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

Optimal result

Integrand size = 37, antiderivative size = 180 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {e (8 b d-3 a e) x \sqrt {a d+(b d+a e) x^2+b e x^4}}{8 b^2 \sqrt {d+e x^2}}+\frac {e^2 x^3 \sqrt {a d+(b d+a e) x^2+b e x^4}}{4 b \sqrt {d+e x^2}}+\frac {\left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{8 b^{5/2}} \] Output:

1/8*e*(-3*a*e+8*b*d)*x*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)/b^2/(e*x^2+d)^(1/ 
2)+1/4*e^2*x^3*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2)/b/(e*x^2+d)^(1/2)+1/8*(3* 
a^2*e^2-8*a*b*d*e+8*b^2*d^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 = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.71 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\sqrt {d+e x^2} \left (\sqrt {b} e x \left (a+b x^2\right ) \left (8 b d-3 a e+2 b e x^2\right )+\left (-8 b^2 d^2+8 a b d e-3 a^2 e^2\right ) \sqrt {a+b x^2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{8 b^{5/2} \sqrt {\left (a+b x^2\right ) \left (d+e x^2\right )}} \] Input:

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

Output:

(Sqrt[d + e*x^2]*(Sqrt[b]*e*x*(a + b*x^2)*(8*b*d - 3*a*e + 2*b*e*x^2) + (- 
8*b^2*d^2 + 8*a*b*d*e - 3*a^2*e^2)*Sqrt[a + b*x^2]*Log[-(Sqrt[b]*x) + Sqrt 
[a + b*x^2]]))/(8*b^(5/2)*Sqrt[(a + b*x^2)*(d + e*x^2)])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1395, 318, 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.

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

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 299

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(Sqrt[a + b*x^2]*Sqrt[d + e*x^2]*((e*x*Sqrt[a + b*x^2]*(d + e*x^2))/(4*b) 
+ ((3*e*(2*b*d - a*e)*x*Sqrt[a + b*x^2])/(2*b) + ((8*b^2*d^2 - 8*a*b*d*e + 
 3*a^2*e^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/(4*b)))/Sqr 
t[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 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 318

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S 
imp[1/(b*(2*(p + q) + 1))   Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b 
*c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 
 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G 
tQ[q, 1] && NeQ[2*(p + q) + 1, 0] &&  !IGtQ[p, 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 = 137, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {e x \left (-2 b e \,x^{2}+3 a e -8 b d \right ) \left (b \,x^{2}+a \right ) \sqrt {e \,x^{2}+d}}{8 b^{2} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}+\frac {\left (3 a^{2} e^{2}-8 a b d e +8 b^{2} d^{2}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {e \,x^{2}+d}}{8 b^{\frac {5}{2}} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}\)	\(137\)
default	\(\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \left (2 b^{\frac {3}{2}} e^{2} x^{3} \sqrt {b \,x^{2}+a}-3 a \,e^{2} x \sqrt {b}\, \sqrt {b \,x^{2}+a}+8 b^{\frac {3}{2}} d e x \sqrt {b \,x^{2}+a}+3 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2} e^{2}-8 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a b d e +8 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) b^{2} d^{2}\right )}{8 b^{\frac {5}{2}} \sqrt {e \,x^{2}+d}\, \sqrt {b \,x^{2}+a}}\)	\(168\)

Input:

int((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.22 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\left [\frac {{\left (8 \, b^{2} d^{3} - 8 \, a b d^{2} e + 3 \, a^{2} d e^{2} + {\left (8 \, b^{2} d^{2} e - 8 \, a b d e^{2} + 3 \, a^{2} e^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (\frac {2 \, b e x^{4} + {\left (2 \, b d + a e\right )} x^{2} + 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d} \sqrt {b} x + a d}{e x^{2} + d}\right ) + 2 \, {\left (2 \, b^{2} e^{2} x^{3} + {\left (8 \, b^{2} d e - 3 \, a b e^{2}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{16 \, {\left (b^{3} e x^{2} + b^{3} d\right )}}, -\frac {{\left (8 \, b^{2} d^{3} - 8 \, a b d^{2} e + 3 \, a^{2} d e^{2} + {\left (8 \, b^{2} d^{2} e - 8 \, a b d e^{2} + 3 \, a^{2} e^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-b} x}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}}\right ) - {\left (2 \, b^{2} e^{2} x^{3} + {\left (8 \, b^{2} d e - 3 \, a b e^{2}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{8 \, {\left (b^{3} e x^{2} + b^{3} d\right )}}\right ] \] Input:

integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
fricas")
 

Output:

[1/16*((8*b^2*d^3 - 8*a*b*d^2*e + 3*a^2*d*e^2 + (8*b^2*d^2*e - 8*a*b*d*e^2 
 + 3*a^2*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*(2*b^2*e^2*x^3 + (8*b^2*d*e - 3*a*b*e^2)*x)*sqrt(b*e*x^4 + (b*d + a* 
e)*x^2 + a*d)*sqrt(e*x^2 + d))/(b^3*e*x^2 + b^3*d), -1/8*((8*b^2*d^3 - 8*a 
*b*d^2*e + 3*a^2*d*e^2 + (8*b^2*d^2*e - 8*a*b*d*e^2 + 3*a^2*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)) - (2*b^2*e^2*x^3 + (8*b^2*d*e - 3*a*b*e^2)*x)*sqrt(b*e*x^4 + (b*d + a* 
e)*x^2 + a*d)*sqrt(e*x^2 + d))/(b^3*e*x^2 + b^3*d)]
 

Sympy [F]

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

integrate((e*x**2+d)**(5/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(1/2),x)
 

Output:

Integral((d + e*x**2)**(5/2)/sqrt((a + b*x**2)*(d + e*x**2)), x)
 

Maxima [F]

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

integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
maxima")
 

Output:

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

Giac [F]

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

integrate((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

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

int((d + e*x^2)^(5/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2),x)
 

Output:

int((d + e*x^2)^(5/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.79 \[ \int \frac {\left (d+e x^2\right )^{5/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {-3 \sqrt {b \,x^{2}+a}\, a b \,e^{2} x +8 \sqrt {b \,x^{2}+a}\, b^{2} d e x +2 \sqrt {b \,x^{2}+a}\, b^{2} e^{2} x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} e^{2}-8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b d e +8 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} d^{2}}{8 b^{3}} \] Input:

int((e*x^2+d)^(5/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x)
 

Output:

( - 3*sqrt(a + b*x**2)*a*b*e**2*x + 8*sqrt(a + b*x**2)*b**2*d*e*x + 2*sqrt 
(a + b*x**2)*b**2*e**2*x**3 + 3*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x) 
/sqrt(a))*a**2*e**2 - 8*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a) 
)*a*b*d*e + 8*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**2*d** 
2)/(8*b**3)