3.411 \(\int \frac{x^{11} (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\)
Optimal. Leaf size=134 \[ \frac{1664 c^3 \sqrt{c+d x^3}}{d^4}-\frac{4992 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}+\frac{3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac{2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}+\frac{x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )} \]
[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
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Rubi [A] time = 0.110103, antiderivative size = 134, normalized size of antiderivative = 1.,
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} \[ \frac{1664 c^3 \sqrt{c+d x^3}}{d^4}-\frac{4992 c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{d^4}+\frac{3 x^6 \left (c+d x^3\right )^{3/2}}{7 d^2}+\frac{2 c \left (c+d x^3\right )^{3/2} \left (694 c+51 d x^3\right )}{21 d^4}+\frac{x^9 \left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )} \]
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 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 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 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 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 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^{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*}
Mathematica [A] time = 0.0690827, size = 111, normalized size = 0.83 \[ \frac{2 \left (\sqrt{c+d x^3} \left (301 c^2 d^2 x^6+12206 c^3 d x^3-145328 c^4+16 c d^3 x^9+d^4 x^{12}\right )+52416 c^{7/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )\right )}{21 d^4 \left (d x^3-8 c\right )} \]
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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Maple [C] time = 0.039, size = 998, normalized size = 7.5 \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+c)^(3/2)/(-d*x^3+8*c)^2,x)
[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/d*c^2*x^3*(d*x^3+c)^(1/2)-4/105/d^2*c^
3*(d*x^3+c)^(1/2))+32/15*c/d*(d*x^3+c)^(5/2))+192*c^2/d^3*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*
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)))
+512*c^3/d^3*(-3/d*c*(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((-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^11*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.89089, size = 562, normalized size = 4.19 \begin{align*} \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{align*}
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)] 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+c)**(3/2)/(-d*x**3+8*c)**2,x)
[Out]
Timed out
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Giac [A] time = 1.12225, size = 171, normalized size = 1.28 \begin{align*} \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{align*}
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