Integrand size = 27, antiderivative size = 76 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2}{27 c^2 \sqrt {c+d x^3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{324 c^{5/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{12 c^{5/2}} \]
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)
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\frac {24 \sqrt {c}}{\sqrt {c+d x^3}}+\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-27 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{324 c^{5/2}} \]
((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]])/(324*c^(5/2))
Time = 0.24 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {948, 96, 25, 27, 174, 73, 219, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^3 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3\) |
\(\Big \downarrow \) 96 |
\(\displaystyle \frac {1}{3} \left (\frac {2}{9 c^2 \sqrt {c+d x^3}}-\frac {\int -\frac {d \left (9 c-d x^3\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2 d}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {d \left (9 c-d x^3\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2 d}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {9 c-d x^3}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {9}{8} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {1}{8} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {1}{4} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {9 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{4 d}}{9 c^2}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {9 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{4 d}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{12 \sqrt {c}}}{9 c^2}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{12 \sqrt {c}}-\frac {9 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{4 \sqrt {c}}}{9 c^2}+\frac {2}{9 c^2 \sqrt {c+d x^3}}\right )\) |
(2/(9*c^2*Sqrt[c + d*x^3]) + (ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(12*Sqr t[c]) - (9*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(4*Sqrt[c]))/(9*c^2))/3
3.4.29.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_] :> Simp[f*((e + f*x)^(p + 1)/((p + 1)*(b*e - a*f)*(d*e - c*f))), x] + S imp[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]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(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]
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] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[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]]
Time = 4.43 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.93
method | result | size |
pseudoelliptic | \(\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}-27 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) \sqrt {d \,x^{3}+c}+24 \sqrt {c}}{324 c^{\frac {5}{2}} \sqrt {d \,x^{3}+c}}\) | \(71\) |
default | \(\frac {\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{8 c}-\frac {-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}}{3}+\sqrt {c}}{108 c^{\frac {5}{2}} \sqrt {d \,x^{3}+c}}\) | \(89\) |
elliptic | \(\text {Expression too large to display}\) | \(1526\) |
1/324/c^(5/2)*(arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))*(d*x^3+c)^(1/2)-27*arc tanh((d*x^3+c)^(1/2)/c^(1/2))*(d*x^3+c)^(1/2)+24*c^(1/2))/(d*x^3+c)^(1/2)
Time = 0.27 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.80 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {{\left (d x^{3} + c\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 27 \, {\left (d x^{3} + c\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 48 \, \sqrt {d x^{3} + c} c}{648 \, {\left (c^{3} d x^{3} + c^{4}\right )}}, \frac {27 \, {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - {\left (d x^{3} + c\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 24 \, \sqrt {d x^{3} + c} c}{324 \, {\left (c^{3} d x^{3} + c^{4}\right )}}\right ] \]
[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)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sq rt(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)]
Time = 4.40 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.28 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {d}{27 c^{2} \sqrt {c + d x^{3}}} - \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{648 c^{2} \sqrt {- c}} + \frac {d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {- c}} \right )}}{24 c^{2} \sqrt {- c}}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\log {\left (x^{3} \right )}}{24 c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(d/(27*c**2*sqrt(c + d*x**3)) - d*atan(sqrt(c + d*x**3)/(3*sq rt(-c)))/(648*c**2*sqrt(-c)) + d*atan(sqrt(c + d*x**3)/sqrt(-c))/(24*c**2* sqrt(-c)))/d, Ne(d, 0)), (log(x**3)/(24*c**(5/2)), True))
\[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x} \,d x } \]
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{12 \, \sqrt {-c} c^{2}} - \frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{324 \, \sqrt {-c} c^{2}} + \frac {2}{27 \, \sqrt {d x^{3} + c} c^{2}} \]
1/12*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/324*arctan(1/3*sq rt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) + 2/27/(sqrt(d*x^3 + c)*c^2)
Time = 7.94 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2}{27\,c^2\,\sqrt {d\,x^3+c}}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )}{12\,\sqrt {c^5}}+\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )}{324\,\sqrt {c^5}} \]