\(\int \frac {(d+e x^2)^{11/2}}{(a d+(b d+a e) x^2+b e x^4)^{5/2}} \, dx\) [15]

Optimal result

Integrand size = 37, antiderivative size = 236 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {(b d-a e)^3 x \left (d+e x^2\right )^{3/2}}{3 a b^3 \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}}+\frac {(b d-a e)^2 (2 b d+7 a e) x \sqrt {d+e x^2}}{3 a^2 b^3 \sqrt {a d+(b d+a e) x^2+b e x^4}}+\frac {e^3 x \sqrt {a d+(b d+a e) x^2+b e x^4}}{2 b^3 \sqrt {d+e x^2}}+\frac {e^2 (6 b d-5 a e) \text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{2 b^{7/2}} \] Output:

1/3*(-a*e+b*d)^3*x*(e*x^2+d)^(3/2)/a/b^3/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(3/2) 
+1/3*(-a*e+b*d)^2*(7*a*e+2*b*d)*x*(e*x^2+d)^(1/2)/a^2/b^3/(a*d+(a*e+b*d)*x 
^2+b*e*x^4)^(1/2)+1/2*e^3*x*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)/b^3/(e*x^2+d 
)^(1/2)+1/2*e^2*(-5*a*e+6*b*d)*arctanh(b^(1/2)*x*(e*x^2+d)^(1/2)/(a*d+(a*e 
+b*d)*x^2+b*e*x^4)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.75 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\left (d+e x^2\right )^{3/2} \left (\sqrt {b} x \left (15 a^4 e^3+4 b^4 d^3 x^2+3 a^2 b^2 e^2 x^2 \left (-8 d+e x^2\right )+6 a b^3 d^2 \left (d+e x^2\right )+2 a^3 b e^2 \left (-9 d+10 e x^2\right )\right )+3 a^2 e^2 (-6 b d+5 a e) \left (a+b x^2\right )^{3/2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{6 a^2 b^{7/2} \left (\left (a+b x^2\right ) \left (d+e x^2\right )\right )^{3/2}} \] Input:

Integrate[(d + e*x^2)^(11/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(5/2),x]
 

Output:

((d + e*x^2)^(3/2)*(Sqrt[b]*x*(15*a^4*e^3 + 4*b^4*d^3*x^2 + 3*a^2*b^2*e^2* 
x^2*(-8*d + e*x^2) + 6*a*b^3*d^2*(d + e*x^2) + 2*a^3*b*e^2*(-9*d + 10*e*x^ 
2)) + 3*a^2*e^2*(-6*b*d + 5*a*e)*(a + b*x^2)^(3/2)*Log[-(Sqrt[b]*x) + Sqrt 
[a + b*x^2]]))/(6*a^2*b^(7/2)*((a + b*x^2)*(d + e*x^2))^(3/2))
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1395, 315, 401, 27, 299, 224, 219}

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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^{11/2}}{\left (x^2 (a e+b d)+a d+b e x^4\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 401

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {x \left (d+e x^2\right ) \left (\frac {2 b d^2}{a}-\frac {5 a e^2}{b}+3 d e\right )}{\sqrt {a+b x^2}}-\frac {\int \frac {e \left (\left (4 b^2 d^2+8 a b e d-15 a^2 e^2\right ) x^2+a d (2 b d-5 a e)\right )}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}+\frac {x \left (d+e x^2\right )^2 (b d-a e)}{3 a b \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {x \left (d+e x^2\right ) \left (\frac {2 b d^2}{a}-\frac {5 a e^2}{b}+3 d e\right )}{\sqrt {a+b x^2}}-\frac {e \int \frac {\left (4 b^2 d^2+8 a b e d-15 a^2 e^2\right ) x^2+a d (2 b d-5 a e)}{\sqrt {b x^2+a}}dx}{a b}}{3 a b}+\frac {x \left (d+e x^2\right )^2 (b d-a e)}{3 a b \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {x \left (d+e x^2\right ) \left (\frac {2 b d^2}{a}-\frac {5 a e^2}{b}+3 d e\right )}{\sqrt {a+b x^2}}-\frac {e \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 e^2+8 a b d e+4 b^2 d^2\right )}{2 b}-\frac {3 a^2 e (6 b d-5 a e) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{a b}}{3 a b}+\frac {x \left (d+e x^2\right )^2 (b d-a e)}{3 a b \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {x \left (d+e x^2\right ) \left (\frac {2 b d^2}{a}-\frac {5 a e^2}{b}+3 d e\right )}{\sqrt {a+b x^2}}-\frac {e \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 e^2+8 a b d e+4 b^2 d^2\right )}{2 b}-\frac {3 a^2 e (6 b d-5 a e) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{a b}}{3 a b}+\frac {x \left (d+e x^2\right )^2 (b d-a e)}{3 a b \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {x \left (d+e x^2\right ) \left (\frac {2 b d^2}{a}-\frac {5 a e^2}{b}+3 d e\right )}{\sqrt {a+b x^2}}-\frac {e \left (\frac {x \sqrt {a+b x^2} \left (-15 a^2 e^2+8 a b d e+4 b^2 d^2\right )}{2 b}-\frac {3 a^2 e \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (6 b d-5 a e)}{2 b^{3/2}}\right )}{a b}}{3 a b}+\frac {x \left (d+e x^2\right )^2 (b d-a e)}{3 a b \left (a+b x^2\right )^{3/2}}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

Input:

Int[(d + e*x^2)^(11/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(5/2),x]
 

Output:

(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*(((b*d - a*e)*x*(d + e*x^2)^2)/(3*a*b*(a 
+ b*x^2)^(3/2)) + ((((2*b*d^2)/a + 3*d*e - (5*a*e^2)/b)*x*(d + e*x^2))/Sqr 
t[a + b*x^2] - (e*(((4*b^2*d^2 + 8*a*b*d*e - 15*a^2*e^2)*x*Sqrt[a + b*x^2] 
)/(2*b) - (3*a^2*e*(6*b*d - 5*a*e)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/( 
2*b^(3/2))))/(a*b))/(3*a*b)))/Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4]
 

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 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 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 315

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

rule 401

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/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

rule 1395

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.24

method	result	size
default	\(-\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \left (-3 b^{\frac {5}{2}} a^{2} e^{3} x^{5}+15 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{3} b \,e^{3} x^{2} \sqrt {b \,x^{2}+a}-18 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2} b^{2} d \,e^{2} x^{2} \sqrt {b \,x^{2}+a}-20 b^{\frac {3}{2}} a^{3} e^{3} x^{3}+24 b^{\frac {5}{2}} a^{2} d \,e^{2} x^{3}-6 b^{\frac {7}{2}} a \,d^{2} e \,x^{3}-4 b^{\frac {9}{2}} d^{3} x^{3}+15 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{4} e^{3} \sqrt {b \,x^{2}+a}-18 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{3} b d \,e^{2} \sqrt {b \,x^{2}+a}-15 \sqrt {b}\, a^{4} e^{3} x +18 b^{\frac {3}{2}} a^{3} d \,e^{2} x -6 b^{\frac {7}{2}} a \,d^{3} x \right )}{6 b^{\frac {7}{2}} \sqrt {e \,x^{2}+d}\, \left (b \,x^{2}+a \right )^{2} a^{2}}\)	\(292\)
risch	\(\frac {e^{3} x \left (b \,x^{2}+a \right ) \sqrt {e \,x^{2}+d}}{2 b^{3} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}-\frac {\left (\frac {e^{2} \left (5 a e -6 b d \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}-\frac {\left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (-\frac {\sqrt {b \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {b \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x -\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b a}-\frac {\left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (\frac {\sqrt {b \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {b \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{3 a \left (x +\frac {\sqrt {-a b}}{b}\right )}\right )}{2 b a}-\frac {\left (5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \sqrt {b \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}+2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}}{2 a^{2} b \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {\left (5 a^{3} e^{3}-9 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \sqrt {b \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}-2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}}{2 a^{2} b \left (x +\frac {\sqrt {-a b}}{b}\right )}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {e \,x^{2}+d}}{2 b^{3} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}\)	\(648\)

Input:

int((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/6*((e*x^2+d)*(b*x^2+a))^(1/2)/b^(7/2)*(-3*b^(5/2)*a^2*e^3*x^5+15*ln(b^( 
1/2)*x+(b*x^2+a)^(1/2))*a^3*b*e^3*x^2*(b*x^2+a)^(1/2)-18*ln(b^(1/2)*x+(b*x 
^2+a)^(1/2))*a^2*b^2*d*e^2*x^2*(b*x^2+a)^(1/2)-20*b^(3/2)*a^3*e^3*x^3+24*b 
^(5/2)*a^2*d*e^2*x^3-6*b^(7/2)*a*d^2*e*x^3-4*b^(9/2)*d^3*x^3+15*ln(b^(1/2) 
*x+(b*x^2+a)^(1/2))*a^4*e^3*(b*x^2+a)^(1/2)-18*ln(b^(1/2)*x+(b*x^2+a)^(1/2 
))*a^3*b*d*e^2*(b*x^2+a)^(1/2)-15*b^(1/2)*a^4*e^3*x+18*b^(3/2)*a^3*d*e^2*x 
-6*b^(7/2)*a*d^3*x)/(e*x^2+d)^(1/2)/(b*x^2+a)^2/a^2
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 777, normalized size of antiderivative = 3.29 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm= 
"fricas")
 

Output:

[-1/12*(3*(6*a^4*b*d^2*e^2 - 5*a^5*d*e^3 + (6*a^2*b^3*d*e^3 - 5*a^3*b^2*e^ 
4)*x^6 + (6*a^2*b^3*d^2*e^2 + 7*a^3*b^2*d*e^3 - 10*a^4*b*e^4)*x^4 + (12*a^ 
3*b^2*d^2*e^2 - 4*a^4*b*d*e^3 - 5*a^5*e^4)*x^2)*sqrt(b)*log((2*b*e*x^4 + ( 
2*b*d + a*e)*x^2 - 2*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e*x^2 + d) 
*sqrt(b)*x + a*d)/(e*x^2 + d)) - 2*(3*a^2*b^3*e^3*x^5 + 2*(2*b^5*d^3 + 3*a 
*b^4*d^2*e - 12*a^2*b^3*d*e^2 + 10*a^3*b^2*e^3)*x^3 + 3*(2*a*b^4*d^3 - 6*a 
^3*b^2*d*e^2 + 5*a^4*b*e^3)*x)*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt( 
e*x^2 + d))/(a^2*b^6*e*x^6 + a^4*b^4*d + (a^2*b^6*d + 2*a^3*b^5*e)*x^4 + ( 
2*a^3*b^5*d + a^4*b^4*e)*x^2), -1/6*(3*(6*a^4*b*d^2*e^2 - 5*a^5*d*e^3 + (6 
*a^2*b^3*d*e^3 - 5*a^3*b^2*e^4)*x^6 + (6*a^2*b^3*d^2*e^2 + 7*a^3*b^2*d*e^3 
 - 10*a^4*b*e^4)*x^4 + (12*a^3*b^2*d^2*e^2 - 4*a^4*b*d*e^3 - 5*a^5*e^4)*x^ 
2)*sqrt(-b)*arctan(sqrt(e*x^2 + d)*sqrt(-b)*x/sqrt(b*e*x^4 + (b*d + a*e)*x 
^2 + a*d)) - (3*a^2*b^3*e^3*x^5 + 2*(2*b^5*d^3 + 3*a*b^4*d^2*e - 12*a^2*b^ 
3*d*e^2 + 10*a^3*b^2*e^3)*x^3 + 3*(2*a*b^4*d^3 - 6*a^3*b^2*d*e^2 + 5*a^4*b 
*e^3)*x)*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e*x^2 + d))/(a^2*b^6*e 
*x^6 + a^4*b^4*d + (a^2*b^6*d + 2*a^3*b^5*e)*x^4 + (2*a^3*b^5*d + a^4*b^4* 
e)*x^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**(11/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {11}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm= 
"maxima")
 

Output:

integrate((e*x^2 + d)^(11/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.68 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\sqrt {b x^{2} + a} e^{3} x}{2 \, b^{3}} + \frac {x {\left (\frac {{\left (2 \, b^{7} d^{3} + 3 \, a b^{6} d^{2} e - 12 \, a^{2} b^{5} d e^{2} + 7 \, a^{3} b^{4} e^{3}\right )} x^{2}}{a^{2} b^{6}} + \frac {3 \, {\left (a b^{6} d^{3} - 3 \, a^{3} b^{4} d e^{2} + 2 \, a^{4} b^{3} e^{3}\right )}}{a^{2} b^{6}}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {{\left (6 \, b d e^{2} - 5 \, a e^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \] Input:

integrate((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm= 
"giac")
 

Output:

1/2*sqrt(b*x^2 + a)*e^3*x/b^3 + 1/3*x*((2*b^7*d^3 + 3*a*b^6*d^2*e - 12*a^2 
*b^5*d*e^2 + 7*a^3*b^4*e^3)*x^2/(a^2*b^6) + 3*(a*b^6*d^3 - 3*a^3*b^4*d*e^2 
 + 2*a^4*b^3*e^3)/(a^2*b^6))/(b*x^2 + a)^(3/2) - 1/2*(6*b*d*e^2 - 5*a*e^3) 
*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{11/2}}{{\left (b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d\right )}^{5/2}} \,d x \] Input:

int((d + e*x^2)^(11/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(5/2),x)
 

Output:

int((d + e*x^2)^(11/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 519, normalized size of antiderivative = 2.20 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {30 \sqrt {b \,x^{2}+a}\, a^{4} b \,e^{3} x -36 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} d \,e^{2} x +40 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} e^{3} x^{3}-48 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d \,e^{2} x^{3}+6 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} e^{3} x^{5}+12 \sqrt {b \,x^{2}+a}\, a \,b^{4} d^{3} x +12 \sqrt {b \,x^{2}+a}\, a \,b^{4} d^{2} e \,x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{5} d^{3} x^{3}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5} e^{3}+36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} b d \,e^{2}-60 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} b \,e^{3} x^{2}+72 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{2} d \,e^{2} x^{2}-30 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{2} e^{3} x^{4}+36 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{3} d \,e^{2} x^{4}-5 \sqrt {b}\, a^{5} e^{3}-10 \sqrt {b}\, a^{4} b \,e^{3} x^{2}+12 \sqrt {b}\, a^{3} b^{2} d^{2} e -5 \sqrt {b}\, a^{3} b^{2} e^{3} x^{4}-8 \sqrt {b}\, a^{2} b^{3} d^{3}+24 \sqrt {b}\, a^{2} b^{3} d^{2} e \,x^{2}-16 \sqrt {b}\, a \,b^{4} d^{3} x^{2}+12 \sqrt {b}\, a \,b^{4} d^{2} e \,x^{4}-8 \sqrt {b}\, b^{5} d^{3} x^{4}}{12 a^{2} b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x)
 

Output:

(30*sqrt(a + b*x**2)*a**4*b*e**3*x - 36*sqrt(a + b*x**2)*a**3*b**2*d*e**2* 
x + 40*sqrt(a + b*x**2)*a**3*b**2*e**3*x**3 - 48*sqrt(a + b*x**2)*a**2*b** 
3*d*e**2*x**3 + 6*sqrt(a + b*x**2)*a**2*b**3*e**3*x**5 + 12*sqrt(a + b*x** 
2)*a*b**4*d**3*x + 12*sqrt(a + b*x**2)*a*b**4*d**2*e*x**3 + 8*sqrt(a + b*x 
**2)*b**5*d**3*x**3 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a 
))*a**5*e**3 + 36*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4 
*b*d*e**2 - 60*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b* 
e**3*x**2 + 72*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b* 
*2*d*e**2*x**2 - 30*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a* 
*3*b**2*e**3*x**4 + 36*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a)) 
*a**2*b**3*d*e**2*x**4 - 5*sqrt(b)*a**5*e**3 - 10*sqrt(b)*a**4*b*e**3*x**2 
 + 12*sqrt(b)*a**3*b**2*d**2*e - 5*sqrt(b)*a**3*b**2*e**3*x**4 - 8*sqrt(b) 
*a**2*b**3*d**3 + 24*sqrt(b)*a**2*b**3*d**2*e*x**2 - 16*sqrt(b)*a*b**4*d** 
3*x**2 + 12*sqrt(b)*a*b**4*d**2*e*x**4 - 8*sqrt(b)*b**5*d**3*x**4)/(12*a** 
2*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4))