\(\int \frac {1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [186]

Optimal result

Integrand size = 28, antiderivative size = 173 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {35 e^3}{8 (b d-a e)^4 \sqrt {d+e x}}-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}-\frac {35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt {d+e x}}+\frac {35 \sqrt {b} e^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{9/2}} \] Output:

-35/8*e^3/(-a*e+b*d)^4/(e*x+d)^(1/2)-1/3/(-a*e+b*d)/(b*x+a)^3/(e*x+d)^(1/2 
)+7/12*e/(-a*e+b*d)^2/(b*x+a)^2/(e*x+d)^(1/2)-35/24*e^2/(-a*e+b*d)^3/(b*x+ 
a)/(e*x+d)^(1/2)+35/8*b^(1/2)*e^3*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d) 
^(1/2))/(-a*e+b*d)^(9/2)
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-48 a^3 e^3-3 a^2 b e^2 (29 d+77 e x)-2 a b^2 e \left (-19 d^2+49 d e x+140 e^2 x^2\right )-b^3 \left (8 d^3-14 d^2 e x+35 d e^2 x^2+105 e^3 x^3\right )}{24 (b d-a e)^4 (a+b x)^3 \sqrt {d+e x}}-\frac {35 \sqrt {b} e^3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 (-b d+a e)^{9/2}} \] Input:

Integrate[1/((d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
 

Output:

(-48*a^3*e^3 - 3*a^2*b*e^2*(29*d + 77*e*x) - 2*a*b^2*e*(-19*d^2 + 49*d*e*x 
 + 140*e^2*x^2) - b^3*(8*d^3 - 14*d^2*e*x + 35*d*e^2*x^2 + 105*e^3*x^3))/( 
24*(b*d - a*e)^4*(a + b*x)^3*Sqrt[d + e*x]) - (35*Sqrt[b]*e^3*ArcTan[(Sqrt 
[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(8*(-(b*d) + a*e)^(9/2))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1098, 27, 52, 52, 52, 61, 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.

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

\(\Big \downarrow \) 1098

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 e \int \frac {1}{(a+b x)^3 (d+e x)^{3/2}}dx}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 e \left (-\frac {5 e \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}}dx}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) \sqrt {d+e x} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 \sqrt {d+e x} (b d-a e)}\right )}{6 (b d-a e)}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)}\)

Input:

Int[1/((d + e*x)^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]
 

Output:

-1/3*1/((b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x]) - (7*e*(-1/2*1/((b*d - a*e) 
*(a + b*x)^2*Sqrt[d + e*x]) - (5*e*(-(1/((b*d - a*e)*(a + b*x)*Sqrt[d + e* 
x])) - (3*e*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*S 
qrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(3/2)))/(2*(b*d - a*e))))/(4*( 
b*d - a*e))))/(6*(b*d - a*e))
 

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 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 61

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

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 = 1.60 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(2 e^{3} \left (-\frac {1}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {b \left (\frac {\frac {19 \left (e x +d \right )^{\frac {5}{2}} b^{2}}{16}+\frac {17 b \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {29}{16} e^{2} a^{2}-\frac {29}{8} a b d e +\frac {29}{16} b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {35 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \sqrt {b \left (a e -b d \right )}}\right )}{\left (a e -b d \right )^{4}}\right )\)	\(156\)
default	\(2 e^{3} \left (-\frac {1}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {b \left (\frac {\frac {19 \left (e x +d \right )^{\frac {5}{2}} b^{2}}{16}+\frac {17 b \left (a e -b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {29}{16} e^{2} a^{2}-\frac {29}{8} a b d e +\frac {29}{16} b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {35 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \sqrt {b \left (a e -b d \right )}}\right )}{\left (a e -b d \right )^{4}}\right )\)	\(156\)
pseudoelliptic	\(-\frac {2 \left (\frac {35 b \,e^{3} \sqrt {e x +d}\, \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16}+\sqrt {b \left (a e -b d \right )}\, \left (\left (\frac {35}{16} b^{3} x^{3}+\frac {35}{6} a \,b^{2} x^{2}+\frac {77}{16} a^{2} b x +a^{3}\right ) e^{3}+\frac {29 b \left (\frac {35}{87} b^{2} x^{2}+\frac {98}{87} a b x +a^{2}\right ) d \,e^{2}}{16}-\frac {19 b^{2} d^{2} \left (\frac {7 b x}{19}+a \right ) e}{24}+\frac {b^{3} d^{3}}{6}\right )\right )}{\sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right )^{4} \left (b x +a \right )^{3}}\)	\(175\)

Input:

int(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
 

Output:

2*e^3*(-1/(a*e-b*d)^4/(e*x+d)^(1/2)-1/(a*e-b*d)^4*b*((19/16*(e*x+d)^(5/2)* 
b^2+17/6*b*(a*e-b*d)*(e*x+d)^(3/2)+(29/16*e^2*a^2-29/8*a*b*d*e+29/16*b^2*d 
^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^3+35/16/(b*(a*e-b*d))^(1/2)*arctan( 
b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (145) = 290\).

Time = 0.14 (sec) , antiderivative size = 1184, normalized size of antiderivative = 6.84 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx =\text {Too large to display} \] Input:

integrate(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
 

Output:

[1/48*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 3*a*b^2*e^4)*x^3 + 3*(a 
*b^2*d*e^3 + a^2*b*e^4)*x^2 + (3*a^2*b*d*e^3 + a^3*e^4)*x)*sqrt(b/(b*d - a 
*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - 
a*e)))/(b*x + a)) - 2*(105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a 
^2*b*d*e^2 + 48*a^3*e^3 + 35*(b^3*d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2* 
e - 14*a*b^2*d*e^2 - 33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3*b^4*d^5 - 4*a^4* 
b^3*d^4*e + 6*a^5*b^2*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 
 4*a*b^6*d^3*e^2 + 6*a^2*b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 
+ (b^7*d^5 - a*b^6*d^4*e - 6*a^2*b^5*d^3*e^2 + 14*a^3*b^4*d^2*e^3 - 11*a^4 
*b^3*d*e^4 + 3*a^5*b^2*e^5)*x^3 + 3*(a*b^6*d^5 - 3*a^2*b^5*d^4*e + 2*a^3*b 
^4*d^3*e^2 + 2*a^4*b^3*d^2*e^3 - 3*a^5*b^2*d*e^4 + a^6*b*e^5)*x^2 + (3*a^2 
*b^5*d^5 - 11*a^3*b^4*d^4*e + 14*a^4*b^3*d^3*e^2 - 6*a^5*b^2*d^2*e^3 - a^6 
*b*d*e^4 + a^7*e^5)*x), -1/24*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 
 3*a*b^2*e^4)*x^3 + 3*(a*b^2*d*e^3 + a^2*b*e^4)*x^2 + (3*a^2*b*d*e^3 + a^3 
*e^4)*x)*sqrt(-b/(b*d - a*e))*arctan(sqrt(e*x + d)*sqrt(-b/(b*d - a*e))) + 
 (105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a^2*b*d*e^2 + 48*a^3*e 
^3 + 35*(b^3*d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2*e - 14*a*b^2*d*e^2 - 
33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3*b^4*d^5 - 4*a^4*b^3*d^4*e + 6*a^5*b^2 
*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6* 
a^2*b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 + (b^7*d^5 - a*b^6...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate(1/(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. \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (145) = 290\).

Time = 0.14 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.88 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {35 \, b e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, e^{3}}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {e x + d}} - \frac {57 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} e^{3} - 136 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d e^{3} + 87 \, \sqrt {e x + d} b^{3} d^{2} e^{3} + 136 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} e^{4} - 174 \, \sqrt {e x + d} a b^{2} d e^{4} + 87 \, \sqrt {e x + d} a^{2} b e^{5}}{24 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \] Input:

integrate(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
 

Output:

-35/8*b*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^4 - 4*a*b 
^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b* 
e)) - 2*e^3/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 
+ a^4*e^4)*sqrt(e*x + d)) - 1/24*(57*(e*x + d)^(5/2)*b^3*e^3 - 136*(e*x + 
d)^(3/2)*b^3*d*e^3 + 87*sqrt(e*x + d)*b^3*d^2*e^3 + 136*(e*x + d)^(3/2)*a* 
b^2*e^4 - 174*sqrt(e*x + d)*a*b^2*d*e^4 + 87*sqrt(e*x + d)*a^2*b*e^5)/((b^ 
4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*((e*x 
 + d)*b - b*d + a*e)^3)
 

Mupad [B] (verification not implemented)

Time = 9.53 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.70 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {2\,e^3}{a\,e-b\,d}+\frac {35\,b^2\,e^3\,{\left (d+e\,x\right )}^2}{3\,{\left (a\,e-b\,d\right )}^3}+\frac {35\,b^3\,e^3\,{\left (d+e\,x\right )}^3}{8\,{\left (a\,e-b\,d\right )}^4}+\frac {77\,b\,e^3\,\left (d+e\,x\right )}{8\,{\left (a\,e-b\,d\right )}^2}}{\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{7/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{5/2}+{\left (d+e\,x\right )}^{3/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}-\frac {35\,\sqrt {b}\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}}\right )}{8\,{\left (a\,e-b\,d\right )}^{9/2}} \] Input:

int(1/((d + e*x)^(3/2)*(a^2 + b^2*x^2 + 2*a*b*x)^2),x)
 

Output:

- ((2*e^3)/(a*e - b*d) + (35*b^2*e^3*(d + e*x)^2)/(3*(a*e - b*d)^3) + (35* 
b^3*e^3*(d + e*x)^3)/(8*(a*e - b*d)^4) + (77*b*e^3*(d + e*x))/(8*(a*e - b* 
d)^2))/((d + e*x)^(1/2)*(a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2 
) + b^3*(d + e*x)^(7/2) - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^(5/2) + (d + e*x 
)^(3/2)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e)) - (35*b^(1/2)*e^3*atan((b 
^(1/2)*(d + e*x)^(1/2)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^ 
3*e - 4*a^3*b*d*e^3))/(a*e - b*d)^(9/2)))/(8*(a*e - b*d)^(9/2))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 702, normalized size of antiderivative = 4.06 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {-105 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{3} e^{3}-315 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} b \,e^{3} x -315 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a \,b^{2} e^{3} x^{2}-105 \sqrt {b}\, \sqrt {e x +d}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{3} e^{3} x^{3}-48 a^{4} e^{4}-39 a^{3} b d \,e^{3}-231 a^{3} b \,e^{4} x +125 a^{2} b^{2} d^{2} e^{2}+133 a^{2} b^{2} d \,e^{3} x -280 a^{2} b^{2} e^{4} x^{2}-46 a \,b^{3} d^{3} e +112 a \,b^{3} d^{2} e^{2} x +245 a \,b^{3} d \,e^{3} x^{2}-105 a \,b^{3} e^{4} x^{3}+8 b^{4} d^{4}-14 b^{4} d^{3} e x +35 b^{4} d^{2} e^{2} x^{2}+105 b^{4} d \,e^{3} x^{3}}{24 \sqrt {e x +d}\, \left (a^{5} b^{3} e^{5} x^{3}-5 a^{4} b^{4} d \,e^{4} x^{3}+10 a^{3} b^{5} d^{2} e^{3} x^{3}-10 a^{2} b^{6} d^{3} e^{2} x^{3}+5 a \,b^{7} d^{4} e \,x^{3}-b^{8} d^{5} x^{3}+3 a^{6} b^{2} e^{5} x^{2}-15 a^{5} b^{3} d \,e^{4} x^{2}+30 a^{4} b^{4} d^{2} e^{3} x^{2}-30 a^{3} b^{5} d^{3} e^{2} x^{2}+15 a^{2} b^{6} d^{4} e \,x^{2}-3 a \,b^{7} d^{5} x^{2}+3 a^{7} b \,e^{5} x -15 a^{6} b^{2} d \,e^{4} x +30 a^{5} b^{3} d^{2} e^{3} x -30 a^{4} b^{4} d^{3} e^{2} x +15 a^{3} b^{5} d^{4} e x -3 a^{2} b^{6} d^{5} x +a^{8} e^{5}-5 a^{7} b d \,e^{4}+10 a^{6} b^{2} d^{2} e^{3}-10 a^{5} b^{3} d^{3} e^{2}+5 a^{4} b^{4} d^{4} e -a^{3} b^{5} d^{5}\right )} \] Input:

int(1/(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)
 

Output:

( - 105*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt 
(b)*sqrt(a*e - b*d)))*a**3*e**3 - 315*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d 
)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**2*b*e**3*x - 315*sq 
rt(b)*sqrt(d + e*x)*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 - 105*sqrt(b)*sqrt(d + e*x)*sqrt(a*e - b*d)*a 
tan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*b**3*e**3*x**3 - 48*a**4* 
e**4 - 39*a**3*b*d*e**3 - 231*a**3*b*e**4*x + 125*a**2*b**2*d**2*e**2 + 13 
3*a**2*b**2*d*e**3*x - 280*a**2*b**2*e**4*x**2 - 46*a*b**3*d**3*e + 112*a* 
b**3*d**2*e**2*x + 245*a*b**3*d*e**3*x**2 - 105*a*b**3*e**4*x**3 + 8*b**4* 
d**4 - 14*b**4*d**3*e*x + 35*b**4*d**2*e**2*x**2 + 105*b**4*d*e**3*x**3)/( 
24*sqrt(d + e*x)*(a**8*e**5 - 5*a**7*b*d*e**4 + 3*a**7*b*e**5*x + 10*a**6* 
b**2*d**2*e**3 - 15*a**6*b**2*d*e**4*x + 3*a**6*b**2*e**5*x**2 - 10*a**5*b 
**3*d**3*e**2 + 30*a**5*b**3*d**2*e**3*x - 15*a**5*b**3*d*e**4*x**2 + a**5 
*b**3*e**5*x**3 + 5*a**4*b**4*d**4*e - 30*a**4*b**4*d**3*e**2*x + 30*a**4* 
b**4*d**2*e**3*x**2 - 5*a**4*b**4*d*e**4*x**3 - a**3*b**5*d**5 + 15*a**3*b 
**5*d**4*e*x - 30*a**3*b**5*d**3*e**2*x**2 + 10*a**3*b**5*d**2*e**3*x**3 - 
 3*a**2*b**6*d**5*x + 15*a**2*b**6*d**4*e*x**2 - 10*a**2*b**6*d**3*e**2*x* 
*3 - 3*a*b**7*d**5*x**2 + 5*a*b**7*d**4*e*x**3 - b**8*d**5*x**3))