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

Optimal. Leaf size=85 \[ \frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{32 \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 \sqrt{c}} \]

[Out]

(3*Sqrt[c + d*x^3])/(8*(8*c - d*x^3)) - (3*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(32*Sqrt[c]) - ArcTanh[Sqrt[c
+ d*x^3]/Sqrt[c]]/(96*Sqrt[c])

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Rubi [A]  time = 0.0790129, antiderivative size = 85, 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, 98, 156, 63, 208, 206} \[ \frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{32 \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[c + d*x^3])/(8*(8*c - d*x^3)) - (3*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(32*Sqrt[c]) - ArcTanh[Sqrt[c
+ d*x^3]/Sqrt[c]]/(96*Sqrt[c])

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 98

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

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{\left (c+d x^3\right )^{3/2}}{x \left (8 c-d x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x (8 c-d x)^2} \, dx,x,x^3\right )\\ &=\frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac{\operatorname{Subst}\left (\int \frac{-c^2 d+\frac{7}{2} c d^2 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{24 c d}\\ &=\frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}+\frac{1}{192} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )-\frac{1}{64} (9 d) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )\\ &=\frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac{9}{32} \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{96 d}\\ &=\frac{3 \sqrt{c+d x^3}}{8 \left (8 c-d x^3\right )}-\frac{3 \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )}{32 \sqrt{c}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0451007, size = 100, normalized size = 1.18 \[ \frac{36 \sqrt{c} \sqrt{c+d x^3}+\left (9 d x^3-72 c\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )+\left (d x^3-8 c\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )}{96 \sqrt{c} \left (8 c-d x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*Sqrt[c]*Sqrt[c + d*x^3] + (-72*c + 9*d*x^3)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] + (-8*c + d*x^3)*ArcTanh[
Sqrt[c + d*x^3]/Sqrt[c]])/(96*Sqrt[c]*(8*c - d*x^3))

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Maple [C]  time = 0.013, size = 956, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*d/c^2*(2/9*x^3*(d*x^3+c)^(1/2)+56/9*c*(d*x^3+c)^(1/2)/d+3*I/d^3*c*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)))+1/8*d/c*(-3/d*c*(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((-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))*Ellip
ticPi(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)),_alp
ha=RootOf(_Z^3*d-8*c)))+1/64/c^2*(2/9*d*x^3*(d*x^3+c)^(1/2)+8/9*c*(d*x^3+c)^(1/2)-2/3*c^(3/2)*arctanh((d*x^3+c
)^(1/2)/c^(1/2)))

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

[Out]

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

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Fricas [A]  time = 1.63902, size = 524, normalized size = 6.16 \begin{align*} \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 ) +{\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 ) - 72 \, \sqrt{d x^{3} + c} c}{192 \,{\left (c d x^{3} - 8 \, c^{2}\right )}}, \frac{{\left (d x^{3} - 8 \, c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) + 9 \,{\left (d x^{3} - 8 \, c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) - 36 \, \sqrt{d x^{3} + c} c}{96 \,{\left (c d x^{3} - 8 \, c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/192*(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)) + (d*x^3 - 8*c)*
sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 72*sqrt(d*x^3 + c)*c)/(c*d*x^3 - 8*c^2), 1/96*((d
*x^3 - 8*c)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + 9*(d*x^3 - 8*c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*
sqrt(-c)/c) - 36*sqrt(d*x^3 + c)*c)/(c*d*x^3 - 8*c^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.12132, size = 95, normalized size = 1.12 \begin{align*} \frac{\arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{96 \, \sqrt{-c}} + \frac{3 \, \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{32 \, \sqrt{-c}} - \frac{3 \, \sqrt{d x^{3} + c}}{8 \,{\left (d x^{3} - 8 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) + 3/32*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) - 3/8*sqrt
(d*x^3 + c)/(d*x^3 - 8*c)