Integrand size = 24, antiderivative size = 222 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=\frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \sqrt {c+d x^2}}{16 c}+\frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \left (c+d x^2\right )^{3/2}}{48 c^2}-\frac {\left (8 b^2 c^2+a d (12 b c+a d)\right ) \left (c+d x^2\right )^{5/2}}{16 c^2 x^2}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{6 c x^6}-\frac {a (12 b c+a d) \left (c+d x^2\right )^{7/2}}{24 c^2 x^4}-\frac {5 d \left (8 b^2 c^2+a d (12 b c+a d)\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}} \]
5/48*d*(8*b^2*c^2+a*d*(a*d+12*b*c))*(d*x^2+c)^(3/2)/c^2-1/16*(8*b^2*c^2+a* d*(a*d+12*b*c))*(d*x^2+c)^(5/2)/c^2/x^2-1/6*a^2*(d*x^2+c)^(7/2)/c/x^6-1/24 *a*(a*d+12*b*c)*(d*x^2+c)^(7/2)/c^2/x^4-5/16*d*(8*b^2*c^2+a*d*(a*d+12*b*c) )*arctanh((d*x^2+c)^(1/2)/c^(1/2))/c^(1/2)+5/16*d*(8*b^2*c^2+a*d*(a*d+12*b *c))*(d*x^2+c)^(1/2)/c
Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.68 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=-\frac {\sqrt {c+d x^2} \left (12 a b x^2 \left (2 c^2+9 c d x^2-8 d^2 x^4\right )-8 b^2 x^4 \left (-3 c^2+14 c d x^2+2 d^2 x^4\right )+a^2 \left (8 c^2+26 c d x^2+33 d^2 x^4\right )\right )}{48 x^6}-\frac {5 d \left (8 b^2 c^2+12 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )}{16 \sqrt {c}} \]
-1/48*(Sqrt[c + d*x^2]*(12*a*b*x^2*(2*c^2 + 9*c*d*x^2 - 8*d^2*x^4) - 8*b^2 *x^4*(-3*c^2 + 14*c*d*x^2 + 2*d^2*x^4) + a^2*(8*c^2 + 26*c*d*x^2 + 33*d^2* x^4)))/x^6 - (5*d*(8*b^2*c^2 + 12*a*b*c*d + a^2*d^2)*ArcTanh[Sqrt[c + d*x^ 2]/Sqrt[c]])/(16*Sqrt[c])
Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {354, 100, 27, 87, 51, 60, 60, 73, 221}
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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^2 \left (d x^2+c\right )^{5/2}}{x^8}dx^2\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\left (6 b^2 c x^2+a (12 b c+a d)\right ) \left (d x^2+c\right )^{5/2}}{2 x^6}dx^2}{3 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \left (\frac {\int \frac {\left (6 b^2 c x^2+a (12 b c+a d)\right ) \left (d x^2+c\right )^{5/2}}{x^6}dx^2}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \int \frac {\left (d x^2+c\right )^{5/2}}{x^4}dx^2}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \left (\frac {5}{2} d \int \frac {\left (d x^2+c\right )^{3/2}}{x^2}dx^2-\frac {\left (c+d x^2\right )^{5/2}}{x^2}\right )}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \left (\frac {5}{2} d \left (c \int \frac {\sqrt {d x^2+c}}{x^2}dx^2+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )-\frac {\left (c+d x^2\right )^{5/2}}{x^2}\right )}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \left (\frac {5}{2} d \left (c \left (c \int \frac {1}{x^2 \sqrt {d x^2+c}}dx^2+2 \sqrt {c+d x^2}\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )-\frac {\left (c+d x^2\right )^{5/2}}{x^2}\right )}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \left (\frac {5}{2} d \left (c \left (\frac {2 c \int \frac {1}{\frac {x^4}{d}-\frac {c}{d}}d\sqrt {d x^2+c}}{d}+2 \sqrt {c+d x^2}\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )-\frac {\left (c+d x^2\right )^{5/2}}{x^2}\right )}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (\frac {\frac {3 \left (a d (a d+12 b c)+8 b^2 c^2\right ) \left (\frac {5}{2} d \left (c \left (2 \sqrt {c+d x^2}-2 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^2}}{\sqrt {c}}\right )\right )+\frac {2}{3} \left (c+d x^2\right )^{3/2}\right )-\frac {\left (c+d x^2\right )^{5/2}}{x^2}\right )}{4 c}-\frac {a \left (c+d x^2\right )^{7/2} (a d+12 b c)}{2 c x^4}}{6 c}-\frac {a^2 \left (c+d x^2\right )^{7/2}}{3 c x^6}\right )\) |
(-1/3*(a^2*(c + d*x^2)^(7/2))/(c*x^6) + (-1/2*(a*(12*b*c + a*d)*(c + d*x^2 )^(7/2))/(c*x^4) + (3*(8*b^2*c^2 + a*d*(12*b*c + a*d))*(-((c + d*x^2)^(5/2 )/x^2) + (5*d*((2*(c + d*x^2)^(3/2))/3 + c*(2*Sqrt[c + d*x^2] - 2*Sqrt[c]* ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]])))/2))/(4*c))/(6*c))/2
3.7.35.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.94 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.65
| method | result | size |
| pseudoelliptic | \(-\frac {11 \left (\frac {5 d \,x^{6} \left (a^{2} d^{2}+12 a b c d +8 b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{\sqrt {c}}\right )}{11}+\sqrt {d \,x^{2}+c}\, \left (\frac {26 \left (-\frac {56}{13} b^{2} x^{4}+\frac {54}{13} a b \,x^{2}+a^{2}\right ) x^{2} d \,c^{\frac {3}{2}}}{33}+\frac {8 \left (b^{2} x^{4}+a b \,x^{2}+\frac {1}{3} a^{2}\right ) c^{\frac {5}{2}}}{11}+d^{2} x^{4} \sqrt {c}\, \left (-\frac {16}{33} b^{2} x^{4}-\frac {32}{11} a b \,x^{2}+a^{2}\right )\right )\right )}{16 \sqrt {c}\, x^{6}}\) | \(144\) |
| risch | \(-\frac {\sqrt {d \,x^{2}+c}\, \left (33 a^{2} d^{2} x^{4}+108 x^{4} a b c d +24 b^{2} c^{2} x^{4}+26 a^{2} c d \,x^{2}+24 a b \,c^{2} x^{2}+8 a^{2} c^{2}\right )}{48 x^{6}}+\frac {d \left (16 b^{2} d^{2} \left (\frac {x^{2} \sqrt {d \,x^{2}+c}}{3 d}-\frac {2 c \sqrt {d \,x^{2}+c}}{3 d^{2}}\right )+32 a b d \sqrt {d \,x^{2}+c}+48 b^{2} c \sqrt {d \,x^{2}+c}-\frac {\left (5 a^{2} d^{2}+60 a b c d +40 b^{2} c^{2}\right ) \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )}{\sqrt {c}}\right )}{16}\) | \(201\) |
| default | \(a^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{6 c \,x^{6}}+\frac {d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{4 c \,x^{4}}+\frac {3 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )}{4 c}\right )}{6 c}\right )+b^{2} \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )+2 a b \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{4 c \,x^{4}}+\frac {3 d \left (-\frac {\left (d \,x^{2}+c \right )^{\frac {7}{2}}}{2 c \,x^{2}}+\frac {5 d \left (\frac {\left (d \,x^{2}+c \right )^{\frac {5}{2}}}{5}+c \left (\frac {\left (d \,x^{2}+c \right )^{\frac {3}{2}}}{3}+c \left (\sqrt {d \,x^{2}+c}-\sqrt {c}\, \ln \left (\frac {2 c +2 \sqrt {c}\, \sqrt {d \,x^{2}+c}}{x}\right )\right )\right )\right )}{2 c}\right )}{4 c}\right )\) | \(356\) |
-11/16*(5/11*d*x^6*(a^2*d^2+12*a*b*c*d+8*b^2*c^2)*arctanh((d*x^2+c)^(1/2)/ c^(1/2))+(d*x^2+c)^(1/2)*(26/33*(-56/13*b^2*x^4+54/13*a*b*x^2+a^2)*x^2*d*c ^(3/2)+8/11*(b^2*x^4+a*b*x^2+1/3*a^2)*c^(5/2)+d^2*x^4*c^(1/2)*(-16/33*b^2* x^4-32/11*a*b*x^2+a^2)))/c^(1/2)/x^6
Time = 0.28 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.56 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=\left [\frac {15 \, {\left (8 \, b^{2} c^{2} d + 12 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {c} x^{6} \log \left (-\frac {d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {c} + 2 \, c}{x^{2}}\right ) + 2 \, {\left (16 \, b^{2} c d^{2} x^{8} + 16 \, {\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \, {\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \, {\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{96 \, c x^{6}}, \frac {15 \, {\left (8 \, b^{2} c^{2} d + 12 \, a b c d^{2} + a^{2} d^{3}\right )} \sqrt {-c} x^{6} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{2} + c}}\right ) + {\left (16 \, b^{2} c d^{2} x^{8} + 16 \, {\left (7 \, b^{2} c^{2} d + 6 \, a b c d^{2}\right )} x^{6} - 8 \, a^{2} c^{3} - 3 \, {\left (8 \, b^{2} c^{3} + 36 \, a b c^{2} d + 11 \, a^{2} c d^{2}\right )} x^{4} - 2 \, {\left (12 \, a b c^{3} + 13 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt {d x^{2} + c}}{48 \, c x^{6}}\right ] \]
[1/96*(15*(8*b^2*c^2*d + 12*a*b*c*d^2 + a^2*d^3)*sqrt(c)*x^6*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(16*b^2*c*d^2*x^8 + 16*(7*b^2*c ^2*d + 6*a*b*c*d^2)*x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a^2 *c*d^2)*x^4 - 2*(12*a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c*x^6), 1/48*(15*(8*b^2*c^2*d + 12*a*b*c*d^2 + a^2*d^3)*sqrt(-c)*x^6*arctan(sqrt( -c)/sqrt(d*x^2 + c)) + (16*b^2*c*d^2*x^8 + 16*(7*b^2*c^2*d + 6*a*b*c*d^2)* x^6 - 8*a^2*c^3 - 3*(8*b^2*c^3 + 36*a*b*c^2*d + 11*a^2*c*d^2)*x^4 - 2*(12* a*b*c^3 + 13*a^2*c^2*d)*x^2)*sqrt(d*x^2 + c))/(c*x^6)]
Time = 89.31 (sec) , antiderivative size = 484, normalized size of antiderivative = 2.18 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=- \frac {a^{2} c^{3}}{6 \sqrt {d} x^{7} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {17 a^{2} c^{2} \sqrt {d}}{24 x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {35 a^{2} c d^{\frac {3}{2}}}{48 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {a^{2} d^{\frac {5}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} - \frac {3 a^{2} d^{\frac {5}{2}}}{16 x \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 a^{2} d^{3} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{16 \sqrt {c}} - \frac {15 a b \sqrt {c} d^{2} \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{4} - \frac {a b c^{3}}{2 \sqrt {d} x^{5} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {3 a b c^{2} \sqrt {d}}{4 x^{3} \sqrt {\frac {c}{d x^{2}} + 1}} - \frac {2 a b c d^{\frac {3}{2}} \sqrt {\frac {c}{d x^{2}} + 1}}{x} + \frac {7 a b c d^{\frac {3}{2}}}{4 x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 a b d^{\frac {5}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} - \frac {5 b^{2} c^{\frac {3}{2}} d \operatorname {asinh}{\left (\frac {\sqrt {c}}{\sqrt {d} x} \right )}}{2} - \frac {b^{2} c^{2} \sqrt {d} \sqrt {\frac {c}{d x^{2}} + 1}}{2 x} + \frac {2 b^{2} c^{2} \sqrt {d}}{x \sqrt {\frac {c}{d x^{2}} + 1}} + \frac {2 b^{2} c d^{\frac {3}{2}} x}{\sqrt {\frac {c}{d x^{2}} + 1}} + b^{2} d^{2} \left (\begin {cases} \frac {c \sqrt {c + d x^{2}}}{3 d} + \frac {x^{2} \sqrt {c + d x^{2}}}{3} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \]
-a**2*c**3/(6*sqrt(d)*x**7*sqrt(c/(d*x**2) + 1)) - 17*a**2*c**2*sqrt(d)/(2 4*x**5*sqrt(c/(d*x**2) + 1)) - 35*a**2*c*d**(3/2)/(48*x**3*sqrt(c/(d*x**2) + 1)) - a**2*d**(5/2)*sqrt(c/(d*x**2) + 1)/(2*x) - 3*a**2*d**(5/2)/(16*x* sqrt(c/(d*x**2) + 1)) - 5*a**2*d**3*asinh(sqrt(c)/(sqrt(d)*x))/(16*sqrt(c) ) - 15*a*b*sqrt(c)*d**2*asinh(sqrt(c)/(sqrt(d)*x))/4 - a*b*c**3/(2*sqrt(d) *x**5*sqrt(c/(d*x**2) + 1)) - 3*a*b*c**2*sqrt(d)/(4*x**3*sqrt(c/(d*x**2) + 1)) - 2*a*b*c*d**(3/2)*sqrt(c/(d*x**2) + 1)/x + 7*a*b*c*d**(3/2)/(4*x*sqr t(c/(d*x**2) + 1)) + 2*a*b*d**(5/2)*x/sqrt(c/(d*x**2) + 1) - 5*b**2*c**(3/ 2)*d*asinh(sqrt(c)/(sqrt(d)*x))/2 - b**2*c**2*sqrt(d)*sqrt(c/(d*x**2) + 1) /(2*x) + 2*b**2*c**2*sqrt(d)/(x*sqrt(c/(d*x**2) + 1)) + 2*b**2*c*d**(3/2)* x/sqrt(c/(d*x**2) + 1) + b**2*d**2*Piecewise((c*sqrt(c + d*x**2)/(3*d) + x **2*sqrt(c + d*x**2)/3, Ne(d, 0)), (sqrt(c)*x**2/2, True))
Time = 0.21 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.59 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=-\frac {5}{2} \, b^{2} c^{\frac {3}{2}} d \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {15}{4} \, a b \sqrt {c} d^{2} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right ) - \frac {5 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {c}{\sqrt {c d} {\left | x \right |}}\right )}{16 \, \sqrt {c}} + \frac {5}{6} \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} d}{2 \, c} + \frac {5}{2} \, \sqrt {d x^{2} + c} b^{2} c d + \frac {15}{4} \, \sqrt {d x^{2} + c} a b d^{2} + \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b d^{2}}{4 \, c^{2}} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b d^{2}}{4 \, c} + \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{3}}{16 \, c^{3}} + \frac {5 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} d^{3}}{48 \, c^{2}} + \frac {5 \, \sqrt {d x^{2} + c} a^{2} d^{3}}{16 \, c} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} b^{2}}{2 \, c x^{2}} - \frac {3 \, {\left (d x^{2} + c\right )}^{\frac {7}{2}} a b d}{4 \, c^{2} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d^{2}}{16 \, c^{3} x^{2}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a b}{2 \, c x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2} d}{24 \, c^{2} x^{4}} - \frac {{\left (d x^{2} + c\right )}^{\frac {7}{2}} a^{2}}{6 \, c x^{6}} \]
-5/2*b^2*c^(3/2)*d*arcsinh(c/(sqrt(c*d)*abs(x))) - 15/4*a*b*sqrt(c)*d^2*ar csinh(c/(sqrt(c*d)*abs(x))) - 5/16*a^2*d^3*arcsinh(c/(sqrt(c*d)*abs(x)))/s qrt(c) + 5/6*(d*x^2 + c)^(3/2)*b^2*d + 1/2*(d*x^2 + c)^(5/2)*b^2*d/c + 5/2 *sqrt(d*x^2 + c)*b^2*c*d + 15/4*sqrt(d*x^2 + c)*a*b*d^2 + 3/4*(d*x^2 + c)^ (5/2)*a*b*d^2/c^2 + 5/4*(d*x^2 + c)^(3/2)*a*b*d^2/c + 1/16*(d*x^2 + c)^(5/ 2)*a^2*d^3/c^3 + 5/48*(d*x^2 + c)^(3/2)*a^2*d^3/c^2 + 5/16*sqrt(d*x^2 + c) *a^2*d^3/c - 1/2*(d*x^2 + c)^(7/2)*b^2/(c*x^2) - 3/4*(d*x^2 + c)^(7/2)*a*b *d/(c^2*x^2) - 1/16*(d*x^2 + c)^(7/2)*a^2*d^2/(c^3*x^2) - 1/2*(d*x^2 + c)^ (7/2)*a*b/(c*x^4) - 1/24*(d*x^2 + c)^(7/2)*a^2*d/(c^2*x^4) - 1/6*(d*x^2 + c)^(7/2)*a^2/(c*x^6)
Time = 0.32 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.29 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=\frac {16 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} d^{2} + 96 \, \sqrt {d x^{2} + c} b^{2} c d^{2} + 96 \, \sqrt {d x^{2} + c} a b d^{3} + \frac {15 \, {\left (8 \, b^{2} c^{2} d^{2} + 12 \, a b c d^{3} + a^{2} d^{4}\right )} \arctan \left (\frac {\sqrt {d x^{2} + c}}{\sqrt {-c}}\right )}{\sqrt {-c}} - \frac {24 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} b^{2} c^{2} d^{2} - 48 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} b^{2} c^{3} d^{2} + 24 \, \sqrt {d x^{2} + c} b^{2} c^{4} d^{2} + 108 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a b c d^{3} - 192 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a b c^{2} d^{3} + 84 \, \sqrt {d x^{2} + c} a b c^{3} d^{3} + 33 \, {\left (d x^{2} + c\right )}^{\frac {5}{2}} a^{2} d^{4} - 40 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}} a^{2} c d^{4} + 15 \, \sqrt {d x^{2} + c} a^{2} c^{2} d^{4}}{d^{3} x^{6}}}{48 \, d} \]
1/48*(16*(d*x^2 + c)^(3/2)*b^2*d^2 + 96*sqrt(d*x^2 + c)*b^2*c*d^2 + 96*sqr t(d*x^2 + c)*a*b*d^3 + 15*(8*b^2*c^2*d^2 + 12*a*b*c*d^3 + a^2*d^4)*arctan( sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) - (24*(d*x^2 + c)^(5/2)*b^2*c^2*d^2 - 4 8*(d*x^2 + c)^(3/2)*b^2*c^3*d^2 + 24*sqrt(d*x^2 + c)*b^2*c^4*d^2 + 108*(d* x^2 + c)^(5/2)*a*b*c*d^3 - 192*(d*x^2 + c)^(3/2)*a*b*c^2*d^3 + 84*sqrt(d*x ^2 + c)*a*b*c^3*d^3 + 33*(d*x^2 + c)^(5/2)*a^2*d^4 - 40*(d*x^2 + c)^(3/2)* a^2*c*d^4 + 15*sqrt(d*x^2 + c)*a^2*c^2*d^4)/(d^3*x^6))/d
Time = 7.17 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.36 \[ \int \frac {\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}}{x^7} \, dx=\frac {\sqrt {d\,x^2+c}\,\left (\frac {5\,a^2\,c^2\,d^3}{16}+\frac {7\,a\,b\,c^3\,d^2}{4}+\frac {b^2\,c^4\,d}{2}\right )-{\left (d\,x^2+c\right )}^{3/2}\,\left (\frac {5\,a^2\,c\,d^3}{6}+4\,a\,b\,c^2\,d^2+b^2\,c^3\,d\right )+{\left (d\,x^2+c\right )}^{5/2}\,\left (\frac {11\,a^2\,d^3}{16}+\frac {9\,a\,b\,c\,d^2}{4}+\frac {b^2\,c^2\,d}{2}\right )}{3\,c\,{\left (d\,x^2+c\right )}^2-3\,c^2\,\left (d\,x^2+c\right )-{\left (d\,x^2+c\right )}^3+c^3}+\left (2\,b\,d\,\left (a\,d-b\,c\right )+4\,b^2\,c\,d\right )\,\sqrt {d\,x^2+c}+\frac {b^2\,d\,{\left (d\,x^2+c\right )}^{3/2}}{3}+\frac {d\,\mathrm {atan}\left (\frac {d\,\sqrt {d\,x^2+c}\,\left (a^2\,d^2+12\,a\,b\,c\,d+8\,b^2\,c^2\right )\,5{}\mathrm {i}}{8\,\sqrt {c}\,\left (\frac {5\,a^2\,d^3}{8}+\frac {15\,a\,b\,c\,d^2}{2}+5\,b^2\,c^2\,d\right )}\right )\,\left (a^2\,d^2+12\,a\,b\,c\,d+8\,b^2\,c^2\right )\,5{}\mathrm {i}}{16\,\sqrt {c}} \]
((c + d*x^2)^(1/2)*((b^2*c^4*d)/2 + (5*a^2*c^2*d^3)/16 + (7*a*b*c^3*d^2)/4 ) - (c + d*x^2)^(3/2)*((5*a^2*c*d^3)/6 + b^2*c^3*d + 4*a*b*c^2*d^2) + (c + d*x^2)^(5/2)*((11*a^2*d^3)/16 + (b^2*c^2*d)/2 + (9*a*b*c*d^2)/4))/(3*c*(c + d*x^2)^2 - 3*c^2*(c + d*x^2) - (c + d*x^2)^3 + c^3) + (2*b*d*(a*d - b*c ) + 4*b^2*c*d)*(c + d*x^2)^(1/2) + (b^2*d*(c + d*x^2)^(3/2))/3 + (d*atan(( d*(c + d*x^2)^(1/2)*(a^2*d^2 + 8*b^2*c^2 + 12*a*b*c*d)*5i)/(8*c^(1/2)*((5* a^2*d^3)/8 + 5*b^2*c^2*d + (15*a*b*c*d^2)/2)))*(a^2*d^2 + 8*b^2*c^2 + 12*a *b*c*d)*5i)/(16*c^(1/2))