\(\int \frac {(d+e x^2)^{9/2}}{(a d+(b d+a e) x^2+b e x^4)^{5/2}} \, dx\) [16]

Optimal result

Integrand size = 37, antiderivative size = 174 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {(b d-a e)^2 x \left (d+e x^2\right )^{3/2}}{3 a b^2 \left (a d+(b d+a e) x^2+b e x^4\right )^{3/2}}+\frac {2 (b d-a e) (b d+2 a e) x \sqrt {d+e x^2}}{3 a^2 b^2 \sqrt {a d+(b d+a e) x^2+b e x^4}}+\frac {e^2 \text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.73 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {\left (d+e x^2\right )^{3/2} \left (\sqrt {b} (b d-a e) x \left (3 a^2 e+2 b^2 d x^2+a b \left (3 d+4 e x^2\right )\right )-3 a^2 e^2 \left (a+b x^2\right )^{3/2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{3 a^2 b^{5/2} \left (\left (a+b x^2\right ) \left (d+e x^2\right )\right )^{3/2}} \] Input:

Integrate[(d + e*x^2)^(9/2)/(a*d + (b*d + a*e)*x^2 + b*e*x^4)^(5/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1395, 315, 298, 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.

\(\displaystyle \int \frac {\left (d+e x^2\right )^{9/2}}{\left (x^2 (a e+b d)+a d+b e x^4\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[(d + e*x^2)^(9/2)/(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))/(3*a*b*(a + 
b*x^2)^(3/2)) + (((b*d - a*e)*(2*b*d + 3*a*e)*x)/(a*b*Sqrt[a + b*x^2]) + ( 
3*a*e^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/b^(3/2))/(3*a*b)))/Sqrt[a*d 
+ (b*d + a*e)*x^2 + b*e*x^4]
 

Defintions of rubi rules used

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 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 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 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 = 173, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \left (3 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2} b \,e^{2} x^{2} \sqrt {b \,x^{2}+a}-4 b^{\frac {3}{2}} a^{2} e^{2} x^{3}+2 b^{\frac {5}{2}} a d e \,x^{3}+2 b^{\frac {7}{2}} d^{2} x^{3}+3 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{3} e^{2} \sqrt {b \,x^{2}+a}-3 \sqrt {b}\, a^{3} e^{2} x +3 b^{\frac {5}{2}} a \,d^{2} x \right )}{3 b^{\frac {5}{2}} \sqrt {e \,x^{2}+d}\, \left (b \,x^{2}+a \right )^{2} a^{2}}\)	\(173\)

Input:

int((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x,method=_RETURNVERB 
OSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 575, normalized size of antiderivative = 3.30 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (a^{2} b^{2} e^{3} x^{6} + a^{4} d e^{2} + {\left (a^{2} b^{2} d e^{2} + 2 \, a^{3} b e^{3}\right )} x^{4} + {\left (2 \, a^{3} b d e^{2} + a^{4} e^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (\frac {2 \, b e x^{4} + {\left (2 \, b d + a e\right )} x^{2} + 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d} \sqrt {b} x + a d}{e x^{2} + d}\right ) + 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} {\left (2 \, {\left (b^{4} d^{2} + a b^{3} d e - 2 \, a^{2} b^{2} e^{2}\right )} x^{3} + 3 \, {\left (a b^{3} d^{2} - a^{3} b e^{2}\right )} x\right )} \sqrt {e x^{2} + d}}{6 \, {\left (a^{2} b^{5} e x^{6} + a^{4} b^{3} d + {\left (a^{2} b^{5} d + 2 \, a^{3} b^{4} e\right )} x^{4} + {\left (2 \, a^{3} b^{4} d + a^{4} b^{3} e\right )} x^{2}\right )}}, -\frac {3 \, {\left (a^{2} b^{2} e^{3} x^{6} + a^{4} d e^{2} + {\left (a^{2} b^{2} d e^{2} + 2 \, a^{3} b e^{3}\right )} x^{4} + {\left (2 \, a^{3} b d e^{2} + a^{4} e^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-b} x}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}}\right ) - \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} {\left (2 \, {\left (b^{4} d^{2} + a b^{3} d e - 2 \, a^{2} b^{2} e^{2}\right )} x^{3} + 3 \, {\left (a b^{3} d^{2} - a^{3} b e^{2}\right )} x\right )} \sqrt {e x^{2} + d}}{3 \, {\left (a^{2} b^{5} e x^{6} + a^{4} b^{3} d + {\left (a^{2} b^{5} d + 2 \, a^{3} b^{4} e\right )} x^{4} + {\left (2 \, a^{3} b^{4} d + a^{4} b^{3} e\right )} x^{2}\right )}}\right ] \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" 
fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x**2+d)**(9/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int { \frac {{\left (e x^{2} + d\right )}^{\frac {9}{2}}}{{\left (b e x^{4} + {\left (b d + a e\right )} x^{2} + a d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" 
maxima")
 

Output:

integrate((e*x^2 + d)^(9/2)/(b*e*x^4 + (b*d + a*e)*x^2 + a*d)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.60 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {x {\left (\frac {2 \, {\left (b^{5} d^{2} + a b^{4} d e - 2 \, a^{2} b^{3} e^{2}\right )} x^{2}}{a^{2} b^{4}} + \frac {3 \, {\left (a b^{4} d^{2} - a^{3} b^{2} e^{2}\right )}}{a^{2} b^{4}}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {e^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} \] Input:

integrate((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\int \frac {{\left (e\,x^2+d\right )}^{9/2}}{{\left (b\,e\,x^4+\left (a\,e+b\,d\right )\,x^2+a\,d\right )}^{5/2}} \,d x \] Input:

int((d + e*x^2)^(9/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(5/2),x)
 

Output:

int((d + e*x^2)^(9/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.72 \[ \int \frac {\left (d+e x^2\right )^{9/2}}{\left (a d+(b d+a e) x^2+b e x^4\right )^{5/2}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a^{3} b \,e^{2} x -4 \sqrt {b \,x^{2}+a}\, a^{2} b^{2} e^{2} x^{3}+3 \sqrt {b \,x^{2}+a}\, a \,b^{3} d^{2} x +2 \sqrt {b \,x^{2}+a}\, a \,b^{3} d e \,x^{3}+2 \sqrt {b \,x^{2}+a}\, b^{4} d^{2} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} e^{2}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b \,e^{2} x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b^{2} e^{2} x^{4}+2 \sqrt {b}\, a^{3} b d e -2 \sqrt {b}\, a^{2} b^{2} d^{2}+4 \sqrt {b}\, a^{2} b^{2} d e \,x^{2}-4 \sqrt {b}\, a \,b^{3} d^{2} x^{2}+2 \sqrt {b}\, a \,b^{3} d e \,x^{4}-2 \sqrt {b}\, b^{4} d^{2} x^{4}}{3 a^{2} b^{3} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

int((e*x^2+d)^(9/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(5/2),x)
 

Output:

( - 3*sqrt(a + b*x**2)*a**3*b*e**2*x - 4*sqrt(a + b*x**2)*a**2*b**2*e**2*x 
**3 + 3*sqrt(a + b*x**2)*a*b**3*d**2*x + 2*sqrt(a + b*x**2)*a*b**3*d*e*x** 
3 + 2*sqrt(a + b*x**2)*b**4*d**2*x**3 + 3*sqrt(b)*log((sqrt(a + b*x**2) + 
sqrt(b)*x)/sqrt(a))*a**4*e**2 + 6*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)* 
x)/sqrt(a))*a**3*b*e**2*x**2 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x 
)/sqrt(a))*a**2*b**2*e**2*x**4 + 2*sqrt(b)*a**3*b*d*e - 2*sqrt(b)*a**2*b** 
2*d**2 + 4*sqrt(b)*a**2*b**2*d*e*x**2 - 4*sqrt(b)*a*b**3*d**2*x**2 + 2*sqr 
t(b)*a*b**3*d*e*x**4 - 2*sqrt(b)*b**4*d**2*x**4)/(3*a**2*b**3*(a**2 + 2*a* 
b*x**2 + b**2*x**4))