Integrand size = 24, antiderivative size = 235 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {c^2 \left (16 a^2 d^2+3 b c (b c-4 a d)\right ) x \sqrt {c+d x^2}}{256 d^3}+\frac {c \left (16 a^2 d^2+3 b c (b c-4 a d)\right ) x^3 \sqrt {c+d x^2}}{128 d^2}+\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) x^3 \left (c+d x^2\right )^{3/2}}{96 d^2}-\frac {b (b c-4 a d) x^3 \left (c+d x^2\right )^{5/2}}{16 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}-\frac {c^3 \left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{256 d^{7/2}} \]
1/96*(16*a^2*d^2+3*b*c*(-4*a*d+b*c))*x^3*(d*x^2+c)^(3/2)/d^2-1/16*b*(-4*a* d+b*c)*x^3*(d*x^2+c)^(5/2)/d^2+1/10*b^2*x^5*(d*x^2+c)^(5/2)/d-1/256*c^3*(1 6*a^2*d^2+3*b*c*(-4*a*d+b*c))*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(7/2)+1 /256*c^2*(16*a^2*d^2+3*b*c*(-4*a*d+b*c))*x*(d*x^2+c)^(1/2)/d^3+1/128*c*(16 *a^2*d^2+3*b*c*(-4*a*d+b*c))*x^3*(d*x^2+c)^(1/2)/d^2
Time = 0.81 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (80 a^2 d^2 \left (3 c^2+14 c d x^2+8 d^2 x^4\right )+60 a b d \left (-3 c^3+2 c^2 d x^2+24 c d^2 x^4+16 d^3 x^6\right )+3 b^2 \left (15 c^4-10 c^3 d x^2+8 c^2 d^2 x^4+176 c d^3 x^6+128 d^4 x^8\right )\right )+30 c^3 \left (3 b^2 c^2-12 a b c d+16 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{3840 d^{7/2}} \]
(Sqrt[d]*x*Sqrt[c + d*x^2]*(80*a^2*d^2*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4) + 60*a*b*d*(-3*c^3 + 2*c^2*d*x^2 + 24*c*d^2*x^4 + 16*d^3*x^6) + 3*b^2*(15*c^ 4 - 10*c^3*d*x^2 + 8*c^2*d^2*x^4 + 176*c*d^3*x^6 + 128*d^4*x^8)) + 30*c^3* (3*b^2*c^2 - 12*a*b*c*d + 16*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/(Sqrt[c] - Sqrt[ c + d*x^2])])/(3840*d^(7/2))
Time = 0.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.80, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {367, 27, 363, 248, 248, 262, 224, 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.
| \(\displaystyle \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 367 |
| \(\displaystyle \frac {\int 5 x^2 \left (d x^2+c\right )^{3/2} \left (2 a^2 d-b (b c-4 a d) x^2\right )dx}{10 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int x^2 \left (d x^2+c\right )^{3/2} \left (2 a^2 d-b (b c-4 a d) x^2\right )dx}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \int x^2 \left (d x^2+c\right )^{3/2}dx}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \left (\frac {1}{2} c \int x^2 \sqrt {d x^2+c}dx+\frac {1}{6} x^3 \left (c+d x^2\right )^{3/2}\right )}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \int \frac {x^2}{\sqrt {d x^2+c}}dx+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )+\frac {1}{6} x^3 \left (c+d x^2\right )^{3/2}\right )}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )+\frac {1}{6} x^3 \left (c+d x^2\right )^{3/2}\right )}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )+\frac {1}{6} x^3 \left (c+d x^2\right )^{3/2}\right )}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+3 b c (b c-4 a d)\right ) \left (\frac {1}{2} c \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )+\frac {1}{6} x^3 \left (c+d x^2\right )^{3/2}\right )}{8 d}-\frac {b x^3 \left (c+d x^2\right )^{5/2} (b c-4 a d)}{8 d}}{2 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{5/2}}{10 d}\) |
(b^2*x^5*(c + d*x^2)^(5/2))/(10*d) + (-1/8*(b*(b*c - 4*a*d)*x^3*(c + d*x^2 )^(5/2))/d + ((16*a^2*d^2 + 3*b*c*(b*c - 4*a*d))*((x^3*(c + d*x^2)^(3/2))/ 6 + (c*((x^3*Sqrt[c + d*x^2])/4 + (c*((x*Sqrt[c + d*x^2])/(2*d) - (c*ArcTa nh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*d^(3/2))))/4))/2))/(8*d))/(2*d)
3.7.15.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[d^2*(e*x)^(m + 3)*((a + b*x^2)^(p + 1)/(b*e^3*(m + 2*p + 5))), x] + Simp[1/(b*(m + 2*p + 5)) Int[(e*x)^m*(a + b*x^2)^p*Simp[b*c^2* (m + 2*p + 5) - d*(a*d*(m + 3) - 2*b*c*(m + 2*p + 5))*x^2, x], x], x] /; Fr eeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + 2*p + 5, 0]
Time = 2.98 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(\frac {\left (-a^{2} c^{3} d^{2}+\frac {3}{4} a b \,c^{4} d -\frac {3}{16} b^{2} c^{5}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \left (c^{2} \left (\frac {1}{10} b^{2} x^{4}+\frac {1}{2} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {14 x^{2} \left (\frac {33}{70} b^{2} x^{4}+\frac {9}{7} a b \,x^{2}+a^{2}\right ) c \,d^{\frac {7}{2}}}{3}+\left (\frac {8}{5} b^{2} x^{8}+4 a b \,x^{6}+\frac {8}{3} a^{2} x^{4}\right ) d^{\frac {9}{2}}-\frac {3 b \left (\left (\frac {b \,x^{2}}{6}+a \right ) d^{\frac {3}{2}}-\frac {b \sqrt {d}\, c}{4}\right ) c^{3}}{4}\right ) \sqrt {d \,x^{2}+c}}{16 d^{\frac {7}{2}}}\) | \(173\) |
| risch | \(\frac {x \left (384 b^{2} x^{8} d^{4}+960 a b \,d^{4} x^{6}+528 b^{2} c \,d^{3} x^{6}+640 a^{2} d^{4} x^{4}+1440 c a b \,x^{4} d^{3}+24 b^{2} c^{2} d^{2} x^{4}+1120 a^{2} c \,d^{3} x^{2}+120 a b \,c^{2} d^{2} x^{2}-30 b^{2} c^{3} d \,x^{2}+240 a^{2} c^{2} d^{2}-180 a b \,c^{3} d +45 b^{2} c^{4}\right ) \sqrt {d \,x^{2}+c}}{3840 d^{3}}-\frac {c^{3} \left (16 a^{2} d^{2}-12 a b c d +3 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{256 d^{\frac {7}{2}}}\) | \(198\) |
| default | \(b^{2} \left (\frac {x^{5} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{10 d}-\frac {c \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{8 d}-\frac {3 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6 d}\right )}{8 d}\right )}{2 d}\right )+a^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6 d}\right )+2 a b \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{8 d}-\frac {3 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6 d}\right )}{8 d}\right )\) | \(305\) |
1/16/d^(7/2)*((-a^2*c^3*d^2+3/4*a*b*c^4*d-3/16*b^2*c^5)*arctanh((d*x^2+c)^ (1/2)/x/d^(1/2))+x*(c^2*(1/10*b^2*x^4+1/2*a*b*x^2+a^2)*d^(5/2)+14/3*x^2*(3 3/70*b^2*x^4+9/7*a*b*x^2+a^2)*c*d^(7/2)+(8/5*b^2*x^8+4*a*b*x^6+8/3*a^2*x^4 )*d^(9/2)-3/4*b*((1/6*b*x^2+a)*d^(3/2)-1/4*b*d^(1/2)*c)*c^3)*(d*x^2+c)^(1/ 2))
Time = 0.33 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.78 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\left [\frac {15 \, {\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (11 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (3 \, b^{2} c^{2} d^{3} + 180 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} - 10 \, {\left (3 \, b^{2} c^{3} d^{2} - 12 \, a b c^{2} d^{3} - 112 \, a^{2} c d^{4}\right )} x^{3} + 15 \, {\left (3 \, b^{2} c^{4} d - 12 \, a b c^{3} d^{2} + 16 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{7680 \, d^{4}}, \frac {15 \, {\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (11 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (3 \, b^{2} c^{2} d^{3} + 180 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} - 10 \, {\left (3 \, b^{2} c^{3} d^{2} - 12 \, a b c^{2} d^{3} - 112 \, a^{2} c d^{4}\right )} x^{3} + 15 \, {\left (3 \, b^{2} c^{4} d - 12 \, a b c^{3} d^{2} + 16 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{3840 \, d^{4}}\right ] \]
[1/7680*(15*(3*b^2*c^5 - 12*a*b*c^4*d + 16*a^2*c^3*d^2)*sqrt(d)*log(-2*d*x ^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(384*b^2*d^5*x^9 + 48*(11*b^2*c* d^4 + 20*a*b*d^5)*x^7 + 8*(3*b^2*c^2*d^3 + 180*a*b*c*d^4 + 80*a^2*d^5)*x^5 - 10*(3*b^2*c^3*d^2 - 12*a*b*c^2*d^3 - 112*a^2*c*d^4)*x^3 + 15*(3*b^2*c^4 *d - 12*a*b*c^3*d^2 + 16*a^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^4, 1/3840*(15* (3*b^2*c^5 - 12*a*b*c^4*d + 16*a^2*c^3*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqr t(d*x^2 + c)) + (384*b^2*d^5*x^9 + 48*(11*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8* (3*b^2*c^2*d^3 + 180*a*b*c*d^4 + 80*a^2*d^5)*x^5 - 10*(3*b^2*c^3*d^2 - 12* a*b*c^2*d^3 - 112*a^2*c*d^4)*x^3 + 15*(3*b^2*c^4*d - 12*a*b*c^3*d^2 + 16*a ^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^4]
Time = 0.45 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.84 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\begin {cases} - \frac {c \left (a^{2} c^{2} - \frac {3 c \left (2 a^{2} c d + 2 a b c^{2} - \frac {5 c \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {7 c \left (2 a b d^{2} + \frac {11 b^{2} c d}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \sqrt {c + d x^{2}} \left (\frac {b^{2} d x^{9}}{10} + \frac {x^{7} \cdot \left (2 a b d^{2} + \frac {11 b^{2} c d}{10}\right )}{8 d} + \frac {x^{5} \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {7 c \left (2 a b d^{2} + \frac {11 b^{2} c d}{10}\right )}{8 d}\right )}{6 d} + \frac {x^{3} \cdot \left (2 a^{2} c d + 2 a b c^{2} - \frac {5 c \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {7 c \left (2 a b d^{2} + \frac {11 b^{2} c d}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d} + \frac {x \left (a^{2} c^{2} - \frac {3 c \left (2 a^{2} c d + 2 a b c^{2} - \frac {5 c \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {7 c \left (2 a b d^{2} + \frac {11 b^{2} c d}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (\frac {a^{2} x^{3}}{3} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-c*(a**2*c**2 - 3*c*(2*a**2*c*d + 2*a*b*c**2 - 5*c*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 7*c*(2*a*b*d**2 + 11*b**2*c*d/10)/(8*d))/(6*d))/( 4*d))*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0) ), (x*log(x)/sqrt(d*x**2), True))/(2*d) + sqrt(c + d*x**2)*(b**2*d*x**9/10 + x**7*(2*a*b*d**2 + 11*b**2*c*d/10)/(8*d) + x**5*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 7*c*(2*a*b*d**2 + 11*b**2*c*d/10)/(8*d))/(6*d) + x**3*(2*a** 2*c*d + 2*a*b*c**2 - 5*c*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 7*c*(2*a*b*d **2 + 11*b**2*c*d/10)/(8*d))/(6*d))/(4*d) + x*(a**2*c**2 - 3*c*(2*a**2*c*d + 2*a*b*c**2 - 5*c*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 7*c*(2*a*b*d**2 + 11*b**2*c*d/10)/(8*d))/(6*d))/(4*d))/(2*d)), Ne(d, 0)), (c**(3/2)*(a**2*x **3/3 + 2*a*b*x**5/5 + b**2*x**7/7), True))
Time = 0.21 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.27 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x^{5}}{10 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c x^{3}}{16 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b x^{3}}{4 \, d} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} x}{32 \, d^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} x}{128 \, d^{3}} - \frac {3 \, \sqrt {d x^{2} + c} b^{2} c^{4} x}{256 \, d^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c x}{8 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} x}{32 \, d^{2}} + \frac {3 \, \sqrt {d x^{2} + c} a b c^{3} x}{64 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} x}{6 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x}{24 \, d} - \frac {\sqrt {d x^{2} + c} a^{2} c^{2} x}{16 \, d} - \frac {3 \, b^{2} c^{5} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{256 \, d^{\frac {7}{2}}} + \frac {3 \, a b c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{64 \, d^{\frac {5}{2}}} - \frac {a^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, d^{\frac {3}{2}}} \]
1/10*(d*x^2 + c)^(5/2)*b^2*x^5/d - 1/16*(d*x^2 + c)^(5/2)*b^2*c*x^3/d^2 + 1/4*(d*x^2 + c)^(5/2)*a*b*x^3/d + 1/32*(d*x^2 + c)^(5/2)*b^2*c^2*x/d^3 - 1 /128*(d*x^2 + c)^(3/2)*b^2*c^3*x/d^3 - 3/256*sqrt(d*x^2 + c)*b^2*c^4*x/d^3 - 1/8*(d*x^2 + c)^(5/2)*a*b*c*x/d^2 + 1/32*(d*x^2 + c)^(3/2)*a*b*c^2*x/d^ 2 + 3/64*sqrt(d*x^2 + c)*a*b*c^3*x/d^2 + 1/6*(d*x^2 + c)^(5/2)*a^2*x/d - 1 /24*(d*x^2 + c)^(3/2)*a^2*c*x/d - 1/16*sqrt(d*x^2 + c)*a^2*c^2*x/d - 3/256 *b^2*c^5*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 3/64*a*b*c^4*arcsinh(d*x/sqrt(c* d))/d^(5/2) - 1/16*a^2*c^3*arcsinh(d*x/sqrt(c*d))/d^(3/2)
Time = 0.30 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.93 \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b^{2} d x^{2} + \frac {11 \, b^{2} c d^{8} + 20 \, a b d^{9}}{d^{8}}\right )} x^{2} + \frac {3 \, b^{2} c^{2} d^{7} + 180 \, a b c d^{8} + 80 \, a^{2} d^{9}}{d^{8}}\right )} x^{2} - \frac {5 \, {\left (3 \, b^{2} c^{3} d^{6} - 12 \, a b c^{2} d^{7} - 112 \, a^{2} c d^{8}\right )}}{d^{8}}\right )} x^{2} + \frac {15 \, {\left (3 \, b^{2} c^{4} d^{5} - 12 \, a b c^{3} d^{6} + 16 \, a^{2} c^{2} d^{7}\right )}}{d^{8}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (3 \, b^{2} c^{5} - 12 \, a b c^{4} d + 16 \, a^{2} c^{3} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{256 \, d^{\frac {7}{2}}} \]
1/3840*(2*(4*(6*(8*b^2*d*x^2 + (11*b^2*c*d^8 + 20*a*b*d^9)/d^8)*x^2 + (3*b ^2*c^2*d^7 + 180*a*b*c*d^8 + 80*a^2*d^9)/d^8)*x^2 - 5*(3*b^2*c^3*d^6 - 12* a*b*c^2*d^7 - 112*a^2*c*d^8)/d^8)*x^2 + 15*(3*b^2*c^4*d^5 - 12*a*b*c^3*d^6 + 16*a^2*c^2*d^7)/d^8)*sqrt(d*x^2 + c)*x + 1/256*(3*b^2*c^5 - 12*a*b*c^4* d + 16*a^2*c^3*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)
Timed out. \[ \int x^2 \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\int x^2\,{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2} \,d x \]