3.5.11 \(\int \frac {x^{11} (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\) [411]

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

[Out]

3/7*x^6*(d*x^3+c)^(3/2)/d^2+1/3*x^9*(d*x^3+c)^(3/2)/d/(-d*x^3+8*c)+2/21*c*(d*x^3+c)^(3/2)*(51*d*x^3+694*c)/d^4
-4992*c^(7/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4+1664*c^3*(d*x^3+c)^(1/2)/d^4

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Rubi [A]
time = 0.08, 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 = {457, 99, 158, 152, 52, 65, 212} \begin {gather*} -\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 )} \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 52

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 65

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 99

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 152

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 158

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 457

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} \text {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 {\text {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 \text {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 ) \text {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 ) \text {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 ) \text {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 = 0.15, size = 103, normalized size = 0.77 \begin {gather*} \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 (-8 c+d x^3\right )}-\frac {4992 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^4} \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))/(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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.37, size = 999, normalized size = 7.46

method	result	size
elliptic	\(\frac {1536 c^{4} \sqrt {d \,x^{3}+c}}{d^{4} \left (-d \,x^{3}+8 c \right )}+\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 {32300 c^{3} \sqrt {d \,x^{3}+c}}{21 d^{4}}+\frac {832 i c^{3} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{18 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \sqrt {d \,x^{3}+c}}\right )}{d^{6}}\)	\(515\)
risch	\(\text {Expression too large to display}\)	\(912\)
default	\(\text {Expression too large to display}\)	\(999\)

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,method=_RETURNVERBOSE)

[Out]

1/d^3*(d*(2/21*d*x^9*(d*x^3+c)^(1/2)+16/105*c*x^6*(d*x^3+c)^(1/2)+2/105*c^2/d*x^3*(d*x^3+c)^(1/2)-4/105*c^3/d^
2*(d*x^3+c)^(1/2))+32/15*c/d*(d*x^3+c)^(5/2))+192/d^3*c^2*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*
I*c/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3
))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3
^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*
d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x
+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1
/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/
2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))
+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*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-
3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))
)/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2
*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d
*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/
3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^
(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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Maxima [A]
time = 0.48, 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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Fricas [A]
time = 3.08, 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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Sympy [F(-1)] Timed out
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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Giac [A]
time = 0.60, 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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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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