\(\int \frac {(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [183]

Optimal result

Integrand size = 28, antiderivative size = 136 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {e \sqrt {d+e x}}{4 b^2 (a+b x)^2}-\frac {e^2 \sqrt {d+e x}}{8 b^2 (b d-a e) (a+b x)}-\frac {(d+e x)^{3/2}}{3 b (a+b x)^3}+\frac {e^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{5/2} (b d-a e)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\sqrt {d+e x} \left (-3 a^2 e^2-2 a b e (d+4 e x)+b^2 \left (8 d^2+14 d e x+3 e^2 x^2\right )\right )}{24 b^2 (-b d+a e) (a+b x)^3}+\frac {e^3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{5/2} (-b d+a e)^{3/2}} \] Input:

Integrate[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
 

Output:

(Sqrt[d + e*x]*(-3*a^2*e^2 - 2*a*b*e*(d + 4*e*x) + b^2*(8*d^2 + 14*d*e*x + 
 3*e^2*x^2)))/(24*b^2*(-(b*d) + a*e)*(a + b*x)^3) + (e^3*ArcTan[(Sqrt[b]*S 
qrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*b^(5/2)*(-(b*d) + a*e)^(3/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1098, 27, 51, 51, 52, 73, 221}

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.

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

\(\Big \downarrow \) 1098

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
 

Output:

-1/3*(d + e*x)^(3/2)/(b*(a + b*x)^3) + (e*(-1/2*Sqrt[d + e*x]/(b*(a + b*x) 
^2) + (e*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTanh[(Sqrt[b]*S 
qrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/(4*b)))/(2*b 
)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 1098

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[ 
{a, b, c, d, e, m}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.51 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {e^{3} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )-\left (\left (3 e x +2 d \right ) b +a e \right ) \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, \left (\frac {\left (-e x -4 d \right ) b}{3}+a e \right )}{8 \sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right ) b^{2} \left (b x +a \right )^{3}}\)	\(119\)
derivativedivides	\(2 e^{3} \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}}}{16 a e -16 b d}-\frac {\left (e x +d \right )^{\frac {3}{2}}}{6 b}-\frac {\left (a e -b d \right ) \sqrt {e x +d}}{16 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \left (a e -b d \right ) b^{2} \sqrt {b \left (a e -b d \right )}}\right )\)	\(126\)
default	\(2 e^{3} \left (\frac {\frac {\left (e x +d \right )^{\frac {5}{2}}}{16 a e -16 b d}-\frac {\left (e x +d \right )^{\frac {3}{2}}}{6 b}-\frac {\left (a e -b d \right ) \sqrt {e x +d}}{16 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \left (a e -b d \right ) b^{2} \sqrt {b \left (a e -b d \right )}}\right )\)	\(126\)

Input:

int((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (112) = 224\).

Time = 0.11 (sec) , antiderivative size = 666, normalized size of antiderivative = 4.90 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\left [-\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}, -\frac {3 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (8 \, b^{4} d^{3} - 10 \, a b^{3} d^{2} e - a^{2} b^{2} d e^{2} + 3 \, a^{3} b e^{3} + 3 \, {\left (b^{4} d e^{2} - a b^{3} e^{3}\right )} x^{2} + 2 \, {\left (7 \, b^{4} d^{2} e - 11 \, a b^{3} d e^{2} + 4 \, a^{2} b^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{5} d^{2} - 2 \, a^{4} b^{4} d e + a^{5} b^{3} e^{2} + {\left (b^{8} d^{2} - 2 \, a b^{7} d e + a^{2} b^{6} e^{2}\right )} x^{3} + 3 \, {\left (a b^{7} d^{2} - 2 \, a^{2} b^{6} d e + a^{3} b^{5} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{2} - 2 \, a^{3} b^{5} d e + a^{4} b^{4} e^{2}\right )} x\right )}}\right ] \] Input:

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
 

Output:

[-1/48*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(b 
^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + 
d))/(b*x + a)) + 2*(8*b^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b*e 
^3 + 3*(b^4*d*e^2 - a*b^3*e^3)*x^2 + 2*(7*b^4*d^2*e - 11*a*b^3*d*e^2 + 4*a 
^2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + 
 (b^8*d^2 - 2*a*b^7*d*e + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*d*e 
+ a^3*b^5*e^2)*x^2 + 3*(a^2*b^6*d^2 - 2*a^3*b^5*d*e + a^4*b^4*e^2)*x), -1/ 
24*(3*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*sqrt(-b^2* 
d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (8*b 
^4*d^3 - 10*a*b^3*d^2*e - a^2*b^2*d*e^2 + 3*a^3*b*e^3 + 3*(b^4*d*e^2 - a*b 
^3*e^3)*x^2 + 2*(7*b^4*d^2*e - 11*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt(e*x 
 + d))/(a^3*b^5*d^2 - 2*a^4*b^4*d*e + a^5*b^3*e^2 + (b^8*d^2 - 2*a*b^7*d*e 
 + a^2*b^6*e^2)*x^3 + 3*(a*b^7*d^2 - 2*a^2*b^6*d*e + a^3*b^5*e^2)*x^2 + 3* 
(a^2*b^6*d^2 - 2*a^3*b^5*d*e + a^4*b^4*e^2)*x)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for m 
ore detail
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{3} d - a b^{2} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} + 8 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {e x + d} b^{2} d^{2} e^{3} - 8 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {e x + d} a b d e^{4} - 3 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left (b^{3} d - a b^{2} e\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \] Input:

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
 

Output:

-1/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d - a*b^2*e)*s 
qrt(-b^2*d + a*b*e)) - 1/24*(3*(e*x + d)^(5/2)*b^2*e^3 + 8*(e*x + d)^(3/2) 
*b^2*d*e^3 - 3*sqrt(e*x + d)*b^2*d^2*e^3 - 8*(e*x + d)^(3/2)*a*b*e^4 + 6*s 
qrt(e*x + d)*a*b*d*e^4 - 3*sqrt(e*x + d)*a^2*e^5)/((b^3*d - a*b^2*e)*((e*x 
 + d)*b - b*d + a*e)^3)
 

Mupad [B] (verification not implemented)

Time = 9.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.54 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{8\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,b}-\frac {e^3\,{\left (d+e\,x\right )}^{5/2}}{8\,\left (a\,e-b\,d\right )}+\frac {e^3\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{8\,b^2}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2} \] Input:

int((d + e*x)^(3/2)/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 470, normalized size of antiderivative = 3.46 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{3} e^{3}+9 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} b \,e^{3} x +9 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a \,b^{2} e^{3} x^{2}+3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{3} e^{3} x^{3}-3 \sqrt {e x +d}\, a^{3} b \,e^{3}+\sqrt {e x +d}\, a^{2} b^{2} d \,e^{2}-8 \sqrt {e x +d}\, a^{2} b^{2} e^{3} x +10 \sqrt {e x +d}\, a \,b^{3} d^{2} e +22 \sqrt {e x +d}\, a \,b^{3} d \,e^{2} x +3 \sqrt {e x +d}\, a \,b^{3} e^{3} x^{2}-8 \sqrt {e x +d}\, b^{4} d^{3}-14 \sqrt {e x +d}\, b^{4} d^{2} e x -3 \sqrt {e x +d}\, b^{4} d \,e^{2} x^{2}}{24 b^{3} \left (a^{2} b^{3} e^{2} x^{3}-2 a \,b^{4} d e \,x^{3}+b^{5} d^{2} x^{3}+3 a^{3} b^{2} e^{2} x^{2}-6 a^{2} b^{3} d e \,x^{2}+3 a \,b^{4} d^{2} x^{2}+3 a^{4} b \,e^{2} x -6 a^{3} b^{2} d e x +3 a^{2} b^{3} d^{2} x +a^{5} e^{2}-2 a^{4} b d e +a^{3} b^{2} d^{2}\right )} \] Input:

int((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Output:

(3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d) 
))*a**3*e**3 + 9*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*s 
qrt(a*e - b*d)))*a**2*b*e**3*x + 9*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + 
e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**2*e**3*x**2 + 3*sqrt(b)*sqrt(a*e - 
 b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**3*e**3*x**3 - 3 
*sqrt(d + e*x)*a**3*b*e**3 + sqrt(d + e*x)*a**2*b**2*d*e**2 - 8*sqrt(d + e 
*x)*a**2*b**2*e**3*x + 10*sqrt(d + e*x)*a*b**3*d**2*e + 22*sqrt(d + e*x)*a 
*b**3*d*e**2*x + 3*sqrt(d + e*x)*a*b**3*e**3*x**2 - 8*sqrt(d + e*x)*b**4*d 
**3 - 14*sqrt(d + e*x)*b**4*d**2*e*x - 3*sqrt(d + e*x)*b**4*d*e**2*x**2)/( 
24*b**3*(a**5*e**2 - 2*a**4*b*d*e + 3*a**4*b*e**2*x + a**3*b**2*d**2 - 6*a 
**3*b**2*d*e*x + 3*a**3*b**2*e**2*x**2 + 3*a**2*b**3*d**2*x - 6*a**2*b**3* 
d*e*x**2 + a**2*b**3*e**2*x**3 + 3*a*b**4*d**2*x**2 - 2*a*b**4*d*e*x**3 + 
b**5*d**2*x**3))