Integrand size = 20, antiderivative size = 204 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=-\frac {\left (b^2-4 a c\right ) \left (7 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{1536 c^4}+\frac {\left (7 b^2-4 a c\right ) \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{576 c^3}-\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{180 c^2}+\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{18 c}+\frac {\left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{3072 c^{9/2}} \]
1/576*(-4*a*c+7*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(3/2)/c^3-7/180*b*(c*x^6+ b*x^3+a)^(5/2)/c^2+1/18*x^3*(c*x^6+b*x^3+a)^(5/2)/c+1/3072*(-4*a*c+b^2)^2* (-4*a*c+7*b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1/2)/(c*x^6+b*x^3+a)^(1/2))/c^(9 /2)-1/1536*(-4*a*c+b^2)*(-4*a*c+7*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c ^4
Time = 0.51 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.95 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {2 \sqrt {c} \sqrt {a+b x^3+c x^6} \left (-105 b^5+70 b^4 c x^3+8 b^3 c \left (95 a-7 c x^6\right )+48 b^2 c^2 x^3 \left (-9 a+c x^6\right )+160 c^3 x^3 \left (3 a^2+14 a c x^6+8 c^2 x^{12}\right )+16 b c^2 \left (-81 a^2+18 a c x^6+104 c^2 x^{12}\right )\right )-15 \left (b^2-4 a c\right )^2 \left (7 b^2-4 a c\right ) \log \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )}{46080 c^{9/2}} \]
(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]*(-105*b^5 + 70*b^4*c*x^3 + 8*b^3*c*(95* a - 7*c*x^6) + 48*b^2*c^2*x^3*(-9*a + c*x^6) + 160*c^3*x^3*(3*a^2 + 14*a*c *x^6 + 8*c^2*x^12) + 16*b*c^2*(-81*a^2 + 18*a*c*x^6 + 104*c^2*x^12)) - 15* (b^2 - 4*a*c)^2*(7*b^2 - 4*a*c)*Log[b + 2*c*x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]])/(46080*c^(9/2))
Time = 0.35 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1693, 1166, 27, 1160, 1087, 1087, 1092, 219}
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.
\(\displaystyle \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1693 |
\(\displaystyle \frac {1}{3} \int x^6 \left (c x^6+b x^3+a\right )^{3/2}dx^3\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {1}{3} \left (\frac {\int -\frac {1}{2} \left (7 b x^3+2 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{6 c}+\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\int \left (7 b x^3+2 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{12 c}\right )\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \int \left (c x^6+b x^3+a\right )^{3/2}dx^3}{2 c}}{12 c}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^6+b x^3+a}dx^3}{16 c}\right )}{2 c}}{12 c}\right )\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^6+b x^3+a}}dx^3}{8 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\) |
\(\Big \downarrow \) 1092 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^6}d\frac {2 c x^3+b}{\sqrt {c x^6+b x^3+a}}}{4 c}\right )}{16 c}\right )}{2 c}}{12 c}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {x^3 \left (a+b x^3+c x^6\right )^{5/2}}{6 c}-\frac {\frac {7 b \left (a+b x^3+c x^6\right )^{5/2}}{5 c}-\frac {\left (7 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{2 c}}{12 c}\right )\) |
((x^3*(a + b*x^3 + c*x^6)^(5/2))/(6*c) - ((7*b*(a + b*x^3 + c*x^6)^(5/2))/ (5*c) - ((7*b^2 - 4*a*c)*(((b + 2*c*x^3)*(a + b*x^3 + c*x^6)^(3/2))/(8*c) - (3*(b^2 - 4*a*c)*(((b + 2*c*x^3)*Sqrt[a + b*x^3 + c*x^6])/(4*c) - ((b^2 - 4*a*c)*ArcTanh[(b + 2*c*x^3)/(2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6])])/(8*c^ (3/2))))/(16*c)))/(2*c))/(12*c))/3
3.3.3.3.1 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_)^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_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && IntegerQ [Simplify[(m + 1)/n]]
\[\int x^{8} \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}d x\]
Time = 0.28 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.21 \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\left [-\frac {15 \, {\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} + 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (1280 \, c^{6} x^{15} + 1664 \, b c^{5} x^{12} + 16 \, {\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{9} - 8 \, {\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} + 2 \, {\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{92160 \, c^{5}}, -\frac {15 \, {\left (7 \, b^{6} - 60 \, a b^{4} c + 144 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) - 2 \, {\left (1280 \, c^{6} x^{15} + 1664 \, b c^{5} x^{12} + 16 \, {\left (3 \, b^{2} c^{4} + 140 \, a c^{5}\right )} x^{9} - 8 \, {\left (7 \, b^{3} c^{3} - 36 \, a b c^{4}\right )} x^{6} - 105 \, b^{5} c + 760 \, a b^{3} c^{2} - 1296 \, a^{2} b c^{3} + 2 \, {\left (35 \, b^{4} c^{2} - 216 \, a b^{2} c^{3} + 240 \, a^{2} c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{46080 \, c^{5}}\right ] \]
[-1/92160*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^2*c^2 - 64*a^3*c^3)*sqrt(c)* log(-8*c^2*x^6 - 8*b*c*x^3 - b^2 + 4*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b) *sqrt(c) - 4*a*c) - 4*(1280*c^6*x^15 + 1664*b*c^5*x^12 + 16*(3*b^2*c^4 + 1 40*a*c^5)*x^9 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^6 - 105*b^5*c + 760*a*b^3*c^2 - 1296*a^2*b*c^3 + 2*(35*b^4*c^2 - 216*a*b^2*c^3 + 240*a^2*c^4)*x^3)*sqrt (c*x^6 + b*x^3 + a))/c^5, -1/46080*(15*(7*b^6 - 60*a*b^4*c + 144*a^2*b^2*c ^2 - 64*a^3*c^3)*sqrt(-c)*arctan(1/2*sqrt(c*x^6 + b*x^3 + a)*(2*c*x^3 + b) *sqrt(-c)/(c^2*x^6 + b*c*x^3 + a*c)) - 2*(1280*c^6*x^15 + 1664*b*c^5*x^12 + 16*(3*b^2*c^4 + 140*a*c^5)*x^9 - 8*(7*b^3*c^3 - 36*a*b*c^4)*x^6 - 105*b^ 5*c + 760*a*b^3*c^2 - 1296*a^2*b*c^3 + 2*(35*b^4*c^2 - 216*a*b^2*c^3 + 240 *a^2*c^4)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^5]
\[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^{8} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
\[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{8} \,d x } \]
Timed out. \[ \int x^8 \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^8\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]