3.5.24 \(\int \frac {x^{11}}{(8 c-d x^3)^2 \sqrt {c+d x^3}} \, dx\) [424]

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

[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 = 0.06, 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 = {457, 100, 152, 65, 212} \begin {gather*} -\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 )} \end {gather*}

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 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 100

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 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 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 (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx &=\frac {1}{3} \text {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 {\text {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 ) \text {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 ) \text {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 = 0.11, size = 81, normalized size = 0.85 \begin {gather*} \frac {2 \left (\frac {3 \sqrt {c+d x^3} \left (-1360 c^2+114 c d x^3+3 d^2 x^6\right )}{-8 c+d x^3}-1472 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )\right )}{81 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((3*Sqrt[c + d*x^3]*(-1360*c^2 + 114*c*d*x^3 + 3*d^2*x^6))/(-8*c + d*x^3) - 1472*c^(3/2)*ArcTanh[Sqrt[c + d
*x^3]/(3*Sqrt[c])]))/(81*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 = 917, normalized size = 9.65

method	result	size
elliptic	\(\frac {512 c^{2} \sqrt {d \,x^{3}+c}}{27 d^{4} \left (-d \,x^{3}+8 c \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 d^{3}}+\frac {92 c \sqrt {d \,x^{3}+c}}{9 d^{4}}+\frac {1472 i c \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 )}{243 d^{6}}\)	\(473\)
risch	\(\text {Expression too large to display}\)	\(890\)
default	\(\text {Expression too large to display}\)	\(917\)

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,method=_RETURNVERBOSE)

[Out]

1/d^3*(d*(2/9/d*x^3*(d*x^3+c)^(1/2)-4/9*c*(d*x^3+c)^(1/2)/d^2)+32/3*c*(d*x^3+c)^(1/2)/d)+512*c^3/d^3*(1/27*(d*
x^3+c)^(1/2)/c/d/(-d*x^3+8*c)-1/486*I/d^3/c^2*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)))+64/9*I/d^6*c*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 = 93, normalized size = 0.98 \begin {gather*} \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}} \end {gather*}

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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Fricas [A]
time = 3.21, size = 195, normalized size = 2.05 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \end {gather*}

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]

Integral(x**11/((-8*c + d*x**3)**2*sqrt(c + d*x**3)), x)

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Giac [A]
time = 1.35, size = 93, normalized size = 0.98 \begin {gather*} \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}} \end {gather*}

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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Mupad [B]
time = 4.06, size = 107, normalized size = 1.13 \begin {gather*} \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 )} \end {gather*}

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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