3.3.80 \(\int \frac {x^{11} (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\)

Optimal. Leaf size=134 \[ -\frac {4992 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}+\frac {1664 c^3 \sqrt {c+d x^3}}{d^4}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}+\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {446, 97, 153, 147, 50, 63, 206} \begin {gather*} \frac {1664 c^3 \sqrt {c+d x^3}}{d^4}-\frac {4992 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}+\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^11*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]

[Out]

(1664*c^3*Sqrt[c + d*x^3])/d^4 + (3*x^6*(c + d*x^3)^(3/2))/(7*d^2) + (x^9*(c + d*x^3)^(3/2))/(3*d*(8*c - d*x^3
)) + (2*c*(c + d*x^3)^(3/2)*(694*c + 51*d*x^3))/(21*d^4) - (4992*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])
/d^4

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 97

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(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^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

Rule 153

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegerQ[m]

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^{11} \left (c+d x^3\right )^{3/2}}{\left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3 (c+d x)^{3/2}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {x^2 \sqrt {c+d x} \left (3 c+\frac {9 d x}{2}\right )}{8 c-d x} \, dx,x,x^3\right )}{3 d}\\ &=\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {2 \operatorname {Subst}\left (\int \frac {x \sqrt {c+d x} \left (-72 c^2 d-\frac {255}{2} c d^2 x\right )}{8 c-d x} \, dx,x,x^3\right )}{21 d^3}\\ &=\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}-\frac {\left (832 c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d^3}\\ &=\frac {1664 c^3 \sqrt {c+d x^3}}{d^4}+\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}-\frac {\left (7488 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{d^3}\\ &=\frac {1664 c^3 \sqrt {c+d x^3}}{d^4}+\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}-\frac {\left (14976 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^4}\\ &=\frac {1664 c^3 \sqrt {c+d x^3}}{d^4}+\frac {3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac {x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}-\frac {4992 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 111, normalized size = 0.83 \begin {gather*} \frac {2 \left (52416 c^{7/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+\sqrt {c+d x^3} \left (-145328 c^4+12206 c^3 d x^3+301 c^2 d^2 x^6+16 c d^3 x^9+d^4 x^{12}\right )\right )}{21 d^4 \left (d x^3-8 c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^11*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]

[Out]

(2*(Sqrt[c + d*x^3]*(-145328*c^4 + 12206*c^3*d*x^3 + 301*c^2*d^2*x^6 + 16*c*d^3*x^9 + d^4*x^12) + 52416*c^(7/2
)*(8*c - d*x^3)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]))/(21*d^4*(-8*c + d*x^3))

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IntegrateAlgebraic [A]  time = 0.10, size = 104, normalized size = 0.78 \begin {gather*} -\frac {4992 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4}-\frac {2 \sqrt {c+d x^3} \left (145328 c^4-12206 c^3 d x^3-301 c^2 d^2 x^6-16 c d^3 x^9-d^4 x^{12}\right )}{21 d^4 \left (d x^3-8 c\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^11*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x]

[Out]

(-2*Sqrt[c + d*x^3]*(145328*c^4 - 12206*c^3*d*x^3 - 301*c^2*d^2*x^6 - 16*c*d^3*x^9 - d^4*x^12))/(21*d^4*(-8*c
+ d*x^3)) - (4992*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^4

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fricas [A]  time = 0.44, size = 239, normalized size = 1.78 \begin {gather*} \left [\frac {2 \, {\left (26208 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\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^{4} x^{12} + 16 \, c d^{3} x^{9} + 301 \, c^{2} d^{2} x^{6} + 12206 \, c^{3} d x^{3} - 145328 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{21 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}, \frac {2 \, {\left (52416 \, {\left (c^{3} d x^{3} - 8 \, c^{4}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (d^{4} x^{12} + 16 \, c d^{3} x^{9} + 301 \, c^{2} d^{2} x^{6} + 12206 \, c^{3} d x^{3} - 145328 \, c^{4}\right )} \sqrt {d x^{3} + c}\right )}}{21 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="fricas")

[Out]

[2/21*(26208*(c^3*d*x^3 - 8*c^4)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + (d^4*
x^12 + 16*c*d^3*x^9 + 301*c^2*d^2*x^6 + 12206*c^3*d*x^3 - 145328*c^4)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4), 2/
21*(52416*(c^3*d*x^3 - 8*c^4)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + (d^4*x^12 + 16*c*d^3*x^9 + 301
*c^2*d^2*x^6 + 12206*c^3*d*x^3 - 145328*c^4)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4)]

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giac [A]  time = 0.17, size = 127, normalized size = 0.95 \begin {gather*} \frac {4992 \, c^{4} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{4}} - \frac {1536 \, \sqrt {d x^{3} + c} c^{4}}{{\left (d x^{3} - 8 \, c\right )} d^{4}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {7}{2}} d^{24} + 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c d^{24} + 448 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} d^{24} + 15680 \, \sqrt {d x^{3} + c} c^{3} d^{24}\right )}}{21 \, d^{28}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")

[Out]

4992*c^4*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 1536*sqrt(d*x^3 + c)*c^4/((d*x^3 - 8*c)*d^4) +
2/21*((d*x^3 + c)^(7/2)*d^24 + 21*(d*x^3 + c)^(5/2)*c*d^24 + 448*(d*x^3 + c)^(3/2)*c^2*d^24 + 15680*sqrt(d*x^3
+ c)*c^3*d^24)/d^28

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maple [C]  time = 0.28, size = 998, normalized size = 7.45

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x)

[Out]

1/d^3*(d*(2/21*(d*x^3+c)^(1/2)*d*x^9+16/105*(d*x^3+c)^(1/2)*c*x^6+2/105*(d*x^3+c)^(1/2)*c^2/d*x^3-4/105*(d*x^3
+c)^(1/2)*c^3/d^2)+32/15*c/d*(d*x^3+c)^(5/2))+512*c^3/d^3*(-3*c/d*(d*x^3+c)^(1/2)/(d*x^3-8*c)+2/3*(d*x^3+c)^(1
/2)/d+1/2*I/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)
))+192*c^2/d^3*(2/9*(d*x^3+c)^(1/2)*x^3+56/9*(d*x^3+c)^(1/2)*c/d+3*I*c/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 = 1.18, size = 119, normalized size = 0.89 \begin {gather*} \frac {2 \, {\left (26208 \, c^{\frac {7}{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 {7}{2}} + 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c + 448 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} + 15680 \, \sqrt {d x^{3} + c} c^{3} - \frac {16128 \, \sqrt {d x^{3} + c} c^{4}}{d x^{3} - 8 \, c}\right )}}{21 \, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")

[Out]

2/21*(26208*c^(7/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + (d*x^3 + c)^(7/2) + 21*
(d*x^3 + c)^(5/2)*c + 448*(d*x^3 + c)^(3/2)*c^2 + 15680*sqrt(d*x^3 + c)*c^3 - 16128*sqrt(d*x^3 + c)*c^4/(d*x^3
- 8*c))/d^4

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mupad [B]  time = 4.10, size = 147, normalized size = 1.10 \begin {gather*} \frac {2496\,c^{7/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^4}+\frac {32300\,c^3\,\sqrt {d\,x^3+c}}{21\,d^4}+\frac {2\,x^9\,\sqrt {d\,x^3+c}}{21\,d}+\frac {16\,c\,x^6\,\sqrt {d\,x^3+c}}{7\,d^2}+\frac {986\,c^2\,x^3\,\sqrt {d\,x^3+c}}{21\,d^3}+\frac {1536\,c^4\,\sqrt {d\,x^3+c}}{d^4\,\left (8\,c-d\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^11*(c + d*x^3)^(3/2))/(8*c - d*x^3)^2,x)

[Out]

(2496*c^(7/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/d^4 + (32300*c^3*(c + d*x^3)^(1
/2))/(21*d^4) + (2*x^9*(c + d*x^3)^(1/2))/(21*d) + (16*c*x^6*(c + d*x^3)^(1/2))/(7*d^2) + (986*c^2*x^3*(c + d*
x^3)^(1/2))/(21*d^3) + (1536*c^4*(c + d*x^3)^(1/2))/(d^4*(8*c - d*x^3))

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sympy [F(-1)]  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**11*(d*x**3+c)**(3/2)/(-d*x**3+8*c)**2,x)

[Out]

Timed out

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