3.4.29 \(\int \frac {1}{x (8 c-d x^3) (c+d x^3)^{3/2}} \, dx\) [329]

Optimal. Leaf size=76 \[ \frac {2}{27 c^2 \sqrt {c+d x^3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{324 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{12 c^{5/2}} \]

[Out]

1/324*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-1/12*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+2/27/c^2/(d*x
^3+c)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 76, normalized size of antiderivative = 1.00, 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 = {457, 87, 162, 65, 214, 212} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{324 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{12 c^{5/2}}+\frac {2}{27 c^2 \sqrt {c+d x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(27*c^2*Sqrt[c + d*x^3]) + ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(324*c^(5/2)) - ArcTanh[Sqrt[c + d*x^3]/Sqrt
[c]]/(12*c^(5/2))

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 87

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

Rule 162

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 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 214

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

Rule 457

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]]

Rubi steps

\begin {align*} \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {1}{x (8 c-d x) (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac {2}{27 c^2 \sqrt {c+d x^3}}-\frac {\text {Subst}\left (\int \frac {-9 c d+d^2 x}{x (8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{27 c^2 d}\\ &=\frac {2}{27 c^2 \sqrt {c+d x^3}}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {c+d x}} \, dx,x,x^3\right )}{24 c^2}+\frac {d \text {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{216 c^2}\\ &=\frac {2}{27 c^2 \sqrt {c+d x^3}}+\frac {\text {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{108 c^2}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {c}{d}+\frac {x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{12 c^2 d}\\ &=\frac {2}{27 c^2 \sqrt {c+d x^3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{324 c^{5/2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{12 c^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 69, normalized size = 0.91 \begin {gather*} \frac {\frac {24 \sqrt {c}}{\sqrt {c+d x^3}}+\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 )}{324 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((24*Sqrt[c])/Sqrt[c + d*x^3] + ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 27*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(3
24*c^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.39, size = 485, normalized size = 6.38

method	result	size
default	\(-\frac {d \left (\frac {2}{27 d c \sqrt {\left (x^{3}+\frac {c}{d}\right ) 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 )}{243 d^{3} c^{2}}\right )}{8 c}+\frac {\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \arctanh \left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{8 c}\)	\(485\)
elliptic	\(\text {Expression too large to display}\)	\(1526\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*d/c*(2/27/d/c/((x^3+c/d)*d)^(1/2)+1/243*I/d^3/c^2*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)*_alpha+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)))+1/8/c*(2/3/c/((x^3+c/d)*d)^(1/2)-2/3*arctanh((d*x^3+c)^(1/2)/c
^(1/2))/c^(3/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.19, size = 213, normalized size = 2.80 \begin {gather*} \left [\frac {{\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 ) + 27 \, {\left (d x^{3} + c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 48 \, \sqrt {d x^{3} + c} c}{648 \, {\left (c^{3} d x^{3} + c^{4}\right )}}, \frac {27 \, {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 24 \, \sqrt {d x^{3} + c} c}{324 \, {\left (c^{3} d x^{3} + c^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/648*((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)) + 27*(d*x^3 + c)*sqr
t(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 48*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 + c^4), 1/324*(27*(
d*x^3 + c)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - (d*x^3 + c)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqrt(
-c)/c) + 24*sqrt(d*x^3 + c)*c)/(c^3*d*x^3 + c^4)]

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Sympy [A]
time = 6.20, size = 78, normalized size = 1.03 \begin {gather*} \frac {2}{27 c^{2} \sqrt {c + d x^{3}}} - \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{324 c^{2} \sqrt {- c}} + \frac {\operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {- c}} \right )}}{12 c^{2} \sqrt {- c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(27*c**2*sqrt(c + d*x**3)) - atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(324*c**2*sqrt(-c)) + atan(sqrt(c + d*x**3)
/sqrt(-c))/(12*c**2*sqrt(-c))

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Giac [A]
time = 0.90, size = 68, normalized size = 0.89 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{12 \, \sqrt {-c} c^{2}} - \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{324 \, \sqrt {-c} c^{2}} + \frac {2}{27 \, \sqrt {d x^{3} + c} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/324*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^
2) + 2/27/(sqrt(d*x^3 + c)*c^2)

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Mupad [B]
time = 3.66, size = 68, normalized size = 0.89 \begin {gather*} \frac {2}{27\,c^2\,\sqrt {d\,x^3+c}}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )}{12\,\sqrt {c^5}}+\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )}{324\,\sqrt {c^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/(27*c^2*(c + d*x^3)^(1/2)) - atanh((c^2*(c + d*x^3)^(1/2))/(c^5)^(1/2))/(12*(c^5)^(1/2)) + atanh((c^2*(c + d
*x^3)^(1/2))/(3*(c^5)^(1/2)))/(324*(c^5)^(1/2))

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