\(\int \frac {x^4}{(a+b x^2)^2 (c+d x^2)^{5/2}} \, dx\) [1032]

Optimal result

Integrand size = 24, antiderivative size = 174 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {(2 b c+3 a d) x}{6 b (b c-a d)^2 \left (c+d x^2\right )^{3/2}}+\frac {a x}{2 b (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2}}+\frac {(4 b c+11 a d) x}{6 (b c-a d)^3 \sqrt {c+d x^2}}-\frac {\sqrt {a} (3 b c+2 a d) \arctan \left (\frac {\sqrt {b c-a d} x}{\sqrt {a} \sqrt {c+d x^2}}\right )}{2 (b c-a d)^{7/2}} \] Output:

1/6*(3*a*d+2*b*c)*x/b/(-a*d+b*c)^2/(d*x^2+c)^(3/2)+1/2*a*x/b/(-a*d+b*c)/(b 
*x^2+a)/(d*x^2+c)^(3/2)+1/6*(11*a*d+4*b*c)*x/(-a*d+b*c)^3/(d*x^2+c)^(1/2)- 
1/2*a^(1/2)*(2*a*d+3*b*c)*arctan((-a*d+b*c)^(1/2)*x/a^(1/2)/(d*x^2+c)^(1/2 
))/(-a*d+b*c)^(7/2)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1715\) vs. \(2(174)=348\).

Time = 15.47 (sec) , antiderivative size = 1715, normalized size of antiderivative = 9.86 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[x^4/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]
 

Output:

((x*(-65536*c^(17/2) - 278528*c^(15/2)*d*x^2 - 487424*c^(13/2)*d^2*x^4 - 4 
52608*c^(11/2)*d^3*x^6 - 239360*c^(9/2)*d^4*x^8 - 71808*c^(7/2)*d^5*x^10 - 
 11424*c^(5/2)*d^6*x^12 - 816*c^(3/2)*d^7*x^14 - 17*Sqrt[c]*d^8*x^16 + 655 
36*c^8*Sqrt[c + d*x^2] + 245760*c^7*d*x^2*Sqrt[c + d*x^2] + 372736*c^6*d^2 
*x^4*Sqrt[c + d*x^2] + 292864*c^5*d^3*x^6*Sqrt[c + d*x^2] + 126720*c^4*d^4 
*x^8*Sqrt[c + d*x^2] + 29568*c^3*d^5*x^10*Sqrt[c + d*x^2] + 3360*c^2*d^6*x 
^12*Sqrt[c + d*x^2] + 144*c*d^7*x^14*Sqrt[c + d*x^2] + d^8*x^16*Sqrt[c + d 
*x^2])*(2*b^2*c*x^2*(3*c + 2*d*x^2) + 2*a^2*d*(3*c + 4*d*x^2) + a*b*(9*c^2 
 + 16*c*d*x^2 + 11*d^2*x^4)))/(3*(b*c - a*d)^3*(a + b*x^2)*(c + d*x^2)*(65 
536*c^9 + 311296*c^8*d*x^2 + 618496*c^7*d^2*x^4 + 665600*c^6*d^3*x^6 + 419 
584*c^5*d^4*x^8 + 156288*c^4*d^5*x^10 + 32928*c^3*d^6*x^12 + 3504*c^2*d^7* 
x^14 + 145*c*d^8*x^16 + d^9*x^18 - 65536*c^(17/2)*Sqrt[c + d*x^2] - 278528 
*c^(15/2)*d*x^2*Sqrt[c + d*x^2] - 487424*c^(13/2)*d^2*x^4*Sqrt[c + d*x^2] 
- 452608*c^(11/2)*d^3*x^6*Sqrt[c + d*x^2] - 239360*c^(9/2)*d^4*x^8*Sqrt[c 
+ d*x^2] - 71808*c^(7/2)*d^5*x^10*Sqrt[c + d*x^2] - 11424*c^(5/2)*d^6*x^12 
*Sqrt[c + d*x^2] - 816*c^(3/2)*d^7*x^14*Sqrt[c + d*x^2] - 17*Sqrt[c]*d^8*x 
^16*Sqrt[c + d*x^2])) + (3*Sqrt[a]*b*c*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)^3*Sqrt[2*b*c - a*d - 2*Sqrt[b]*Sqrt[c]*Sqrt[b*c - a*d]]) + (3*Sqr 
t[a]*b^(3/2)*c^(3/2)*ArcTan[(Sqrt[2*b*c - a*d + 2*Sqrt[b]*Sqrt[c]*Sqrt[...
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {372, 402, 27, 402, 27, 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.

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 402

\(\displaystyle \frac {a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\frac {b \left (\frac {\int \frac {3 a c (3 b c+2 a d)}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{c (b c-a d)}-\frac {x (11 a d+4 b c)}{\sqrt {c+d x^2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {x (3 a d+2 b c)}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 b (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\frac {b \left (\frac {3 a (2 a d+3 b c) \int \frac {1}{\left (b x^2+a\right ) \sqrt {d x^2+c}}dx}{b c-a d}-\frac {x (11 a d+4 b c)}{\sqrt {c+d x^2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {x (3 a d+2 b c)}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 b (b c-a d)}\)

\(\Big \downarrow \) 291

\(\displaystyle \frac {a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\frac {b \left (\frac {3 a (2 a d+3 b c) \int \frac {1}{a-\frac {(a d-b c) x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{b c-a d}-\frac {x (11 a d+4 b c)}{\sqrt {c+d x^2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {x (3 a d+2 b c)}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 b (b c-a d)}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a x}{2 b \left (a+b x^2\right ) \left (c+d x^2\right )^{3/2} (b c-a d)}-\frac {\frac {b \left (\frac {3 \sqrt {a} (2 a d+3 b c) \arctan \left (\frac {x \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^2}}\right )}{(b c-a d)^{3/2}}-\frac {x (11 a d+4 b c)}{\sqrt {c+d x^2} (b c-a d)}\right )}{3 (b c-a d)}-\frac {x (3 a d+2 b c)}{3 \left (c+d x^2\right )^{3/2} (b c-a d)}}{2 b (b c-a d)}\)

Input:

Int[x^4/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x]
 

Output:

(a*x)/(2*b*(b*c - a*d)*(a + b*x^2)*(c + d*x^2)^(3/2)) - (-1/3*((2*b*c + 3* 
a*d)*x)/((b*c - a*d)*(c + d*x^2)^(3/2)) + (b*(-(((4*b*c + 11*a*d)*x)/((b*c 
 - a*d)*Sqrt[c + d*x^2])) + (3*Sqrt[a]*(3*b*c + 2*a*d)*ArcTan[(Sqrt[b*c - 
a*d]*x)/(Sqrt[a]*Sqrt[c + d*x^2])])/(b*c - a*d)^(3/2)))/(3*(b*c - a*d)))/( 
2*b*(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 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 402

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

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {a \left (\frac {\left (b \,x^{2}+a \right ) \left (a d +\frac {3 b c}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x^{2} d +c}\, a}{x \sqrt {\left (a d -b c \right ) a}}\right )}{\sqrt {\left (a d -b c \right ) a}}-\frac {\sqrt {x^{2} d +c}\, b x}{2}\right )}{\left (b \,x^{2}+a \right ) \left (a d -b c \right )^{3}}-\frac {\left (a d +b c \right ) x}{\left (a d -b c \right )^{3} \sqrt {x^{2} d +c}}-\frac {d \,x^{3}}{3 \left (a d -b c \right )^{2} \left (x^{2} d +c \right )^{\frac {3}{2}}}\)	\(146\)
default	\(\text {Expression too large to display}\)	\(3522\)

Input:

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

a*((b*x^2+a)*(a*d+3/2*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))-1/2*(d*x^2+c)^(1/2)*b*x)/(b*x^2+a)/(a*d-b*c)^3-(a*d+ 
b*c)*x/(a*d-b*c)^3/(d*x^2+c)^(1/2)-1/3*d/(a*d-b*c)^2/(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. 458 vs. \(2 (150) = 300\).

Time = 0.53 (sec) , antiderivative size = 1008, normalized size of antiderivative = 5.79 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="fricas")
 

Output:

[-1/24*(3*((3*b^2*c*d^2 + 2*a*b*d^3)*x^6 + 3*a*b*c^3 + 2*a^2*c^2*d + (6*b^ 
2*c^2*d + 7*a*b*c*d^2 + 2*a^2*d^3)*x^4 + (3*b^2*c^3 + 8*a*b*c^2*d + 4*a^2* 
c*d^2)*x^2)*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)) - 4*((4*b^2*c*d + 11*a*b*d^2)*x^5 + 2*(3*b^ 
2*c^2 + 8*a*b*c*d + 4*a^2*d^2)*x^3 + 3*(3*a*b*c^2 + 2*a^2*c*d)*x)*sqrt(d*x 
^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b*c^3*d^2 - a^4*c^2*d^3 + (b 
^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - a^3*b*d^5)*x^6 + (2*b^4*c 
^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3*b*c*d^4 - a^4*d^5)*x^4 + 
(b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + 5*a^3*b*c^2*d^3 - 2*a^4*c*d^4 
)*x^2), 1/12*(3*((3*b^2*c*d^2 + 2*a*b*d^3)*x^6 + 3*a*b*c^3 + 2*a^2*c^2*d + 
 (6*b^2*c^2*d + 7*a*b*c*d^2 + 2*a^2*d^3)*x^4 + (3*b^2*c^3 + 8*a*b*c^2*d + 
4*a^2*c*d^2)*x^2)*sqrt(a/(b*c - a*d))*arctan(-1/2*((b*c - 2*a*d)*x^2 - a*c 
)*sqrt(d*x^2 + c)*sqrt(a/(b*c - a*d))/(a*d*x^3 + a*c*x)) + 2*((4*b^2*c*d + 
 11*a*b*d^2)*x^5 + 2*(3*b^2*c^2 + 8*a*b*c*d + 4*a^2*d^2)*x^3 + 3*(3*a*b*c^ 
2 + 2*a^2*c*d)*x)*sqrt(d*x^2 + c))/(a*b^3*c^5 - 3*a^2*b^2*c^4*d + 3*a^3*b* 
c^3*d^2 - a^4*c^2*d^3 + (b^4*c^3*d^2 - 3*a*b^3*c^2*d^3 + 3*a^2*b^2*c*d^4 - 
 a^3*b*d^5)*x^6 + (2*b^4*c^4*d - 5*a*b^3*c^3*d^2 + 3*a^2*b^2*c^2*d^3 + a^3 
*b*c*d^4 - a^4*d^5)*x^4 + (b^4*c^5 - a*b^3*c^4*d - 3*a^2*b^2*c^3*d^2 + ...
 

Sympy [F]

\[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x^{2}\right )^{2} \left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4/(b*x**2+a)**2/(d*x**2+c)**(5/2),x)
 

Output:

Integral(x**4/((a + b*x**2)**2*(c + d*x**2)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {x^{4}}{{\left (b x^{2} + a\right )}^{2} {\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="maxima")
 

Output:

integrate(x^4/((b*x^2 + a)^2*(d*x^2 + c)^(5/2)), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (150) = 300\).

Time = 0.41 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.41 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\frac {{\left (\frac {2 \, {\left (b^{4} c^{5} d^{2} - a b^{3} c^{4} d^{3} - 3 \, a^{2} b^{2} c^{3} d^{4} + 5 \, a^{3} b c^{2} d^{5} - 2 \, a^{4} c d^{6}\right )} x^{2}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}} + \frac {3 \, {\left (b^{4} c^{6} d - 2 \, a b^{3} c^{5} d^{2} + 2 \, a^{3} b c^{3} d^{4} - a^{4} c^{2} d^{5}\right )}}{b^{6} c^{7} d - 6 \, a b^{5} c^{6} d^{2} + 15 \, a^{2} b^{4} c^{5} d^{3} - 20 \, a^{3} b^{3} c^{4} d^{4} + 15 \, a^{4} b^{2} c^{3} d^{5} - 6 \, a^{5} b c^{2} d^{6} + a^{6} c d^{7}}\right )} x}{3 \, {\left (d x^{2} + c\right )}^{\frac {3}{2}}} + \frac {{\left (3 \, a b c \sqrt {d} + 2 \, a^{2} d^{\frac {3}{2}}\right )} \arctan \left (\frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{2 \, {\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} - \frac {{\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a b c \sqrt {d} - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a^{2} d^{\frac {3}{2}} - a b c^{2} \sqrt {d}}{{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left ({\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x - \sqrt {d x^{2} + c}\right )}^{2} a d + b c^{2}\right )}} \] Input:

integrate(x^4/(b*x^2+a)^2/(d*x^2+c)^(5/2),x, algorithm="giac")
 

Output:

1/3*(2*(b^4*c^5*d^2 - a*b^3*c^4*d^3 - 3*a^2*b^2*c^3*d^4 + 5*a^3*b*c^2*d^5 
- 2*a^4*c*d^6)*x^2/(b^6*c^7*d - 6*a*b^5*c^6*d^2 + 15*a^2*b^4*c^5*d^3 - 20* 
a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3*d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7) + 3*(b 
^4*c^6*d - 2*a*b^3*c^5*d^2 + 2*a^3*b*c^3*d^4 - a^4*c^2*d^5)/(b^6*c^7*d - 6 
*a*b^5*c^6*d^2 + 15*a^2*b^4*c^5*d^3 - 20*a^3*b^3*c^4*d^4 + 15*a^4*b^2*c^3* 
d^5 - 6*a^5*b*c^2*d^6 + a^6*c*d^7))*x/(d*x^2 + c)^(3/2) + 1/2*(3*a*b*c*sqr 
t(d) + 2*a^2*d^(3/2))*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))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^ 
2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) - ((sqrt(d)*x - sqrt(d*x^2 + c))^2*a 
*b*c*sqrt(d) - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*a^2*d^(3/2) - a*b*c^2*sqr 
t(d))/((b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*((sqrt(d)*x - s 
qrt(d*x^2 + c))^4*b - 2*(sqrt(d)*x - sqrt(d*x^2 + c))^2*b*c + 4*(sqrt(d)*x 
 - sqrt(d*x^2 + c))^2*a*d + b*c^2))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx=\int \frac {x^4}{{\left (b\,x^2+a\right )}^2\,{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(5/2)),x)
 

Output:

int(x^4/((a + b*x^2)^2*(c + d*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 3695, normalized size of antiderivative = 21.24 \[ \int \frac {x^4}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(b*x^2+a)^2/(d*x^2+c)^(5/2),x)
 

Output:

(24*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 
**3 + 48*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**4*x**2 + 24*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**5*x**4 + 42*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**2*b*c**3*d**2 + 108*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**2*b*c**2*d**3*x**2 + 90*sqrt(a)*sqrt(a*d - b*c)*log( - sq 
rt(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**2*b*c*d**4*x**4 + 24*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)*sqr 
t(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a**2*b*d**5*x**6 + 9*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*b**2*c**4*d + 60*sqrt(a)*sqrt(a 
*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2*a*d + b*c) + s 
qrt(b)*sqrt(c + d*x**2) + sqrt(d)*sqrt(b)*x)*a*b**2*c**3*d**2*x**2 + 93*sq 
rt(a)*sqrt(a*d - b*c)*log( - sqrt(2*sqrt(d)*sqrt(a)*sqrt(a*d - b*c) - 2...