Integrand size = 20, antiderivative size = 223 \[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {b \left (b^2-4 a c\right ) \left (3 b^2-4 a c\right ) \left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{1024 c^5}-\frac {b \left (3 b^2-4 a c\right ) \left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{384 c^4}+\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{21 c}+\frac {\left (21 b^2-16 a c-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{840 c^3}-\frac {b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{2048 c^{11/2}} \]
-1/384*b*(-4*a*c+3*b^2)*(2*c*x^3+b)*(c*x^6+b*x^3+a)^(3/2)/c^4+1/21*x^6*(c* x^6+b*x^3+a)^(5/2)/c+1/840*(-30*b*c*x^3-16*a*c+21*b^2)*(c*x^6+b*x^3+a)^(5/ 2)/c^3-1/2048*b*(-4*a*c+b^2)^2*(-4*a*c+3*b^2)*arctanh(1/2*(2*c*x^3+b)/c^(1 /2)/(c*x^6+b*x^3+a)^(1/2))/c^(11/2)+1/1024*b*(-4*a*c+b^2)*(-4*a*c+3*b^2)*( 2*c*x^3+b)*(c*x^6+b*x^3+a)^(1/2)/c^5
Time = 0.66 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99 \[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\frac {\sqrt {a+b x^3+c x^6} \left (315 b^6-210 b^5 c x^3+16 b^3 c^2 x^3 \left (91 a-9 c x^6\right )+168 b^4 c \left (-15 a+c x^6\right )+1024 c^3 \left (a+c x^6\right )^2 \left (-2 a+5 c x^6\right )+16 b^2 c^2 \left (343 a^2-62 a c x^6+8 c^2 x^{12}\right )+32 b c^3 x^3 \left (-73 a^2+22 a c x^6+200 c^2 x^{12}\right )\right )}{107520 c^5}+\frac {b \left (b^2-4 a c\right )^2 \left (3 b^2-4 a c\right ) \log \left (b+2 c x^3-2 \sqrt {c} \sqrt {a+b x^3+c x^6}\right )}{2048 c^{11/2}} \]
(Sqrt[a + b*x^3 + c*x^6]*(315*b^6 - 210*b^5*c*x^3 + 16*b^3*c^2*x^3*(91*a - 9*c*x^6) + 168*b^4*c*(-15*a + c*x^6) + 1024*c^3*(a + c*x^6)^2*(-2*a + 5*c *x^6) + 16*b^2*c^2*(343*a^2 - 62*a*c*x^6 + 8*c^2*x^12) + 32*b*c^3*x^3*(-73 *a^2 + 22*a*c*x^6 + 200*c^2*x^12)))/(107520*c^5) + (b*(b^2 - 4*a*c)^2*(3*b ^2 - 4*a*c)*Log[b + 2*c*x^3 - 2*Sqrt[c]*Sqrt[a + b*x^3 + c*x^6]])/(2048*c^ (11/2))
Time = 0.38 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.02, 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, 1225, 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.
| \(\displaystyle \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1693 |
| \(\displaystyle \frac {1}{3} \int x^9 \left (c x^6+b x^3+a\right )^{3/2}dx^3\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {1}{3} \left (\frac {\int -\frac {1}{2} x^3 \left (9 b x^3+4 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{7 c}+\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\int x^3 \left (9 b x^3+4 a\right ) \left (c x^6+b x^3+a\right )^{3/2}dx^3}{14 c}\right )\) |
| \(\Big \downarrow \) 1225 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\frac {7 b \left (3 b^2-4 a c\right ) \int \left (c x^6+b x^3+a\right )^{3/2}dx^3}{8 c^2}-\frac {\left (-16 a c+21 b^2-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{20 c^2}}{14 c}\right )\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\frac {7 b \left (3 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \int \sqrt {c x^6+b x^3+a}dx^3}{16 c}\right )}{8 c^2}-\frac {\left (-16 a c+21 b^2-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{20 c^2}}{14 c}\right )\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\frac {7 b \left (3 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\sqrt {c x^6+b x^3+a}}dx^3}{8 c}\right )}{16 c}\right )}{8 c^2}-\frac {\left (-16 a c+21 b^2-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{20 c^2}}{14 c}\right )\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\frac {7 b \left (3 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 c-x^6}d\frac {2 c x^3+b}{\sqrt {c x^6+b x^3+a}}}{4 c}\right )}{16 c}\right )}{8 c^2}-\frac {\left (-16 a c+21 b^2-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{20 c^2}}{14 c}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {x^6 \left (a+b x^3+c x^6\right )^{5/2}}{7 c}-\frac {\frac {7 b \left (3 b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \left (a+b x^3+c x^6\right )^{3/2}}{8 c}-\frac {3 \left (b^2-4 a c\right ) \left (\frac {\left (b+2 c x^3\right ) \sqrt {a+b x^3+c x^6}}{4 c}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {b+2 c x^3}{2 \sqrt {c} \sqrt {a+b x^3+c x^6}}\right )}{8 c^{3/2}}\right )}{16 c}\right )}{8 c^2}-\frac {\left (-16 a c+21 b^2-30 b c x^3\right ) \left (a+b x^3+c x^6\right )^{5/2}}{20 c^2}}{14 c}\right )\) |
((x^6*(a + b*x^3 + c*x^6)^(5/2))/(7*c) - (-1/20*((21*b^2 - 16*a*c - 30*b*c *x^3)*(a + b*x^3 + c*x^6)^(5/2))/c^2 + (7*b*(3*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)))/(8*c^2))/(14*c))/3
3.3.2.3.1 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_))^(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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
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]]
\[\int x^{11} \left (c \,x^{6}+b \,x^{3}+a \right )^{\frac {3}{2}}d x\]
Time = 0.28 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.40 \[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\left [-\frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{6} - 8 \, b c x^{3} - b^{2} - 4 \, \sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (5120 \, c^{7} x^{18} + 6400 \, b c^{6} x^{15} + 128 \, {\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{12} - 16 \, {\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{9} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} + 8 \, {\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{6} - 2 \, {\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{430080 \, c^{6}}, \frac {105 \, {\left (3 \, b^{7} - 28 \, a b^{5} c + 80 \, a^{2} b^{3} c^{2} - 64 \, a^{3} b c^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{6} + b x^{3} + a} {\left (2 \, c x^{3} + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{6} + b c x^{3} + a c\right )}}\right ) + 2 \, {\left (5120 \, c^{7} x^{18} + 6400 \, b c^{6} x^{15} + 128 \, {\left (b^{2} c^{5} + 64 \, a c^{6}\right )} x^{12} - 16 \, {\left (9 \, b^{3} c^{4} - 44 \, a b c^{5}\right )} x^{9} + 315 \, b^{6} c - 2520 \, a b^{4} c^{2} + 5488 \, a^{2} b^{2} c^{3} - 2048 \, a^{3} c^{4} + 8 \, {\left (21 \, b^{4} c^{3} - 124 \, a b^{2} c^{4} + 128 \, a^{2} c^{5}\right )} x^{6} - 2 \, {\left (105 \, b^{5} c^{2} - 728 \, a b^{3} c^{3} + 1168 \, a^{2} b c^{4}\right )} x^{3}\right )} \sqrt {c x^{6} + b x^{3} + a}}{215040 \, c^{6}}\right ] \]
[-1/430080*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3*c^2 - 64*a^3*b*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*(5120*c^7*x^18 + 6400*b*c^6*x^15 + 128*(b^2*c^5 + 64*a*c^6)*x^12 - 16*(9*b^3*c^4 - 44*a*b*c^5)*x^9 + 315*b^6*c - 2520*a*b^4 *c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 1 28*a^2*c^5)*x^6 - 2*(105*b^5*c^2 - 728*a*b^3*c^3 + 1168*a^2*b*c^4)*x^3)*sq rt(c*x^6 + b*x^3 + a))/c^6, 1/215040*(105*(3*b^7 - 28*a*b^5*c + 80*a^2*b^3 *c^2 - 64*a^3*b*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*(5120*c^7*x^18 + 6400*b*c^6*x ^15 + 128*(b^2*c^5 + 64*a*c^6)*x^12 - 16*(9*b^3*c^4 - 44*a*b*c^5)*x^9 + 31 5*b^6*c - 2520*a*b^4*c^2 + 5488*a^2*b^2*c^3 - 2048*a^3*c^4 + 8*(21*b^4*c^3 - 124*a*b^2*c^4 + 128*a^2*c^5)*x^6 - 2*(105*b^5*c^2 - 728*a*b^3*c^3 + 116 8*a^2*b*c^4)*x^3)*sqrt(c*x^6 + b*x^3 + a))/c^6]
\[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^{11} \left (a + b x^{3} + c x^{6}\right )^{\frac {3}{2}}\, dx \]
Exception generated. \[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int { {\left (c x^{6} + b x^{3} + a\right )}^{\frac {3}{2}} x^{11} \,d x } \]
Timed out. \[ \int x^{11} \left (a+b x^3+c x^6\right )^{3/2} \, dx=\int x^{11}\,{\left (c\,x^6+b\,x^3+a\right )}^{3/2} \,d x \]