Integrand size = 21, antiderivative size = 196 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {c \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \sqrt {c+d x^2}}{128 d^2}+\frac {\left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{192 d^2}-\frac {b (3 b c-10 a d) x \left (c+d x^2\right )^{5/2}}{48 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}+\frac {c^2 \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{5/2}} \]
1/192*(48*a^2*d^2-16*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(3/2)/d^2-1/48*b*(-10* a*d+3*b*c)*x*(d*x^2+c)^(5/2)/d^2+1/8*b*x*(b*x^2+a)*(d*x^2+c)^(5/2)/d+1/128 *c^2*(48*a^2*d^2-16*a*b*c*d+3*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/ d^(5/2)+1/128*c*(48*a^2*d^2-16*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(1/2)/d^2
Time = 0.21 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (5 c+2 d x^2\right )+16 a b d \left (3 c^2+14 c d x^2+8 d^2 x^4\right )+b^2 \left (-9 c^3+6 c^2 d x^2+72 c d^2 x^4+48 d^3 x^6\right )\right )-3 c^2 \left (3 b^2 c^2-16 a b c d+48 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{384 d^{5/2}} \]
(Sqrt[d]*x*Sqrt[c + d*x^2]*(48*a^2*d^2*(5*c + 2*d*x^2) + 16*a*b*d*(3*c^2 + 14*c*d*x^2 + 8*d^2*x^4) + b^2*(-9*c^3 + 6*c^2*d*x^2 + 72*c*d^2*x^4 + 48*d ^3*x^6)) - 3*c^2*(3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(384*d^(5/2))
Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {318, 25, 299, 211, 211, 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.
| \(\displaystyle \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\int -\left (d x^2+c\right )^{3/2} \left (b (3 b c-10 a d) x^2+a (b c-8 a d)\right )dx}{8 d}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\int \left (d x^2+c\right )^{3/2} \left (b (3 b c-10 a d) x^2+a (b c-8 a d)\right )dx}{8 d}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\frac {b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{6 d}-\frac {\left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \int \left (d x^2+c\right )^{3/2}dx}{6 d}}{8 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\frac {b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{6 d}-\frac {\left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \left (\frac {3}{4} c \int \sqrt {d x^2+c}dx+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{6 d}}{8 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\frac {b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{6 d}-\frac {\left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{6 d}}{8 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\frac {b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{6 d}-\frac {\left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{6 d}}{8 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}}{8 d}-\frac {\frac {b x \left (c+d x^2\right )^{5/2} (3 b c-10 a d)}{6 d}-\frac {\left (48 a^2 d^2-16 a b c d+3 b^2 c^2\right ) \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )}{6 d}}{8 d}\) |
(b*x*(a + b*x^2)*(c + d*x^2)^(5/2))/(8*d) - ((b*(3*b*c - 10*a*d)*x*(c + d* x^2)^(5/2))/(6*d) - ((3*b^2*c^2 - 16*a*b*c*d + 48*a^2*d^2)*((x*(c + d*x^2) ^(3/2))/4 + (3*c*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d])))/4))/(6*d))/(8*d)
3.7.17.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Time = 2.94 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.73
| method | result | size |
| pseudoelliptic | \(\frac {\frac {\left (3 a^{2} c^{2} d^{2}-a b \,c^{3} d +\frac {3}{16} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{8}+\frac {5 x \sqrt {d \,x^{2}+c}\, \left (c \left (\frac {3}{10} b^{2} x^{4}+\frac {14}{15} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {\left (b^{2} x^{6}+\frac {8}{3} a b \,x^{4}+2 a^{2} x^{2}\right ) d^{\frac {7}{2}}}{5}+\frac {b \,c^{2} \left (\left (\frac {b \,x^{2}}{8}+a \right ) d^{\frac {3}{2}}-\frac {3 b \sqrt {d}\, c}{16}\right )}{5}\right )}{8}}{d^{\frac {5}{2}}}\) | \(144\) |
| risch | \(\frac {x \left (48 b^{2} d^{3} x^{6}+128 a b \,d^{3} x^{4}+72 b^{2} c \,d^{2} x^{4}+96 a^{2} d^{3} x^{2}+224 a b c \,d^{2} x^{2}+6 b^{2} c^{2} d \,x^{2}+240 c \,a^{2} d^{2}+48 a b \,c^{2} d -9 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{384 d^{2}}+\frac {c^{2} \left (48 a^{2} d^{2}-16 a b c d +3 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {5}{2}}}\) | \(157\) |
| default | \(a^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )+b^{2} \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{8 d}-\frac {3 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6 d}\right )}{8 d}\right )+2 a b \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6 d}\right )\) | \(235\) |
5/8*(1/5*(3*a^2*c^2*d^2-a*b*c^3*d+3/16*b^2*c^4)*arctanh((d*x^2+c)^(1/2)/x/ d^(1/2))+x*(d*x^2+c)^(1/2)*(c*(3/10*b^2*x^4+14/15*a*b*x^2+a^2)*d^(5/2)+1/5 *(b^2*x^6+8/3*a*b*x^4+2*a^2*x^2)*d^(7/2)+1/5*b*c^2*((1/8*b*x^2+a)*d^(3/2)- 3/16*b*d^(1/2)*c)))/d^(5/2)
Time = 0.30 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.76 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\left [\frac {3 \, {\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (9 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (3 \, b^{2} c^{2} d^{2} + 112 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (3 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} - 80 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{3}}, -\frac {3 \, {\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (48 \, b^{2} d^{4} x^{7} + 8 \, {\left (9 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (3 \, b^{2} c^{2} d^{2} + 112 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (3 \, b^{2} c^{3} d - 16 \, a b c^{2} d^{2} - 80 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, d^{3}}\right ] \]
[1/768*(3*(3*b^2*c^4 - 16*a*b*c^3*d + 48*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*(9*b^2*c*d^3 + 16*a*b*d^4)*x^5 + 2*(3*b^2*c^2*d^2 + 112*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3* (3*b^2*c^3*d - 16*a*b*c^2*d^2 - 80*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^3, -1/ 384*(3*(3*b^2*c^4 - 16*a*b*c^3*d + 48*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*(9*b^2*c*d^3 + 16*a*b*d^4)*x^5 + 2*(3*b^2*c^2*d^2 + 112*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3*(3*b^2*c^3*d - 16 *a*b*c^2*d^2 - 80*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^3]
Time = 0.39 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.68 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {c + d x^{2}} \left (\frac {b^{2} d x^{7}}{8} + \frac {x^{5} \cdot \left (2 a b d^{2} + \frac {9 b^{2} c d}{8}\right )}{6 d} + \frac {x^{3} \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {5 c \left (2 a b d^{2} + \frac {9 b^{2} c d}{8}\right )}{6 d}\right )}{4 d} + \frac {x \left (2 a^{2} c d + 2 a b c^{2} - \frac {3 c \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {5 c \left (2 a b d^{2} + \frac {9 b^{2} c d}{8}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) + \left (a^{2} c^{2} - \frac {c \left (2 a^{2} c d + 2 a b c^{2} - \frac {3 c \left (a^{2} d^{2} + 4 a b c d + b^{2} c^{2} - \frac {5 c \left (2 a b d^{2} + \frac {9 b^{2} c d}{8}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(c + d*x**2)*(b**2*d*x**7/8 + x**5*(2*a*b*d**2 + 9*b**2*c*d /8)/(6*d) + x**3*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 5*c*(2*a*b*d**2 + 9* b**2*c*d/8)/(6*d))/(4*d) + x*(2*a**2*c*d + 2*a*b*c**2 - 3*c*(a**2*d**2 + 4 *a*b*c*d + b**2*c**2 - 5*c*(2*a*b*d**2 + 9*b**2*c*d/8)/(6*d))/(4*d))/(2*d) ) + (a**2*c**2 - c*(2*a**2*c*d + 2*a*b*c**2 - 3*c*(a**2*d**2 + 4*a*b*c*d + b**2*c**2 - 5*c*(2*a*b*d**2 + 9*b**2*c*d/8)/(6*d))/(4*d))/(2*d))*Piecewis e((log(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/s qrt(d*x**2), True)), Ne(d, 0)), (c**(3/2)*(a**2*x + 2*a*b*x**3/3 + b**2*x* *5/5), True))
Time = 0.19 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.16 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} x^{3}}{8 \, d} + \frac {1}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} x + \frac {3}{8} \, \sqrt {d x^{2} + c} a^{2} c x - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c x}{16 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{64 \, d^{2}} + \frac {3 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b x}{3 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x}{12 \, d} - \frac {\sqrt {d x^{2} + c} a b c^{2} x}{8 \, d} + \frac {3 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {5}{2}}} - \frac {a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} + \frac {3 \, a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} \]
1/8*(d*x^2 + c)^(5/2)*b^2*x^3/d + 1/4*(d*x^2 + c)^(3/2)*a^2*x + 3/8*sqrt(d *x^2 + c)*a^2*c*x - 1/16*(d*x^2 + c)^(5/2)*b^2*c*x/d^2 + 1/64*(d*x^2 + c)^ (3/2)*b^2*c^2*x/d^2 + 3/128*sqrt(d*x^2 + c)*b^2*c^3*x/d^2 + 1/3*(d*x^2 + c )^(5/2)*a*b*x/d - 1/12*(d*x^2 + c)^(3/2)*a*b*c*x/d - 1/8*sqrt(d*x^2 + c)*a *b*c^2*x/d + 3/128*b^2*c^4*arcsinh(d*x/sqrt(c*d))/d^(5/2) - 1/8*a*b*c^3*ar csinh(d*x/sqrt(c*d))/d^(3/2) + 3/8*a^2*c^2*arcsinh(d*x/sqrt(c*d))/sqrt(d)
Time = 0.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} d x^{2} + \frac {9 \, b^{2} c d^{6} + 16 \, a b d^{7}}{d^{6}}\right )} x^{2} + \frac {3 \, b^{2} c^{2} d^{5} + 112 \, a b c d^{6} + 48 \, a^{2} d^{7}}{d^{6}}\right )} x^{2} - \frac {3 \, {\left (3 \, b^{2} c^{3} d^{4} - 16 \, a b c^{2} d^{5} - 80 \, a^{2} c d^{6}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{4} - 16 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, d^{\frac {5}{2}}} \]
1/384*(2*(4*(6*b^2*d*x^2 + (9*b^2*c*d^6 + 16*a*b*d^7)/d^6)*x^2 + (3*b^2*c^ 2*d^5 + 112*a*b*c*d^6 + 48*a^2*d^7)/d^6)*x^2 - 3*(3*b^2*c^3*d^4 - 16*a*b*c ^2*d^5 - 80*a^2*c*d^6)/d^6)*sqrt(d*x^2 + c)*x - 1/128*(3*b^2*c^4 - 16*a*b* c^3*d + 48*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)
Timed out. \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{3/2} \, dx=\int {\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{3/2} \,d x \]