\(\int \frac {(d+e x^2)^{9/2}}{(a d+(b d+a e) x^2+b e x^4)^{7/2}} \, dx\) [24]

Optimal result

Integrand size = 37, antiderivative size = 166 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {(b d-a e) x \left (d+e x^2\right )^{5/2}}{5 a b \left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}}+\frac {(4 b d+a e) x \left (d+e x^2\right )^{3/2}}{15 a^2 b \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}}+\frac {2 (4 b d+a e) x \sqrt {d+e x^2}}{15 a^3 b \sqrt {a d+(b d+a e) x^2+b e x^4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.47 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\left (d+e x^2\right )^{5/2} \left (8 b^2 d x^5+2 a b x^3 \left (10 d+e x^2\right )+5 a^2 \left (3 d x+e x^3\right )\right )}{15 a^3 \left (\left (a+b x^2\right ) \left (d+e x^2\right )\right )^{5/2}} \] Input:

Integrate[(d + e*x^2)^(9/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(7/2),x]
 

Output:

((d + e*x^2)^(5/2)*(8*b^2*d*x^5 + 2*a*b*x^3*(10*d + e*x^2) + 5*a^2*(3*d*x 
+ e*x^3)))/(15*a^3*((a + b*x^2)*(d + e*x^2))^(5/2))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.82, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {1395, 298, 209, 208}

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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^{9/2}}{\left (x^2 (a e+b d)+a d+b e x^4\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 1395

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \int \frac {e x^2+d}{\left (b x^2+a\right )^{7/2}}dx}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

Int[(d + e*x^2)^(9/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(7/2),x]
 

Output:

(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*(((b*d - a*e)*x)/(5*a*b*(a + b*x^2)^(5/2) 
) + ((4*b*d + a*e)*(x/(3*a*(a + b*x^2)^(3/2)) + (2*x)/(3*a^2*Sqrt[a + b*x^ 
2])))/(5*a*b)))/Sqrt[a*d + (b*d + a*e)*x^2 + b*e*x^4]
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

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

rule 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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 = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.50

method	result	size
default	\(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, x \left (2 a b e \,x^{4}+8 b^{2} d \,x^{4}+5 a^{2} e \,x^{2}+20 a b d \,x^{2}+15 a^{2} d \right )}{15 \sqrt {e \,x^{2}+d}\, \left (b \,x^{2}+a \right )^{3} a^{3}}\)	\(83\)
orering	\(\frac {x \left (2 a b e \,x^{4}+8 b^{2} d \,x^{4}+5 a^{2} e \,x^{2}+20 a b d \,x^{2}+15 a^{2} d \right ) \left (e \,x^{2}+d \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )}{15 a^{3} \left (a d +\left (a e +b d \right ) x^{2}+b e \,x^{4}\right )^{\frac {7}{2}}}\)	\(87\)
gosper	\(\frac {\left (b \,x^{2}+a \right ) x \left (2 a b e \,x^{4}+8 b^{2} d \,x^{4}+5 a^{2} e \,x^{2}+20 a b d \,x^{2}+15 a^{2} d \right ) \left (e \,x^{2}+d \right )^{\frac {7}{2}}}{15 a^{3} \left (b e \,x^{4}+a e \,x^{2}+b d \,x^{2}+a d \right )^{\frac {7}{2}}}\)	\(88\)

Input:

int((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/15/(e*x^2+d)^(1/2)*((e*x^2+d)*(b*x^2+a))^(1/2)*x*(2*a*b*e*x^4+8*b^2*d*x^ 
4+5*a^2*e*x^2+20*a*b*d*x^2+15*a^2*d)/(b*x^2+a)^3/a^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.92 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} {\left (2 \, {\left (4 \, b^{2} d + a b e\right )} x^{5} + 15 \, a^{2} d x + 5 \, {\left (4 \, a b d + a^{2} e\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{15 \, {\left (a^{3} b^{3} e x^{8} + a^{6} d + {\left (a^{3} b^{3} d + 3 \, a^{4} b^{2} e\right )} x^{6} + 3 \, {\left (a^{4} b^{2} d + a^{5} b e\right )} x^{4} + {\left (3 \, a^{5} b d + a^{6} e\right )} x^{2}\right )}} \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x, algorithm=" 
fricas")
 

Output:

1/15*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*(2*(4*b^2*d + a*b*e)*x^5 + 15*a 
^2*d*x + 5*(4*a*b*d + a^2*e)*x^3)*sqrt(e*x^2 + d)/(a^3*b^3*e*x^8 + a^6*d + 
 (a^3*b^3*d + 3*a^4*b^2*e)*x^6 + 3*(a^4*b^2*d + a^5*b*e)*x^4 + (3*a^5*b*d 
+ a^6*e)*x^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**(9/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {9}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x, algorithm=" 
maxima")
 

Output:

integrate((e*x^2 + d)^(9/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(7/2), x)
 

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {9}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x, algorithm=" 
giac")
 

Output:

integrate((e*x^2 + d)^(9/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 17.54 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\sqrt {b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d}\,\left (\frac {d\,x\,\sqrt {e\,x^2+d}}{a\,b^3\,e}+\frac {x^3\,\sqrt {e\,x^2+d}\,\left (\frac {e\,a^2}{3}+\frac {4\,b\,d\,a}{3}\right )}{a^3\,b^3\,e}+\frac {x^5\,\sqrt {e\,x^2+d}\,\left (\frac {8\,d\,b^2}{15}+\frac {2\,a\,e\,b}{15}\right )}{a^3\,b^3\,e}\right )}{x^8+\frac {a^3\,d}{b^3\,e}+\frac {x^6\,\left (3\,a\,e+b\,d\right )}{b\,e}+\frac {x^2\,\left (e\,a^6+3\,b\,d\,a^5\right )}{a^3\,b^3\,e}+\frac {3\,a\,x^4\,\left (a\,e+b\,d\right )}{b^2\,e}} \] Input:

int((d + e*x^2)^(9/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(7/2),x)
 

Output:

((a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2)*((d*x*(d + e*x^2)^(1/2))/(a*b^3*e 
) + (x^3*(d + e*x^2)^(1/2)*((a^2*e)/3 + (4*a*b*d)/3))/(a^3*b^3*e) + (x^5*( 
d + e*x^2)^(1/2)*((8*b^2*d)/15 + (2*a*b*e)/15))/(a^3*b^3*e)))/(x^8 + (a^3* 
d)/(b^3*e) + (x^6*(3*a*e + b*d))/(b*e) + (x^2*(a^6*e + 3*a^5*b*d))/(a^3*b^ 
3*e) + (3*a*x^4*(a*e + b*d))/(b^2*e))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.34 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} d x +5 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} e \,x^{3}+20 \sqrt {b \,x^{2}+a}\, a \,b^{3} d \,x^{3}+2 \sqrt {b \,x^{2}+a}\, a \,b^{3} e \,x^{5}+8 \sqrt {b \,x^{2}+a}\, b^{4} d \,x^{5}-2 \sqrt {b}\, a^{4} e -8 \sqrt {b}\, a^{3} b d -6 \sqrt {b}\, a^{3} b e \,x^{2}-24 \sqrt {b}\, a^{2} b^{2} d \,x^{2}-6 \sqrt {b}\, a^{2} b^{2} e \,x^{4}-24 \sqrt {b}\, a \,b^{3} d \,x^{4}-2 \sqrt {b}\, a \,b^{3} e \,x^{6}-8 \sqrt {b}\, b^{4} d \,x^{6}}{15 a^{3} b^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

int((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b**2*d*x + 5*sqrt(a + b*x**2)*a**2*b**2*e*x**3 + 
 20*sqrt(a + b*x**2)*a*b**3*d*x**3 + 2*sqrt(a + b*x**2)*a*b**3*e*x**5 + 8* 
sqrt(a + b*x**2)*b**4*d*x**5 - 2*sqrt(b)*a**4*e - 8*sqrt(b)*a**3*b*d - 6*s 
qrt(b)*a**3*b*e*x**2 - 24*sqrt(b)*a**2*b**2*d*x**2 - 6*sqrt(b)*a**2*b**2*e 
*x**4 - 24*sqrt(b)*a*b**3*d*x**4 - 2*sqrt(b)*a*b**3*e*x**6 - 8*sqrt(b)*b** 
4*d*x**6)/(15*a**3*b**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6) 
)