Integrand size = 27, antiderivative size = 81 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{24 c x^3}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 c^{3/2}}-\frac {5 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{3/2}} \]
1/32*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)-5/96*d*arctanh((d*x^3+ c)^(1/2)/c^(1/2))/c^(3/2)-1/24*(d*x^3+c)^(1/2)/c/x^3
Time = 0.16 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{24 c x^3}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{32 c^{3/2}}-\frac {5 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{3/2}} \]
-1/24*Sqrt[c + d*x^3]/(c*x^3) + (d*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/( 32*c^(3/2)) - (5*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(96*c^(3/2))
Time = 0.23 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {948, 110, 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.
\(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt {d x^3+c}}{x^6 \left (8 c-d x^3\right )}dx^3\) |
\(\Big \downarrow \) 110 |
\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {d \left (d x^3+10 c\right )}{2 x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{8 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\frac {d \int \frac {d x^3+10 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {5}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {9}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {9}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {5 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {5 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (\frac {d \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3}\right )\) |
(-1/8*Sqrt[c + d*x^3]/(c*x^3) + (d*((3*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c]) ])/(2*Sqrt[c]) - (5*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c])))/(16*c) )/3
3.3.87.3.1 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[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f)) Int[(a + b*x)^(m + 1) *(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
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 = 4.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.80
method | result | size |
risch | \(-\frac {\sqrt {d \,x^{3}+c}}{24 c \,x^{3}}+\frac {d \left (-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{6 \sqrt {c}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{16 c}\) | \(65\) |
pseudoelliptic | \(\frac {-5 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d \,x^{3}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) d \,x^{3}-4 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{96 c^{\frac {3}{2}} x^{3}}\) | \(65\) |
default | \(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}}{8 c}+\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) \sqrt {c}}{3}\right )}{64 c^{2}}+\frac {d \left (-2 \sqrt {d \,x^{3}+c}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {c}\right )}{192 c^{2}}\) | \(117\) |
elliptic | \(\text {Expression too large to display}\) | \(1523\) |
-1/24*(d*x^3+c)^(1/2)/c/x^3+1/16*d/c*(-5/6*arctanh((d*x^3+c)^(1/2)/c^(1/2) )/c^(1/2)+1/2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(1/2))
Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.30 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=\left [\frac {3 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 5 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, \sqrt {d x^{3} + c} c}{192 \, c^{2} x^{3}}, \frac {5 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 3 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 4 \, \sqrt {d x^{3} + c} c}{96 \, c^{2} x^{3}}\right ] \]
[1/192*(3*sqrt(c)*d*x^3*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d* x^3 - 8*c)) + 5*sqrt(c)*d*x^3*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c )/x^3) - 8*sqrt(d*x^3 + c)*c)/(c^2*x^3), 1/96*(5*sqrt(-c)*d*x^3*arctan(sqr t(d*x^3 + c)*sqrt(-c)/c) - 3*sqrt(-c)*d*x^3*arctan(1/3*sqrt(d*x^3 + c)*sqr t(-c)/c) - 4*sqrt(d*x^3 + c)*c)/(c^2*x^3)]
\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=- \int \frac {\sqrt {c + d x^{3}}}{- 8 c x^{4} + d x^{7}}\, dx \]
\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=\int { -\frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{4}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=\frac {5 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c} c} - \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{32 \, \sqrt {-c} c} - \frac {\sqrt {d x^{3} + c}}{24 \, c x^{3}} \]
5/96*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/32*d*arctan(1/3*s qrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/24*sqrt(d*x^3 + c)/(c*x^3)
Time = 7.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )} \, dx=\frac {d\,\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^3}}\right )}{32\,\sqrt {c^3}}-\frac {5\,d\,\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{\sqrt {c^3}}\right )}{96\,\sqrt {c^3}}-\frac {\sqrt {d\,x^3+c}}{24\,c\,x^3} \]