Integrand size = 27, antiderivative size = 124 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {d \sqrt {c+d x^3}}{96 c^2 \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 c x^3 \left (8 c-d x^3\right )}+\frac {7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1152 c^{5/2}}-\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{128 c^{5/2}} \]
7/1152*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-1/128*d*arctanh((d*x ^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/96*d*(d*x^3+c)^(1/2)/c^2/(-d*x^3+8*c)-1/24* (d*x^3+c)^(1/2)/c/x^3/(-d*x^3+8*c)
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {12 \sqrt {c} \left (4 c-d x^3\right ) \sqrt {c+d x^3}}{-8 c x^3+d x^6}+7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-9 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{1152 c^{5/2}} \]
((12*Sqrt[c]*(4*c - d*x^3)*Sqrt[c + d*x^3])/(-8*c*x^3 + d*x^6) + 7*d*ArcTa nh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 9*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(1 152*c^(5/2))
Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.15, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 110, 27, 168, 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.
| \(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {\sqrt {d x^3+c}}{x^6 \left (8 c-d x^3\right )^2}dx^3\) |
| \(\Big \downarrow \) 110 |
| \(\displaystyle \frac {1}{3} \left (\frac {\int \frac {3 d \left (d x^3+4 c\right )}{2 x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \int \frac {d x^3+4 c}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}-\frac {\int -\frac {6 c d \left (d x^3+6 c\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{72 c^2 d}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\int \frac {d x^3+6 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{12 c}+\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\frac {3}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {7}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{12 c}+\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\frac {7}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{12 c}+\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\frac {3 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}}{12 c}+\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (\frac {3 d \left (\frac {\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{12 c}+\frac {\sqrt {c+d x^3}}{6 c \left (8 c-d x^3\right )}\right )}{16 c}-\frac {\sqrt {c+d x^3}}{8 c x^3 \left (8 c-d x^3\right )}\right )\) |
(-1/8*Sqrt[c + d*x^3]/(c*x^3*(8*c - d*x^3)) + (3*d*(Sqrt[c + d*x^3]/(6*c*( 8*c - d*x^3)) + ((7*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(6*Sqrt[c]) - (3 *ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(12*c)))/(16*c))/3
3.5.3.3.1 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[((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] && ILtQ[m, -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 = 4.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.79
| method | result | size |
| pseudoelliptic | \(\frac {d \left (-\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d \,x^{3}+2 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{2 d \,x^{3} c^{\frac {5}{2}}}+\frac {\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}}{c^{2}}\right )}{192}\) | \(98\) |
| risch | \(-\frac {\sqrt {d \,x^{3}+c}}{192 c^{2} x^{3}}-\frac {d \left (\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{9 \sqrt {c}}-\frac {2 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{3}\right )}{128 c^{2}}\) | \(113\) |
| default | \(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}}{64 c^{2}}+\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) \sqrt {c}}{3}\right )}{256 c^{3}}+\frac {d \left (\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 \sqrt {c}}\right )}{192 c^{2}}+\frac {d \left (-2 \sqrt {d \,x^{3}+c}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {c}\right )}{768 c^{3}}\) | \(166\) |
| elliptic | \(\text {Expression too large to display}\) | \(1550\) |
1/192*d*(-1/2*(3*arctanh((d*x^3+c)^(1/2)/c^(1/2))*d*x^3+2*(d*x^3+c)^(1/2)* c^(1/2))/d/x^3/c^(5/2)+((d*x^3+c)^(1/2)/(-d*x^3+8*c)+7/6*arctanh(1/3*(d*x^ 3+c)^(1/2)/c^(1/2))/c^(1/2))/c^2)
Time = 0.33 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.24 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\left [\frac {7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, {\left (c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{2304 \, {\left (c^{3} d x^{6} - 8 \, c^{4} x^{3}\right )}}, \frac {9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 12 \, {\left (c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{1152 \, {\left (c^{3} d x^{6} - 8 \, c^{4} x^{3}\right )}}\right ] \]
[1/2304*(7*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sq rt(c) + 10*c)/(d*x^3 - 8*c)) + 9*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/(c^3*d*x^6 - 8*c^4*x^3), 1/1152*(9*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*ar ctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 7*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan (1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 12*(c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/( c^3*d*x^6 - 8*c^4*x^3)]
\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\int \frac {\sqrt {c + d x^{3}}}{x^{4} \left (- 8 c + d x^{3}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{4}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.91 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{128 \, \sqrt {-c} c^{2}} - \frac {7 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{1152 \, \sqrt {-c} c^{2}} - \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}} d - 5 \, \sqrt {d x^{3} + c} c d}{96 \, {\left ({\left (d x^{3} + c\right )}^{2} - 10 \, {\left (d x^{3} + c\right )} c + 9 \, c^{2}\right )} c^{2}} \]
1/128*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 7/1152*d*arctan( 1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/96*((d*x^3 + c)^(3/2)*d - 5*sqrt(d*x^3 + c)*c*d)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c^2)
Time = 8.43 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {5\,d\,\sqrt {d\,x^3+c}}{32\,c}-\frac {d\,{\left (d\,x^3+c\right )}^{3/2}}{32\,c^2}}{3\,{\left (d\,x^3+c\right )}^2-30\,c\,\left (d\,x^3+c\right )+27\,c^2}+\frac {d\,\left (\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{\sqrt {c^5}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c^2\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^5}}\right )\,7{}\mathrm {i}}{9}\right )\,1{}\mathrm {i}}{128\,\sqrt {c^5}} \]