3.442 \(\int \frac{x^{11}}{(8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=95 \[ \frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}+\frac{2 \left (38 c+39 d x^3\right )}{81 d^4 \sqrt{c+d x^3}}-\frac{640 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{243 d^4} \]
[Out]
(8*x^6)/(27*d^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (2*(38*c + 39*d*x^3))/(81*d^4*Sqrt[c + d*x^3]) - (640*Sqrt[c]
*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(243*d^4)
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Rubi [A] time = 0.0757577, antiderivative size = 95, normalized size of antiderivative = 1.,
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, 98, 146, 63, 206} \[ \frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}+\frac{2 \left (38 c+39 d x^3\right )}{81 d^4 \sqrt{c+d x^3}}-\frac{640 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{243 d^4} \]
Antiderivative was successfully verified.
[In]
Int[x^11/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(8*x^6)/(27*d^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (2*(38*c + 39*d*x^3))/(81*d^4*Sqrt[c + d*x^3]) - (640*Sqrt[c]
*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(243*d^4)
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 98
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])
Rule 146
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]
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^{11}}{\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^3}{(8 c-d x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}-\frac{\operatorname{Subst}\left (\int \frac{x \left (16 c^2+13 c d x\right )}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{27 c d^2}\\ &=\frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}+\frac{2 \left (38 c+39 d x^3\right )}{81 d^4 \sqrt{c+d x^3}}-\frac{(320 c) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{81 d^3}\\ &=\frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}+\frac{2 \left (38 c+39 d x^3\right )}{81 d^4 \sqrt{c+d x^3}}-\frac{(640 c) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{81 d^4}\\ &=\frac{8 x^6}{27 d^2 \left (8 c-d x^3\right ) \sqrt{c+d x^3}}+\frac{2 \left (38 c+39 d x^3\right )}{81 d^4 \sqrt{c+d x^3}}-\frac{640 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{243 d^4}\\ \end{align*}
Mathematica [C] time = 0.0368726, size = 88, normalized size = 0.93 \[ \frac{6 \left (752 c^2-198 c d x^3+9 d^2 x^6\right )-640 c \left (8 c-d x^3\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{d x^3+c}{9 c}\right )}{81 d^4 \left (d x^3-8 c\right ) \sqrt{c+d x^3}} \]
Antiderivative was successfully verified.
[In]
Integrate[x^11/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(6*(752*c^2 - 198*c*d*x^3 + 9*d^2*x^6) - 640*c*(8*c - d*x^3)*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)
])/(81*d^4*(-8*c + d*x^3)*Sqrt[c + d*x^3])
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Maple [C] time = 0.046, size = 970, normalized size = 10.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
[Out]
1/d^3*(d*(2/3/d^2*c/((x^3+1/d*c)*d)^(1/2)+2/3*(d*x^3+c)^(1/2)/d^2)-32/3*c/d/(d*x^3+c)^(1/2))+192*c^2/d^3*(2/27
/d/c/((x^3+1/d*c)*d)^(1/2)+1/243*I/d^3/c^2*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)))+512*c^3/d^3*(-2/243/d/c^2/((x^3+1/d*c)*d)^(1/2)-1/243/d/c^2*(d*x^3+c)^(1/2
)/(d*x^3-8*c)-1/1458*I/d^3/c^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))*Ellip
ticPi(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)),_alp
ha=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^11/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.86252, size = 532, normalized size = 5.6 \begin{align*} \left [\frac{2 \,{\left (160 \,{\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 (27 \, d^{2} x^{6} - 274 \, c d x^{3} - 304 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{243 \,{\left (d^{6} x^{6} - 7 \, c d^{5} x^{3} - 8 \, c^{2} d^{4}\right )}}, \frac{2 \,{\left (320 \,{\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 (27 \, d^{2} x^{6} - 274 \, c d x^{3} - 304 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{243 \,{\left (d^{6} x^{6} - 7 \, c d^{5} x^{3} - 8 \, c^{2} d^{4}\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
[2/243*(160*(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*(27*d^2*x^6 - 274*c*d*x^3 - 304*c^2)*sqrt(d*x^3 + c))/(d^6*x^6 - 7*c*d^5*x^3 - 8*c^2*d^4), 2/243*(320*(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*(27*d^2*x^6 - 274*c*d*x^3 - 30
4*c^2)*sqrt(d*x^3 + c))/(d^6*x^6 - 7*c*d^5*x^3 - 8*c^2*d^4)]
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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**11/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
[Out]
Timed out
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Giac [A] time = 1.15573, size = 119, normalized size = 1.25 \begin{align*} \frac{640 \, c \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{243 \, \sqrt{-c} d^{4}} + \frac{2 \, \sqrt{d x^{3} + c}}{3 \, d^{4}} - \frac{2 \,{\left (85 \,{\left (d x^{3} + c\right )} c + 3 \, c^{2}\right )}}{81 \,{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} - 9 \, \sqrt{d x^{3} + c} c\right )} d^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
640/243*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) + 2/3*sqrt(d*x^3 + c)/d^4 - 2/81*(85*(d*x^3 + c)
*c + 3*c^2)/(((d*x^3 + c)^(3/2) - 9*sqrt(d*x^3 + c)*c)*d^4)