Integrand size = 21, antiderivative size = 196 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\frac {a \left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) x \sqrt {a+b x^2}}{128 b^2}+\frac {\left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{192 b^2}+\frac {d (10 b c-3 a d) x \left (a+b x^2\right )^{5/2}}{48 b^2}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}+\frac {a^2 \left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{128 b^{5/2}} \]
1/192*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a)^(3/2)/b^2+1/48*d*(-3*a *d+10*b*c)*x*(b*x^2+a)^(5/2)/b^2+1/8*d*x*(b*x^2+a)^(5/2)*(d*x^2+c)/b+1/128 *a^2*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/ b^(5/2)+1/128*a*(3*a^2*d^2-16*a*b*c*d+48*b^2*c^2)*x*(b*x^2+a)^(1/2)/b^2
Time = 0.23 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (-9 a^3 d^2+6 a^2 b d \left (8 c+d x^2\right )+16 b^3 x^2 \left (6 c^2+8 c d x^2+3 d^2 x^4\right )+8 a b^2 \left (30 c^2+28 c d x^2+9 d^2 x^4\right )\right )-3 a^2 \left (48 b^2 c^2-16 a b c d+3 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{384 b^{5/2}} \]
(Sqrt[b]*x*Sqrt[a + b*x^2]*(-9*a^3*d^2 + 6*a^2*b*d*(8*c + d*x^2) + 16*b^3* x^2*(6*c^2 + 8*c*d*x^2 + 3*d^2*x^4) + 8*a*b^2*(30*c^2 + 28*c*d*x^2 + 9*d^2 *x^4)) - 3*a^2*(48*b^2*c^2 - 16*a*b*c*d + 3*a^2*d^2)*Log[-(Sqrt[b]*x) + Sq rt[a + b*x^2]])/(384*b^(5/2))
Time = 0.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {318, 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.
| \(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\int \left (b x^2+a\right )^{3/2} \left (d (10 b c-3 a d) x^2+c (8 b c-a d)\right )dx}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \int \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) \left (3 a^2 d^2-16 a b c d+48 b^2 c^2\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2} (10 b c-3 a d)}{6 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{8 b}\) |
(d*x*(a + b*x^2)^(5/2)*(c + d*x^2))/(8*b) + ((d*(10*b*c - 3*a*d)*x*(a + b* x^2)^(5/2))/(6*b) + ((48*b^2*c^2 - 16*a*b*c*d + 3*a^2*d^2)*((x*(a + b*x^2) ^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/(6*b))/(8*b)
3.1.54.3.1 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 = 2.40 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.72
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 a^{2} \left (a^{2} d^{2}-\frac {16}{3} a b c d +16 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{128}-\frac {3 x \sqrt {b \,x^{2}+a}\, \left (-\frac {80 \left (\frac {3}{10} d^{2} x^{4}+\frac {14}{15} c d \,x^{2}+c^{2}\right ) a \,b^{\frac {5}{2}}}{3}-\frac {32 x^{2} \left (\frac {1}{2} d^{2} x^{4}+\frac {4}{3} c d \,x^{2}+c^{2}\right ) b^{\frac {7}{2}}}{3}+\left (\frac {2 \left (-d \,x^{2}-8 c \right ) b^{\frac {3}{2}}}{3}+a d \sqrt {b}\right ) d \,a^{2}\right )}{128}}{b^{\frac {5}{2}}}\) | \(142\) |
| risch | \(-\frac {x \left (-48 b^{3} d^{2} x^{6}-72 a \,b^{2} d^{2} x^{4}-128 b^{3} d c \,x^{4}-6 x^{2} a^{2} b \,d^{2}-224 x^{2} a \,b^{2} c d -96 x^{2} b^{3} c^{2}+9 a^{3} d^{2}-48 a^{2} b c d -240 a \,b^{2} c^{2}\right ) \sqrt {b \,x^{2}+a}}{384 b^{2}}+\frac {a^{2} \left (3 a^{2} d^{2}-16 a b c d +48 b^{2} c^{2}\right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{128 b^{\frac {5}{2}}}\) | \(157\) |
| default | \(c^{2} \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+2 c d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(235\) |
3/128/b^(5/2)*(a^2*(a^2*d^2-16/3*a*b*c*d+16*b^2*c^2)*arctanh((b*x^2+a)^(1/ 2)/x/b^(1/2))-x*(b*x^2+a)^(1/2)*(-80/3*(3/10*d^2*x^4+14/15*c*d*x^2+c^2)*a* b^(5/2)-32/3*x^2*(1/2*d^2*x^4+4/3*c*d*x^2+c^2)*b^(7/2)+(2/3*(-d*x^2-8*c)*b ^(3/2)+a*d*b^(1/2))*d*a^2))
Time = 0.30 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.76 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\left [\frac {3 \, {\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (48 \, b^{4} d^{2} x^{7} + 8 \, {\left (16 \, b^{4} c d + 9 \, a b^{3} d^{2}\right )} x^{5} + 2 \, {\left (48 \, b^{4} c^{2} + 112 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (80 \, a b^{3} c^{2} + 16 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{768 \, b^{3}}, -\frac {3 \, {\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (48 \, b^{4} d^{2} x^{7} + 8 \, {\left (16 \, b^{4} c d + 9 \, a b^{3} d^{2}\right )} x^{5} + 2 \, {\left (48 \, b^{4} c^{2} + 112 \, a b^{3} c d + 3 \, a^{2} b^{2} d^{2}\right )} x^{3} + 3 \, {\left (80 \, a b^{3} c^{2} + 16 \, a^{2} b^{2} c d - 3 \, a^{3} b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{384 \, b^{3}}\right ] \]
[1/768*(3*(48*a^2*b^2*c^2 - 16*a^3*b*c*d + 3*a^4*d^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(48*b^4*d^2*x^7 + 8*(16*b^4*c*d + 9*a*b^3*d^2)*x^5 + 2*(48*b^4*c^2 + 112*a*b^3*c*d + 3*a^2*b^2*d^2)*x^3 + 3* (80*a*b^3*c^2 + 16*a^2*b^2*c*d - 3*a^3*b*d^2)*x)*sqrt(b*x^2 + a))/b^3, -1/ 384*(3*(48*a^2*b^2*c^2 - 16*a^3*b*c*d + 3*a^4*d^2)*sqrt(-b)*arctan(sqrt(-b )*x/sqrt(b*x^2 + a)) - (48*b^4*d^2*x^7 + 8*(16*b^4*c*d + 9*a*b^3*d^2)*x^5 + 2*(48*b^4*c^2 + 112*a*b^3*c*d + 3*a^2*b^2*d^2)*x^3 + 3*(80*a*b^3*c^2 + 1 6*a^2*b^2*c*d - 3*a^3*b*d^2)*x)*sqrt(b*x^2 + a))/b^3]
Time = 0.44 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.68 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {b d^{2} x^{7}}{8} + \frac {x^{5} \cdot \left (\frac {9 a b d^{2}}{8} + 2 b^{2} c d\right )}{6 b} + \frac {x^{3} \left (a^{2} d^{2} + 4 a b c d - \frac {5 a \left (\frac {9 a b d^{2}}{8} + 2 b^{2} c d\right )}{6 b} + b^{2} c^{2}\right )}{4 b} + \frac {x \left (2 a^{2} c d + 2 a b c^{2} - \frac {3 a \left (a^{2} d^{2} + 4 a b c d - \frac {5 a \left (\frac {9 a b d^{2}}{8} + 2 b^{2} c d\right )}{6 b} + b^{2} c^{2}\right )}{4 b}\right )}{2 b}\right ) + \left (a^{2} c^{2} - \frac {a \left (2 a^{2} c d + 2 a b c^{2} - \frac {3 a \left (a^{2} d^{2} + 4 a b c d - \frac {5 a \left (\frac {9 a b d^{2}}{8} + 2 b^{2} c d\right )}{6 b} + b^{2} c^{2}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (c^{2} x + \frac {2 c d x^{3}}{3} + \frac {d^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(b*d**2*x**7/8 + x**5*(9*a*b*d**2/8 + 2*b**2*c *d)/(6*b) + x**3*(a**2*d**2 + 4*a*b*c*d - 5*a*(9*a*b*d**2/8 + 2*b**2*c*d)/ (6*b) + b**2*c**2)/(4*b) + x*(2*a**2*c*d + 2*a*b*c**2 - 3*a*(a**2*d**2 + 4 *a*b*c*d - 5*a*(9*a*b*d**2/8 + 2*b**2*c*d)/(6*b) + b**2*c**2)/(4*b))/(2*b) ) + (a**2*c**2 - a*(2*a**2*c*d + 2*a*b*c**2 - 3*a*(a**2*d**2 + 4*a*b*c*d - 5*a*(9*a*b*d**2/8 + 2*b**2*c*d)/(6*b) + b**2*c**2)/(4*b))/(2*b))*Piecewis e((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/s qrt(b*x**2), True)), Ne(b, 0)), (a**(3/2)*(c**2*x + 2*c*d*x**3/3 + d**2*x* *5/5), True))
Time = 0.20 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.16 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d^{2} x^{3}}{8 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{2} x + \frac {3}{8} \, \sqrt {b x^{2} + a} a c^{2} x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} c d x}{3 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a c d x}{12 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} c d x}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a d^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} a^{3} d^{2} x}{128 \, b^{2}} + \frac {3 \, a^{2} c^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {a^{3} c d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} + \frac {3 \, a^{4} d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} \]
1/8*(b*x^2 + a)^(5/2)*d^2*x^3/b + 1/4*(b*x^2 + a)^(3/2)*c^2*x + 3/8*sqrt(b *x^2 + a)*a*c^2*x + 1/3*(b*x^2 + a)^(5/2)*c*d*x/b - 1/12*(b*x^2 + a)^(3/2) *a*c*d*x/b - 1/8*sqrt(b*x^2 + a)*a^2*c*d*x/b - 1/16*(b*x^2 + a)^(5/2)*a*d^ 2*x/b^2 + 1/64*(b*x^2 + a)^(3/2)*a^2*d^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*a^3 *d^2*x/b^2 + 3/8*a^2*c^2*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/8*a^3*c*d*arcs inh(b*x/sqrt(a*b))/b^(3/2) + 3/128*a^4*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.31 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b d^{2} x^{2} + \frac {16 \, b^{7} c d + 9 \, a b^{6} d^{2}}{b^{6}}\right )} x^{2} + \frac {48 \, b^{7} c^{2} + 112 \, a b^{6} c d + 3 \, a^{2} b^{5} d^{2}}{b^{6}}\right )} x^{2} + \frac {3 \, {\left (80 \, a b^{6} c^{2} + 16 \, a^{2} b^{5} c d - 3 \, a^{3} b^{4} d^{2}\right )}}{b^{6}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (48 \, a^{2} b^{2} c^{2} - 16 \, a^{3} b c d + 3 \, a^{4} d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{128 \, b^{\frac {5}{2}}} \]
1/384*(2*(4*(6*b*d^2*x^2 + (16*b^7*c*d + 9*a*b^6*d^2)/b^6)*x^2 + (48*b^7*c ^2 + 112*a*b^6*c*d + 3*a^2*b^5*d^2)/b^6)*x^2 + 3*(80*a*b^6*c^2 + 16*a^2*b^ 5*c*d - 3*a^3*b^4*d^2)/b^6)*sqrt(b*x^2 + a)*x - 1/128*(48*a^2*b^2*c^2 - 16 *a^3*b*c*d + 3*a^4*d^2)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^2 \, dx=\int {\left (b\,x^2+a\right )}^{3/2}\,{\left (d\,x^2+c\right )}^2 \,d x \]