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

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

[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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Rubi [A]  time = 0.0457229, antiderivative size = 52, normalized size of antiderivative = 1., 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} \[ \frac{2}{27 d^2 \sqrt{c+d x^3}}+\frac{16 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{81 \sqrt{c} d^2} \]

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 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]]

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 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])

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*}

Mathematica [A]  time = 0.030077, size = 49, normalized size = 0.94 \[ \frac{2 \left (\frac{3}{\sqrt{c+d x^3}}+\frac{8 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{\sqrt{c}}\right )}{81 d^2} \]

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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Maple [C]  time = 0.01, size = 456, normalized size = 8.8 \begin{align*}{\frac{2}{3\,{d}^{2}}{\frac{1}{\sqrt{d{x}^{3}+c}}}}-8\,{\frac{c}{d} \left ({\frac{2}{27\,cd}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{c}{d}} \right ) d}}}}+{\frac{{\frac{i}{243}}\sqrt{2}}{{d}^{3}{c}^{2}}\sum _{{\it \_alpha}={\it RootOf} \left ({{\it \_Z}}^{3}d-8\,c \right ) }{\frac{\sqrt [3]{-{d}^{2}c} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{2/3} \right ) }{\sqrt{d{x}^{3}+c}}\sqrt{{\frac{i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{-i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{d}{-3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c}} \left ( x-{\frac{\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{-i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}{\it EllipticPi} \left ( 1/3\,\sqrt{3}\sqrt{{\frac{i\sqrt{3}d}{\sqrt [3]{-{d}^{2}c}} \left ( x+1/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}-{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) }},-1/18\,{\frac{2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd}{cd}},\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d} \left ( -3/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}+{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) ^{-1}}} \right ) }} \right ) } \end{align*}

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/d*c*(2/27/d/c/((x^3+1/d*c)*d)^(1/2)+1/243*I/d^3/c^2*2^(1/2)*sum((-d^2*c)^(1/3)*(1/2*
I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(
-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(
-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^
2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-
d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*
3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/
3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.54801, size = 344, normalized size = 6.62 \begin{align*} \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{align*}

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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Sympy [A]  time = 19.0056, size = 58, normalized size = 1.12 \begin{align*} \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{align*}

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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Giac [A]  time = 1.12979, size = 63, normalized size = 1.21 \begin{align*} -\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{align*}

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