3.428 \(\int \frac{1}{x (8 c-d x^3)^2 \sqrt{c+d x^3}} \, dx\)

Optimal. Leaf size=88 \[ \frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac{13 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{5/2}} \]

[Out]

Sqrt[c + d*x^3]/(216*c^2*(8*c - d*x^3)) + (13*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2592*c^(5/2)) - ArcTanh[S
qrt[c + d*x^3]/Sqrt[c]]/(96*c^(5/2))

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Rubi [A]  time = 0.075345, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {446, 103, 156, 63, 208, 206} \[ \frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac{13 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]

[Out]

Sqrt[c + d*x^3]/(216*c^2*(8*c - d*x^3)) + (13*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2592*c^(5/2)) - ArcTanh[S
qrt[c + d*x^3]/Sqrt[c]]/(96*c^(5/2))

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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

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{1}{x \left (8 c-d x^3\right )^2 \sqrt{c+d x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x (8 c-d x)^2 \sqrt{c+d x}} \, dx,x,x^3\right )\\ &=\frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{-9 c d-\frac{d^2 x}{2}}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{216 c^2 d}\\ &=\frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )}{192 c^2}+\frac{(13 d) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{1728 c^2}\\ &=\frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac{13 \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )}{864 c^2}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{96 c^2 d}\\ &=\frac{\sqrt{c+d x^3}}{216 c^2 \left (8 c-d x^3\right )}+\frac{13 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{2592 c^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0731727, size = 83, normalized size = 0.94 \[ \frac{\frac{12 \sqrt{c} \sqrt{c+d x^3}}{8 c-d x^3}+13 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-27 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{2592 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]

[Out]

((12*Sqrt[c]*Sqrt[c + d*x^3])/(8*c - d*x^3) + 13*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 27*ArcTanh[Sqrt[c + d*
x^3]/Sqrt[c]])/(2592*c^(5/2))

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Maple [C]  time = 0.014, size = 880, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1728*I/d^2/c^3*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)*_alph
a*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))+1/8*d/c*(-1/27/d/c*(d*x^3+c)^(1/2)/(d*x^3-8*c)-1/486*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)))-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{d x^{3} + c}{\left (d x^{3} - 8 \, c\right )}^{2} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85705, size = 544, normalized size = 6.18 \begin{align*} \left [\frac{13 \,{\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 ) + 27 \,{\left (d x^{3} - 8 \, c\right )} \sqrt{c} \log \left (\frac{d x^{3} - 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right ) - 24 \, \sqrt{d x^{3} + c} c}{5184 \,{\left (c^{3} d x^{3} - 8 \, c^{4}\right )}}, \frac{27 \,{\left (d x^{3} - 8 \, c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) - 13 \,{\left (d x^{3} - 8 \, c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) - 12 \, \sqrt{d x^{3} + c} c}{2592 \,{\left (c^{3} d x^{3} - 8 \, c^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/5184*(13*(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)) + 27*(d*x^3 -
8*c)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 - 8*c^4), 1
/2592*(27*(d*x^3 - 8*c)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 13*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqr
t(d*x^3 + c)*sqrt(-c)/c) - 12*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 - 8*c^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (- 8 c + d x^{3}\right )^{2} \sqrt{c + d x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12233, size = 107, normalized size = 1.22 \begin{align*} \frac{\arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{96 \, \sqrt{-c} c^{2}} - \frac{13 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{2592 \, \sqrt{-c} c^{2}} - \frac{\sqrt{d x^{3} + c}}{216 \,{\left (d x^{3} - 8 \, c\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 13/2592*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*
c^2) - 1/216*sqrt(d*x^3 + c)/((d*x^3 - 8*c)*c^2)