3.5.14 \(\int \frac {x^2 (c+d x^3)^{3/2}}{(8 c-d x^3)^2} \, dx\) [414]

Optimal. Leaf size=77 \[ \frac {\sqrt {c+d x^3}}{d}+\frac {\left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 77, 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 = {455, 43, 52, 65, 212} \begin {gather*} \frac {\left (c+d x^3\right )^{3/2}}{3 d \left (8 c-d x^3\right )}+\frac {\sqrt {c+d x^3}}{d}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

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

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 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 455

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

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Mathematica [A]
time = 0.09, size = 72, normalized size = 0.94 \begin {gather*} \frac {\left (25 c-2 d x^3\right ) \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {3 \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.36, size = 452, normalized size = 5.87

method	result	size
default	\(\frac {3 c \sqrt {d \,x^{3}+c}}{d \left (-d \,x^{3}+8 c \right )}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \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 )}{2 d^{3}}\)	\(452\)
elliptic	\(\frac {3 c \sqrt {d \,x^{3}+c}}{d \left (-d \,x^{3}+8 c \right )}+\frac {2 \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \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 )}{2 d^{3}}\)	\(452\)
risch	\(\text {Expression too large to display}\)	\(876\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(d*x^3+c)^(3/2)/(-d*x^3+8*c)^2,x,method=_RETURNVERBOSE)

[Out]

3*c/d*(d*x^3+c)^(1/2)/(-d*x^3+8*c)+2/3*(d*x^3+c)^(1/2)/d+1/2*I/d^3*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)*_al
pha+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 = 0.49, size = 79, normalized size = 1.03 \begin {gather*} \frac {9 \, \sqrt {c} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 4 \, \sqrt {d x^{3} + c} - \frac {18 \, \sqrt {d x^{3} + c} c}{d x^{3} - 8 \, c}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(9*sqrt(c)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 4*sqrt(d*x^3 + c) - 18*sqrt(
d*x^3 + c)*c/(d*x^3 - 8*c))/d

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Fricas [A]
time = 3.95, size = 162, normalized size = 2.10 \begin {gather*} \left [\frac {9 \, {\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 ) + 2 \, {\left (2 \, d x^{3} - 25 \, c\right )} \sqrt {d x^{3} + c}}{6 \, {\left (d^{2} x^{3} - 8 \, c d\right )}}, \frac {9 \, {\left (d x^{3} - 8 \, c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (2 \, d x^{3} - 25 \, c\right )} \sqrt {d x^{3} + c}}{3 \, {\left (d^{2} x^{3} - 8 \, c d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(9*(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)) + 2*(2*d*x^3 - 25*
c)*sqrt(d*x^3 + c))/(d^2*x^3 - 8*c*d), 1/3*(9*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) +
(2*d*x^3 - 25*c)*sqrt(d*x^3 + c))/(d^2*x^3 - 8*c*d)]

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

[Out]

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

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Giac [A]
time = 0.63, size = 69, normalized size = 0.90 \begin {gather*} \frac {3 \, c \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d} + \frac {2 \, \sqrt {d x^{3} + c}}{3 \, d} - \frac {3 \, \sqrt {d x^{3} + c} c}{{\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)^(3/2)/(-d*x^3+8*c)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 3.99, size = 87, normalized size = 1.13 \begin {gather*} \frac {2\,\sqrt {d\,x^3+c}}{3\,d}+\frac {3\,\sqrt {c}\,\ln \left (\frac {10\,c+d\,x^3-6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{2\,d}+\frac {3\,c\,\sqrt {d\,x^3+c}}{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)^(3/2))/(8*c - d*x^3)^2,x)

[Out]

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

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