Integrand size = 27, antiderivative size = 107 \[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{48 c^2 x^6}+\frac {5 d \sqrt {c+d x^3}}{192 c^3 x^3}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2304 c^{7/2}}-\frac {7 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{7/2}} \]
1/2304*d^2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)-7/256*d^2*arctanh( (d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)-1/48*(d*x^3+c)^(1/2)/c^2/x^6+5/192*d*(d*x ^3+c)^(1/2)/c^3/x^3
Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {\sqrt {c+d x^3} \left (-4 c+5 d x^3\right )}{192 c^3 x^6}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2304 c^{7/2}}-\frac {7 d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{7/2}} \]
(Sqrt[c + d*x^3]*(-4*c + 5*d*x^3))/(192*c^3*x^6) + (d^2*ArcTanh[Sqrt[c + d *x^3]/(3*Sqrt[c])])/(2304*c^(7/2)) - (7*d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c ]])/(256*c^(7/2))
Time = 0.26 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14, 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, 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.
| \(\displaystyle \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {1}{x^9 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\) |
| \(\Big \downarrow \) 114 |
| \(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {d \left (20 c-3 d x^3\right )}{2 x^6 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{16 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \int \frac {20 c-3 d x^3}{x^6 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {\int \frac {2 c d \left (42 c-5 d x^3\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{8 c^2}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {d \int \frac {42 c-5 d x^3}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {d \left (\frac {21}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {1}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\right )}{4 c}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {d \left (\frac {1}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {21 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}\right )}{4 c}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {d \left (\frac {21 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}\right )}{4 c}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (-\frac {d \left (-\frac {d \left (\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}-\frac {21 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{4 c}-\frac {5 \sqrt {c+d x^3}}{2 c x^3}\right )}{32 c^2}-\frac {\sqrt {c+d x^3}}{16 c^2 x^6}\right )\) |
(-1/16*Sqrt[c + d*x^3]/(c^2*x^6) - (d*((-5*Sqrt[c + d*x^3])/(2*c*x^3) - (d *(ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/(6*Sqrt[c]) - (21*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c])))/(4*c)))/(32*c^2))/3
3.4.14.3.1 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] && 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 = 4.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72
| method | result | size |
| risch | \(-\frac {\sqrt {d \,x^{3}+c}\, \left (-5 d \,x^{3}+4 c \right )}{192 c^{3} x^{6}}+\frac {d^{2} \left (-\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{18 \sqrt {c}}\right )}{128 c^{3}}\) | \(77\) |
| pseudoelliptic | \(\frac {-63 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d^{2} x^{6}+\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) d^{2} x^{6}+60 d \,x^{3} \sqrt {d \,x^{3}+c}\, \sqrt {c}-48 \sqrt {d \,x^{3}+c}\, c^{\frac {3}{2}}}{2304 c^{\frac {7}{2}} x^{6}}\) | \(86\) |
| default | \(\frac {-\frac {\sqrt {d \,x^{3}+c}}{6 c \,x^{6}}+\frac {d \sqrt {d \,x^{3}+c}}{4 c^{2} x^{3}}-\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{4 c^{\frac {5}{2}}}}{8 c}+\frac {d \left (-\frac {\sqrt {d \,x^{3}+c}}{3 c \,x^{3}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{64 c^{2}}-\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{768 c^{\frac {7}{2}}}+\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{2304 c^{\frac {7}{2}}}\) | \(154\) |
| elliptic | \(\text {Expression too large to display}\) | \(1551\) |
-1/192*(d*x^3+c)^(1/2)*(-5*d*x^3+4*c)/c^3/x^6+1/128*d^2/c^3*(-7/2*arctanh( (d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)+1/18*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2)) /c^(1/2))
Time = 0.29 (sec) , antiderivative size = 217, normalized size of antiderivative = 2.03 \[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {\sqrt {c} d^{2} x^{6} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 63 \, \sqrt {c} d^{2} x^{6} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 24 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{4608 \, c^{4} x^{6}}, \frac {63 \, \sqrt {-c} d^{2} x^{6} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - \sqrt {-c} d^{2} x^{6} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + 12 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{2304 \, c^{4} x^{6}}\right ] \]
[1/4608*(sqrt(c)*d^2*x^6*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d *x^3 - 8*c)) + 63*sqrt(c)*d^2*x^6*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 24*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/(c^4*x^6), 1/2304*(63 *sqrt(-c)*d^2*x^6*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) - sqrt(-c)*d^2*x^6*ar ctan(1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) + 12*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/(c^4*x^6)]
\[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=- \int \frac {1}{- 8 c x^{7} \sqrt {c + d x^{3}} + d x^{10} \sqrt {c + d x^{3}}}\, dx \]
\[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{7}} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {7 \, d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{256 \, \sqrt {-c} c^{3}} - \frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{2304 \, \sqrt {-c} c^{3}} + \frac {5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} - 9 \, \sqrt {d x^{3} + c} c d^{2}}{192 \, c^{3} d^{2} x^{6}} \]
7/256*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/2304*d^2*arc tan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) + 1/192*(5*(d*x^3 + c)^(3 /2)*d^2 - 9*sqrt(d*x^3 + c)*c*d^2)/(c^3*d^2*x^6)
Time = 7.67 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^7 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {d^2\,\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^7}}\right )}{2304\,\sqrt {c^7}}-\frac {7\,d^2\,\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{\sqrt {c^7}}\right )}{256\,\sqrt {c^7}}-\frac {3\,\sqrt {d\,x^3+c}}{64\,c^2\,x^6}+\frac {5\,{\left (d\,x^3+c\right )}^{3/2}}{192\,c^3\,x^6} \]