Integrand size = 20, antiderivative size = 143 \[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=-\frac {\sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (-\frac {2}{3},\frac {3}{2},\frac {3}{2},\frac {1}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 a x^2 \sqrt {a+b x^3+c x^6}} \]
-1/2*AppellF1(-2/3,3/2,3/2,1/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2)),-2*c*x^3/(b +(-4*a*c+b^2)^(1/2)))*(1+2*c*x^3/(b-(-4*a*c+b^2)^(1/2)))^(1/2)*(1+2*c*x^3/ (b+(-4*a*c+b^2)^(1/2)))^(1/2)/a/x^2/(c*x^6+b*x^3+a)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(405\) vs. \(2(143)=286\).
Time = 10.44 (sec) , antiderivative size = 405, normalized size of antiderivative = 2.83 \[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\frac {-48 a^2 c+28 b^2 x^3 \left (b+c x^3\right )+4 a \left (3 b^2-24 b c x^3-20 c^2 x^6\right )+2 b \left (7 b^2-36 a c\right ) x^3 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{2},\frac {1}{2},\frac {4}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+c \left (-7 b^2+20 a c\right ) x^6 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{2},\frac {1}{2},\frac {7}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )}{24 a^2 \left (-b^2+4 a c\right ) x^2 \sqrt {a+b x^3+c x^6}} \]
(-48*a^2*c + 28*b^2*x^3*(b + c*x^3) + 4*a*(3*b^2 - 24*b*c*x^3 - 20*c^2*x^6 ) + 2*b*(7*b^2 - 36*a*c)*x^3*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - S qrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[1/3, 1/2, 1/2, 4/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + c*(-7*b^2 + 20*a*c)*x^6*Sqrt[(b - Sq rt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4 *a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[4/3, 1/2, 1/2, 7/3, (-2 *c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])/(24* a^2*(-b^2 + 4*a*c)*x^2*Sqrt[a + b*x^3 + c*x^6])
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1721, 1012}
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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1721 |
\(\displaystyle \frac {\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} \int \frac {1}{x^3 \left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}}dx}{a \sqrt {a+b x^3+c x^6}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1} \operatorname {AppellF1}\left (-\frac {2}{3},\frac {3}{2},\frac {3}{2},\frac {1}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 a x^2 \sqrt {a+b x^3+c x^6}}\) |
-1/2*(Sqrt[1 + (2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[-2/3, 3/2, 3/2, 1/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(a*x^2*Sqrt[a + b*x^3 + c *x^6])
3.3.47.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x _Symbol] :> Simp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2* c*(x^n/(b + Rt[b^2 - 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4 *a*c, 2])))^FracPart[p])) Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c ])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]
\[\int \frac {1}{x^{3} \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
integral(sqrt(c*x^6 + b*x^3 + a)/(c^2*x^15 + 2*b*c*x^12 + (b^2 + 2*a*c)*x^ 9 + 2*a*b*x^6 + a^2*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int \frac {1}{x^{3} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
\[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 \left (a+b x^3+c x^6\right )^{3/2}} \, dx=\int \frac {1}{x^3\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2}} \,d x \]