Integrand size = 22, antiderivative size = 158 \[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {a (d x)^{1+m} \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {1+m}{3},-\frac {3}{2},-\frac {3}{2},\frac {4+m}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m) \sqrt {1+\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}}} \]
a*(d*x)^(1+m)*AppellF1(1/3+1/3*m,-3/2,-3/2,4/3+1/3*m,-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)/d/(1+m)/ (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. \(357\) vs. \(2(158)=316\).
Time = 1.85 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.26 \[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {x (d x)^m \sqrt {a+b x^3+c x^6} \left (a \left (28+11 m+m^2\right ) \operatorname {AppellF1}\left (\frac {1+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {4+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+(1+m) x^3 \left (b (7+m) \operatorname {AppellF1}\left (\frac {4+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {7+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )+c (4+m) x^3 \operatorname {AppellF1}\left (\frac {7+m}{3},-\frac {1}{2},-\frac {1}{2},\frac {10+m}{3},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}},\frac {2 c x^3}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{(1+m) (4+m) (7+m) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^3}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^3}{b+\sqrt {b^2-4 a c}}}} \]
(x*(d*x)^m*Sqrt[a + b*x^3 + c*x^6]*(a*(28 + 11*m + m^2)*AppellF1[(1 + m)/3 , -1/2, -1/2, (4 + m)/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])] + (1 + m)*x^3*(b*(7 + m)*AppellF1[(4 + m)/3, -1/2, -1/2, (7 + m)/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[ b^2 - 4*a*c])] + c*(4 + m)*x^3*AppellF1[(7 + m)/3, -1/2, -1/2, (10 + m)/3, (-2*c*x^3)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^3)/(-b + Sqrt[b^2 - 4*a*c])])) )/((1 + m)*(4 + m)*(7 + m)*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b - Sqr t[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^3)/(b + Sqrt[b^2 - 4* a*c])])
Time = 0.31 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.
\(\displaystyle \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1721 |
\(\displaystyle \frac {a \sqrt {a+b x^3+c x^6} \int (d x)^m \left (\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1\right )^{3/2} \left (\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}+1\right )^{3/2}dx}{\sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {a (d x)^{m+1} \sqrt {a+b x^3+c x^6} \operatorname {AppellF1}\left (\frac {m+1}{3},-\frac {3}{2},-\frac {3}{2},\frac {m+4}{3},-\frac {2 c x^3}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^3}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1) \sqrt {\frac {2 c x^3}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^3}{\sqrt {b^2-4 a c}+b}+1}}\) |
(a*(d*x)^(1 + m)*Sqrt[a + b*x^3 + c*x^6]*AppellF1[(1 + m)/3, -3/2, -3/2, ( 4 + m)/3, (-2*c*x^3)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^3)/(b + Sqrt[b^2 - 4 *a*c])])/(d*(1 + m)*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])])
3.3.52.3.1 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]
\[\int \left (d x \right )^{m} \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}d x\]
\[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} \left (d x\right )^{m} \,d x } \]
\[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int \left (d x\right )^{m} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]
\[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} \left (d x\right )^{m} \,d x } \]
\[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} \left (d x\right )^{m} \,d x } \]
Timed out. \[ \int (d x)^m \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int {\left (d\,x\right )}^m\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]