3.3.40 \(\int \frac {x^5}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=52 \[ \frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2} \]

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Rubi [A]  time = 0.05, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {446, 78, 63, 206} \begin {gather*} \frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

2/(27*d^2*Sqrt[c + d*x^3]) + (16*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*Sqrt[c]*d^2)

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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^5}{\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}{(8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {8 \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27 d}\\ &=\frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {16 \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{27 d^2}\\ &=\frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.94 \begin {gather*} \frac {2 \left (\frac {3}{\sqrt {c+d x^3}}+\frac {8 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )}{81 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(2*(3/Sqrt[c + d*x^3] + (8*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/Sqrt[c]))/(81*d^2)

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IntegrateAlgebraic [A]  time = 0.05, size = 52, normalized size = 1.00 \begin {gather*} \frac {2}{27 d^2 \sqrt {c+d x^3}}+\frac {16 \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 \sqrt {c} d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^5/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

2/(27*d^2*Sqrt[c + d*x^3]) + (16*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(81*Sqrt[c]*d^2)

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fricas [A]  time = 0.51, size = 149, normalized size = 2.87 \begin {gather*} \left [\frac {2 \, {\left (4 \, {\left (d x^{3} + c\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 \, \sqrt {d x^{3} + c} c\right )}}{81 \, {\left (c d^{3} x^{3} + c^{2} d^{2}\right )}}, -\frac {2 \, {\left (8 \, {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 3 \, \sqrt {d x^{3} + c} c\right )}}{81 \, {\left (c d^{3} x^{3} + c^{2} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[2/81*(4*(d*x^3 + c)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 3*sqrt(d*x^3 + c)
*c)/(c*d^3*x^3 + c^2*d^2), -2/81*(8*(d*x^3 + c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 3*sqrt(d*x^3
+ c)*c)/(c*d^3*x^3 + c^2*d^2)]

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giac [A]  time = 0.17, size = 47, normalized size = 0.90 \begin {gather*} -\frac {2 \, {\left (\frac {8 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} - \frac {3}{\sqrt {d x^{3} + c} d}\right )}}{81 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

-2/81*(8*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) - 3/(sqrt(d*x^3 + c)*d))/d

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maple [C]  time = 0.18, size = 456, normalized size = 8.77 \begin {gather*} -\frac {8 \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}{d}+\frac {2}{3 \sqrt {d \,x^{3}+c}\, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)

[Out]

2/3/d^2/(d*x^3+c)^(1/2)-8*c/d*(2/27/((x^3+c/d)*d)^(1/2)/c/d+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 = 1.21, size = 56, normalized size = 1.08 \begin {gather*} -\frac {2 \, {\left (\frac {4 \, \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {3}{\sqrt {d x^{3} + c}}\right )}}{81 \, d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.68, size = 60, normalized size = 1.15 \begin {gather*} \frac {2}{27\,d^2\,\sqrt {d\,x^3+c}}+\frac {8\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{81\,\sqrt {c}\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/((c + d*x^3)^(3/2)*(8*c - d*x^3)),x)

[Out]

2/(27*d^2*(c + d*x^3)^(1/2)) + (8*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/(81*c^(1/2)
*d^2)

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sympy [A]  time = 27.57, size = 58, normalized size = 1.12 \begin {gather*} \begin {cases} \frac {2 \left (\frac {1}{27 d \sqrt {c + d x^{3}}} - \frac {8 \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{81 d \sqrt {- c}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {x^{6}}{48 c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)

[Out]

Piecewise((2*(1/(27*d*sqrt(c + d*x**3)) - 8*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(81*d*sqrt(-c)))/d, Ne(d, 0)),
(x**6/(48*c**(5/2)), True))

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