Integrand size = 24, antiderivative size = 217 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=-\frac {5 c \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \sqrt {c+d x^2}}{128 d}-\frac {5 \left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{3/2}}{192 d}-\frac {\left (b^2 c^2-16 a d (b c+3 a d)\right ) x \left (c+d x^2\right )^{5/2}}{48 c d}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}+\frac {b^2 x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {5 c^2 \left (b^2 c^2-16 a d (b c+3 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{3/2}} \]
-5/192*(b^2*c^2-16*a*d*(3*a*d+b*c))*x*(d*x^2+c)^(3/2)/d-1/48*(b^2*c^2-16*a *d*(3*a*d+b*c))*x*(d*x^2+c)^(5/2)/c/d-a^2*(d*x^2+c)^(7/2)/c/x+1/8*b^2*x*(d *x^2+c)^(7/2)/d-5/128*c^2*(b^2*c^2-16*a*d*(3*a*d+b*c))*arctanh(x*d^(1/2)/( d*x^2+c)^(1/2))/d^(3/2)-5/128*c*(b^2*c^2-16*a*d*(3*a*d+b*c))*x*(d*x^2+c)^( 1/2)/d
Time = 0.30 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {d} \sqrt {c+d x^2} \left (48 a^2 d \left (-8 c^2+9 c d x^2+2 d^2 x^4\right )+16 a b d x^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )+b^2 x^2 \left (15 c^3+118 c^2 d x^2+136 c d^2 x^4+48 d^3 x^6\right )\right )+15 c^2 \left (b^2 c^2-16 a b c d-48 a^2 d^2\right ) x \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{384 d^{3/2} x} \]
(Sqrt[d]*Sqrt[c + d*x^2]*(48*a^2*d*(-8*c^2 + 9*c*d*x^2 + 2*d^2*x^4) + 16*a *b*d*x^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4) + b^2*x^2*(15*c^3 + 118*c^2*d*x ^2 + 136*c*d^2*x^4 + 48*d^3*x^6)) + 15*c^2*(b^2*c^2 - 16*a*b*c*d - 48*a^2* d^2)*x*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]])/(384*d^(3/2)*x)
Time = 0.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.77, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {365, 299, 211, 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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\int \left (b^2 c x^2+2 a (b c+3 a d)\right ) \left (d x^2+c\right )^{5/2}dx}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \int \left (d x^2+c\right )^{5/2}dx}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \left (\frac {5}{6} c \int \left (d x^2+c\right )^{3/2}dx+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \int \sqrt {d x^2+c}dx+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {b^2 c x \left (c+d x^2\right )^{7/2}}{8 d}-\frac {\left (b^2 c^2-16 a d (3 a d+b c)\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{c x}\) |
-((a^2*(c + d*x^2)^(7/2))/(c*x)) + ((b^2*c*x*(c + d*x^2)^(7/2))/(8*d) - (( b^2*c^2 - 16*a*d*(b*c + 3*a*d))*((x*(c + d*x^2)^(5/2))/6 + (5*c*((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))/(8*d))/c
3.7.30.3.1 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[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x _Symbol] :> Simp[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Time = 2.98 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(\frac {\frac {15 x \,c^{2} \left (a^{2} d^{2}+\frac {1}{3} a b c d -\frac {1}{48} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{8}+\frac {9 \sqrt {d \,x^{2}+c}\, \left (-\frac {8 \left (-\frac {59}{192} b^{2} x^{4}-\frac {11}{8} a b \,x^{2}+a^{2}\right ) c^{2} d^{\frac {3}{2}}}{9}+x^{2} \left (c \left (\frac {17}{54} b^{2} x^{4}+\frac {26}{27} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {\left (b^{2} x^{6}+\frac {8}{3} a b \,x^{4}+2 a^{2} x^{2}\right ) d^{\frac {7}{2}}}{9}+\frac {5 b^{2} c^{3} \sqrt {d}}{144}\right )\right )}{8}}{x \,d^{\frac {3}{2}}}\) | \(161\) |
| risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (-48 b^{2} d^{3} x^{8}-128 a b \,d^{3} x^{6}-136 b^{2} c \,d^{2} x^{6}-96 a^{2} d^{3} x^{4}-416 a b c \,d^{2} x^{4}-118 b^{2} c^{2} d \,x^{4}-432 a^{2} c \,d^{2} x^{2}-528 a b \,c^{2} d \,x^{2}-15 b^{2} c^{3} x^{2}+384 a^{2} c^{2} d \right )}{384 d x}+\frac {5 c^{2} \left (48 a^{2} d^{2}+16 a b c d -b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {3}{2}}}\) | \(177\) |
| default | \(b^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{8 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{8 d}\right )+a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{c x}+\frac {6 d \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{c}\right )+2 a b \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )\) | \(261\) |
9/8/d^(3/2)*(5/3*x*c^2*(a^2*d^2+1/3*a*b*c*d-1/48*b^2*c^2)*arctanh((d*x^2+c )^(1/2)/x/d^(1/2))+(d*x^2+c)^(1/2)*(-8/9*(-59/192*b^2*x^4-11/8*a*b*x^2+a^2 )*c^2*d^(3/2)+x^2*(c*(17/54*b^2*x^4+26/27*a*b*x^2+a^2)*d^(5/2)+1/9*(b^2*x^ 6+8/3*a*b*x^4+2*a^2*x^2)*d^(7/2)+5/144*b^2*c^3*d^(1/2))))/x
Time = 0.30 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.73 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\left [-\frac {15 \, {\left (b^{2} c^{4} - 16 \, a b c^{3} d - 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (48 \, b^{2} d^{4} x^{8} + 8 \, {\left (17 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{6} - 384 \, a^{2} c^{2} d^{2} + 2 \, {\left (59 \, b^{2} c^{2} d^{2} + 208 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 176 \, a b c^{2} d^{2} + 144 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{768 \, d^{2} x}, \frac {15 \, {\left (b^{2} c^{4} - 16 \, a b c^{3} d - 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (48 \, b^{2} d^{4} x^{8} + 8 \, {\left (17 \, b^{2} c d^{3} + 16 \, a b d^{4}\right )} x^{6} - 384 \, a^{2} c^{2} d^{2} + 2 \, {\left (59 \, b^{2} c^{2} d^{2} + 208 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{4} + 3 \, {\left (5 \, b^{2} c^{3} d + 176 \, a b c^{2} d^{2} + 144 \, a^{2} c d^{3}\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{384 \, d^{2} x}\right ] \]
[-1/768*(15*(b^2*c^4 - 16*a*b*c^3*d - 48*a^2*c^2*d^2)*sqrt(d)*x*log(-2*d*x ^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 2*(48*b^2*d^4*x^8 + 8*(17*b^2*c*d^ 3 + 16*a*b*d^4)*x^6 - 384*a^2*c^2*d^2 + 2*(59*b^2*c^2*d^2 + 208*a*b*c*d^3 + 48*a^2*d^4)*x^4 + 3*(5*b^2*c^3*d + 176*a*b*c^2*d^2 + 144*a^2*c*d^3)*x^2) *sqrt(d*x^2 + c))/(d^2*x), 1/384*(15*(b^2*c^4 - 16*a*b*c^3*d - 48*a^2*c^2* d^2)*sqrt(-d)*x*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (48*b^2*d^4*x^8 + 8*( 17*b^2*c*d^3 + 16*a*b*d^4)*x^6 - 384*a^2*c^2*d^2 + 2*(59*b^2*c^2*d^2 + 208 *a*b*c*d^3 + 48*a^2*d^4)*x^4 + 3*(5*b^2*c^3*d + 176*a*b*c^2*d^2 + 144*a^2* c*d^3)*x^2)*sqrt(d*x^2 + c))/(d^2*x)]
Time = 2.58 (sec) , antiderivative size = 915, normalized size of antiderivative = 4.22 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\text {Too large to display} \]
-a**2*c**(5/2)/(x*sqrt(1 + d*x**2/c)) - a**2*c**(3/2)*d*x/sqrt(1 + d*x**2/ c) + a**2*c**2*sqrt(d)*asinh(sqrt(d)*x/sqrt(c)) + 2*a**2*c*d*Piecewise((c* 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 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sqrt(c)* x, True)) + a**2*d**2*Piecewise((-c**2*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))/(8*d) + c*x*sqrt(c + d*x**2)/(8*d) + x**3*sqrt(c + d*x**2)/4, Ne(d, 0)), (sqrt(c) *x**3/3, True)) + 2*a*b*c**2*Piecewise((c*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 + x*sqrt(c + d*x**2)/2, Ne(d, 0)), (sqrt(c)*x, True)) + 4*a*b*c*d*Piecewise( (-c**2*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))/(8*d) + c*x*sqrt(c + d*x**2)/(8*d) + x* *3*sqrt(c + d*x**2)/4, Ne(d, 0)), (sqrt(c)*x**3/3, True)) + 2*a*b*d**2*Pie cewise((c**3*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))/(16*d**2) - c**2*x*sqrt(c + d*x** 2)/(16*d**2) + c*x**3*sqrt(c + d*x**2)/(24*d) + x**5*sqrt(c + d*x**2)/6, N e(d, 0)), (sqrt(c)*x**5/5, True)) + b**2*c**2*Piecewise((-c**2*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))/(8*d) + c*x*sqrt(c + d*x**2)/(8*d) + x**3*sqrt(c + d*x**2 )/4, Ne(d, 0)), (sqrt(c)*x**3/3, True)) + 2*b**2*c*d*Piecewise((c**3*Pi...
Time = 0.22 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\frac {1}{3} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b x + \frac {5}{12} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c x + \frac {5}{8} \, \sqrt {d x^{2} + c} a b c^{2} x + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x}{8 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c x}{48 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{2} x}{192 \, d} - \frac {5 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d} + \frac {5}{4} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d x + \frac {15}{8} \, \sqrt {d x^{2} + c} a^{2} c d x - \frac {5 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {3}{2}}} + \frac {5 \, a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {d}} + \frac {15}{8} \, a^{2} c^{2} \sqrt {d} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right ) - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2}}{x} \]
1/3*(d*x^2 + c)^(5/2)*a*b*x + 5/12*(d*x^2 + c)^(3/2)*a*b*c*x + 5/8*sqrt(d* x^2 + c)*a*b*c^2*x + 1/8*(d*x^2 + c)^(7/2)*b^2*x/d - 1/48*(d*x^2 + c)^(5/2 )*b^2*c*x/d - 5/192*(d*x^2 + c)^(3/2)*b^2*c^2*x/d - 5/128*sqrt(d*x^2 + c)* b^2*c^3*x/d + 5/4*(d*x^2 + c)^(3/2)*a^2*d*x + 15/8*sqrt(d*x^2 + c)*a^2*c*d *x - 5/128*b^2*c^4*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 5/8*a*b*c^3*arcsinh(d* x/sqrt(c*d))/sqrt(d) + 15/8*a^2*c^2*sqrt(d)*arcsinh(d*x/sqrt(c*d)) - (d*x^ 2 + c)^(5/2)*a^2/x
Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\frac {2 \, a^{2} c^{3} \sqrt {d}}{{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} - c} + \frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (6 \, b^{2} d^{2} x^{2} + \frac {17 \, b^{2} c d^{7} + 16 \, a b d^{8}}{d^{6}}\right )} x^{2} + \frac {59 \, b^{2} c^{2} d^{6} + 208 \, a b c d^{7} + 48 \, a^{2} d^{8}}{d^{6}}\right )} x^{2} + \frac {3 \, {\left (5 \, b^{2} c^{3} d^{5} + 176 \, a b c^{2} d^{6} + 144 \, a^{2} c d^{7}\right )}}{d^{6}}\right )} \sqrt {d x^{2} + c} x + \frac {5 \, {\left (b^{2} c^{4} - 16 \, a b c^{3} d - 48 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{256 \, d^{\frac {3}{2}}} \]
2*a^2*c^3*sqrt(d)/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c) + 1/384*(2*(4*(6*b ^2*d^2*x^2 + (17*b^2*c*d^7 + 16*a*b*d^8)/d^6)*x^2 + (59*b^2*c^2*d^6 + 208* a*b*c*d^7 + 48*a^2*d^8)/d^6)*x^2 + 3*(5*b^2*c^3*d^5 + 176*a*b*c^2*d^6 + 14 4*a^2*c*d^7)/d^6)*sqrt(d*x^2 + c)*x + 5/256*(b^2*c^4 - 16*a*b*c^3*d - 48*a ^2*c^2*d^2)*log((sqrt(d)*x - sqrt(d*x^2 + c))^2)/d^(3/2)
Timed out. \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^2} \, dx=\int \frac {{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}}{x^2} \,d x \]