Integrand size = 24, antiderivative size = 191 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {c \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) x \sqrt {c+d x^2}}{128 d^3}+\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) x^3 \sqrt {c+d x^2}}{64 d^2}-\frac {b (5 b c-16 a d) x^3 \left (c+d x^2\right )^{3/2}}{48 d^2}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}-\frac {c^2 \left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{7/2}} \]
-1/48*b*(-16*a*d+5*b*c)*x^3*(d*x^2+c)^(3/2)/d^2+1/8*b^2*x^5*(d*x^2+c)^(3/2 )/d-1/128*c^2*(16*a^2*d^2+b*c*(-16*a*d+5*b*c))*arctanh(x*d^(1/2)/(d*x^2+c) ^(1/2))/d^(7/2)+1/128*c*(16*a^2*d^2+b*c*(-16*a*d+5*b*c))*x*(d*x^2+c)^(1/2) /d^3+1/64*(16*a^2*d^2+b*c*(-16*a*d+5*b*c))*x^3*(d*x^2+c)^(1/2)/d^2
Time = 0.59 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.86 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (c+2 d x^2\right )+16 a b d \left (-3 c^2+2 c d x^2+8 d^2 x^4\right )+b^2 \left (15 c^3-10 c^2 d x^2+8 c d^2 x^4+48 d^3 x^6\right )\right )+6 c^2 \left (5 b^2 c^2-16 a b c d+16 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{384 d^{7/2}} \]
(Sqrt[d]*x*Sqrt[c + d*x^2]*(48*a^2*d^2*(c + 2*d*x^2) + 16*a*b*d*(-3*c^2 + 2*c*d*x^2 + 8*d^2*x^4) + b^2*(15*c^3 - 10*c^2*d*x^2 + 8*c*d^2*x^4 + 48*d^3 *x^6)) + 6*c^2*(5*b^2*c^2 - 16*a*b*c*d + 16*a^2*d^2)*ArcTanh[(Sqrt[d]*x)/( Sqrt[c] - Sqrt[c + d*x^2])])/(384*d^(7/2))
Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {367, 363, 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.
| \(\displaystyle \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx\) |
| \(\Big \downarrow \) 367 |
| \(\displaystyle \frac {\int x^2 \sqrt {d x^2+c} \left (8 a^2 d-b (5 b c-16 a d) x^2\right )dx}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \int x^2 \sqrt {d x^2+c}dx}{2 d}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
| \(\Big \downarrow \) 248 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \left (\frac {1}{4} c \int \frac {x^2}{\sqrt {d x^2+c}}dx+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )}{2 d}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )}{2 d}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )}{2 d}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (16 a^2 d^2+b c (5 b c-16 a d)\right ) \left (\frac {1}{4} c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )+\frac {1}{4} x^3 \sqrt {c+d x^2}\right )}{2 d}-\frac {b x^3 \left (c+d x^2\right )^{3/2} (5 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^5 \left (c+d x^2\right )^{3/2}}{8 d}\) |
(b^2*x^5*(c + d*x^2)^(3/2))/(8*d) + (-1/6*(b*(5*b*c - 16*a*d)*x^3*(c + d*x ^2)^(3/2))/d + ((16*a^2*d^2 + b*c*(5*b*c - 16*a*d))*((x^3*Sqrt[c + d*x^2]) /4 + (c*((x*Sqrt[c + d*x^2])/(2*d) - (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2 ]])/(2*d^(3/2))))/4))/(2*d))/(8*d)
3.7.5.3.1 Defintions of rubi rules used
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 = 3.01 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(\frac {\left (-a^{2} c^{2} d^{2}+a b \,c^{3} d -\frac {5}{16} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+x \sqrt {d \,x^{2}+c}\, \left (c \left (\frac {1}{6} b^{2} x^{4}+\frac {2}{3} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\left (b^{2} x^{6}+\frac {8}{3} a b \,x^{4}+2 a^{2} x^{2}\right ) d^{\frac {7}{2}}-\left (\left (\frac {5 b \,x^{2}}{24}+a \right ) d^{\frac {3}{2}}-\frac {5 b \sqrt {d}\, c}{16}\right ) b \,c^{2}\right )}{8 d^{\frac {7}{2}}}\) | \(141\) |
| risch | \(\frac {x \left (48 b^{2} d^{3} x^{6}+128 a b \,d^{3} x^{4}+8 b^{2} c \,d^{2} x^{4}+96 a^{2} d^{3} x^{2}+32 a b c \,d^{2} x^{2}-10 b^{2} c^{2} d \,x^{2}+48 c \,a^{2} d^{2}-48 a b \,c^{2} d +15 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{384 d^{3}}-\frac {c^{2} \left (16 a^{2} d^{2}-16 a b c d +5 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {7}{2}}}\) | \(157\) |
| default | \(b^{2} \left (\frac {x^{5} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{8 d}-\frac {5 c \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )}{2 d}\right )}{8 d}\right )+a^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )+2 a b \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4 d}-\frac {c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4 d}\right )}{2 d}\right )\) | \(257\) |
1/8/d^(7/2)*((-a^2*c^2*d^2+a*b*c^3*d-5/16*b^2*c^4)*arctanh((d*x^2+c)^(1/2) /x/d^(1/2))+x*(d*x^2+c)^(1/2)*(c*(1/6*b^2*x^4+2/3*a*b*x^2+a^2)*d^(5/2)+(b^ 2*x^6+8/3*a*b*x^4+2*a^2*x^2)*d^(7/2)-((5/24*b*x^2+a)*d^(3/2)-5/16*b*d^(1/2 )*c)*b*c^2))
Time = 0.30 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.79 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\left [\frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{4}}, \frac {3 \, {\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} - 2 \, {\left (5 \, b^{2} c^{2} d^{2} - 16 \, a b c d^{3} - 48 \, a^{2} d^{4}\right )} x^{3} + 3 \, {\left (5 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} + 16 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, d^{4}}\right ] \]
[1/768*(3*(5*b^2*c^4 - 16*a*b*c^3*d + 16*a^2*c^2*d^2)*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(48*b^2*d^4*x^7 + 8*(b^2*c*d^3 + 1 6*a*b*d^4)*x^5 - 2*(5*b^2*c^2*d^2 - 16*a*b*c*d^3 - 48*a^2*d^4)*x^3 + 3*(5* b^2*c^3*d - 16*a*b*c^2*d^2 + 16*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^4, 1/384* (3*(5*b^2*c^4 - 16*a*b*c^3*d + 16*a^2*c^2*d^2)*sqrt(-d)*arctan(sqrt(-d)*x/ sqrt(d*x^2 + c)) + (48*b^2*d^4*x^7 + 8*(b^2*c*d^3 + 16*a*b*d^4)*x^5 - 2*(5 *b^2*c^2*d^2 - 16*a*b*c*d^3 - 48*a^2*d^4)*x^3 + 3*(5*b^2*c^3*d - 16*a*b*c^ 2*d^2 + 16*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^4]
Time = 0.38 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.31 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\begin {cases} - \frac {c \left (a^{2} c - \frac {3 c \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{2 d} + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{7}}{8} + \frac {x^{5} \cdot \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d} + \frac {x^{3} \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d} + \frac {x \left (a^{2} c - \frac {3 c \left (a^{2} d + 2 a b c - \frac {5 c \left (2 a b d + \frac {b^{2} c}{8}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) & \text {for}\: d \neq 0 \\\sqrt {c} \left (\frac {a^{2} x^{3}}{3} + \frac {2 a b x^{5}}{5} + \frac {b^{2} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((-c*(a**2*c - 3*c*(a**2*d + 2*a*b*c - 5*c*(2*a*b*d + b**2*c/8)/( 6*d))/(4*d))*Piecewise((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), N e(c, 0)), (x*log(x)/sqrt(d*x**2), True))/(2*d) + sqrt(c + d*x**2)*(b**2*x* *7/8 + x**5*(2*a*b*d + b**2*c/8)/(6*d) + x**3*(a**2*d + 2*a*b*c - 5*c*(2*a *b*d + b**2*c/8)/(6*d))/(4*d) + x*(a**2*c - 3*c*(a**2*d + 2*a*b*c - 5*c*(2 *a*b*d + b**2*c/8)/(6*d))/(4*d))/(2*d)), Ne(d, 0)), (sqrt(c)*(a**2*x**3/3 + 2*a*b*x**5/5 + b**2*x**7/7), True))
Time = 0.20 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.24 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} x^{5}}{8 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c x^{3}}{48 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b x^{3}}{3 \, d} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{64 \, d^{3}} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{3}} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x}{4 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b c^{2} x}{8 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x}{4 \, d} - \frac {\sqrt {d x^{2} + c} a^{2} c x}{8 \, d} - \frac {5 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {7}{2}}} + \frac {a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} - \frac {a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} \]
1/8*(d*x^2 + c)^(3/2)*b^2*x^5/d - 5/48*(d*x^2 + c)^(3/2)*b^2*c*x^3/d^2 + 1 /3*(d*x^2 + c)^(3/2)*a*b*x^3/d + 5/64*(d*x^2 + c)^(3/2)*b^2*c^2*x/d^3 - 5/ 128*sqrt(d*x^2 + c)*b^2*c^3*x/d^3 - 1/4*(d*x^2 + c)^(3/2)*a*b*c*x/d^2 + 1/ 8*sqrt(d*x^2 + c)*a*b*c^2*x/d^2 + 1/4*(d*x^2 + c)^(3/2)*a^2*x/d - 1/8*sqrt (d*x^2 + c)*a^2*c*x/d - 5/128*b^2*c^4*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 1/8 *a*b*c^3*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/8*a^2*c^2*arcsinh(d*x/sqrt(c*d ))/d^(3/2)
Time = 0.29 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.91 \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} x^{2} + \frac {b^{2} c d^{5} + 16 \, a b d^{6}}{d^{6}}\right )} x^{2} - \frac {5 \, b^{2} c^{2} d^{4} - 16 \, a b c d^{5} - 48 \, a^{2} d^{6}}{d^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{3} - 16 \, a b c^{2} d^{4} + 16 \, a^{2} c d^{5}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (5 \, b^{2} c^{4} - 16 \, a b c^{3} d + 16 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, d^{\frac {7}{2}}} \]
1/384*(2*(4*(6*b^2*x^2 + (b^2*c*d^5 + 16*a*b*d^6)/d^6)*x^2 - (5*b^2*c^2*d^ 4 - 16*a*b*c*d^5 - 48*a^2*d^6)/d^6)*x^2 + 3*(5*b^2*c^3*d^3 - 16*a*b*c^2*d^ 4 + 16*a^2*c*d^5)/d^6)*sqrt(d*x^2 + c)*x + 1/128*(5*b^2*c^4 - 16*a*b*c^3*d + 16*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(7/2)
Timed out. \[ \int x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2} \, dx=\int x^2\,{\left (b\,x^2+a\right )}^2\,\sqrt {d\,x^2+c} \,d x \]