Integrand size = 21, antiderivative size = 240 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {c^2 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \sqrt {c+d x^2}}{256 d^2}+\frac {c \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{3/2}}{384 d^2}+\frac {\left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) x \left (c+d x^2\right )^{5/2}}{480 d^2}-\frac {3 b (b c-4 a d) x \left (c+d x^2\right )^{7/2}}{80 d^2}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}+\frac {c^3 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{256 d^{5/2}} \]
1/384*c*(80*a^2*d^2-20*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(3/2)/d^2+1/480*(80* a^2*d^2-20*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(5/2)/d^2-3/80*b*(-4*a*d+b*c)*x* (d*x^2+c)^(7/2)/d^2+1/10*b*x*(b*x^2+a)*(d*x^2+c)^(7/2)/d+1/256*c^3*(80*a^2 *d^2-20*a*b*c*d+3*b^2*c^2)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/d^(5/2)+1/25 6*c^2*(80*a^2*d^2-20*a*b*c*d+3*b^2*c^2)*x*(d*x^2+c)^(1/2)/d^2
Time = 0.28 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.80 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (80 a^2 d^2 \left (33 c^2+26 c d x^2+8 d^2 x^4\right )+20 a b d \left (15 c^3+118 c^2 d x^2+136 c d^2 x^4+48 d^3 x^6\right )+b^2 \left (-45 c^4+30 c^3 d x^2+744 c^2 d^2 x^4+1008 c d^3 x^6+384 d^4 x^8\right )\right )-15 c^3 \left (3 b^2 c^2-20 a b c d+80 a^2 d^2\right ) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )}{3840 d^{5/2}} \]
(Sqrt[d]*x*Sqrt[c + d*x^2]*(80*a^2*d^2*(33*c^2 + 26*c*d*x^2 + 8*d^2*x^4) + 20*a*b*d*(15*c^3 + 118*c^2*d*x^2 + 136*c*d^2*x^4 + 48*d^3*x^6) + b^2*(-45 *c^4 + 30*c^3*d*x^2 + 744*c^2*d^2*x^4 + 1008*c*d^3*x^6 + 384*d^4*x^8)) - 1 5*c^3*(3*b^2*c^2 - 20*a*b*c*d + 80*a^2*d^2)*Log[-(Sqrt[d]*x) + Sqrt[c + d* x^2]])/(3840*d^(5/2))
Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.77, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {318, 25, 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.
| \(\displaystyle \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\int -\left (d x^2+c\right )^{5/2} \left (3 b (b c-4 a d) x^2+a (b c-10 a d)\right )dx}{10 d}+\frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\int \left (d x^2+c\right )^{5/2} \left (3 b (b c-4 a d) x^2+a (b c-10 a d)\right )dx}{10 d}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \int \left (d x^2+c\right )^{5/2}dx}{8 d}}{10 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \left (\frac {5}{6} c \int \left (d x^2+c\right )^{3/2}dx+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{10 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \int \sqrt {d x^2+c}dx+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{10 d}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{10 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{10 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {b x \left (a+b x^2\right ) \left (c+d x^2\right )^{7/2}}{10 d}-\frac {\frac {3 b x \left (c+d x^2\right )^{7/2} (b c-4 a d)}{8 d}-\frac {\left (80 a^2 d^2-20 a b c d+3 b^2 c^2\right ) \left (\frac {5}{6} c \left (\frac {3}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {1}{4} x \left (c+d x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c+d x^2\right )^{5/2}\right )}{8 d}}{10 d}\) |
(b*x*(a + b*x^2)*(c + d*x^2)^(7/2))/(10*d) - ((3*b*(b*c - 4*a*d)*x*(c + d* x^2)^(7/2))/(8*d) - ((3*b^2*c^2 - 20*a*b*c*d + 80*a^2*d^2)*((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))/(10* d)
3.7.28.3.1 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 = 3.00 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {\frac {5 c^{3} \left (a^{2} d^{2}-\frac {1}{4} a b c d +\frac {3}{80} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{16}+\frac {11 x \left (c^{2} \left (\frac {31}{110} b^{2} x^{4}+\frac {59}{66} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {26 x^{2} \left (\frac {63}{130} b^{2} x^{4}+\frac {17}{13} a b \,x^{2}+a^{2}\right ) c \,d^{\frac {7}{2}}}{33}+\frac {8 x^{4} \left (\frac {3}{5} b^{2} x^{4}+\frac {3}{2} a b \,x^{2}+a^{2}\right ) d^{\frac {9}{2}}}{33}+\frac {5 \left (\left (\frac {b \,x^{2}}{10}+a \right ) d^{\frac {3}{2}}-\frac {3 b \sqrt {d}\, c}{20}\right ) b \,c^{3}}{44}\right ) \sqrt {d \,x^{2}+c}}{16}}{d^{\frac {5}{2}}}\) | \(170\) |
| risch | \(\frac {x \left (384 b^{2} x^{8} d^{4}+960 a b \,d^{4} x^{6}+1008 b^{2} c \,d^{3} x^{6}+640 a^{2} d^{4} x^{4}+2720 c a b \,x^{4} d^{3}+744 b^{2} c^{2} d^{2} x^{4}+2080 a^{2} c \,d^{3} x^{2}+2360 a b \,c^{2} d^{2} x^{2}+30 b^{2} c^{3} d \,x^{2}+2640 a^{2} c^{2} d^{2}+300 a b \,c^{3} d -45 b^{2} c^{4}\right ) \sqrt {d \,x^{2}+c}}{3840 d^{2}}+\frac {c^{3} \left (80 a^{2} d^{2}-20 a b c d +3 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{256 d^{\frac {5}{2}}}\) | \(198\) |
| default | \(a^{2} \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )+b^{2} \left (\frac {x^{3} \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{10 d}-\frac {3 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{8 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{8 d}\right )}{10 d}\right )+2 a b \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {7}{2}}}{8 d}-\frac {c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {5}{2}}}{6}+\frac {5 c \left (\frac {x \left (d \,x^{2}+c \right )^{\frac {3}{2}}}{4}+\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2}+\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 \sqrt {d}}\right )}{4}\right )}{6}\right )}{8 d}\right )\) | \(283\) |
11/16/d^(5/2)*(5/11*c^3*(a^2*d^2-1/4*a*b*c*d+3/80*b^2*c^2)*arctanh((d*x^2+ c)^(1/2)/x/d^(1/2))+x*(c^2*(31/110*b^2*x^4+59/66*a*b*x^2+a^2)*d^(5/2)+26/3 3*x^2*(63/130*b^2*x^4+17/13*a*b*x^2+a^2)*c*d^(7/2)+8/33*x^4*(3/5*b^2*x^4+3 /2*a*b*x^2+a^2)*d^(9/2)+5/44*((1/10*b*x^2+a)*d^(3/2)-3/20*b*d^(1/2)*c)*b*c ^3)*(d*x^2+c)^(1/2))
Time = 0.33 (sec) , antiderivative size = 420, normalized size of antiderivative = 1.75 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\left [\frac {15 \, {\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (21 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (93 \, b^{2} c^{2} d^{3} + 340 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} + 10 \, {\left (3 \, b^{2} c^{3} d^{2} + 236 \, a b c^{2} d^{3} + 208 \, a^{2} c d^{4}\right )} x^{3} - 15 \, {\left (3 \, b^{2} c^{4} d - 20 \, a b c^{3} d^{2} - 176 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{7680 \, d^{3}}, -\frac {15 \, {\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (384 \, b^{2} d^{5} x^{9} + 48 \, {\left (21 \, b^{2} c d^{4} + 20 \, a b d^{5}\right )} x^{7} + 8 \, {\left (93 \, b^{2} c^{2} d^{3} + 340 \, a b c d^{4} + 80 \, a^{2} d^{5}\right )} x^{5} + 10 \, {\left (3 \, b^{2} c^{3} d^{2} + 236 \, a b c^{2} d^{3} + 208 \, a^{2} c d^{4}\right )} x^{3} - 15 \, {\left (3 \, b^{2} c^{4} d - 20 \, a b c^{3} d^{2} - 176 \, a^{2} c^{2} d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{3840 \, d^{3}}\right ] \]
[1/7680*(15*(3*b^2*c^5 - 20*a*b*c^4*d + 80*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*(21*b^2*c* d^4 + 20*a*b*d^5)*x^7 + 8*(93*b^2*c^2*d^3 + 340*a*b*c*d^4 + 80*a^2*d^5)*x^ 5 + 10*(3*b^2*c^3*d^2 + 236*a*b*c^2*d^3 + 208*a^2*c*d^4)*x^3 - 15*(3*b^2*c ^4*d - 20*a*b*c^3*d^2 - 176*a^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^3, -1/3840* (15*(3*b^2*c^5 - 20*a*b*c^4*d + 80*a^2*c^3*d^2)*sqrt(-d)*arctan(sqrt(-d)*x /sqrt(d*x^2 + c)) - (384*b^2*d^5*x^9 + 48*(21*b^2*c*d^4 + 20*a*b*d^5)*x^7 + 8*(93*b^2*c^2*d^3 + 340*a*b*c*d^4 + 80*a^2*d^5)*x^5 + 10*(3*b^2*c^3*d^2 + 236*a*b*c^2*d^3 + 208*a^2*c*d^4)*x^3 - 15*(3*b^2*c^4*d - 20*a*b*c^3*d^2 - 176*a^2*c^2*d^3)*x)*sqrt(d*x^2 + c))/d^3]
Leaf count of result is larger than twice the leaf count of optimal. 520 vs. \(2 (236) = 472\).
Time = 0.48 (sec) , antiderivative size = 520, normalized size of antiderivative = 2.17 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {c + d x^{2}} \left (\frac {b^{2} d^{2} x^{9}}{10} + \frac {x^{7} \cdot \left (2 a b d^{3} + \frac {21 b^{2} c d^{2}}{10}\right )}{8 d} + \frac {x^{5} \left (a^{2} d^{3} + 6 a b c d^{2} + 3 b^{2} c^{2} d - \frac {7 c \left (2 a b d^{3} + \frac {21 b^{2} c d^{2}}{10}\right )}{8 d}\right )}{6 d} + \frac {x^{3} \cdot \left (3 a^{2} c d^{2} + 6 a b c^{2} d + b^{2} c^{3} - \frac {5 c \left (a^{2} d^{3} + 6 a b c d^{2} + 3 b^{2} c^{2} d - \frac {7 c \left (2 a b d^{3} + \frac {21 b^{2} c d^{2}}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d} + \frac {x \left (3 a^{2} c^{2} d + 2 a b c^{3} - \frac {3 c \left (3 a^{2} c d^{2} + 6 a b c^{2} d + b^{2} c^{3} - \frac {5 c \left (a^{2} d^{3} + 6 a b c d^{2} + 3 b^{2} c^{2} d - \frac {7 c \left (2 a b d^{3} + \frac {21 b^{2} c d^{2}}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) + \left (a^{2} c^{3} - \frac {c \left (3 a^{2} c^{2} d + 2 a b c^{3} - \frac {3 c \left (3 a^{2} c d^{2} + 6 a b c^{2} d + b^{2} c^{3} - \frac {5 c \left (a^{2} d^{3} + 6 a b c d^{2} + 3 b^{2} c^{2} d - \frac {7 c \left (2 a b d^{3} + \frac {21 b^{2} c d^{2}}{10}\right )}{8 d}\right )}{6 d}\right )}{4 d}\right )}{2 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d \neq 0 \\c^{\frac {5}{2}} \left (a^{2} x + \frac {2 a b x^{3}}{3} + \frac {b^{2} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(c + d*x**2)*(b**2*d**2*x**9/10 + x**7*(2*a*b*d**3 + 21*b** 2*c*d**2/10)/(8*d) + x**5*(a**2*d**3 + 6*a*b*c*d**2 + 3*b**2*c**2*d - 7*c* (2*a*b*d**3 + 21*b**2*c*d**2/10)/(8*d))/(6*d) + x**3*(3*a**2*c*d**2 + 6*a* b*c**2*d + b**2*c**3 - 5*c*(a**2*d**3 + 6*a*b*c*d**2 + 3*b**2*c**2*d - 7*c *(2*a*b*d**3 + 21*b**2*c*d**2/10)/(8*d))/(6*d))/(4*d) + x*(3*a**2*c**2*d + 2*a*b*c**3 - 3*c*(3*a**2*c*d**2 + 6*a*b*c**2*d + b**2*c**3 - 5*c*(a**2*d* *3 + 6*a*b*c*d**2 + 3*b**2*c**2*d - 7*c*(2*a*b*d**3 + 21*b**2*c*d**2/10)/( 8*d))/(6*d))/(4*d))/(2*d)) + (a**2*c**3 - c*(3*a**2*c**2*d + 2*a*b*c**3 - 3*c*(3*a**2*c*d**2 + 6*a*b*c**2*d + b**2*c**3 - 5*c*(a**2*d**3 + 6*a*b*c*d **2 + 3*b**2*c**2*d - 7*c*(2*a*b*d**3 + 21*b**2*c*d**2/10)/(8*d))/(6*d))/( 4*d))/(2*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)), Ne(d, 0)), (c**(5/2)*(a**2*x + 2 *a*b*x**3/3 + b**2*x**5/5), True))
Time = 0.21 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.19 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} x^{3}}{10 \, d} + \frac {1}{6} \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} x + \frac {5}{24} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c x + \frac {5}{16} \, \sqrt {d x^{2} + c} a^{2} c^{2} x - \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2} c x}{80 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} x}{160 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} x}{128 \, d^{2}} + \frac {3 \, \sqrt {d x^{2} + c} b^{2} c^{4} x}{256 \, d^{2}} + \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a b x}{4 \, d} - \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c x}{24 \, d} - \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} x}{96 \, d} - \frac {5 \, \sqrt {d x^{2} + c} a b c^{3} x}{64 \, d} + \frac {3 \, b^{2} c^{5} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{256 \, d^{\frac {5}{2}}} - \frac {5 \, a b c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{64 \, d^{\frac {3}{2}}} + \frac {5 \, a^{2} c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{16 \, \sqrt {d}} \]
1/10*(d*x^2 + c)^(7/2)*b^2*x^3/d + 1/6*(d*x^2 + c)^(5/2)*a^2*x + 5/24*(d*x ^2 + c)^(3/2)*a^2*c*x + 5/16*sqrt(d*x^2 + c)*a^2*c^2*x - 3/80*(d*x^2 + c)^ (7/2)*b^2*c*x/d^2 + 1/160*(d*x^2 + c)^(5/2)*b^2*c^2*x/d^2 + 1/128*(d*x^2 + c)^(3/2)*b^2*c^3*x/d^2 + 3/256*sqrt(d*x^2 + c)*b^2*c^4*x/d^2 + 1/4*(d*x^2 + c)^(7/2)*a*b*x/d - 1/24*(d*x^2 + c)^(5/2)*a*b*c*x/d - 5/96*(d*x^2 + c)^ (3/2)*a*b*c^2*x/d - 5/64*sqrt(d*x^2 + c)*a*b*c^3*x/d + 3/256*b^2*c^5*arcsi nh(d*x/sqrt(c*d))/d^(5/2) - 5/64*a*b*c^4*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 5/16*a^2*c^3*arcsinh(d*x/sqrt(c*d))/sqrt(d)
Time = 0.34 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.92 \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, b^{2} d^{2} x^{2} + \frac {21 \, b^{2} c d^{9} + 20 \, a b d^{10}}{d^{8}}\right )} x^{2} + \frac {93 \, b^{2} c^{2} d^{8} + 340 \, a b c d^{9} + 80 \, a^{2} d^{10}}{d^{8}}\right )} x^{2} + \frac {5 \, {\left (3 \, b^{2} c^{3} d^{7} + 236 \, a b c^{2} d^{8} + 208 \, a^{2} c d^{9}\right )}}{d^{8}}\right )} x^{2} - \frac {15 \, {\left (3 \, b^{2} c^{4} d^{6} - 20 \, a b c^{3} d^{7} - 176 \, a^{2} c^{2} d^{8}\right )}}{d^{8}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{5} - 20 \, a b c^{4} d + 80 \, a^{2} c^{3} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{256 \, d^{\frac {5}{2}}} \]
1/3840*(2*(4*(6*(8*b^2*d^2*x^2 + (21*b^2*c*d^9 + 20*a*b*d^10)/d^8)*x^2 + ( 93*b^2*c^2*d^8 + 340*a*b*c*d^9 + 80*a^2*d^10)/d^8)*x^2 + 5*(3*b^2*c^3*d^7 + 236*a*b*c^2*d^8 + 208*a^2*c*d^9)/d^8)*x^2 - 15*(3*b^2*c^4*d^6 - 20*a*b*c ^3*d^7 - 176*a^2*c^2*d^8)/d^8)*sqrt(d*x^2 + c)*x - 1/256*(3*b^2*c^5 - 20*a *b*c^4*d + 80*a^2*c^3*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(5/2)
Timed out. \[ \int \left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2} \, dx=\int {\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2} \,d x \]