Integrand size = 28, antiderivative size = 129 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2 a^6 (d x)^{7/2}}{7 d}+\frac {12 a^5 b (d x)^{11/2}}{11 d^3}+\frac {2 a^4 b^2 (d x)^{15/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{19/2}}{19 d^7}+\frac {30 a^2 b^4 (d x)^{23/2}}{23 d^9}+\frac {4 a b^5 (d x)^{27/2}}{9 d^{11}}+\frac {2 b^6 (d x)^{31/2}}{31 d^{13}} \]
2/7*a^6*(d*x)^(7/2)/d+12/11*a^5*b*(d*x)^(11/2)/d^3+2*a^4*b^2*(d*x)^(15/2)/ d^5+40/19*a^3*b^3*(d*x)^(19/2)/d^7+30/23*a^2*b^4*(d*x)^(23/2)/d^9+4/9*a*b^ 5*(d*x)^(27/2)/d^11+2/31*b^6*(d*x)^(31/2)/d^13
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.60 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2 x (d x)^{5/2} \left (1341153 a^6+5120766 a^5 b x^2+9388071 a^4 b^2 x^4+9882180 a^3 b^3 x^6+6122655 a^2 b^4 x^8+2086238 a b^5 x^{10}+302841 b^6 x^{12}\right )}{9388071} \]
(2*x*(d*x)^(5/2)*(1341153*a^6 + 5120766*a^5*b*x^2 + 9388071*a^4*b^2*x^4 + 9882180*a^3*b^3*x^6 + 6122655*a^2*b^4*x^8 + 2086238*a*b^5*x^10 + 302841*b^ 6*x^12))/9388071
Time = 0.26 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1380, 27, 244, 2009}
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.
\(\displaystyle \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {\int b^6 (d x)^{5/2} \left (b x^2+a\right )^6dx}{b^6}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int (d x)^{5/2} \left (a+b x^2\right )^6dx\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \int \left (a^6 (d x)^{5/2}+\frac {6 a^5 b (d x)^{9/2}}{d^2}+\frac {15 a^4 b^2 (d x)^{13/2}}{d^4}+\frac {20 a^3 b^3 (d x)^{17/2}}{d^6}+\frac {15 a^2 b^4 (d x)^{21/2}}{d^8}+\frac {6 a b^5 (d x)^{25/2}}{d^{10}}+\frac {b^6 (d x)^{29/2}}{d^{12}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2 a^6 (d x)^{7/2}}{7 d}+\frac {12 a^5 b (d x)^{11/2}}{11 d^3}+\frac {2 a^4 b^2 (d x)^{15/2}}{d^5}+\frac {40 a^3 b^3 (d x)^{19/2}}{19 d^7}+\frac {30 a^2 b^4 (d x)^{23/2}}{23 d^9}+\frac {4 a b^5 (d x)^{27/2}}{9 d^{11}}+\frac {2 b^6 (d x)^{31/2}}{31 d^{13}}\) |
(2*a^6*(d*x)^(7/2))/(7*d) + (12*a^5*b*(d*x)^(11/2))/(11*d^3) + (2*a^4*b^2* (d*x)^(15/2))/d^5 + (40*a^3*b^3*(d*x)^(19/2))/(19*d^7) + (30*a^2*b^4*(d*x) ^(23/2))/(23*d^9) + (4*a*b^5*(d*x)^(27/2))/(9*d^11) + (2*b^6*(d*x)^(31/2)) /(31*d^13)
3.7.79.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.81 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(\frac {2 x \left (302841 b^{6} x^{12}+2086238 a \,b^{5} x^{10}+6122655 a^{2} b^{4} x^{8}+9882180 a^{3} b^{3} x^{6}+9388071 a^{4} b^{2} x^{4}+5120766 a^{5} b \,x^{2}+1341153 a^{6}\right ) \left (d x \right )^{\frac {5}{2}}}{9388071}\) | \(74\) |
pseudoelliptic | \(\frac {2 \left (\frac {7}{31} b^{6} x^{12}+\frac {14}{9} a \,b^{5} x^{10}+\frac {105}{23} a^{2} b^{4} x^{8}+\frac {140}{19} a^{3} b^{3} x^{6}+7 a^{4} b^{2} x^{4}+\frac {42}{11} a^{5} b \,x^{2}+a^{6}\right ) \sqrt {d x}\, d^{2} x^{3}}{7}\) | \(77\) |
trager | \(\frac {2 d^{2} x^{3} \left (302841 b^{6} x^{12}+2086238 a \,b^{5} x^{10}+6122655 a^{2} b^{4} x^{8}+9882180 a^{3} b^{3} x^{6}+9388071 a^{4} b^{2} x^{4}+5120766 a^{5} b \,x^{2}+1341153 a^{6}\right ) \sqrt {d x}}{9388071}\) | \(79\) |
risch | \(\frac {2 d^{3} x^{4} \left (302841 b^{6} x^{12}+2086238 a \,b^{5} x^{10}+6122655 a^{2} b^{4} x^{8}+9882180 a^{3} b^{3} x^{6}+9388071 a^{4} b^{2} x^{4}+5120766 a^{5} b \,x^{2}+1341153 a^{6}\right )}{9388071 \sqrt {d x}}\) | \(79\) |
derivativedivides | \(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {31}{2}}}{31}+\frac {4 a \,b^{5} d^{2} \left (d x \right )^{\frac {27}{2}}}{9}+\frac {30 a^{2} d^{4} b^{4} \left (d x \right )^{\frac {23}{2}}}{23}+\frac {40 a^{3} d^{6} b^{3} \left (d x \right )^{\frac {19}{2}}}{19}+2 a^{4} d^{8} b^{2} \left (d x \right )^{\frac {15}{2}}+\frac {12 a^{5} d^{10} b \left (d x \right )^{\frac {11}{2}}}{11}+\frac {2 a^{6} d^{12} \left (d x \right )^{\frac {7}{2}}}{7}}{d^{13}}\) | \(105\) |
default | \(\frac {\frac {2 b^{6} \left (d x \right )^{\frac {31}{2}}}{31}+\frac {4 a \,b^{5} d^{2} \left (d x \right )^{\frac {27}{2}}}{9}+\frac {30 a^{2} d^{4} b^{4} \left (d x \right )^{\frac {23}{2}}}{23}+\frac {40 a^{3} d^{6} b^{3} \left (d x \right )^{\frac {19}{2}}}{19}+2 a^{4} d^{8} b^{2} \left (d x \right )^{\frac {15}{2}}+\frac {12 a^{5} d^{10} b \left (d x \right )^{\frac {11}{2}}}{11}+\frac {2 a^{6} d^{12} \left (d x \right )^{\frac {7}{2}}}{7}}{d^{13}}\) | \(105\) |
2/9388071*x*(302841*b^6*x^12+2086238*a*b^5*x^10+6122655*a^2*b^4*x^8+988218 0*a^3*b^3*x^6+9388071*a^4*b^2*x^4+5120766*a^5*b*x^2+1341153*a^6)*(d*x)^(5/ 2)
Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.74 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2}{9388071} \, {\left (302841 \, b^{6} d^{2} x^{15} + 2086238 \, a b^{5} d^{2} x^{13} + 6122655 \, a^{2} b^{4} d^{2} x^{11} + 9882180 \, a^{3} b^{3} d^{2} x^{9} + 9388071 \, a^{4} b^{2} d^{2} x^{7} + 5120766 \, a^{5} b d^{2} x^{5} + 1341153 \, a^{6} d^{2} x^{3}\right )} \sqrt {d x} \]
2/9388071*(302841*b^6*d^2*x^15 + 2086238*a*b^5*d^2*x^13 + 6122655*a^2*b^4* d^2*x^11 + 9882180*a^3*b^3*d^2*x^9 + 9388071*a^4*b^2*d^2*x^7 + 5120766*a^5 *b*d^2*x^5 + 1341153*a^6*d^2*x^3)*sqrt(d*x)
Time = 0.79 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.99 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2 a^{6} x \left (d x\right )^{\frac {5}{2}}}{7} + \frac {12 a^{5} b x^{3} \left (d x\right )^{\frac {5}{2}}}{11} + 2 a^{4} b^{2} x^{5} \left (d x\right )^{\frac {5}{2}} + \frac {40 a^{3} b^{3} x^{7} \left (d x\right )^{\frac {5}{2}}}{19} + \frac {30 a^{2} b^{4} x^{9} \left (d x\right )^{\frac {5}{2}}}{23} + \frac {4 a b^{5} x^{11} \left (d x\right )^{\frac {5}{2}}}{9} + \frac {2 b^{6} x^{13} \left (d x\right )^{\frac {5}{2}}}{31} \]
2*a**6*x*(d*x)**(5/2)/7 + 12*a**5*b*x**3*(d*x)**(5/2)/11 + 2*a**4*b**2*x** 5*(d*x)**(5/2) + 40*a**3*b**3*x**7*(d*x)**(5/2)/19 + 30*a**2*b**4*x**9*(d* x)**(5/2)/23 + 4*a*b**5*x**11*(d*x)**(5/2)/9 + 2*b**6*x**13*(d*x)**(5/2)/3 1
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.81 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2 \, {\left (302841 \, \left (d x\right )^{\frac {31}{2}} b^{6} + 2086238 \, \left (d x\right )^{\frac {27}{2}} a b^{5} d^{2} + 6122655 \, \left (d x\right )^{\frac {23}{2}} a^{2} b^{4} d^{4} + 9882180 \, \left (d x\right )^{\frac {19}{2}} a^{3} b^{3} d^{6} + 9388071 \, \left (d x\right )^{\frac {15}{2}} a^{4} b^{2} d^{8} + 5120766 \, \left (d x\right )^{\frac {11}{2}} a^{5} b d^{10} + 1341153 \, \left (d x\right )^{\frac {7}{2}} a^{6} d^{12}\right )}}{9388071 \, d^{13}} \]
2/9388071*(302841*(d*x)^(31/2)*b^6 + 2086238*(d*x)^(27/2)*a*b^5*d^2 + 6122 655*(d*x)^(23/2)*a^2*b^4*d^4 + 9882180*(d*x)^(19/2)*a^3*b^3*d^6 + 9388071* (d*x)^(15/2)*a^4*b^2*d^8 + 5120766*(d*x)^(11/2)*a^5*b*d^10 + 1341153*(d*x) ^(7/2)*a^6*d^12)/d^13
Time = 0.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.96 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2}{31} \, \sqrt {d x} b^{6} d^{2} x^{15} + \frac {4}{9} \, \sqrt {d x} a b^{5} d^{2} x^{13} + \frac {30}{23} \, \sqrt {d x} a^{2} b^{4} d^{2} x^{11} + \frac {40}{19} \, \sqrt {d x} a^{3} b^{3} d^{2} x^{9} + 2 \, \sqrt {d x} a^{4} b^{2} d^{2} x^{7} + \frac {12}{11} \, \sqrt {d x} a^{5} b d^{2} x^{5} + \frac {2}{7} \, \sqrt {d x} a^{6} d^{2} x^{3} \]
2/31*sqrt(d*x)*b^6*d^2*x^15 + 4/9*sqrt(d*x)*a*b^5*d^2*x^13 + 30/23*sqrt(d* x)*a^2*b^4*d^2*x^11 + 40/19*sqrt(d*x)*a^3*b^3*d^2*x^9 + 2*sqrt(d*x)*a^4*b^ 2*d^2*x^7 + 12/11*sqrt(d*x)*a^5*b*d^2*x^5 + 2/7*sqrt(d*x)*a^6*d^2*x^3
Time = 0.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.80 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx=\frac {2\,a^6\,{\left (d\,x\right )}^{7/2}}{7\,d}+\frac {2\,b^6\,{\left (d\,x\right )}^{31/2}}{31\,d^{13}}+\frac {2\,a^4\,b^2\,{\left (d\,x\right )}^{15/2}}{d^5}+\frac {40\,a^3\,b^3\,{\left (d\,x\right )}^{19/2}}{19\,d^7}+\frac {30\,a^2\,b^4\,{\left (d\,x\right )}^{23/2}}{23\,d^9}+\frac {12\,a^5\,b\,{\left (d\,x\right )}^{11/2}}{11\,d^3}+\frac {4\,a\,b^5\,{\left (d\,x\right )}^{27/2}}{9\,d^{11}} \]