\(\int \frac {(d+e x^2)^{11/2}}{(a d+(b d+a e) x^2+b e x^4)^{7/2}} \, dx\) [23]

Optimal result

Integrand size = 37, antiderivative size = 210 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\left (8 b^2 d^2+4 a b d e+3 a^2 e^2\right ) x \sqrt {d+e x^2}}{15 a^3 b^2 \sqrt {a d+(b d+a e) x^2+b e x^4}}+\frac {(b d-a e)^2 x \sqrt {d+e x^2}}{5 a b^2 \left (a+b x^2\right )^2 \sqrt {a d+(b d+a e) x^2+b e x^4}}+\frac {2 (b d-a e) (2 b d+3 a e) x \sqrt {d+e x^2}}{15 a^2 b^2 \left (a+b x^2\right ) \sqrt {a d+(b d+a e) x^2+b e x^4}} \] Output:

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

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.48 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\sqrt {d+e x^2} \left (8 b^2 d^2 x^5+4 a b d x^3 \left (5 d+e x^2\right )+a^2 \left (15 d^2 x+10 d e x^3+3 e^2 x^5\right )\right )}{15 a^3 \left (a+b x^2\right )^2 \sqrt {\left (a+b x^2\right ) \left (d+e x^2\right )}} \] Input:

Integrate[(d + e*x^2)^(11/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(7/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {1395, 292, 292, 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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^{11/2}}{\left (x^2 (a e+b d)+a d+b e x^4\right )^{7/2}} \, dx\)

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 292

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

\(\Big \downarrow \) 208

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

Input:

Int[(d + e*x^2)^(11/2)/(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]*((x*(d + e*x^2)^2)/(5*a*(a + b*x^2)^(5/2) 
) + (4*d*((2*d*x)/(3*a^2*Sqrt[a + b*x^2]) + (x*(d + e*x^2))/(3*a*(a + b*x^ 
2)^(3/2))))/(5*a)))/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 292

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

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 = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, x \left (3 a^{2} e^{2} x^{4}+4 a b d e \,x^{4}+8 b^{2} d^{2} x^{4}+10 a^{2} d e \,x^{2}+20 a \,d^{2} b \,x^{2}+15 a^{2} d^{2}\right )}{15 \sqrt {e \,x^{2}+d}\, \left (b \,x^{2}+a \right )^{3} a^{3}}\)	\(102\)
orering	\(\frac {x \left (3 a^{2} e^{2} x^{4}+4 a b d e \,x^{4}+8 b^{2} d^{2} x^{4}+10 a^{2} d e \,x^{2}+20 a \,d^{2} b \,x^{2}+15 a^{2} d^{2}\right ) \left (e \,x^{2}+d \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )}{15 a^{3} \left (a d +\left (a e +b d \right ) x^{2}+b e \,x^{4}\right )^{\frac {7}{2}}}\)	\(106\)
gosper	\(\frac {\left (b \,x^{2}+a \right ) x \left (3 a^{2} e^{2} x^{4}+4 a b d e \,x^{4}+8 b^{2} d^{2} x^{4}+10 a^{2} d e \,x^{2}+20 a \,d^{2} b \,x^{2}+15 a^{2} d^{2}\right ) \left (e \,x^{2}+d \right )^{\frac {7}{2}}}{15 a^{3} \left (b e \,x^{4}+a e \,x^{2}+b d \,x^{2}+a d \right )^{\frac {7}{2}}}\)	\(107\)

Input:

int((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x,method=_RETURNVER 
BOSE)
 

Output:

1/15/(e*x^2+d)^(1/2)*((e*x^2+d)*(b*x^2+a))^(1/2)*x*(3*a^2*e^2*x^4+4*a*b*d* 
e*x^4+8*b^2*d^2*x^4+10*a^2*d*e*x^2+20*a*b*d^2*x^2+15*a^2*d^2)/(b*x^2+a)^3/ 
a^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.80 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} {\left ({\left (8 \, b^{2} d^{2} + 4 \, a b d e + 3 \, a^{2} e^{2}\right )} x^{5} + 15 \, a^{2} d^{2} x + 10 \, {\left (2 \, a b d^{2} + a^{2} d e\right )} x^{3}\right )} \sqrt {e x^{2} + d}}{15 \, {\left (a^{3} b^{3} e x^{8} + a^{6} d + {\left (a^{3} b^{3} d + 3 \, a^{4} b^{2} e\right )} x^{6} + 3 \, {\left (a^{4} b^{2} d + a^{5} b e\right )} x^{4} + {\left (3 \, a^{5} b d + a^{6} e\right )} x^{2}\right )}} \] Input:

integrate((e*x^2+d)^(11/2)/(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)*((8*b^2*d^2 + 4*a*b*d*e + 3*a^2 
*e^2)*x^5 + 15*a^2*d^2*x + 10*(2*a*b*d^2 + a^2*d*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. \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**(11/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {11}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x, algorithm= 
"maxima")
 

Output:

integrate((e*x^2 + d)^(11/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(7/2), x)
 

Giac [F]

\[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {11}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(11/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(7/2),x, algorithm= 
"giac")
 

Output:

integrate((e*x^2 + d)^(11/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 17.79 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {\sqrt {b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d}\,\left (\frac {x^5\,\sqrt {e\,x^2+d}\,\left (\frac {a^2\,e^2}{5}+\frac {4\,a\,b\,d\,e}{15}+\frac {8\,b^2\,d^2}{15}\right )}{a^3\,b^3\,e}+\frac {d^2\,x\,\sqrt {e\,x^2+d}}{a\,b^3\,e}+\frac {2\,d\,x^3\,\sqrt {e\,x^2+d}\,\left (a\,e+2\,b\,d\right )}{3\,a^2\,b^3\,e}\right )}{x^8+\frac {a^3\,d}{b^3\,e}+\frac {x^6\,\left (3\,a\,e+b\,d\right )}{b\,e}+\frac {x^2\,\left (e\,a^6+3\,b\,d\,a^5\right )}{a^3\,b^3\,e}+\frac {3\,a\,x^4\,\left (a\,e+b\,d\right )}{b^2\,e}} \] Input:

int((d + e*x^2)^(11/2)/(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)*((x^5*(d + e*x^2)^(1/2)*((a^2*e^2 
)/5 + (8*b^2*d^2)/15 + (4*a*b*d*e)/15))/(a^3*b^3*e) + (d^2*x*(d + e*x^2)^( 
1/2))/(a*b^3*e) + (2*d*x^3*(d + e*x^2)^(1/2)*(a*e + 2*b*d))/(3*a^2*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 = 326, normalized size of antiderivative = 1.55 \[ \int \frac {\left (d+e x^2\right )^{11/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{7/2}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d^{2} x +10 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d e \,x^{3}+3 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} e^{2} x^{5}+20 \sqrt {b \,x^{2}+a}\, a \,b^{4} d^{2} x^{3}+4 \sqrt {b \,x^{2}+a}\, a \,b^{4} d e \,x^{5}+8 \sqrt {b \,x^{2}+a}\, b^{5} d^{2} x^{5}+3 \sqrt {b}\, a^{5} e^{2}-4 \sqrt {b}\, a^{4} b d e +9 \sqrt {b}\, a^{4} b \,e^{2} x^{2}-8 \sqrt {b}\, a^{3} b^{2} d^{2}-12 \sqrt {b}\, a^{3} b^{2} d e \,x^{2}+9 \sqrt {b}\, a^{3} b^{2} e^{2} x^{4}-24 \sqrt {b}\, a^{2} b^{3} d^{2} x^{2}-12 \sqrt {b}\, a^{2} b^{3} d e \,x^{4}+3 \sqrt {b}\, a^{2} b^{3} e^{2} x^{6}-24 \sqrt {b}\, a \,b^{4} d^{2} x^{4}-4 \sqrt {b}\, a \,b^{4} d e \,x^{6}-8 \sqrt {b}\, b^{5} d^{2} x^{6}}{15 a^{3} b^{3} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:

int((e*x^2+d)^(11/2)/(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**3*d**2*x + 10*sqrt(a + b*x**2)*a**2*b**3*d*e* 
x**3 + 3*sqrt(a + b*x**2)*a**2*b**3*e**2*x**5 + 20*sqrt(a + b*x**2)*a*b**4 
*d**2*x**3 + 4*sqrt(a + b*x**2)*a*b**4*d*e*x**5 + 8*sqrt(a + b*x**2)*b**5* 
d**2*x**5 + 3*sqrt(b)*a**5*e**2 - 4*sqrt(b)*a**4*b*d*e + 9*sqrt(b)*a**4*b* 
e**2*x**2 - 8*sqrt(b)*a**3*b**2*d**2 - 12*sqrt(b)*a**3*b**2*d*e*x**2 + 9*s 
qrt(b)*a**3*b**2*e**2*x**4 - 24*sqrt(b)*a**2*b**3*d**2*x**2 - 12*sqrt(b)*a 
**2*b**3*d*e*x**4 + 3*sqrt(b)*a**2*b**3*e**2*x**6 - 24*sqrt(b)*a*b**4*d**2 
*x**4 - 4*sqrt(b)*a*b**4*d*e*x**6 - 8*sqrt(b)*b**5*d**2*x**6)/(15*a**3*b** 
3*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))