Integrand size = 18, antiderivative size = 141 \[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {a x^2 \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {2}{3},-\frac {3}{2},-\frac {3}{2},\frac {5}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \]
1/2*a*x^2*AppellF1(2/3,-3/2,-3/2,5/3,-2*c*x^3/(b-(-4*a*c+b^2)^(1/2)),-2*c* x^3/(b+(-4*a*c+b^2)^(1/2)))*(c*x^6+b*x^3+a)^(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)
Leaf count is larger than twice the leaf count of optimal. \(410\) vs. \(2(141)=282\).
Time = 10.50 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.91 \[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {x^2 \left (10 \left (27 a b^2+448 a^2 c+27 b^3 x^3+698 a b c x^3+277 b^2 c x^6+608 a c^2 x^6+410 b c^2 x^9+160 c^3 x^{12}\right )-270 a \left (b^2-16 a c\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},\frac {1}{2},\frac {5}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )-27 b \left (7 b^2-52 a c\right ) x^3 \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},\frac {1}{2},\frac {8}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )\right )}{17600 c \sqrt {a+b x^3+c x^6}} \]
(x^2*(10*(27*a*b^2 + 448*a^2*c + 27*b^3*x^3 + 698*a*b*c*x^3 + 277*b^2*c*x^ 6 + 608*a*c^2*x^6 + 410*b*c^2*x^9 + 160*c^3*x^12) - 270*a*(b^2 - 16*a*c)*S qrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + S qrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[2/3, 1/2, 1/ 2, 5/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a *c])] - 27*b*(7*b^2 - 52*a*c)*x^3*Sqrt[(b - Sqrt[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[5/3, 1/2, 1/2, 8/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a* c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])]))/(17600*c*Sqrt[a + b*x^3 + c*x^6 ])
Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1721 |
\(\displaystyle \frac {a \sqrt {a+b x^3+c x^6} \int x \left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}dx}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {a x^2 \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {2}{3},-\frac {3}{2},-\frac {3}{2},\frac {5}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\) |
(a*x^2*Sqrt[a + b*x^3 + c*x^6]*AppellF1[2/3, -3/2, -3/2, 5/3, (-2*c*x^3)/( b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c])])/(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 ])])
3.3.15.3.1 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]
\[\int x \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}d x\]
\[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x \,d x } \]
\[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]
\[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x \,d x } \]
\[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x \,d x } \]
Timed out. \[ \int x \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]