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

Optimal. Leaf size=90 \[ \frac {1024 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}+\frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}-\frac {4 c \sqrt {c+d x^3}}{d^4}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^4} \]

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Rubi [A]  time = 0.10, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {446, 87, 43, 63, 206} \begin {gather*} \frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}+\frac {1024 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}-\frac {4 c \sqrt {c+d x^3}}{d^4}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c^2)/(27*d^4*Sqrt[c + d*x^3]) - (4*c*Sqrt[c + d*x^3])/d^4 - (2*(c + d*x^3)^(3/2))/(9*d^4) + (1024*c^(3/2)*A
rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*d^4)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^3}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {c^2}{9 d^3 (c+d x)^{3/2}}-\frac {7 c}{d^3 \sqrt {c+d x}}-\frac {x}{d^2 \sqrt {c+d x}}+\frac {512 c^2}{9 d^3 (8 c-d x) \sqrt {c+d x}}\right ) \, dx,x,x^3\right )\\ &=\frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}-\frac {14 c \sqrt {c+d x^3}}{3 d^4}+\frac {\left (512 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 {\operatorname {Subst}\left (\int \frac {x}{\sqrt {c+d x}} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}-\frac {14 c \sqrt {c+d x^3}}{3 d^4}+\frac {\left (1024 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 {\operatorname {Subst}\left (\int \left (-\frac {c}{d \sqrt {c+d x}}+\frac {\sqrt {c+d x}}{d}\right ) \, dx,x,x^3\right )}{3 d^2}\\ &=\frac {2 c^2}{27 d^4 \sqrt {c+d x^3}}-\frac {4 c \sqrt {c+d x^3}}{d^4}-\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^4}+\frac {1024 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 [C]  time = 0.06, size = 66, normalized size = 0.73 \begin {gather*} -\frac {2 \left (512 c^2 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {d x^3+c}{9 c}\right )-456 c^2+60 c d x^3+3 d^2 x^6\right )}{27 d^4 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-456*c^2 + 60*c*d*x^3 + 3*d^2*x^6 + 512*c^2*Hypergeometric2F1[-1/2, 1, 1/2, (c + d*x^3)/(9*c)]))/(27*d^4*
Sqrt[c + d*x^3])

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IntegrateAlgebraic [A]  time = 0.08, size = 73, normalized size = 0.81 \begin {gather*} \frac {1024 c^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 d^4}-\frac {2 \left (56 c^2+60 c d x^3+3 d^2 x^6\right )}{27 d^4 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(56*c^2 + 60*c*d*x^3 + 3*d^2*x^6))/(27*d^4*Sqrt[c + d*x^3]) + (1024*c^(3/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqr
t[c])])/(81*d^4)

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fricas [A]  time = 0.73, size = 189, normalized size = 2.10 \begin {gather*} \left [\frac {2 \, {\left (256 \, {\left (c d x^{3} + 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} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} + c d^{4}\right )}}, -\frac {2 \, {\left (512 \, {\left (c d x^{3} + 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} + 60 \, c d x^{3} + 56 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{81 \, {\left (d^{5} x^{3} + 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)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[2/81*(256*(c*d*x^3 + 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 + 60*c*d*x^3 + 56*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 + c*d^4), -2/81*(512*(c*d*x^3 + c^2)*sqrt(-c)*arctan(1/3*sq
rt(d*x^3 + c)*sqrt(-c)/c) + 3*(3*d^2*x^6 + 60*c*d*x^3 + 56*c^2)*sqrt(d*x^3 + c))/(d^5*x^3 + c*d^4)]

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giac [A]  time = 0.20, size = 82, normalized size = 0.91 \begin {gather*} -\frac {1024 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{81 \, \sqrt {-c} d^{4}} + \frac {2 \, c^{2}}{27 \, \sqrt {d x^{3} + c} d^{4}} - \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{8} + 18 \, \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)/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-1024/81*c^2*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^4) + 2/27*c^2/(sqrt(d*x^3 + c)*d^4) - 2/9*((d*x^
3 + c)^(3/2)*d^8 + 18*sqrt(d*x^3 + c)*c*d^8)/d^12

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maple [C]  time = 0.30, size = 560, normalized size = 6.22 \begin {gather*} -\frac {512 \left (\frac {2}{27 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, c d}+\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \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 ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \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 ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\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}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\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 ) d}}\right )}{243 c^{2} d^{3} \sqrt {d \,x^{3}+c}}\right ) c^{3}}{d^{3}}-\frac {8 \left (\frac {2 c}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, d^{2}}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d^{2}}\right ) c}{d^{2}}-\frac {\frac {2 \sqrt {d \,x^{3}+c}\, x^{3}}{9 d^{2}}-\frac {2 c^{2}}{3 \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}\, d^{3}}-\frac {10 \sqrt {d \,x^{3}+c}\, c}{9 d^{3}}}{d}+\frac {128 c^{2}}{3 \sqrt {d \,x^{3}+c}\, d^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(-2/3/d^3*c^2/((x^3+c/d)*d)^(1/2)+2/9/d^2*x^3*(d*x^3+c)^(1/2)-10/9*c*(d*x^3+c)^(1/2)/d^3)-8*c/d^2*(2/3/d^
2*c/((x^3+c/d)*d)^(1/2)+2/3*(d*x^3+c)^(1/2)/d^2)+128/3*c^2/d^4/(d*x^3+c)^(1/2)-512*c^3/d^3*(2/27/d/c/((x^3+c/d
)*d)^(1/2)+1/243*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)))

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maxima [A]  time = 1.25, size = 82, normalized size = 0.91 \begin {gather*} -\frac {2 \, {\left (256 \, 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}} + 162 \, \sqrt {d x^{3} + c} c - \frac {3 \, c^{2}}{\sqrt {d x^{3} + 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)/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

-2/81*(256*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) + 16
2*sqrt(d*x^3 + c)*c - 3*c^2/sqrt(d*x^3 + c))/d^4

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mupad [B]  time = 3.78, size = 95, normalized size = 1.06 \begin {gather*} \frac {512\,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 {38\,c\,\sqrt {d\,x^3+c}}{9\,d^4}+\frac {2\,c^2}{27\,d^4\,\sqrt {d\,x^3+c}}-\frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^3} \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)),x)

[Out]

(512*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) - (38*c*(c + d*x^3)^(1/
2))/(9*d^4) + (2*c^2)/(27*d^4*(c + d*x^3)^(1/2)) - (2*x^3*(c + d*x^3)^(1/2))/(9*d^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+8*c)/(d*x**3+c)**(3/2),x)

[Out]

Timed out

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