Integrand size = 27, antiderivative size = 106 \[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {5}{648 c^3 \sqrt {c+d x^3}}+\frac {1}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{7776 c^{7/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{7/2}} \]
7/7776*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)-1/96*arctanh((d*x^3+c) ^(1/2)/c^(1/2))/c^(7/2)+5/648/c^3/(d*x^3+c)^(1/2)+1/216/c^2/(-d*x^3+8*c)/( d*x^3+c)^(1/2)
Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {\frac {12 \sqrt {c} \left (43 c-5 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {c+d x^3}}+7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-81 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{7776 c^{7/2}} \]
((12*Sqrt[c]*(43*c - 5*d*x^3))/((8*c - d*x^3)*Sqrt[c + d*x^3]) + 7*ArcTanh [Sqrt[c + d*x^3]/(3*Sqrt[c])] - 81*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(7776 *c^(7/2))
Time = 0.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.19, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 114, 27, 169, 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.
| \(\displaystyle \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{x^3 \left (8 c-d x^3\right )^2 \left (d x^3+c\right )^{3/2}}dx^3\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\int -\frac {3 d \left (d x^3+6 c\right )}{2 x^3 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3}{72 c^2 d}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {\int \frac {d x^3+6 c}{x^3 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 169 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {2 \int \frac {c d \left (54 c-5 d x^3\right )}{2 x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2 d}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {\int \frac {54 c-5 d x^3}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {27}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {7}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {7}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {27 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{9 c}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {27 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}}{9 c}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (\frac {\frac {\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}-\frac {27 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{9 c}+\frac {10}{9 c \sqrt {c+d x^3}}}{48 c^2}+\frac {1}{72 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\right )\) |
(1/(72*c^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (10/(9*c*Sqrt[c + d*x^3]) + (( 7*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(6*Sqrt[c]) - (27*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(9*c))/(48*c^2))/3
3.5.46.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> 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] + Simp[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] && ILtQ[m, -1] && (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(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] + S imp[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*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n, 2*p]
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 = 4.55 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.87
| method | result | size |
| pseudoelliptic | \(-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{96 c^{\frac {7}{2}}}+\frac {\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \left (d \,x^{3}-8 c \right )}{\sqrt {c}}-4 \sqrt {d \,x^{3}+c}}{7776 \left (d \,x^{3}-8 c \right ) c^{3}}+\frac {2}{243 c^{3} \sqrt {d \,x^{3}+c}}\) | \(92\) |
| default | \(\frac {\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{64 c^{2}}+\frac {-\frac {2}{\sqrt {d \,x^{3}+c}}+\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}}{1944 c^{3}}-\frac {-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {d \,x^{3}+c}}{3}+\sqrt {c}}{864 c^{\frac {7}{2}} \sqrt {d \,x^{3}+c}}\) | \(147\) |
| elliptic | \(\text {Expression too large to display}\) | \(1552\) |
-1/96*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)+1/7776*(7*arctanh(1/3*(d*x^ 3+c)^(1/2)/c^(1/2))/c^(1/2)*(d*x^3-8*c)-4*(d*x^3+c)^(1/2))/(d*x^3-8*c)/c^3 +2/243/c^3/(d*x^3+c)^(1/2)
Time = 0.35 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.98 \[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {7 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 81 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 24 \, {\left (5 \, c d x^{3} - 43 \, c^{2}\right )} \sqrt {d x^{3} + c}}{15552 \, {\left (c^{4} d^{2} x^{6} - 7 \, c^{5} d x^{3} - 8 \, c^{6}\right )}}, \frac {81 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 7 \, {\left (d^{2} x^{6} - 7 \, c d x^{3} - 8 \, c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 12 \, {\left (5 \, c d x^{3} - 43 \, c^{2}\right )} \sqrt {d x^{3} + c}}{7776 \, {\left (c^{4} d^{2} x^{6} - 7 \, c^{5} d x^{3} - 8 \, c^{6}\right )}}\right ] \]
[1/15552*(7*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^ 3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) + 81*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*s qrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 24*(5*c*d*x^3 - 43*c^2)*sqrt(d*x^3 + c))/(c^4*d^2*x^6 - 7*c^5*d*x^3 - 8*c^6), 1/7776*(81 *(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(-c)*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 7*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(-c)*arctan(1/3*sqrt(d*x^3 + c)*sqr t(-c)/c) + 12*(5*c*d*x^3 - 43*c^2)*sqrt(d*x^3 + c))/(c^4*d^2*x^6 - 7*c^5*d *x^3 - 8*c^6)]
\[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x \left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2} x} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c} c^{3}} - \frac {7 \, \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{7776 \, \sqrt {-c} c^{3}} + \frac {5 \, d x^{3} - 43 \, c}{648 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - 9 \, \sqrt {d x^{3} + c} c\right )} c^{3}} \]
1/96*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 7/7776*arctan(1/3*s qrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) + 1/648*(5*d*x^3 - 43*c)/(((d*x^3 + c)^(3/2) - 9*sqrt(d*x^3 + c)*c)*c^3)
Time = 8.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x \left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {\frac {5\,\left (d\,x^3+c\right )}{216\,c^3}-\frac {2}{9\,c^2}}{27\,c\,\sqrt {d\,x^3+c}-3\,{\left (d\,x^3+c\right )}^{3/2}}+\frac {\left (\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{\sqrt {c^7}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^7}}\right )\,7{}\mathrm {i}}{81}\right )\,1{}\mathrm {i}}{96\,\sqrt {c^7}} \]