Integrand size = 19, antiderivative size = 134 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {128 b^4 \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac {64 b^3 \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac {16 b^2 \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac {8 b x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]
128/15015*b^4*(c*x^4+b*x^2)^(5/2)/c^5/x^5-64/3003*b^3*(c*x^4+b*x^2)^(5/2)/ c^4/x^3+16/429*b^2*(c*x^4+b*x^2)^(5/2)/c^3/x-8/143*b*x*(c*x^4+b*x^2)^(5/2) /c^2+1/13*x^3*(c*x^4+b*x^2)^(5/2)/c
Time = 0.05 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.51 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2} \left (128 b^4-320 b^3 c x^2+560 b^2 c^2 x^4-840 b c^3 x^6+1155 c^4 x^8\right )}{15015 c^5 x^5} \]
((x^2*(b + c*x^2))^(5/2)*(128*b^4 - 320*b^3*c*x^2 + 560*b^2*c^2*x^4 - 840* b*c^3*x^6 + 1155*c^4*x^8))/(15015*c^5*x^5)
Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {1421, 1421, 1421, 1399, 1420}
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.
\(\displaystyle \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {8 b \int x^4 \left (c x^4+b x^2\right )^{3/2}dx}{13 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \int x^2 \left (c x^4+b x^2\right )^{3/2}dx}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1421 |
\(\displaystyle \frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \int \left (c x^4+b x^2\right )^{3/2}dx}{9 c}\right )}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1399 |
\(\displaystyle \frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^2}dx}{7 c}\right )}{9 c}\right )}{11 c}\right )}{13 c}\) |
\(\Big \downarrow \) 1420 |
\(\displaystyle \frac {x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {8 b \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}\right )}{9 c}\right )}{11 c}\right )}{13 c}\) |
(x^3*(b*x^2 + c*x^4)^(5/2))/(13*c) - (8*b*((x*(b*x^2 + c*x^4)^(5/2))/(11*c ) - (6*b*((b*x^2 + c*x^4)^(5/2)/(9*c*x) - (4*b*((-2*b*(b*x^2 + c*x^4)^(5/2 ))/(35*c^2*x^5) + (b*x^2 + c*x^4)^(5/2)/(7*c*x^3)))/(9*c)))/(11*c)))/(13*c )
3.3.49.3.1 Defintions of rubi rules used
Int[((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(b*x^2 + c*x^4)^( p + 1)/(c*(4*p + 1)*x^3), x] - Simp[b*((2*p - 1)/(c*(4*p + 1))) Int[(b*x^ 2 + c*x^4)^p/x^2, x], x] /; FreeQ[{b, c, p}, x] && IGtQ[p - 1/2, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && EqQ[m + 2*p - 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b *d^2*((m + 2*p - 1)/(c*(m + 4*p + 1))) Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && IGtQ[Simplify[(m + 2*p - 1)/2], 0] && NeQ[m + 4*p + 1, 0]
Time = 0.45 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.54
method | result | size |
gosper | \(\frac {\left (c \,x^{2}+b \right ) \left (1155 c^{4} x^{8}-840 c^{3} x^{6} b +560 b^{2} c^{2} x^{4}-320 x^{2} c \,b^{3}+128 b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 c^{5} x^{3}}\) | \(72\) |
default | \(\frac {\left (c \,x^{2}+b \right ) \left (1155 c^{4} x^{8}-840 c^{3} x^{6} b +560 b^{2} c^{2} x^{4}-320 x^{2} c \,b^{3}+128 b^{4}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{15015 c^{5} x^{3}}\) | \(72\) |
trager | \(\frac {\left (1155 c^{6} x^{12}+1470 b \,c^{5} x^{10}+35 b^{2} c^{4} x^{8}-40 b^{3} c^{3} x^{6}+48 b^{4} c^{2} x^{4}-64 b^{5} c \,x^{2}+128 b^{6}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{15015 c^{5} x}\) | \(87\) |
risch | \(\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (1155 c^{6} x^{12}+1470 b \,c^{5} x^{10}+35 b^{2} c^{4} x^{8}-40 b^{3} c^{3} x^{6}+48 b^{4} c^{2} x^{4}-64 b^{5} c \,x^{2}+128 b^{6}\right )}{15015 x \,c^{5}}\) | \(87\) |
1/15015*(c*x^2+b)*(1155*c^4*x^8-840*b*c^3*x^6+560*b^2*c^2*x^4-320*b^3*c*x^ 2+128*b^4)*(c*x^4+b*x^2)^(3/2)/c^5/x^3
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (1155 \, c^{6} x^{12} + 1470 \, b c^{5} x^{10} + 35 \, b^{2} c^{4} x^{8} - 40 \, b^{3} c^{3} x^{6} + 48 \, b^{4} c^{2} x^{4} - 64 \, b^{5} c x^{2} + 128 \, b^{6}\right )} \sqrt {c x^{4} + b x^{2}}}{15015 \, c^{5} x} \]
1/15015*(1155*c^6*x^12 + 1470*b*c^5*x^10 + 35*b^2*c^4*x^8 - 40*b^3*c^3*x^6 + 48*b^4*c^2*x^4 - 64*b^5*c*x^2 + 128*b^6)*sqrt(c*x^4 + b*x^2)/(c^5*x)
\[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{6} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.59 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (1155 \, c^{6} x^{12} + 1470 \, b c^{5} x^{10} + 35 \, b^{2} c^{4} x^{8} - 40 \, b^{3} c^{3} x^{6} + 48 \, b^{4} c^{2} x^{4} - 64 \, b^{5} c x^{2} + 128 \, b^{6}\right )} \sqrt {c x^{2} + b}}{15015 \, c^{5}} \]
1/15015*(1155*c^6*x^12 + 1470*b*c^5*x^10 + 35*b^2*c^4*x^8 - 40*b^3*c^3*x^6 + 48*b^4*c^2*x^4 - 64*b^5*c*x^2 + 128*b^6)*sqrt(c*x^2 + b)/c^5
Time = 0.29 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.69 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {128 \, b^{\frac {13}{2}} \mathrm {sgn}\left (x\right )}{15015 \, c^{5}} + \frac {1155 \, {\left (c x^{2} + b\right )}^{\frac {13}{2}} \mathrm {sgn}\left (x\right ) - 5460 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} b \mathrm {sgn}\left (x\right ) + 10010 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} b^{2} \mathrm {sgn}\left (x\right ) - 8580 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} b^{3} \mathrm {sgn}\left (x\right ) + 3003 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b^{4} \mathrm {sgn}\left (x\right )}{15015 \, c^{5}} \]
-128/15015*b^(13/2)*sgn(x)/c^5 + 1/15015*(1155*(c*x^2 + b)^(13/2)*sgn(x) - 5460*(c*x^2 + b)^(11/2)*b*sgn(x) + 10010*(c*x^2 + b)^(9/2)*b^2*sgn(x) - 8 580*(c*x^2 + b)^(7/2)*b^3*sgn(x) + 3003*(c*x^2 + b)^(5/2)*b^4*sgn(x))/c^5
Time = 13.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.54 \[ \int x^6 \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (c\,x^2+b\right )}^2\,\sqrt {c\,x^4+b\,x^2}\,\left (128\,b^4-320\,b^3\,c\,x^2+560\,b^2\,c^2\,x^4-840\,b\,c^3\,x^6+1155\,c^4\,x^8\right )}{15015\,c^5\,x} \]