2 \sqrt{c+d x^3} \left (-488 c^2+41 c d x^3+d^2 x^6\right )+352 c^{3/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{9 d^3 \left (d x^3-8 c\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(x^8*Sqrt[c + d*x^3])/(8*c - d*x^3)^2,x]
[Out]
(2*Sqrt[c + d*x^3]*(-488*c^2 + 41*c*d*x^3 + d^2*x^6) + 352*c^(3/2)*(8*c - d*x^3)*ArcTanh[Sqrt[c + d*x^3]/(3*Sq
rt[c])])/(9*d^3*(-8*c + d*x^3))
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Maple [C] time = 0.014, size = 892, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x)
[Out]
2/9*(d*x^3+c)^(3/2)/d^3+16*c/d^2*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d
*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3
)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3
))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(
1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)
)*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alp
ha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(
1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))+64*c^2/d^2*(-1/3/d*(d*x^3+c)^(1/2)/(d*x^3-8*c)+1/54
*I/d^3/c*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/
3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*
3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)
*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(
x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(
1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1
/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c))
)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.09995, size = 433, normalized size = 4.25 \begin{align*} \left [\frac{2 \,{\left (88 \,{\left (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 ) +{\left (d^{2} x^{6} + 41 \, c d x^{3} - 488 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{9 \,{\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}, \frac{2 \,{\left (176 \,{\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) +{\left (d^{2} x^{6} + 41 \, c d x^{3} - 488 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{9 \,{\left (d^{4} x^{3} - 8 \, c d^{3}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")
[Out]
[2/9*(88*(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)) + (d^2*x^6 +
41*c*d*x^3 - 488*c^2)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3), 2/9*(176*(c*d*x^3 - 8*c^2)*sqrt(-c)*arctan(1/3*sqr
t(d*x^3 + c)*sqrt(-c)/c) + (d^2*x^6 + 41*c*d*x^3 - 488*c^2)*sqrt(d*x^3 + c))/(d^4*x^3 - 8*c*d^3)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**8*(d*x**3+c)**(1/2)/(-d*x**3+8*c)**2,x)
[Out]
Timed out
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Giac [A] time = 1.12101, size = 126, normalized size = 1.24 \begin{align*} \frac{352 \, c^{2} \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{9 \, \sqrt{-c} d^{3}} - \frac{64 \, \sqrt{d x^{3} + c} c^{2}}{3 \,{\left (d x^{3} - 8 \, c\right )} d^{3}} + \frac{2 \,{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} d^{6} + 48 \, \sqrt{d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^8*(d*x^3+c)^(1/2)/(-d*x^3+8*c)^2,x, algorithm="giac")
[Out]
352/9*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 64/3*sqrt(d*x^3 + c)*c^2/((d*x^3 - 8*c)*d^3) +
2/9*((d*x^3 + c)^(3/2)*d^6 + 48*sqrt(d*x^3 + c)*c*d^6)/d^9