Integrand size = 24, antiderivative size = 258 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {\left (19 b^2 c^2-52 a b c d+32 a^2 d^2\right ) x \sqrt {c+d x^2}}{16 b^4}+\frac {d (7 b c-8 a d) x^3 \sqrt {c+d x^2}}{8 b^3}+\frac {2 d x^3 \left (c+d x^2\right )^{3/2}}{3 b^2}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}-\frac {\sqrt {a} (3 b c-8 a d) (b c-a d)^{3/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 b^5}+\frac {\left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{16 b^5 \sqrt {d}} \]
2/3*d*x^3*(d*x^2+c)^(3/2)/b^2-1/2*x^3*(d*x^2+c)^(5/2)/b/(b*x^2+a)-1/2*(-8* a*d+3*b*c)*(-a*d+b*c)^(3/2)*arctan(x*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^2+c)^(1 /2))*a^(1/2)/b^5+1/16*(-64*a^3*d^3+120*a^2*b*c*d^2-60*a*b^2*c^2*d+5*b^3*c^ 3)*arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))/b^5/d^(1/2)+1/16*(32*a^2*d^2-52*a*b* c*d+19*b^2*c^2)*x*(d*x^2+c)^(1/2)/b^4+1/8*d*(-8*a*d+7*b*c)*x^3*(d*x^2+c)^( 1/2)/b^3
Time = 10.20 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {b x \sqrt {c+d x^2} \left (33 b^2 c^2-108 a b c d+72 a^2 d^2+2 b d (13 b c-12 a d) x^2+8 b^2 d^2 x^4+\frac {24 a (b c-a d)^2}{a+b x^2}\right )+24 \sqrt {a} (b c-a d)^{3/2} (-3 b c+8 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )+\frac {3 \left (5 b^3 c^3-60 a b^2 c^2 d+120 a^2 b c d^2-64 a^3 d^3\right ) \log \left (d x+\sqrt {d} \sqrt {c+d x^2}\right )}{\sqrt {d}}}{48 b^5} \]
(b*x*Sqrt[c + d*x^2]*(33*b^2*c^2 - 108*a*b*c*d + 72*a^2*d^2 + 2*b*d*(13*b* c - 12*a*d)*x^2 + 8*b^2*d^2*x^4 + (24*a*(b*c - a*d)^2)/(a + b*x^2)) + 24*S qrt[a]*(b*c - a*d)^(3/2)*(-3*b*c + 8*a*d)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt [a]*Sqrt[c + d*x^2])] + (3*(5*b^3*c^3 - 60*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*Log[d*x + Sqrt[d]*Sqrt[c + d*x^2]])/Sqrt[d])/(48*b^5)
Time = 0.60 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {369, 443, 27, 443, 444, 27, 398, 224, 219, 291, 218}
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.
| \(\displaystyle \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 369 |
| \(\displaystyle \frac {\int \frac {x^2 \left (d x^2+c\right )^{3/2} \left (8 d x^2+3 c\right )}{b x^2+a}dx}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 443 |
| \(\displaystyle \frac {\frac {\int \frac {6 x^2 \sqrt {d x^2+c} \left (d (7 b c-8 a d) x^2+c (3 b c-4 a d)\right )}{b x^2+a}dx}{6 b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {x^2 \sqrt {d x^2+c} \left (d (7 b c-8 a d) x^2+c (3 b c-4 a d)\right )}{b x^2+a}dx}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 443 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {x^2 \left (d \left (19 b^2 c^2-52 a b d c+32 a^2 d^2\right ) x^2+c \left (12 b^2 c^2-37 a b d c+24 a^2 d^2\right )\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 444 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\int \frac {d \left (a c \left (19 b^2 c^2-52 a b d c+32 a^2 d^2\right )-\left (5 b^3 c^3-60 a b^2 d c^2+120 a^2 b d^2 c-64 a^3 d^3\right ) x^2\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{2 b d}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\int \frac {a c \left (19 b^2 c^2-52 a b d c+32 a^2 d^2\right )-\left (5 b^3 c^3-60 a b^2 d c^2+120 a^2 b d^2 c-64 a^3 d^3\right ) x^2}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 398 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\frac {8 a (3 b c-8 a d) (b c-a d)^2 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}-\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \int \frac {1}{\sqrt {d x^2+c}}dx}{b}}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\frac {8 a (3 b c-8 a d) (b c-a d)^2 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}-\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b}}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\frac {8 a (3 b c-8 a d) (b c-a d)^2 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}-\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\frac {8 a (3 b c-8 a d) (b c-a d)^2 \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b}-\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {x \sqrt {c+d x^2} \left (32 a^2 d^2-52 a b c d+19 b^2 c^2\right )}{2 b}-\frac {\frac {8 \sqrt {a} (3 b c-8 a d) (b c-a d)^{3/2} \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b}-\frac {\left (-64 a^3 d^3+120 a^2 b c d^2-60 a b^2 c^2 d+5 b^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}}{2 b}}{4 b}+\frac {d x^3 \sqrt {c+d x^2} (7 b c-8 a d)}{4 b}}{b}+\frac {4 d x^3 \left (c+d x^2\right )^{3/2}}{3 b}}{2 b}-\frac {x^3 \left (c+d x^2\right )^{5/2}}{2 b \left (a+b x^2\right )}\) |
-1/2*(x^3*(c + d*x^2)^(5/2))/(b*(a + b*x^2)) + ((4*d*x^3*(c + d*x^2)^(3/2) )/(3*b) + ((d*(7*b*c - 8*a*d)*x^3*Sqrt[c + d*x^2])/(4*b) + (((19*b^2*c^2 - 52*a*b*c*d + 32*a^2*d^2)*x*Sqrt[c + d*x^2])/(2*b) - ((8*Sqrt[a]*(3*b*c - 8*a*d)*(b*c - a*d)^(3/2)*ArcTan[(Sqrt[b*c - a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^ 2])])/b - ((5*b^3*c^3 - 60*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*Arc Tanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(b*Sqrt[d]))/(2*b))/(4*b))/b)/(2*b)
3.8.49.3.1 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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(2* b*(p + 1))), x] - Simp[e^2/(2*b*(p + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 1)*Simp[c*(m - 1) + d*(m + 2*q - 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 0 ] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*(g*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*g*(m + 2*(p + q + 1) + 1))), x] + Simp[1/(b*(m + 2* (p + q + 1) + 1)) Int[(g*x)^m*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(( b*e - a*f)*(m + 1) + b*e*2*(p + q + 1)) + (d*(b*e - a*f)*(m + 1) + f*2*q*(b *c - a*d) + b*e*d*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , g, m, p}, x] && GtQ[q, 0] && !(EqQ[q, 1] && SimplerQ[e + f*x^2, c + d*x^ 2])
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[f*g*(g*x)^(m - 1)*(a + b*x^2)^ (p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q + 1) + 1))), x] - Simp[g^2/ (b*d*(m + 2*(p + q + 1) + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2) ^q*Simp[a*f*c*(m - 1) + (a*f*d*(m + 2*q + 1) + b*(f*c*(m + 2*p + 1) - e*d*( m + 2*(p + q + 1) + 1)))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && GtQ[m, 1]
Time = 3.21 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.85
| method | result | size |
| pseudoelliptic | \(-\frac {-\frac {b \sqrt {d \,x^{2}+c}\, \left (8 b^{2} d^{2} x^{4}-24 x^{2} a b \,d^{2}+26 x^{2} b^{2} c d +72 a^{2} d^{2}-108 a b c d +33 b^{2} c^{2}\right ) x}{12}+\frac {\left (64 a^{3} d^{3}-120 a^{2} b c \,d^{2}+60 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )}{4 \sqrt {d}}+2 \left (a d -b c \right )^{2} a \left (-\frac {b \sqrt {d \,x^{2}+c}\, x}{b \,x^{2}+a}-\frac {\left (8 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}\right )}{4 b^{5}}\) | \(219\) |
| risch | \(\text {Expression too large to display}\) | \(1101\) |
| default | \(\text {Expression too large to display}\) | \(5359\) |
-1/4/b^5*(-1/12*b*(d*x^2+c)^(1/2)*(8*b^2*d^2*x^4-24*a*b*d^2*x^2+26*b^2*c*d *x^2+72*a^2*d^2-108*a*b*c*d+33*b^2*c^2)*x+1/4*(64*a^3*d^3-120*a^2*b*c*d^2+ 60*a*b^2*c^2*d-5*b^3*c^3)/d^(1/2)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2))+2*(a* d-b*c)^2*a*(-b*(d*x^2+c)^(1/2)*x/(b*x^2+a)-(8*a*d-3*b*c)/((a*d-b*c)*a)^(1/ 2)*arctanh((d*x^2+c)^(1/2)/x*a/((a*d-b*c)*a)^(1/2))))
Time = 1.69 (sec) , antiderivative size = 1697, normalized size of antiderivative = 6.58 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\text {Too large to display} \]
[-1/96*(3*(5*a*b^3*c^3 - 60*a^2*b^2*c^2*d + 120*a^3*b*c*d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2*d + 120*a^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt (d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) - 12*(3*a*b^2*c^2*d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2*d - 11*a*b^2*c*d^2 + 8*a^2*b*d^3)* x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2 *c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt( -a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 2*(8*b^4*d ^3*x^7 + 2*(13*b^4*c*d^2 - 8*a*b^3*d^3)*x^5 + (33*b^4*c^2*d - 82*a*b^3*c*d ^2 + 48*a^2*b^2*d^3)*x^3 + 3*(19*a*b^3*c^2*d - 52*a^2*b^2*c*d^2 + 32*a^3*b *d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a*b^5*d), -1/48*(3*(5*a*b^3*c^3 - 6 0*a^2*b^2*c^2*d + 120*a^3*b*c*d^2 - 64*a^4*d^3 + (5*b^4*c^3 - 60*a*b^3*c^2 *d + 120*a^2*b^2*c*d^2 - 64*a^3*b*d^3)*x^2)*sqrt(-d)*arctan(sqrt(-d)*x/sqr t(d*x^2 + c)) - 6*(3*a*b^2*c^2*d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*c^2 *d - 11*a*b^2*c*d^2 + 8*a^2*b*d^3)*x^2)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d + 8*a^2*d^2)*x^4 + a^2*c^2 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^2 - 4*((b*c - 2*a*d)*x^3 - a*c*x)*sqrt(-a*b*c + a^2*d)*sqrt(d*x^2 + c))/(b^2*x ^4 + 2*a*b*x^2 + a^2)) - (8*b^4*d^3*x^7 + 2*(13*b^4*c*d^2 - 8*a*b^3*d^3)*x ^5 + (33*b^4*c^2*d - 82*a*b^3*c*d^2 + 48*a^2*b^2*d^3)*x^3 + 3*(19*a*b^3*c^ 2*d - 52*a^2*b^2*c*d^2 + 32*a^3*b*d^3)*x)*sqrt(d*x^2 + c))/(b^6*d*x^2 + a* b^5*d), -1/96*(24*(3*a*b^2*c^2*d - 11*a^2*b*c*d^2 + 8*a^3*d^3 + (3*b^3*...
\[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^{4} \left (c + d x^{2}\right )^{\frac {5}{2}}}{\left (a + b x^{2}\right )^{2}}\, dx \]
\[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {5}{2}} x^{4}}{{\left (b x^{2} + a\right )}^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (222) = 444\).
Time = 0.35 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.00 \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, d^{2} x^{2}}{b^{2}} + \frac {13 \, b^{12} c d^{5} - 12 \, a b^{11} d^{6}}{b^{14} d^{4}}\right )} x^{2} + \frac {3 \, {\left (11 \, b^{12} c^{2} d^{4} - 36 \, a b^{11} c d^{5} + 24 \, a^{2} b^{10} d^{6}\right )}}{b^{14} d^{4}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (3 \, a b^{3} c^{3} \sqrt {d} - 14 \, a^{2} b^{2} c^{2} d^{\frac {3}{2}} + 19 \, a^{3} b c d^{\frac {5}{2}} - 8 \, a^{4} d^{\frac {7}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, \sqrt {a b c d - a^{2} d^{2}} b^{5}} - \frac {{\left (5 \, b^{3} c^{3} - 60 \, a b^{2} c^{2} d + 120 \, a^{2} b c d^{2} - 64 \, a^{3} d^{3}\right )} \log \left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2}\right )}{32 \, b^{5} \sqrt {d}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b^{3} c^{3} \sqrt {d} - 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} b^{2} c^{2} d^{\frac {3}{2}} + 5 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{3} b c d^{\frac {5}{2}} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{4} d^{\frac {7}{2}} - a b^{3} c^{4} \sqrt {d} + 2 \, a^{2} b^{2} c^{3} d^{\frac {3}{2}} - a^{3} b c^{2} d^{\frac {5}{2}}}{{\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )} b^{5}} \]
1/48*(2*(4*d^2*x^2/b^2 + (13*b^12*c*d^5 - 12*a*b^11*d^6)/(b^14*d^4))*x^2 + 3*(11*b^12*c^2*d^4 - 36*a*b^11*c*d^5 + 24*a^2*b^10*d^6)/(b^14*d^4))*sqrt( d*x^2 + c)*x + 1/2*(3*a*b^3*c^3*sqrt(d) - 14*a^2*b^2*c^2*d^(3/2) + 19*a^3* b*c*d^(5/2) - 8*a^4*d^(7/2))*arctan(1/2*((sqrt(d)*x - sqrt(d*x^2 + c))^2*b - b*c + 2*a*d)/sqrt(a*b*c*d - a^2*d^2))/(sqrt(a*b*c*d - a^2*d^2)*b^5) - 1 /32*(5*b^3*c^3 - 60*a*b^2*c^2*d + 120*a^2*b*c*d^2 - 64*a^3*d^3)*log((sqrt( d)*x - sqrt(d*x^2 + c))^2)/(b^5*sqrt(d)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^ 2*a*b^3*c^3*sqrt(d) - 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*b^2*c^2*d^(3/2 ) + 5*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^3*b*c*d^(5/2) - 2*(sqrt(d)*x - sqr t(d*x^2 + c))^2*a^4*d^(7/2) - a*b^3*c^4*sqrt(d) + 2*a^2*b^2*c^3*d^(3/2) - a^3*b*c^2*d^(5/2))/(((sqrt(d)*x - sqrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sq rt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a*d + b*c^2)*b^5)
Timed out. \[ \int \frac {x^4 \left (c+d x^2\right )^{5/2}}{\left (a+b x^2\right )^2} \, dx=\int \frac {x^4\,{\left (d\,x^2+c\right )}^{5/2}}{{\left (b\,x^2+a\right )}^2} \,d x \]