\(\int \frac {x^{5/2}}{(a+b x^2)^2 (c+d x^2)^2} \, dx\) [814]

Optimal result

Integrand size = 24, antiderivative size = 465 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=-\frac {d x^{3/2}}{(b c-a d)^2 \left (c+d x^2\right )}-\frac {x^{3/2}}{2 (b c-a d) \left (a+b x^2\right ) \left (c+d x^2\right )}-\frac {\sqrt [4]{b} (3 b c+5 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}+\frac {\sqrt [4]{b} (3 b c+5 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}+\frac {\sqrt [4]{d} (5 b c+3 a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)^3}-\frac {\sqrt [4]{d} (5 b c+3 a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)^3}-\frac {\sqrt [4]{b} (3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)^3}+\frac {\sqrt [4]{d} (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)^3} \] Output:

-d*x^(3/2)/(-a*d+b*c)^2/(d*x^2+c)-1/2*x^(3/2)/(-a*d+b*c)/(b*x^2+a)/(d*x^2+ 
c)-1/8*b^(1/4)*(5*a*d+3*b*c)*arctan(1-2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^( 
1/2)/a^(1/4)/(-a*d+b*c)^3+1/8*b^(1/4)*(5*a*d+3*b*c)*arctan(1+2^(1/2)*b^(1/ 
4)*x^(1/2)/a^(1/4))*2^(1/2)/a^(1/4)/(-a*d+b*c)^3+1/8*d^(1/4)*(3*a*d+5*b*c) 
*arctan(1-2^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/c^(1/4)/(-a*d+b*c)^3-1/ 
8*d^(1/4)*(3*a*d+5*b*c)*arctan(1+2^(1/2)*d^(1/4)*x^(1/2)/c^(1/4))*2^(1/2)/ 
c^(1/4)/(-a*d+b*c)^3-1/8*b^(1/4)*(5*a*d+3*b*c)*arctanh(2^(1/2)*a^(1/4)*b^( 
1/4)*x^(1/2)/(a^(1/2)+b^(1/2)*x))*2^(1/2)/a^(1/4)/(-a*d+b*c)^3+1/8*d^(1/4) 
*(3*a*d+5*b*c)*arctanh(2^(1/2)*c^(1/4)*d^(1/4)*x^(1/2)/(c^(1/2)+d^(1/2)*x) 
)*2^(1/2)/c^(1/4)/(-a*d+b*c)^3
 

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 340, normalized size of antiderivative = 0.73 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\frac {1}{8} \left (-\frac {4 x^{3/2} \left (a d+b \left (c+2 d x^2\right )\right )}{(b c-a d)^2 \left (a+b x^2\right ) \left (c+d x^2\right )}+\frac {\sqrt {2} \sqrt [4]{b} (3 b c+5 a d) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{a} (-b c+a d)^3}+\frac {\sqrt {2} \sqrt [4]{d} (5 b c+3 a d) \arctan \left (\frac {\sqrt {c}-\sqrt {d} x}{\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}\right )}{\sqrt [4]{c} (b c-a d)^3}+\frac {\sqrt {2} \sqrt [4]{b} (3 b c+5 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{a} (-b c+a d)^3}+\frac {\sqrt {2} \sqrt [4]{d} (5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}}{\sqrt {c}+\sqrt {d} x}\right )}{\sqrt [4]{c} (b c-a d)^3}\right ) \] Input:

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

Output:

((-4*x^(3/2)*(a*d + b*(c + 2*d*x^2)))/((b*c - a*d)^2*(a + b*x^2)*(c + d*x^ 
2)) + (Sqrt[2]*b^(1/4)*(3*b*c + 5*a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[ 
2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(a^(1/4)*(-(b*c) + a*d)^3) + (Sqrt[2]*d^(1/4 
)*(5*b*c + 3*a*d)*ArcTan[(Sqrt[c] - Sqrt[d]*x)/(Sqrt[2]*c^(1/4)*d^(1/4)*Sq 
rt[x])])/(c^(1/4)*(b*c - a*d)^3) + (Sqrt[2]*b^(1/4)*(3*b*c + 5*a*d)*ArcTan 
h[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(a^(1/4)*(-(b* 
c) + a*d)^3) + (Sqrt[2]*d^(1/4)*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[2]*c^(1/4)*d 
^(1/4)*Sqrt[x])/(Sqrt[c] + Sqrt[d]*x)])/(c^(1/4)*(b*c - a*d)^3))/8
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 638, normalized size of antiderivative = 1.37, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {368, 971, 1049, 27, 1054, 2009}

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.

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

\(\Big \downarrow \) 368

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

\(\Big \downarrow \) 971

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

\(\Big \downarrow \) 1049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1054

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {\frac {-\frac {\sqrt [4]{b} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (5 a d+3 b c)}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {\sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (5 a d+3 b c)}{2 \sqrt {2} \sqrt [4]{a} (b c-a d)}+\frac {\sqrt [4]{d} (3 a d+5 b c) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}-\frac {\sqrt [4]{d} (3 a d+5 b c) \arctan \left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt {x}}{\sqrt [4]{c}}+1\right )}{2 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {\sqrt [4]{b} (5 a d+3 b c) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {\sqrt [4]{b} (5 a d+3 b c) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} \sqrt [4]{a} (b c-a d)}-\frac {\sqrt [4]{d} (3 a d+5 b c) \log \left (-\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}+\frac {\sqrt [4]{d} (3 a d+5 b c) \log \left (\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt {x}+\sqrt {c}+\sqrt {d} x\right )}{4 \sqrt {2} \sqrt [4]{c} (b c-a d)}}{b c-a d}-\frac {2 d x^{3/2}}{\left (c+d x^2\right ) (b c-a d)}}{4 (b c-a d)}-\frac {x^{3/2}}{4 \left (a+b x^2\right ) \left (c+d x^2\right ) (b c-a d)}\right )\)

Input:

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

Output:

2*(-1/4*x^(3/2)/((b*c - a*d)*(a + b*x^2)*(c + d*x^2)) + ((-2*d*x^(3/2))/(( 
b*c - a*d)*(c + d*x^2)) + (-1/2*(b^(1/4)*(3*b*c + 5*a*d)*ArcTan[1 - (Sqrt[ 
2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*a^(1/4)*(b*c - a*d)) + (b^(1/4)*(3* 
b*c + 5*a*d)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^( 
1/4)*(b*c - a*d)) + (d^(1/4)*(5*b*c + 3*a*d)*ArcTan[1 - (Sqrt[2]*d^(1/4)*S 
qrt[x])/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*(b*c - a*d)) - (d^(1/4)*(5*b*c + 3*a* 
d)*ArcTan[1 + (Sqrt[2]*d^(1/4)*Sqrt[x])/c^(1/4)])/(2*Sqrt[2]*c^(1/4)*(b*c 
- a*d)) + (b^(1/4)*(3*b*c + 5*a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*S 
qrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a^(1/4)*(b*c - a*d)) - (b^(1/4)*(3*b*c + 5 
*a*d)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[ 
2]*a^(1/4)*(b*c - a*d)) - (d^(1/4)*(5*b*c + 3*a*d)*Log[Sqrt[c] - Sqrt[2]*c 
^(1/4)*d^(1/4)*Sqrt[x] + Sqrt[d]*x])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)) + (d^ 
(1/4)*(5*b*c + 3*a*d)*Log[Sqrt[c] + Sqrt[2]*c^(1/4)*d^(1/4)*Sqrt[x] + Sqrt 
[d]*x])/(4*Sqrt[2]*c^(1/4)*(b*c - a*d)))/(b*c - a*d))/(4*(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 368

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

rule 971

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

rule 1049

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m 
 + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p + 1))) 
, x] + Simp[1/(a*n*(b*c - a*d)*(p + 1))   Int[(g*x)^m*(a + b*x^n)^(p + 1)*( 
c + d*x^n)^q*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e 
- a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, 
g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]
 

rule 1054

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n 
_)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a 
+ b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, 
 m, p}, x] && IGtQ[n, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 302, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {2 d \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) x^{\frac {3}{2}}}{x^{2} d +c}+\frac {\left (\frac {5 b c}{4}+\frac {3 a d}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 b \left (\frac {\left (\frac {a d}{4}-\frac {b c}{4}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {5 a d}{4}+\frac {3 b c}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}\)	\(302\)
default	\(\frac {2 d \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) x^{\frac {3}{2}}}{x^{2} d +c}+\frac {\left (\frac {5 b c}{4}+\frac {3 a d}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}{x +\left (\frac {c}{d}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {c}{d}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {c}{d}\right )^{\frac {1}{4}}}-1\right )\right )}{8 d \left (\frac {c}{d}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}-\frac {2 b \left (\frac {\left (\frac {a d}{4}-\frac {b c}{4}\right ) x^{\frac {3}{2}}}{b \,x^{2}+a}+\frac {\left (\frac {5 a d}{4}+\frac {3 b c}{4}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{8 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{\left (a d -b c \right )^{3}}\)	\(302\)

Input:

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

Output:

2*d/(a*d-b*c)^3*((-1/4*a*d+1/4*b*c)*x^(3/2)/(d*x^2+c)+1/8*(5/4*b*c+3/4*a*d 
)/d/(c/d)^(1/4)*2^(1/2)*(ln((x-(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2))/(x 
+(c/d)^(1/4)*x^(1/2)*2^(1/2)+(c/d)^(1/2)))+2*arctan(2^(1/2)/(c/d)^(1/4)*x^ 
(1/2)+1)+2*arctan(2^(1/2)/(c/d)^(1/4)*x^(1/2)-1)))-2*b/(a*d-b*c)^3*((1/4*a 
*d-1/4*b*c)*x^(3/2)/(b*x^2+a)+1/8*(5/4*a*d+3/4*b*c)/b/(a/b)^(1/4)*2^(1/2)* 
(ln((x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/(x+(a/b)^(1/4)*x^(1/2)*2^( 
1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2 
)/(a/b)^(1/4)*x^(1/2)-1)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 12.48 (sec) , antiderivative size = 5884, normalized size of antiderivative = 12.65 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 567, normalized size of antiderivative = 1.22 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/16*(3*b^2*c + 5*a*b*d)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^ 
(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*s 
qrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt( 
b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) - sqrt( 
2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^( 
3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a) 
)/(a^(1/4)*b^(3/4)))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) - 
 1/16*(5*b*c*d + 3*a*d^2)*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d 
^(1/4) + 2*sqrt(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))* 
sqrt(d)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*c^(1/4)*d^(1/4) - 2*sqrt 
(d)*sqrt(x))/sqrt(sqrt(c)*sqrt(d)))/(sqrt(sqrt(c)*sqrt(d))*sqrt(d)) - sqrt 
(2)*log(sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c))/(c^(1/4)*d^ 
(3/4)) + sqrt(2)*log(-sqrt(2)*c^(1/4)*d^(1/4)*sqrt(x) + sqrt(d)*x + sqrt(c 
))/(c^(1/4)*d^(3/4)))/(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3) 
- 1/2*(2*b*d*x^(7/2) + (b*c + a*d)*x^(3/2))/(a*b^2*c^3 - 2*a^2*b*c^2*d + a 
^3*c*d^2 + (b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^4 + (b^3*c^3 - a*b^2* 
c^2*d - a^2*b*c*d^2 + a^3*d^3)*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 952 vs. \(2 (355) = 710\).

Time = 0.28 (sec) , antiderivative size = 952, normalized size of antiderivative = 2.05 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*arctan(1/2*sqrt(2)*(sqrt(2 
)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2 
*b^4*c^2*d + 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1/4*(3*(a*b^ 
3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/ 
4) - 2*sqrt(x))/(a/b)^(1/4))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2*b^4*c^2*d 
+ 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) - 1/4*(5*(c*d^3)^(3/4)*b* 
c + 3*(c*d^3)^(3/4)*a*d)*arctan(1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) + 2*sqrt( 
x))/(c/d)^(1/4))/(sqrt(2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2 
)*a^2*b*c^2*d^4 - sqrt(2)*a^3*c*d^5) - 1/4*(5*(c*d^3)^(3/4)*b*c + 3*(c*d^3 
)^(3/4)*a*d)*arctan(-1/2*sqrt(2)*(sqrt(2)*(c/d)^(1/4) - 2*sqrt(x))/(c/d)^( 
1/4))/(sqrt(2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2)*a^2*b*c^2 
*d^4 - sqrt(2)*a^3*c*d^5) - 1/8*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d 
)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(sqrt(2)*a*b^5*c^3 - 3* 
sqrt(2)*a^2*b^4*c^2*d + 3*sqrt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1 
/8*(3*(a*b^3)^(3/4)*b*c + 5*(a*b^3)^(3/4)*a*d)*log(-sqrt(2)*sqrt(x)*(a/b)^ 
(1/4) + x + sqrt(a/b))/(sqrt(2)*a*b^5*c^3 - 3*sqrt(2)*a^2*b^4*c^2*d + 3*sq 
rt(2)*a^3*b^3*c*d^2 - sqrt(2)*a^4*b^2*d^3) + 1/8*(5*(c*d^3)^(3/4)*b*c + 3* 
(c*d^3)^(3/4)*a*d)*log(sqrt(2)*sqrt(x)*(c/d)^(1/4) + x + sqrt(c/d))/(sqrt( 
2)*b^3*c^4*d^2 - 3*sqrt(2)*a*b^2*c^3*d^3 + 3*sqrt(2)*a^2*b*c^2*d^4 - sqrt( 
2)*a^3*c*d^5) - 1/8*(5*(c*d^3)^(3/4)*b*c + 3*(c*d^3)^(3/4)*a*d)*log(-sq...
 

Mupad [B] (verification not implemented)

Time = 4.51 (sec) , antiderivative size = 30956, normalized size of antiderivative = 66.57 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

2*atan((((-(81*b^5*c^4 + 625*a^4*b*d^4 + 1500*a^3*b^2*c*d^3 + 1350*a^2*b^3 
*c^2*d^2 + 540*a*b^4*c^3*d)/(4096*a^13*d^12 + 4096*a*b^12*c^12 - 49152*a^2 
*b^11*c^11*d + 270336*a^3*b^10*c^10*d^2 - 901120*a^4*b^9*c^9*d^3 + 2027520 
*a^5*b^8*c^8*d^4 - 3244032*a^6*b^7*c^7*d^5 + 3784704*a^7*b^6*c^6*d^6 - 324 
4032*a^8*b^5*c^5*d^7 + 2027520*a^9*b^4*c^4*d^8 - 901120*a^10*b^3*c^3*d^9 + 
 270336*a^11*b^2*c^2*d^10 - 49152*a^12*b*c*d^11))^(3/4)*(((864*a*b^20*c^17 
*d^4 + 864*a^17*b^4*c*d^20 - 5184*a^2*b^19*c^16*d^5 + 3200*a^3*b^18*c^15*d 
^6 + 56640*a^4*b^17*c^14*d^7 - 220800*a^5*b^16*c^13*d^8 + 369088*a^6*b^15* 
c^12*d^9 - 240768*a^7*b^14*c^11*d^10 - 158400*a^8*b^13*c^10*d^11 + 390720* 
a^9*b^12*c^9*d^12 - 158400*a^10*b^11*c^8*d^13 - 240768*a^11*b^10*c^7*d^14 
+ 369088*a^12*b^9*c^6*d^15 - 220800*a^13*b^8*c^5*d^16 + 56640*a^14*b^7*c^4 
*d^17 + 3200*a^15*b^6*c^3*d^18 - 5184*a^16*b^5*c^2*d^19)*1i)/(a^14*d^14 + 
b^14*c^14 + 91*a^2*b^12*c^12*d^2 - 364*a^3*b^11*c^11*d^3 + 1001*a^4*b^10*c 
^10*d^4 - 2002*a^5*b^9*c^9*d^5 + 3003*a^6*b^8*c^8*d^6 - 3432*a^7*b^7*c^7*d 
^7 + 3003*a^8*b^6*c^6*d^8 - 2002*a^9*b^5*c^5*d^9 + 1001*a^10*b^4*c^4*d^10 
- 364*a^11*b^3*c^3*d^11 + 91*a^12*b^2*c^2*d^12 - 14*a*b^13*c^13*d - 14*a^1 
3*b*c*d^13) - (x^(1/2)*(-(81*b^5*c^4 + 625*a^4*b*d^4 + 1500*a^3*b^2*c*d^3 
+ 1350*a^2*b^3*c^2*d^2 + 540*a*b^4*c^3*d)/(4096*a^13*d^12 + 4096*a*b^12*c^ 
12 - 49152*a^2*b^11*c^11*d + 270336*a^3*b^10*c^10*d^2 - 901120*a^4*b^9*c^9 
*d^3 + 2027520*a^5*b^8*c^8*d^4 - 3244032*a^6*b^7*c^7*d^5 + 3784704*a^7*...
 

Reduce [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 2425, normalized size of antiderivative = 5.22 \[ \int \frac {x^{5/2}}{\left (a+b x^2\right )^2 \left (c+d x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(10*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)* 
sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*c**2*d + 10*b**(1/4)*a**(3/4)*s 
qrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**( 
1/4)*sqrt(2)))*a**2*c*d**2*x**2 + 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/ 
4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b* 
c**3 + 16*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sq 
rt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*c**2*d*x**2 + 10*b**(1/4)* 
a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b** 
(1/4)*a**(1/4)*sqrt(2)))*a*b*c*d**2*x**4 + 6*b**(1/4)*a**(3/4)*sqrt(2)*ata 
n((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt( 
2)))*b**2*c**3*x**2 + 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)* 
sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*b**2*c**2*d*x**4 
 - 10*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x 
)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*c**2*d - 10*b**(1/4)*a**(3/4) 
*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a* 
*(1/4)*sqrt(2)))*a**2*c*d**2*x**2 - 6*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**( 
1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a* 
b*c**3 - 16*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2* 
sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b*c**2*d*x**2 - 10*b**(1/4 
)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))...