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

Optimal. Leaf size=95 \[ -\frac {2944 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4}+\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )} \]

[Out]

-2944/81*c^(3/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^4+8/27*x^6*(d*x^3+c)^(1/2)/d^2/(-d*x^3+8*c)+2/27*(7*d*
x^3+170*c)*(d*x^3+c)^(1/2)/d^4

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Rubi [A]  time = 0.07, antiderivative size = 95, normalized size of antiderivative = 1.00, 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, 147, 63, 206} \[ -\frac {2944 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}+\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*x^6*Sqrt[c + d*x^3])/(27*d^2*(8*c - d*x^3)) + (2*Sqrt[c + d*x^3]*(170*c + 7*d*x^3))/(27*d^4) - (2944*c^(3/2
)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*d^4)

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 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 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 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 (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3}{(8 c-d x)^2 \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {x \left (16 c^2+21 c d x\right )}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27 c d^2}\\ &=\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4}-\frac {\left (1472 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27 d^3}\\ &=\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4}-\frac {\left (2944 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{27 d^4}\\ &=\frac {8 x^6 \sqrt {c+d x^3}}{27 d^2 \left (8 c-d x^3\right )}+\frac {2 \sqrt {c+d x^3} \left (170 c+7 d x^3\right )}{27 d^4}-\frac {2944 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 91, normalized size = 0.96 \[ \frac {2944 c^{3/2} \left (8 c-d x^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+6 \sqrt {c+d x^3} \left (-1360 c^2+114 c d x^3+3 d^2 x^6\right )}{81 d^4 \left (d x^3-8 c\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*Sqrt[c + d*x^3]*(-1360*c^2 + 114*c*d*x^3 + 3*d^2*x^6) + 2944*c^(3/2)*(8*c - d*x^3)*ArcTanh[Sqrt[c + d*x^3]/
(3*Sqrt[c])])/(81*d^4*(-8*c + d*x^3))

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fricas [A]  time = 0.64, size = 195, normalized size = 2.05 \[ \left [\frac {2 \, {\left (736 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 3 \, {\left (3 \, d^{2} x^{6} + 114 \, c d x^{3} - 1360 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}, \frac {2 \, {\left (1472 \, {\left (c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 3 \, {\left (3 \, d^{2} x^{6} + 114 \, c d x^{3} - 1360 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} - 8 \, c d^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[2/81*(736*(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*(3*d^2*
x^6 + 114*c*d*x^3 - 1360*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^4), 2/81*(1472*(c*d*x^3 - 8*c^2)*sqrt(-c)*arct
an(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 3*(3*d^2*x^6 + 114*c*d*x^3 - 1360*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 - 8*c*d^
4)]

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giac [A]  time = 0.19, size = 93, normalized size = 0.98 \[ \frac {2944 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{4}} - \frac {512 \, \sqrt {d x^{3} + c} c^{2}}{27 \, {\left (d x^{3} - 8 \, c\right )} d^{4}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{8} + 45 \, \sqrt {d x^{3} + c} c d^{8}\right )}}{9 \, d^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

2944/81*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) - 512/27*sqrt(d*x^3 + c)*c^2/((d*x^3 - 8*c)*d^
4) + 2/9*((d*x^3 + c)^(3/2)*d^8 + 45*sqrt(d*x^3 + c)*c*d^8)/d^12

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maple [C]  time = 0.30, size = 916, normalized size = 9.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x)

[Out]

1/d^3*((2/9*(d*x^3+c)^(1/2)/d*x^3-4/9*(d*x^3+c)^(1/2)*c/d^2)*d+32/3*(d*x^3+c)^(1/2)*c/d)+512*c^3/d^3*(-1/27/c/
d*(d*x^3+c)^(1/2)/(d*x^3-8*c)-1/486*I/c^2/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)))+64/9*I*c/d^6*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 = 1.34, size = 93, normalized size = 0.98 \[ \frac {2 \, {\left (736 \, c^{\frac {3}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 9 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} + 405 \, \sqrt {d x^{3} + c} c - \frac {768 \, \sqrt {d x^{3} + c} c^{2}}{d x^{3} - 8 \, c}\right )}}{81 \, d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

2/81*(736*c^(3/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 9*(d*x^3 + c)^(3/2) + 405
*sqrt(d*x^3 + c)*c - 768*sqrt(d*x^3 + c)*c^2/(d*x^3 - 8*c))/d^4

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mupad [B]  time = 4.06, size = 107, normalized size = 1.13 \[ \frac {92\,c\,\sqrt {d\,x^3+c}}{9\,d^4}+\frac {1472\,c^{3/2}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,d^4}+\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^3}+\frac {512\,c^2\,\sqrt {d\,x^3+c}}{27\,d^4\,\left (8\,c-d\,x^3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/((c + d*x^3)^(1/2)*(8*c - d*x^3)^2),x)

[Out]

(92*c*(c + d*x^3)^(1/2))/(9*d^4) + (1472*c^(3/2)*log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3
)))/(81*d^4) + (2*x^3*(c + d*x^3)^(1/2))/(9*d^3) + (512*c^2*(c + d*x^3)^(1/2))/(27*d^4*(8*c - d*x^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(-d*x**3+8*c)**2/(d*x**3+c)**(1/2),x)

[Out]

Timed out

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