Integrand size = 26, antiderivative size = 88 \[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{12 c^2 x^3}+\frac {d \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{24 \sqrt {3} c^{5/2}}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{8 c^{5/2}} \]
1/8*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/72*d*arctan(1/3*(d*x^3+c) ^(1/2)*3^(1/2)/c^(1/2))/c^(5/2)*3^(1/2)-1/12*(d*x^3+c)^(1/2)/c^2/x^3
Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{12 c^2 x^3}+\frac {d \arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{24 \sqrt {3} c^{5/2}}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{8 c^{5/2}} \]
-1/12*Sqrt[c + d*x^3]/(c^2*x^3) + (d*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[ c])])/(24*Sqrt[3]*c^(5/2)) + (d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(8*c^(5/ 2))
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {948, 114, 27, 174, 73, 216, 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.
\(\displaystyle \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^6 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3\) |
\(\Big \downarrow \) 114 |
\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {d \left (d x^3+6 c\right )}{2 x^3 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3}{4 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (-\frac {d \int \frac {d x^3+6 c}{x^3 \sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
\(\Big \downarrow \) 174 |
\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {3}{2} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3-\frac {1}{2} d \int \frac {1}{\sqrt {d x^3+c} \left (d x^3+4 c\right )}dx^3\right )}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{d}-\int \frac {1}{x^6+3 c}d\sqrt {d x^3+c}\right )}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{d}-\frac {\arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} \sqrt {c}}\right )}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {\arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} \sqrt {c}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}\right )}{8 c^2}-\frac {\sqrt {c+d x^3}}{4 c^2 x^3}\right )\) |
(-1/4*Sqrt[c + d*x^3]/(c^2*x^3) - (d*(-(ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sq rt[c])]/(Sqrt[3]*Sqrt[c])) - (3*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/Sqrt[c]) )/(8*c^2))/3
3.3.74.3.1 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[(((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]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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 = 4.46 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{8 c^{\frac {5}{2}}}+\frac {d \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{72 c^{\frac {5}{2}}}-\frac {\sqrt {d \,x^{3}+c}}{12 c^{2} x^{3}}\) | \(66\) |
pseudoelliptic | \(\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) d \,x^{3}+9 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d \,x^{3}-6 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{72 c^{\frac {5}{2}} x^{3}}\) | \(70\) |
default | \(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c \,x^{3}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{4 c}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{24 c^{\frac {5}{2}}}+\frac {d \arctan \left (\frac {\sqrt {d \,x^{3}+c}\, \sqrt {3}}{3 \sqrt {c}}\right ) \sqrt {3}}{72 c^{\frac {5}{2}}}\) | \(92\) |
elliptic | \(\text {Expression too large to display}\) | \(1523\) |
1/8*d*arctanh((d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/72*d*arctan(1/3*(d*x^3+c) ^(1/2)*3^(1/2)/c^(1/2))/c^(5/2)*3^(1/2)-1/12*(d*x^3+c)^(1/2)/c^2/x^3
Time = 0.26 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.20 \[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\left [\frac {2 \, \sqrt {3} \sqrt {c} d x^{3} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + 9 \, \sqrt {c} d x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 12 \, \sqrt {d x^{3} + c} c}{144 \, c^{3} x^{3}}, -\frac {\sqrt {3} \sqrt {-c} d x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + 18 \, \sqrt {-c} d x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + 12 \, \sqrt {d x^{3} + c} c}{144 \, c^{3} x^{3}}\right ] \]
[1/144*(2*sqrt(3)*sqrt(c)*d*x^3*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c) ) + 9*sqrt(c)*d*x^3*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 1 2*sqrt(d*x^3 + c)*c)/(c^3*x^3), -1/144*(sqrt(3)*sqrt(-c)*d*x^3*log((d*x^3 - 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + 18*sqrt(-c)*d *x^3*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) + 12*sqrt(d*x^3 + c)*c)/(c^3*x^3)]
\[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {1}{x^{4} \sqrt {c + d x^{3}} \cdot \left (4 c + d x^{3}\right )}\, dx \]
\[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int { \frac {1}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} x^{4}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {\sqrt {3} d \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{72 \, c^{\frac {5}{2}}} - \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{8 \, \sqrt {-c} c^{2}} - \frac {\sqrt {d x^{3} + c}}{12 \, c^{2} x^{3}} \]
1/72*sqrt(3)*d*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/c^(5/2) - 1/8*d *arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/12*sqrt(d*x^3 + c)/(c ^2*x^3)
Time = 9.50 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^4 \sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {d\,\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )}{16\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{12\,c^2\,x^3}+\frac {\sqrt {3}\,d\,\ln \left (\frac {\sqrt {3}\,d\,x^3-2\,\sqrt {3}\,c+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,1{}\mathrm {i}}{144\,c^{5/2}} \]