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

Optimal. Leaf size=64 \[ \frac {\sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 \sqrt {c} d} \]

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Rubi [A]  time = 0.05, antiderivative size = 64, 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 = {444, 47, 63, 206} \begin {gather*} \frac {\sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 \sqrt {c} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + d*x^3]/(3*d*(8*c - d*x^3)) - ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(9*Sqrt[c]*d)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{(8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac {\sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )\\ &=\frac {\sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d}\\ &=\frac {\sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 \sqrt {c} d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 61, normalized size = 0.95 \begin {gather*} \frac {\frac {3 \sqrt {c+d x^3}}{8 c-d x^3}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{\sqrt {c}}}{9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*Sqrt[c + d*x^3])/(8*c - d*x^3) - ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/Sqrt[c])/(9*d)

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IntegrateAlgebraic [A]  time = 0.06, size = 63, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {c+d x^3}}{3 d \left (d x^3-8 c\right )}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 \sqrt {c} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*Sqrt[c + d*x^3]/(d*(-8*c + d*x^3)) - ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(9*Sqrt[c]*d)

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fricas [A]  time = 0.49, size = 149, normalized size = 2.33 \begin {gather*} \left [\frac {{\left (d x^{3} - 8 \, 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 ) - 6 \, \sqrt {d x^{3} + c} c}{18 \, {\left (c d^{2} x^{3} - 8 \, c^{2} d\right )}}, \frac {{\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 3 \, \sqrt {d x^{3} + c} c}{9 \, {\left (c d^{2} x^{3} - 8 \, c^{2} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*((d*x^3 - 8*c)*sqrt(c)*log((d*x^3 - 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - 6*sqrt(d*x^3 + c)
*c)/(c*d^2*x^3 - 8*c^2*d), 1/9*((d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 3*sqrt(d*x^3 +
c)*c)/(c*d^2*x^3 - 8*c^2*d)]

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giac [A]  time = 0.16, size = 53, normalized size = 0.83 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{9 \, \sqrt {-c} d} - \frac {\sqrt {d x^{3} + c}}{3 \, {\left (d x^{3} - 8 \, c\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d) - 1/3*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*d)

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maple [C]  time = 0.18, size = 439, normalized size = 6.86 \begin {gather*} -\frac {\sqrt {d \,x^{3}+c}}{3 \left (d \,x^{3}-8 c \right ) 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 )}{54 c \,d^{3} \sqrt {d \,x^{3}+c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(d*x^3+c)^(1/2)/(d*x^3-8*c)/d+1/54*I/d^3/c*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 = 1.39, size = 66, normalized size = 1.03 \begin {gather*} \frac {\frac {\log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right )}{\sqrt {c}} - \frac {6 \, \sqrt {d x^{3} + c}}{d x^{3} - 8 \, c}}{18 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c)))/sqrt(c) - 6*sqrt(d*x^3 + c)/(d*x^3 - 8*
c))/d

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mupad [B]  time = 3.93, size = 72, normalized size = 1.12 \begin {gather*} \frac {\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{18\,\sqrt {c}\,d}+\frac {\sqrt {d\,x^3+c}}{3\,d\,\left (8\,c-d\,x^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((10*c + d*x^3 - 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3))/(18*c^(1/2)*d) + (c + d*x^3)^(1/2)/(3*d*(8*c -
d*x^3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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