\(\int \frac {(d+e x^2)^{7/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx\) [1]

Optimal result

Integrand size = 37, antiderivative size = 261 \[ \int \frac {\left (d+e x^2\right )^{7/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {e \left (24 b^2 d^2-18 a b d e+5 a^2 e^2\right ) x \sqrt {a d+(b d+a e) x^2+b e x^4}}{16 b^3 \sqrt {d+e x^2}}+\frac {e^2 (18 b d-5 a e) x^3 \sqrt {a d+(b d+a e) x^2+b e x^4}}{24 b^2 \sqrt {d+e x^2}}+\frac {e^3 x^5 \sqrt {a d+(b d+a e) x^2+b e x^4}}{6 b \sqrt {d+e x^2}}+\frac {(2 b d-a e) \left (8 b^2 d^2-8 a b d e+5 a^2 e^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x \sqrt {d+e x^2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}}\right )}{16 b^{7/2}} \] Output:

1/16*e*(5*a^2*e^2-18*a*b*d*e+24*b^2*d^2)*x*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/ 
2)/b^3/(e*x^2+d)^(1/2)+1/24*e^2*(-5*a*e+18*b*d)*x^3*(a*d+(a*e+b*d)*x^2+b*e 
*x^4)^(1/2)/b^2/(e*x^2+d)^(1/2)+1/6*e^3*x^5*(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1 
/2)/b/(e*x^2+d)^(1/2)+1/16*(-a*e+2*b*d)*(5*a^2*e^2-8*a*b*d*e+8*b^2*d^2)*ar 
ctanh(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 = 0.22 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.67 \[ \int \frac {\left (d+e x^2\right )^{7/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {\sqrt {d+e x^2} \left (\sqrt {b} e x \left (a+b x^2\right ) \left (15 a^2 e^2-2 a b e \left (27 d+5 e x^2\right )+4 b^2 \left (18 d^2+9 d e x^2+2 e^2 x^4\right )\right )+3 \left (-16 b^3 d^3+24 a b^2 d^2 e-18 a^2 b d e^2+5 a^3 e^3\right ) \sqrt {a+b x^2} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )\right )}{48 b^{7/2} \sqrt {\left (a+b x^2\right ) \left (d+e x^2\right )}} \] Input:

Integrate[(d + e*x^2)^(7/2)/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)*(15*a^2*e^2 - 2*a*b*e*(27*d + 5* 
e*x^2) + 4*b^2*(18*d^2 + 9*d*e*x^2 + 2*e^2*x^4)) + 3*(-16*b^3*d^3 + 24*a*b 
^2*d^2*e - 18*a^2*b*d*e^2 + 5*a^3*e^3)*Sqrt[a + b*x^2]*Log[-(Sqrt[b]*x) + 
Sqrt[a + b*x^2]]))/(48*b^(7/2)*Sqrt[(a + b*x^2)*(d + e*x^2)])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1395, 318, 403, 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.

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

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 318

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

\(\Big \downarrow \) 403

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\int \frac {e \left (44 b^2 d^2-44 a b e d+15 a^2 e^2\right ) x^2+d \left (24 b^2 d^2-14 a b e d+5 a^2 e^2\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {5 e x \sqrt {a+b x^2} \left (d+e x^2\right ) (2 b d-a e)}{4 b}}{6 b}+\frac {e x \sqrt {a+b x^2} \left (d+e x^2\right )^2}{6 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\frac {3 (2 b d-a e) \left (5 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {e x \sqrt {a+b x^2} \left (15 a^2 e^2-44 a b d e+44 b^2 d^2\right )}{2 b}}{4 b}+\frac {5 e x \sqrt {a+b x^2} \left (d+e x^2\right ) (2 b d-a e)}{4 b}}{6 b}+\frac {e x \sqrt {a+b x^2} \left (d+e x^2\right )^2}{6 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\frac {3 (2 b d-a e) \left (5 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {e x \sqrt {a+b x^2} \left (15 a^2 e^2-44 a b d e+44 b^2 d^2\right )}{2 b}}{4 b}+\frac {5 e x \sqrt {a+b x^2} \left (d+e x^2\right ) (2 b d-a e)}{4 b}}{6 b}+\frac {e x \sqrt {a+b x^2} \left (d+e x^2\right )^2}{6 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {a+b x^2} \sqrt {d+e x^2} \left (\frac {\frac {\frac {3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (2 b d-a e) \left (5 a^2 e^2-8 a b d e+8 b^2 d^2\right )}{2 b^{3/2}}+\frac {e x \sqrt {a+b x^2} \left (15 a^2 e^2-44 a b d e+44 b^2 d^2\right )}{2 b}}{4 b}+\frac {5 e x \sqrt {a+b x^2} \left (d+e x^2\right ) (2 b d-a e)}{4 b}}{6 b}+\frac {e x \sqrt {a+b x^2} \left (d+e x^2\right )^2}{6 b}\right )}{\sqrt {x^2 (a e+b d)+a d+b e x^4}}\)

Input:

Int[(d + e*x^2)^(7/2)/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)^2)/(6*b 
) + ((5*e*(2*b*d - a*e)*x*Sqrt[a + b*x^2]*(d + e*x^2))/(4*b) + ((e*(44*b^2 
*d^2 - 44*a*b*d*e + 15*a^2*e^2)*x*Sqrt[a + b*x^2])/(2*b) + (3*(2*b*d - a*e 
)*(8*b^2*d^2 - 8*a*b*d*e + 5*a^2*e^2)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]] 
)/(2*b^(3/2)))/(4*b))/(6*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 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 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 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 = 0.13 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {e x \left (8 e^{2} x^{4} b^{2}-10 a b \,e^{2} x^{2}+36 b^{2} d e \,x^{2}+15 a^{2} e^{2}-54 a b d e +72 b^{2} d^{2}\right ) \left (b \,x^{2}+a \right ) \sqrt {e \,x^{2}+d}}{48 b^{3} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}-\frac {\left (5 a^{3} e^{3}-18 a^{2} b d \,e^{2}+24 a \,b^{2} d^{2} e -16 b^{3} d^{3}\right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) \sqrt {b \,x^{2}+a}\, \sqrt {e \,x^{2}+d}}{16 b^{\frac {7}{2}} \sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}}\)	\(189\)
default	\(-\frac {\sqrt {\left (e \,x^{2}+d \right ) \left (b \,x^{2}+a \right )}\, \left (-8 b^{\frac {5}{2}} e^{3} x^{5} \sqrt {b \,x^{2}+a}+10 a \,b^{\frac {3}{2}} e^{3} x^{3} \sqrt {b \,x^{2}+a}-36 b^{\frac {5}{2}} d \,e^{2} x^{3} \sqrt {b \,x^{2}+a}-15 a^{2} e^{3} x \sqrt {b \,x^{2}+a}\, \sqrt {b}+54 a \,b^{\frac {3}{2}} d \,e^{2} x \sqrt {b \,x^{2}+a}-72 b^{\frac {5}{2}} d^{2} e x \sqrt {b \,x^{2}+a}+15 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{3} e^{3}-54 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a^{2} b d \,e^{2}+72 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) a \,b^{2} d^{2} e -48 \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right ) b^{3} d^{3}\right )}{48 b^{\frac {7}{2}} \sqrt {e \,x^{2}+d}\, \sqrt {b \,x^{2}+a}}\)	\(264\)

Input:

int((e*x^2+d)^(7/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/48*e*x*(8*b^2*e^2*x^4-10*a*b*e^2*x^2+36*b^2*d*e*x^2+15*a^2*e^2-54*a*b*d* 
e+72*b^2*d^2)*(b*x^2+a)/b^3/((e*x^2+d)*(b*x^2+a))^(1/2)*(e*x^2+d)^(1/2)-1/ 
16*(5*a^3*e^3-18*a^2*b*d*e^2+24*a*b^2*d^2*e-16*b^3*d^3)/b^(7/2)*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 = 535, normalized size of antiderivative = 2.05 \[ \int \frac {\left (d+e x^2\right )^{7/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\left [-\frac {3 \, {\left (16 \, b^{3} d^{4} - 24 \, a b^{2} d^{3} e + 18 \, a^{2} b d^{2} e^{2} - 5 \, a^{3} d e^{3} + {\left (16 \, b^{3} d^{3} e - 24 \, a b^{2} d^{2} e^{2} + 18 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x^{2}\right )} \sqrt {b} \log \left (\frac {2 \, b e x^{4} + {\left (2 \, b d + a e\right )} x^{2} - 2 \, \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d} \sqrt {b} x + a d}{e x^{2} + d}\right ) - 2 \, {\left (8 \, b^{3} e^{3} x^{5} + 2 \, {\left (18 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} d^{2} e - 18 \, a b^{2} d e^{2} + 5 \, a^{2} b e^{3}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{96 \, {\left (b^{4} e x^{2} + b^{4} d\right )}}, -\frac {3 \, {\left (16 \, b^{3} d^{4} - 24 \, a b^{2} d^{3} e + 18 \, a^{2} b d^{2} e^{2} - 5 \, a^{3} d e^{3} + {\left (16 \, b^{3} d^{3} e - 24 \, a b^{2} d^{2} e^{2} + 18 \, a^{2} b d e^{3} - 5 \, a^{3} e^{4}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {e x^{2} + d} \sqrt {-b} x}{\sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d}}\right ) - {\left (8 \, b^{3} e^{3} x^{5} + 2 \, {\left (18 \, b^{3} d e^{2} - 5 \, a b^{2} e^{3}\right )} x^{3} + 3 \, {\left (24 \, b^{3} d^{2} e - 18 \, a b^{2} d e^{2} + 5 \, a^{2} b e^{3}\right )} x\right )} \sqrt {b e x^{4} + {\left (b d + a e\right )} x^{2} + a d} \sqrt {e x^{2} + d}}{48 \, {\left (b^{4} e x^{2} + b^{4} d\right )}}\right ] \] Input:

integrate((e*x^2+d)^(7/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
fricas")
 

Output:

[-1/96*(3*(16*b^3*d^4 - 24*a*b^2*d^3*e + 18*a^2*b*d^2*e^2 - 5*a^3*d*e^3 + 
(16*b^3*d^3*e - 24*a*b^2*d^2*e^2 + 18*a^2*b*d*e^3 - 5*a^3*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*(8*b^3*e^3*x^5 + 2* 
(18*b^3*d*e^2 - 5*a*b^2*e^3)*x^3 + 3*(24*b^3*d^2*e - 18*a*b^2*d*e^2 + 5*a^ 
2*b*e^3)*x)*sqrt(b*e*x^4 + (b*d + a*e)*x^2 + a*d)*sqrt(e*x^2 + d))/(b^4*e* 
x^2 + b^4*d), -1/48*(3*(16*b^3*d^4 - 24*a*b^2*d^3*e + 18*a^2*b*d^2*e^2 - 5 
*a^3*d*e^3 + (16*b^3*d^3*e - 24*a*b^2*d^2*e^2 + 18*a^2*b*d*e^3 - 5*a^3*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)) - (8*b^3*e^3*x^5 + 2*(18*b^3*d*e^2 - 5*a*b^2*e^3)*x^3 + 3*( 
24*b^3*d^2*e - 18*a*b^2*d*e^2 + 5*a^2*b*e^3)*x)*sqrt(b*e*x^4 + (b*d + a*e) 
*x^2 + a*d)*sqrt(e*x^2 + d))/(b^4*e*x^2 + b^4*d)]
 

Sympy [F(-1)]

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

integrate((e*x**2+d)**(7/2)/(a*d+(a*e+b*d)*x**2+b*e*x**4)**(1/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*x^2+d)^(7/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
maxima")
 

Output:

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

Giac [F]

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

integrate((e*x^2+d)^(7/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

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

int((d + e*x^2)^(7/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2),x)
 

Output:

int((d + e*x^2)^(7/2)/(a*d + x^2*(a*e + b*d) + b*e*x^4)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.92 \[ \int \frac {\left (d+e x^2\right )^{7/2}}{\sqrt {a d+(b d+a e) x^2+b e x^4}} \, dx=\frac {15 \sqrt {b \,x^{2}+a}\, a^{2} b \,e^{3} x -54 \sqrt {b \,x^{2}+a}\, a \,b^{2} d \,e^{2} x -10 \sqrt {b \,x^{2}+a}\, a \,b^{2} e^{3} x^{3}+72 \sqrt {b \,x^{2}+a}\, b^{3} d^{2} e x +36 \sqrt {b \,x^{2}+a}\, b^{3} d \,e^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} e^{3} x^{5}-15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} e^{3}+54 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b d \,e^{2}-72 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d^{2} e +48 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{3} d^{3}}{48 b^{4}} \] Input:

int((e*x^2+d)^(7/2)/(a*d+(a*e+b*d)*x^2+b*e*x^4)^(1/2),x)
 

Output:

(15*sqrt(a + b*x**2)*a**2*b*e**3*x - 54*sqrt(a + b*x**2)*a*b**2*d*e**2*x - 
 10*sqrt(a + b*x**2)*a*b**2*e**3*x**3 + 72*sqrt(a + b*x**2)*b**3*d**2*e*x 
+ 36*sqrt(a + b*x**2)*b**3*d*e**2*x**3 + 8*sqrt(a + b*x**2)*b**3*e**3*x**5 
 - 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*e**3 + 54*s 
qrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d*e**2 - 72*sqrt 
(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*d**2*e + 48*sqrt(b) 
*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*b**3*d**3)/(48*b**4)