\(\int \frac {x^8}{(a+b x^2) (c+d x^2)^{5/2}} \, dx\) [980]

Optimal result

Integrand size = 24, antiderivative size = 220 \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=-\frac {c x^5}{3 d (b c-a d) \left (c+d x^2\right )^{3/2}}-\frac {c (5 b c-8 a d) x^3}{3 d^2 (b c-a d)^2 \sqrt {c+d x^2}}+\frac {\left (5 b^2 c^2-8 a b c d+a^2 d^2\right ) x \sqrt {c+d x^2}}{2 b d^3 (b c-a d)^2}+\frac {a^{7/2} \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b^2 (b c-a d)^{5/2}}-\frac {(5 b c+2 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 b^2 d^{7/2}} \] Output:

-1/3*c*x^5/d/(-a*d+b*c)/(d*x^2+c)^(3/2)-1/3*c*(-8*a*d+5*b*c)*x^3/d^2/(-a*d 
+b*c)^2/(d*x^2+c)^(1/2)+1/2*(a^2*d^2-8*a*b*c*d+5*b^2*c^2)*x*(d*x^2+c)^(1/2 
)/b/d^3/(-a*d+b*c)^2+a^(7/2)*arctan((-a*d+b*c)^(1/2)*x/a^(1/2)/(d*x^2+c)^( 
1/2))/b^2/(-a*d+b*c)^(5/2)-1/2*(2*a*d+5*b*c)*arctanh(d^(1/2)*x/(d*x^2+c)^( 
1/2))/b^2/d^(7/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(458\) vs. \(2(220)=440\).

Time = 3.27 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.08 \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\frac {\frac {b \sqrt {d} x \left (3 a^2 d^2 \left (c+d x^2\right )^2-2 a b c d \left (12 c^2+16 c d x^2+3 d^2 x^4\right )+b^2 c^2 \left (15 c^2+20 c d x^2+3 d^2 x^4\right )\right )}{(b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {6 a^{5/2} d^{5/2} \left (\sqrt {b} \sqrt {c}+\sqrt {b c-a d}\right ) \sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d-2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2}}+\frac {6 a^{5/2} d^{5/2} \left (-\sqrt {b} \sqrt {c}+\sqrt {b c-a d}\right ) \sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} \arctan \left (\frac {\sqrt {2 b c-a d+2 \sqrt {b} \sqrt {c} \sqrt {b c-a d}} x}{\sqrt {a} \left (\sqrt {c}-\sqrt {c+d x^2}\right )}\right )}{(b c-a d)^{5/2}}+6 (5 b c+2 a d) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c}-\sqrt {c+d x^2}}\right )}{6 b^2 d^{7/2}} \] Input:

Integrate[x^8/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

((b*Sqrt[d]*x*(3*a^2*d^2*(c + d*x^2)^2 - 2*a*b*c*d*(12*c^2 + 16*c*d*x^2 + 
3*d^2*x^4) + b^2*c^2*(15*c^2 + 20*c*d*x^2 + 3*d^2*x^4)))/((b*c - a*d)^2*(c 
 + d*x^2)^(3/2)) + (6*a^(5/2)*d^(5/2)*(Sqrt[b]*Sqrt[c] + Sqrt[b*c - a*d])* 
Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*ArcTan[(Sqrt[2*b*c - 
 a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt[c] - Sqrt[c + 
d*x^2]))])/(b*c - a*d)^(5/2) + (6*a^(5/2)*d^(5/2)*(-(Sqrt[b]*Sqrt[c]) + Sq 
rt[b*c - a*d])*Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*ArcTa 
n[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]*x)/(Sqrt[a]*(Sqrt 
[c] - Sqrt[c + d*x^2]))])/(b*c - a*d)^(5/2) + 6*(5*b*c + 2*a*d)*ArcTanh[(S 
qrt[d]*x)/(Sqrt[c] - Sqrt[c + d*x^2])])/(6*b^2*d^(7/2))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {372, 440, 27, 444, 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.

\(\displaystyle \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 372

\(\displaystyle \frac {\int \frac {x^4 \left ((5 b c-3 a d) x^2+5 a c\right )}{\left (b x^2+a\right ) \left (d x^2+c\right )^{3/2}}dx}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 440

\(\displaystyle \frac {-\frac {\int -\frac {3 x^2 \left (\left (a^2 d^2+b c (5 b c-8 a d)\right ) x^2+a c (5 b c-8 a d)\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 \int \frac {x^2 \left (\left (a^2 d^2+b c (5 b c-8 a d)\right ) x^2+a c (5 b c-8 a d)\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 444

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\int \frac {(b c-a d)^2 (5 b c+2 a d) x^2+a c \left (a^2 d^2+b c (5 b c-8 a d)\right )}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 398

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\frac {(b c-a d)^2 (2 a d+5 b c) \int \frac {1}{\sqrt {d x^2+c}}dx}{b}-\frac {2 a^4 d^3 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\frac {(b c-a d)^2 (2 a d+5 b c) \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b}-\frac {2 a^4 d^3 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\frac {(b c-a d)^2 (2 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}-\frac {2 a^4 d^3 \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b}}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\frac {(b c-a d)^2 (2 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}-\frac {2 a^4 d^3 \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b}}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {3 \left (\frac {1}{2} x \sqrt {c+d x^2} \left (\frac {a^2 d}{b}+\frac {c (5 b c-8 a d)}{d}\right )-\frac {\frac {(b c-a d)^2 (2 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{b \sqrt {d}}-\frac {2 a^{7/2} d^3 \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{b \sqrt {b c-a d}}}{2 b d}\right )}{d (b c-a d)}-\frac {c x^3 (5 b c-8 a d)}{d \sqrt {c+d x^2} (b c-a d)}}{3 d (b c-a d)}-\frac {c x^5}{3 d \left (c+d x^2\right )^{3/2} (b c-a d)}\)

Input:

Int[x^8/((a + b*x^2)*(c + d*x^2)^(5/2)),x]
 

Output:

-1/3*(c*x^5)/(d*(b*c - a*d)*(c + d*x^2)^(3/2)) + (-((c*(5*b*c - 8*a*d)*x^3 
)/(d*(b*c - a*d)*Sqrt[c + d*x^2])) + (3*((((a^2*d)/b + (c*(5*b*c - 8*a*d)) 
/d)*x*Sqrt[c + d*x^2])/2 - ((-2*a^(7/2)*d^3*ArcTan[(Sqrt[b*c - a*d]*x)/(Sq 
rt[a]*Sqrt[c + d*x^2])])/(b*Sqrt[b*c - a*d]) + ((b*c - a*d)^2*(5*b*c + 2*a 
*d)*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(b*Sqrt[d]))/(2*b*d)))/(d*(b*c - 
 a*d)))/(3*d*(b*c - a*d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 219

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])
 

rule 224

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]
 

rule 291

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]
 

rule 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a 
, b, c, d, e, m, 2, p, q, x]
 

rule 398

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]
 

rule 440

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + 
 b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ 
g^2/(2*b*(b*c - a*d)*(p + 1))   Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + 
d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c 
*f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && 
 LtQ[p, -1] && GtQ[m, 1]
 

rule 444

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]
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {c^{2} \left (-\frac {2 a^{4} \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\left (a d -b c \right )^{2} b^{2} c^{2} \sqrt {\left (a d -b c \right ) a}}+\frac {-\sqrt {x^{2} d +c}\, b x +\frac {\left (2 a d +5 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}}{x \sqrt {d}}\right )}{\sqrt {d}}}{b^{2} c^{2} d^{3}}+\frac {2 \left (3 a d -2 b c \right ) x}{\left (a d -b c \right )^{2} d^{3} \sqrt {x^{2} d +c}}+\frac {2 x^{3}}{3 d^{2} \left (a d -b c \right ) \left (x^{2} d +c \right )^{\frac {3}{2}}}\right )}{2}\)	\(182\)
risch	\(\text {Expression too large to display}\)	\(1019\)
default	\(\text {Expression too large to display}\)	\(1657\)

Input:

int(x^8/(b*x^2+a)/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*c^2*(-2/(a*d-b*c)^2*a^4/b^2/c^2/((a*d-b*c)*a)^(1/2)*arctanh((d*x^2+c) 
^(1/2)/x*a/((a*d-b*c)*a)^(1/2))+(-(d*x^2+c)^(1/2)*b*x+(2*a*d+5*b*c)/d^(1/2 
)*arctanh((d*x^2+c)^(1/2)/x/d^(1/2)))/b^2/c^2/d^3+2*(3*a*d-2*b*c)*x/(a*d-b 
*c)^2/d^3/(d*x^2+c)^(1/2)+2/3/d^2/(a*d-b*c)/(d*x^2+c)^(3/2)*x^3)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 497 vs. \(2 (190) = 380\).

Time = 1.14 (sec) , antiderivative size = 2188, normalized size of antiderivative = 9.95 \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^8/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*(5*b^3*c^5 - 8*a*b^2*c^4*d + a^2*b*c^3*d^2 + 2*a^3*c^2*d^3 + (5*b 
^3*c^3*d^2 - 8*a*b^2*c^2*d^3 + a^2*b*c*d^4 + 2*a^3*d^5)*x^4 + 2*(5*b^3*c^4 
*d - 8*a*b^2*c^3*d^2 + a^2*b*c^2*d^3 + 2*a^3*c*d^4)*x^2)*sqrt(d)*log(-2*d* 
x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 3*(a^3*d^6*x^4 + 2*a^3*c*d^5*x^2 
+ a^3*c^2*d^4)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 
 2*a^2*d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a* 
d)))/(b^2*x^4 + 2*a*b*x^2 + a^2)) + 2*(3*(b^3*c^2*d^3 - 2*a*b^2*c*d^4 + a^ 
2*b*d^5)*x^5 + 2*(10*b^3*c^3*d^2 - 16*a*b^2*c^2*d^3 + 3*a^2*b*c*d^4)*x^3 + 
 3*(5*b^3*c^4*d - 8*a*b^2*c^3*d^2 + a^2*b*c^2*d^3)*x)*sqrt(d*x^2 + c))/(b^ 
4*c^4*d^4 - 2*a*b^3*c^3*d^5 + a^2*b^2*c^2*d^6 + (b^4*c^2*d^6 - 2*a*b^3*c*d 
^7 + a^2*b^2*d^8)*x^4 + 2*(b^4*c^3*d^5 - 2*a*b^3*c^2*d^6 + a^2*b^2*c*d^7)* 
x^2), 1/12*(6*(5*b^3*c^5 - 8*a*b^2*c^4*d + a^2*b*c^3*d^2 + 2*a^3*c^2*d^3 + 
 (5*b^3*c^3*d^2 - 8*a*b^2*c^2*d^3 + a^2*b*c*d^4 + 2*a^3*d^5)*x^4 + 2*(5*b^ 
3*c^4*d - 8*a*b^2*c^3*d^2 + a^2*b*c^2*d^3 + 2*a^3*c*d^4)*x^2)*sqrt(-d)*arc 
tan(sqrt(-d)*x/sqrt(d*x^2 + c)) + 3*(a^3*d^6*x^4 + 2*a^3*c*d^5*x^2 + a^3*c 
^2*d^4)*sqrt(-a/(b*c - a*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^2*c^2 - 3*a*b*c*d + 2*a^2* 
d^2)*x^3 - (a*b*c^2 - a^2*c*d)*x)*sqrt(d*x^2 + c)*sqrt(-a/(b*c - a*d)))/(b 
^2*x^4 + 2*a*b*x^2 + a^2)) + 2*(3*(b^3*c^2*d^3 - 2*a*b^2*c*d^4 + a^2*b*...
 

Sympy [F]

\[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{8}}{\left (a + b x^{2}\right ) \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**8/(b*x**2+a)/(d*x**2+c)**(5/2),x)
 

Output:

Integral(x**8/((a + b*x**2)*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{8}}{{\left (b x^{2} + a\right )} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^8/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x^8/((b*x^2 + a)*(d*x^2 + c)^(5/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^8/(b*x^2+a)/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^8}{\left (b\,x^2+a\right )\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^8/((a + b*x^2)*(c + d*x^2)^(5/2)),x)
 

Output:

int(x^8/((a + b*x^2)*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.50 (sec) , antiderivative size = 1708, normalized size of antiderivative = 7.76 \[ \int \frac {x^8}{\left (a+b x^2\right ) \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^8/(b*x^2+a)/(d*x^2+c)^(5/2),x)
 

Output:

(6*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 
 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**3*c**2*d* 
*4 + 12*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b 
*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**3*c* 
d**5*x**2 + 6*sqrt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a 
*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a 
**3*d**6*x**4 + 6*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqrt( 
a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)* 
a**3*c**2*d**4 + 12*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)*sqr 
t(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x 
)*a**3*c*d**5*x**2 + 6*sqrt(a)*sqrt(a*d - b*c)*log(sqrt(2*sqrt(d)*sqrt(a)* 
sqrt(a*d - b*c) - 2*a*d + b*c) + sqrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b 
)*x)*a**3*d**6*x**4 - 6*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt 
(a*d - b*c) + 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**3*c* 
*2*d**4 - 12*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) 
 + 2*sqrt(d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**3*c*d**5*x**2 - 
 6*sqrt(a)*sqrt(a*d - b*c)*log(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) + 2*sqrt( 
d)*sqrt(c + d*x**2)*b*x + 2*a*d + 2*b*d*x**2)*a**3*d**6*x**4 + 6*sqrt(c + 
d*x**2)*a**3*b*c**2*d**4*x + 12*sqrt(c + d*x**2)*a**3*b*c*d**5*x**3 + 6*sq 
rt(c + d*x**2)*a**3*b*d**6*x**5 - 54*sqrt(c + d*x**2)*a**2*b**2*c**3*d*...