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

Optimal. Leaf size=73 \[ -\frac{2}{3} \sqrt{c+d x^3}+\frac{9}{4} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-\frac{1}{12} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right ) \]

[Out]

(-2*Sqrt[c + d*x^3])/3 + (9*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/4 - (Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]
/Sqrt[c]])/12

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Rubi [A]  time = 0.0696853, antiderivative size = 73, 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, 84, 156, 63, 208, 206} \[ -\frac{2}{3} \sqrt{c+d x^3}+\frac{9}{4} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-\frac{1}{12} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3])/3 + (9*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/4 - (Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]
/Sqrt[c]])/12

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 84

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

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 )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x (8 c-d x)} \, dx,x,x^3\right )\\ &=-\frac{2}{3} \sqrt{c+d x^3}-\frac{\operatorname{Subst}\left (\int \frac{-c^2 d-10 c d^2 x}{x (8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 d}\\ &=-\frac{2}{3} \sqrt{c+d x^3}+\frac{1}{24} c \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^3\right )+\frac{1}{8} (27 c d) \operatorname{Subst}\left (\int \frac{1}{(8 c-d x) \sqrt{c+d x}} \, dx,x,x^3\right )\\ &=-\frac{2}{3} \sqrt{c+d x^3}+\frac{1}{4} (27 c) \operatorname{Subst}\left (\int \frac{1}{9 c-x^2} \, dx,x,\sqrt{c+d x^3}\right )+\frac{c \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{12 d}\\ &=-\frac{2}{3} \sqrt{c+d x^3}+\frac{9}{4} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-\frac{1}{12} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0260791, size = 73, normalized size = 1. \[ -\frac{2}{3} \sqrt{c+d x^3}+\frac{9}{4} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{3 \sqrt{c}}\right )-\frac{1}{12} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{c}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[c + d*x^3])/3 + (9*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/4 - (Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]
/Sqrt[c]])/12

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Maple [C]  time = 0.021, size = 500, normalized size = 6.9 \begin{align*} -{\frac{d}{8\,c} \left ({\frac{2\,{x}^{3}}{9}\sqrt{d{x}^{3}+c}}+{\frac{56\,c}{9\,d}\sqrt{d{x}^{3}+c}}+{\frac{3\,ic\sqrt{2}}{{d}^{3}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}-8\,c \right ) }{\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{i\sqrt{3}d \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},-{\frac{1}{18\,cd} \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \right ) }+{\frac{1}{8\,c} \left ({\frac{2\,d{x}^{3}}{9}\sqrt{d{x}^{3}+c}}+{\frac{8\,c}{9}\sqrt{d{x}^{3}+c}}-{\frac{2}{3}{c}^{{\frac{3}{2}}}{\it Artanh} \left ({\sqrt{d{x}^{3}+c}{\frac{1}{\sqrt{c}}}} \right ) } \right ) } \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),x)

[Out]

-1/8*d/c*(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/c*(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 )} 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),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.43051, size = 396, normalized size = 5.42 \begin{align*} \left [\frac{9}{8} \, \sqrt{c} \log \left (\frac{d x^{3} + 6 \, \sqrt{d x^{3} + c} \sqrt{c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + \frac{1}{24} \, \sqrt{c} \log \left (\frac{d x^{3} - 2 \, \sqrt{d x^{3} + c} \sqrt{c} + 2 \, c}{x^{3}}\right ) - \frac{2}{3} \, \sqrt{d x^{3} + c}, \frac{1}{12} \, \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{c}\right ) - \frac{9}{4} \, \sqrt{-c} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-c}}{3 \, c}\right ) - \frac{2}{3} \, \sqrt{d x^{3} + 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),x, algorithm="fricas")

[Out]

[9/8*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 1/24*sqrt(c)*log((d*x^3 - 2*sqrt(
d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 2/3*sqrt(d*x^3 + c), 1/12*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 9/4*s
qrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 2/3*sqrt(d*x^3 + c)]

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Sympy [A]  time = 21.3097, size = 73, normalized size = 1. \begin{align*} - \frac{9 c \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{3 \sqrt{- c}} \right )}}{4 \sqrt{- c}} + \frac{c \operatorname{atan}{\left (\frac{\sqrt{c + d x^{3}}}{\sqrt{- c}} \right )}}{12 \sqrt{- c}} - \frac{2 \sqrt{c + d x^{3}}}{3} \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),x)

[Out]

-9*c*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(4*sqrt(-c)) + c*atan(sqrt(c + d*x**3)/sqrt(-c))/(12*sqrt(-c)) - 2*sq
rt(c + d*x**3)/3

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Giac [A]  time = 1.13432, size = 82, normalized size = 1.12 \begin{align*} \frac{c \arctan \left (\frac{\sqrt{d x^{3} + c}}{\sqrt{-c}}\right )}{12 \, \sqrt{-c}} - \frac{9 \, c \arctan \left (\frac{\sqrt{d x^{3} + c}}{3 \, \sqrt{-c}}\right )}{4 \, \sqrt{-c}} - \frac{2}{3} \, \sqrt{d x^{3} + c} \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),x, algorithm="giac")

[Out]

1/12*c*arctan(sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) - 9/4*c*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/sqrt(-c) - 2/3*s
qrt(d*x^3 + c)