Integrand size = 26, antiderivative size = 168 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {16 b^3 (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{15015 c^5 x^5}-\frac {8 b^2 (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{3003 c^4 x^3}+\frac {2 b (8 b B-13 A c) \left (b x^2+c x^4\right )^{5/2}}{429 c^3 x}-\frac {(8 b B-13 A c) x \left (b x^2+c x^4\right )^{5/2}}{143 c^2}+\frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c} \]
16/15015*b^3*(-13*A*c+8*B*b)*(c*x^4+b*x^2)^(5/2)/c^5/x^5-8/3003*b^2*(-13*A *c+8*B*b)*(c*x^4+b*x^2)^(5/2)/c^4/x^3+2/429*b*(-13*A*c+8*B*b)*(c*x^4+b*x^2 )^(5/2)/c^3/x-1/143*(-13*A*c+8*B*b)*x*(c*x^4+b*x^2)^(5/2)/c^2+1/13*B*x^3*( c*x^4+b*x^2)^(5/2)/c
Time = 0.13 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.67 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {x \left (b+c x^2\right )^3 \left (128 b^4 B+105 c^4 x^6 \left (13 A+11 B x^2\right )-70 b c^3 x^4 \left (13 A+12 B x^2\right )+40 b^2 c^2 x^2 \left (13 A+14 B x^2\right )-16 b^3 c \left (13 A+20 B x^2\right )\right )}{15015 c^5 \sqrt {x^2 \left (b+c x^2\right )}} \]
(x*(b + c*x^2)^3*(128*b^4*B + 105*c^4*x^6*(13*A + 11*B*x^2) - 70*b*c^3*x^4 *(13*A + 12*B*x^2) + 40*b^2*c^2*x^2*(13*A + 14*B*x^2) - 16*b^3*c*(13*A + 2 0*B*x^2)))/(15015*c^5*Sqrt[x^2*(b + c*x^2)])
Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1945, 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.
| \(\displaystyle \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1945 |
| \(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {(8 b B-13 A c) \int x^4 \left (c x^4+b x^2\right )^{3/2}dx}{13 c}\) |
| \(\Big \downarrow \) 1421 |
| \(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {(8 b B-13 A c) \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \int x^2 \left (c x^4+b x^2\right )^{3/2}dx}{11 c}\right )}{13 c}\) |
| \(\Big \downarrow \) 1421 |
| \(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {(8 b B-13 A c) \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \int \left (c x^4+b x^2\right )^{3/2}dx}{9 c}\right )}{11 c}\right )}{13 c}\) |
| \(\Big \downarrow \) 1399 |
| \(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {(8 b B-13 A c) \left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^2}dx}{7 c}\right )}{9 c}\right )}{11 c}\right )}{13 c}\) |
| \(\Big \downarrow \) 1420 |
| \(\displaystyle \frac {B x^3 \left (b x^2+c x^4\right )^{5/2}}{13 c}-\frac {\left (\frac {x \left (b x^2+c x^4\right )^{5/2}}{11 c}-\frac {6 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{9 c x}-\frac {4 b \left (\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}\right )}{9 c}\right )}{11 c}\right ) (8 b B-13 A c)}{13 c}\) |
(B*x^3*(b*x^2 + c*x^4)^(5/2))/(13*c) - ((8*b*B - 13*A*c)*((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)
3.2.19.3.1 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]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[d*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(b*(m + n + p*(j + n) + 1))), x] - Simp[(a*d*(m + j* p + 1) - b*c*(m + n + p*(j + n) + 1))/(b*(m + n + p*(j + n) + 1)) Int[(e* x)^m*(a*x^j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, m, n, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && NeQ[m + n + p *(j + n) + 1, 0] && (GtQ[e, 0] || IntegerQ[j])
Time = 2.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.68
| method | result | size |
| gosper | \(-\frac {\left (c \,x^{2}+b \right ) \left (-1155 B \,x^{8} c^{4}-1365 A \,x^{6} c^{4}+840 B \,x^{6} b \,c^{3}+910 A \,x^{4} b \,c^{3}-560 B \,x^{4} b^{2} c^{2}-520 A \,x^{2} b^{2} c^{2}+320 B \,x^{2} b^{3} c +208 A \,b^{3} c -128 B \,b^{4}\right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{15015 c^{5} x^{3}}\) | \(115\) |
| default | \(-\frac {\left (c \,x^{2}+b \right ) \left (-1155 B \,x^{8} c^{4}-1365 A \,x^{6} c^{4}+840 B \,x^{6} b \,c^{3}+910 A \,x^{4} b \,c^{3}-560 B \,x^{4} b^{2} c^{2}-520 A \,x^{2} b^{2} c^{2}+320 B \,x^{2} b^{3} c +208 A \,b^{3} c -128 B \,b^{4}\right ) \left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}}}{15015 c^{5} x^{3}}\) | \(115\) |
| trager | \(-\frac {\left (-1155 B \,c^{6} x^{12}-1365 A \,c^{6} x^{10}-1470 B b \,c^{5} x^{10}-1820 A b \,c^{5} x^{8}-35 B \,b^{2} c^{4} x^{8}-65 A \,b^{2} c^{4} x^{6}+40 B \,b^{3} c^{3} x^{6}+78 A \,b^{3} c^{3} x^{4}-48 B \,b^{4} c^{2} x^{4}-104 A \,b^{4} c^{2} x^{2}+64 B \,b^{5} c \,x^{2}+208 A \,b^{5} c -128 B \,b^{6}\right ) \sqrt {x^{4} c +b \,x^{2}}}{15015 c^{5} x}\) | \(156\) |
| risch | \(-\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (-1155 B \,c^{6} x^{12}-1365 A \,c^{6} x^{10}-1470 B b \,c^{5} x^{10}-1820 A b \,c^{5} x^{8}-35 B \,b^{2} c^{4} x^{8}-65 A \,b^{2} c^{4} x^{6}+40 B \,b^{3} c^{3} x^{6}+78 A \,b^{3} c^{3} x^{4}-48 B \,b^{4} c^{2} x^{4}-104 A \,b^{4} c^{2} x^{2}+64 B \,b^{5} c \,x^{2}+208 A \,b^{5} c -128 B \,b^{6}\right )}{15015 x \,c^{5}}\) | \(156\) |
-1/15015*(c*x^2+b)*(-1155*B*c^4*x^8-1365*A*c^4*x^6+840*B*b*c^3*x^6+910*A*b *c^3*x^4-560*B*b^2*c^2*x^4-520*A*b^2*c^2*x^2+320*B*b^3*c*x^2+208*A*b^3*c-1 28*B*b^4)*(c*x^4+b*x^2)^(3/2)/c^5/x^3
Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.92 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (1155 \, B c^{6} x^{12} + 105 \, {\left (14 \, B b c^{5} + 13 \, A c^{6}\right )} x^{10} + 35 \, {\left (B b^{2} c^{4} + 52 \, A b c^{5}\right )} x^{8} + 128 \, B b^{6} - 208 \, A b^{5} c - 5 \, {\left (8 \, B b^{3} c^{3} - 13 \, A b^{2} c^{4}\right )} x^{6} + 6 \, {\left (8 \, B b^{4} c^{2} - 13 \, A b^{3} c^{3}\right )} x^{4} - 8 \, {\left (8 \, B b^{5} c - 13 \, A b^{4} c^{2}\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15015 \, c^{5} x} \]
1/15015*(1155*B*c^6*x^12 + 105*(14*B*b*c^5 + 13*A*c^6)*x^10 + 35*(B*b^2*c^ 4 + 52*A*b*c^5)*x^8 + 128*B*b^6 - 208*A*b^5*c - 5*(8*B*b^3*c^3 - 13*A*b^2* c^4)*x^6 + 6*(8*B*b^4*c^2 - 13*A*b^3*c^3)*x^4 - 8*(8*B*b^5*c - 13*A*b^4*c^ 2)*x^2)*sqrt(c*x^4 + b*x^2)/(c^5*x)
\[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\int x^{4} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )\, dx \]
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.89 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {{\left (105 \, c^{5} x^{10} + 140 \, b c^{4} x^{8} + 5 \, b^{2} c^{3} x^{6} - 6 \, b^{3} c^{2} x^{4} + 8 \, b^{4} c x^{2} - 16 \, b^{5}\right )} \sqrt {c x^{2} + b} A}{1155 \, c^{4}} + \frac {{\left (1155 \, c^{6} x^{12} + 1470 \, b c^{5} x^{10} + 35 \, b^{2} c^{4} x^{8} - 40 \, b^{3} c^{3} x^{6} + 48 \, b^{4} c^{2} x^{4} - 64 \, b^{5} c x^{2} + 128 \, b^{6}\right )} \sqrt {c x^{2} + b} B}{15015 \, c^{5}} \]
1/1155*(105*c^5*x^10 + 140*b*c^4*x^8 + 5*b^2*c^3*x^6 - 6*b^3*c^2*x^4 + 8*b ^4*c*x^2 - 16*b^5)*sqrt(c*x^2 + b)*A/c^4 + 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)*B/c^5
Time = 0.29 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.04 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=-\frac {16 \, {\left (8 \, B b^{\frac {13}{2}} - 13 \, A b^{\frac {11}{2}} c\right )} \mathrm {sgn}\left (x\right )}{15015 \, c^{5}} + \frac {1155 \, {\left (c x^{2} + b\right )}^{\frac {13}{2}} B \mathrm {sgn}\left (x\right ) - 5460 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} B b \mathrm {sgn}\left (x\right ) + 10010 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b^{2} \mathrm {sgn}\left (x\right ) - 8580 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{3} \mathrm {sgn}\left (x\right ) + 3003 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{4} \mathrm {sgn}\left (x\right ) + 1365 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} A c \mathrm {sgn}\left (x\right ) - 5005 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A b c \mathrm {sgn}\left (x\right ) + 6435 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b^{2} c \mathrm {sgn}\left (x\right ) - 3003 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{3} c \mathrm {sgn}\left (x\right )}{15015 \, c^{5}} \]
-16/15015*(8*B*b^(13/2) - 13*A*b^(11/2)*c)*sgn(x)/c^5 + 1/15015*(1155*(c*x ^2 + b)^(13/2)*B*sgn(x) - 5460*(c*x^2 + b)^(11/2)*B*b*sgn(x) + 10010*(c*x^ 2 + b)^(9/2)*B*b^2*sgn(x) - 8580*(c*x^2 + b)^(7/2)*B*b^3*sgn(x) + 3003*(c* x^2 + b)^(5/2)*B*b^4*sgn(x) + 1365*(c*x^2 + b)^(11/2)*A*c*sgn(x) - 5005*(c *x^2 + b)^(9/2)*A*b*c*sgn(x) + 6435*(c*x^2 + b)^(7/2)*A*b^2*c*sgn(x) - 300 3*(c*x^2 + b)^(5/2)*A*b^3*c*sgn(x))/c^5
Time = 9.25 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.85 \[ \int x^4 \left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2} \, dx=\frac {\sqrt {c\,x^4+b\,x^2}\,\left (\frac {128\,B\,b^6-208\,A\,b^5\,c}{15015\,c^5}+\frac {x^{10}\,\left (1365\,A\,c^6+1470\,B\,b\,c^5\right )}{15015\,c^5}+\frac {B\,c\,x^{12}}{13}+\frac {b^2\,x^6\,\left (13\,A\,c-8\,B\,b\right )}{3003\,c^2}-\frac {2\,b^3\,x^4\,\left (13\,A\,c-8\,B\,b\right )}{5005\,c^3}+\frac {8\,b^4\,x^2\,\left (13\,A\,c-8\,B\,b\right )}{15015\,c^4}+\frac {b\,x^8\,\left (52\,A\,c+B\,b\right )}{429\,c}\right )}{x} \]