\(\int \frac {(A+B x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [372]

Optimal result

Integrand size = 33, antiderivative size = 292 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {(10 b B d+3 A b e-13 a B e) \sqrt {d+e x}}{40 b^3 (a+b x)^4}-\frac {e (30 b B d+A b e-31 a B e) \sqrt {d+e x}}{80 b^3 (b d-a e) (a+b x)^3}-\frac {e^2 (2 b B d-A b e-a B e) \sqrt {d+e x}}{64 b^3 (b d-a e)^2 (a+b x)^2}+\frac {3 e^3 (2 b B d-A b e-a B e) \sqrt {d+e x}}{128 b^3 (b d-a e)^3 (a+b x)}-\frac {(A b-a B) (d+e x)^{3/2}}{5 b^2 (a+b x)^5}-\frac {3 e^4 (2 b B d-A b e-a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{7/2} (b d-a e)^{7/2}} \] Output:

-1/40*(3*A*b*e-13*B*a*e+10*B*b*d)*(e*x+d)^(1/2)/b^3/(b*x+a)^4-1/80*e*(A*b* 
e-31*B*a*e+30*B*b*d)*(e*x+d)^(1/2)/b^3/(-a*e+b*d)/(b*x+a)^3-1/64*e^2*(-A*b 
*e-B*a*e+2*B*b*d)*(e*x+d)^(1/2)/b^3/(-a*e+b*d)^2/(b*x+a)^2+3/128*e^3*(-A*b 
*e-B*a*e+2*B*b*d)*(e*x+d)^(1/2)/b^3/(-a*e+b*d)^3/(b*x+a)-1/5*(A*b-B*a)*(e* 
x+d)^(3/2)/b^2/(b*x+a)^5-3/128*e^4*(-A*b*e-B*a*e+2*B*b*d)*arctanh(b^(1/2)* 
(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/(-a*e+b*d)^(7/2)
 

Mathematica [A] (verified)

Time = 4.44 (sec) , antiderivative size = 421, normalized size of antiderivative = 1.44 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {\sqrt {b} \sqrt {d+e x} \left (B \left (-15 a^5 e^4+10 a^4 b e^3 (2 d-7 e x)+2 a^3 b^2 e^2 \left (6 d^2+47 d e x-64 e^2 x^2\right )+10 b^5 d x \left (16 d^3+24 d^2 e x+2 d e^2 x^2-3 e^3 x^3\right )+2 a^2 b^3 e \left (-32 d^3+46 d^2 e x+233 d e^2 x^2+35 e^3 x^3\right )+a b^4 \left (32 d^4-336 d^3 e x-668 d^2 e^2 x^2-150 d e^3 x^3+15 e^4 x^4\right )\right )+A b \left (-15 a^4 e^4-10 a^3 b e^3 (d+7 e x)+2 a^2 b^2 e^2 \left (124 d^2+233 d e x+64 e^2 x^2\right )-2 a b^3 e \left (168 d^3+256 d^2 e x+23 d e^2 x^2-35 e^3 x^3\right )+b^4 \left (128 d^4+176 d^3 e x+8 d^2 e^2 x^2-10 d e^3 x^3+15 e^4 x^4\right )\right )\right )}{(-b d+a e)^3 (a+b x)^5}+\frac {15 e^4 (-2 b B d+A b e+a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}}}{640 b^{7/2}} \] Input:

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

Output:

((Sqrt[b]*Sqrt[d + e*x]*(B*(-15*a^5*e^4 + 10*a^4*b*e^3*(2*d - 7*e*x) + 2*a 
^3*b^2*e^2*(6*d^2 + 47*d*e*x - 64*e^2*x^2) + 10*b^5*d*x*(16*d^3 + 24*d^2*e 
*x + 2*d*e^2*x^2 - 3*e^3*x^3) + 2*a^2*b^3*e*(-32*d^3 + 46*d^2*e*x + 233*d* 
e^2*x^2 + 35*e^3*x^3) + a*b^4*(32*d^4 - 336*d^3*e*x - 668*d^2*e^2*x^2 - 15 
0*d*e^3*x^3 + 15*e^4*x^4)) + A*b*(-15*a^4*e^4 - 10*a^3*b*e^3*(d + 7*e*x) + 
 2*a^2*b^2*e^2*(124*d^2 + 233*d*e*x + 64*e^2*x^2) - 2*a*b^3*e*(168*d^3 + 2 
56*d^2*e*x + 23*d*e^2*x^2 - 35*e^3*x^3) + b^4*(128*d^4 + 176*d^3*e*x + 8*d 
^2*e^2*x^2 - 10*d*e^3*x^3 + 15*e^4*x^4))))/((-(b*d) + a*e)^3*(a + b*x)^5) 
+ (15*e^4*(-2*b*B*d + A*b*e + a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[- 
(b*d) + a*e]])/(-(b*d) + a*e)^(7/2))/(640*b^(7/2))
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.89, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1184, 27, 87, 51, 51, 52, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^6 \int \frac {(A+B x) (d+e x)^{3/2}}{b^6 (a+b x)^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {(A+B x) (d+e x)^{3/2}}{(a+b x)^6}dx\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {(-a B e-A b e+2 b B d) \int \frac {(d+e x)^{3/2}}{(a+b x)^5}dx}{2 b (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{5 b (a+b x)^5 (b d-a e)}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {(-a B e-A b e+2 b B d) \left (\frac {3 e \int \frac {\sqrt {d+e x}}{(a+b x)^4}dx}{8 b}-\frac {(d+e x)^{3/2}}{4 b (a+b x)^4}\right )}{2 b (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{5 b (a+b x)^5 (b d-a e)}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/5*((A*b - a*B)*(d + e*x)^(5/2))/(b*(b*d - a*e)*(a + b*x)^5) + ((2*b*B*d 
 - A*b*e - a*B*e)*(-1/4*(d + e*x)^(3/2)/(b*(a + b*x)^4) + (3*e*(-1/3*Sqrt[ 
d + e*x]/(b*(a + b*x)^3) + (e*(-1/2*Sqrt[d + e*x]/((b*d - a*e)*(a + b*x)^2 
) - (3*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*d - a* 
e))))/(6*b)))/(8*b)))/(2*b*(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 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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 1184

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p   Int[(d + e*x)^m*(f + g*x 
)^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E 
qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(2 e^{4} \left (\frac {\frac {3 \left (A b e +B a e -2 B b d \right ) b \left (e x +d \right )^{\frac {9}{2}}}{256 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {7 \left (A b e +B a e -2 B b d \right ) \left (e x +d \right )^{\frac {7}{2}}}{128 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\left (A b -B a \right ) e \left (e x +d \right )^{\frac {5}{2}}}{10 b \left (a e -b d \right )}-\frac {7 \left (A b e +B a e -2 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (A b e +B a e -2 B b d \right ) \left (a e -b d \right ) \sqrt {e x +d}}{256 b^{3}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \left (A b e +B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{256 b^{3} \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(316\)
default	\(2 e^{4} \left (\frac {\frac {3 \left (A b e +B a e -2 B b d \right ) b \left (e x +d \right )^{\frac {9}{2}}}{256 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {7 \left (A b e +B a e -2 B b d \right ) \left (e x +d \right )^{\frac {7}{2}}}{128 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\left (A b -B a \right ) e \left (e x +d \right )^{\frac {5}{2}}}{10 b \left (a e -b d \right )}-\frac {7 \left (A b e +B a e -2 B b d \right ) \left (e x +d \right )^{\frac {3}{2}}}{128 b^{2}}-\frac {3 \left (A b e +B a e -2 B b d \right ) \left (a e -b d \right ) \sqrt {e x +d}}{256 b^{3}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \left (A b e +B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{256 b^{3} \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(316\)
pseudoelliptic	\(-\frac {3 \left (-e^{4} \left (b x +a \right )^{5} \left (\left (A e -2 B d \right ) b +B a e \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )+\sqrt {e x +d}\, \left (\left (-A \,e^{4} x^{4}+\frac {2 d \,x^{3} \left (3 B x +A \right ) e^{3}}{3}-\frac {8 d^{2} x^{2} \left (\frac {5 B x}{2}+A \right ) e^{2}}{15}-\frac {176 d^{3} x \left (\frac {15 B x}{11}+A \right ) e}{15}-\frac {128 d^{4} \left (\frac {5 B x}{4}+A \right )}{15}\right ) b^{5}+\frac {112 \left (-\frac {5 \left (\frac {3 B x}{14}+A \right ) x^{3} e^{4}}{24}+\frac {23 d \left (\frac {75 B x}{23}+A \right ) x^{2} e^{3}}{168}+\frac {32 \left (\frac {167 B x}{128}+A \right ) d^{2} x \,e^{2}}{21}+d^{3} \left (B x +A \right ) e -\frac {2 B \,d^{4}}{21}\right ) a \,b^{4}}{5}-\frac {248 e \,a^{2} \left (\frac {16 \left (\frac {35 B x}{64}+A \right ) x^{2} e^{3}}{31}+\frac {233 d x \left (B x +A \right ) e^{2}}{124}+d^{2} \left (\frac {23 B x}{62}+A \right ) e -\frac {8 B \,d^{3}}{31}\right ) b^{3}}{15}+\frac {2 e^{2} a^{3} \left (\left (\frac {64}{5} B \,x^{2}+7 A x \right ) e^{2}+d \left (-\frac {47 B x}{5}+A \right ) e -\frac {6 B \,d^{2}}{5}\right ) b^{2}}{3}+e^{3} a^{4} \left (\left (\frac {14 B x}{3}+A \right ) e -\frac {4 B d}{3}\right ) b +B \,a^{5} e^{4}\right ) \sqrt {b \left (a e -b d \right )}\right )}{128 \sqrt {b \left (a e -b d \right )}\, \left (b x +a \right )^{5} \left (a e -b d \right )^{3} b^{3}}\)	\(367\)

Input:

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

Output:

2*e^4*((3/256*(A*b*e+B*a*e-2*B*b*d)*b/(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e 
-b^3*d^3)*(e*x+d)^(9/2)+7/128*(A*b*e+B*a*e-2*B*b*d)/(a^2*e^2-2*a*b*d*e+b^2 
*d^2)*(e*x+d)^(7/2)+1/10*(A*b-B*a)*e/b/(a*e-b*d)*(e*x+d)^(5/2)-7/128*(A*b* 
e+B*a*e-2*B*b*d)/b^2*(e*x+d)^(3/2)-3/256*(A*b*e+B*a*e-2*B*b*d)*(a*e-b*d)/b 
^3*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)^5+3/256*(A*b*e+B*a*e-2*B*b*d)/b^3/(a 
^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)/(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. 1168 vs. \(2 (260) = 520\).

Time = 0.20 (sec) , antiderivative size = 2349, normalized size of antiderivative = 8.04 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fric 
as")
 

Output:

[1/1280*(15*(2*B*a^5*b*d*e^4 - (B*a^6 + A*a^5*b)*e^5 + (2*B*b^6*d*e^4 - (B 
*a*b^5 + A*b^6)*e^5)*x^5 + 5*(2*B*a*b^5*d*e^4 - (B*a^2*b^4 + A*a*b^5)*e^5) 
*x^4 + 10*(2*B*a^2*b^4*d*e^4 - (B*a^3*b^3 + A*a^2*b^4)*e^5)*x^3 + 10*(2*B* 
a^3*b^3*d*e^4 - (B*a^4*b^2 + A*a^3*b^3)*e^5)*x^2 + 5*(2*B*a^4*b^2*d*e^4 - 
(B*a^5*b + A*a^4*b^2)*e^5)*x)*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*(32*(B*a*b^6 + 4*A* 
b^7)*d^5 - 16*(6*B*a^2*b^5 + 29*A*a*b^6)*d^4*e + 4*(19*B*a^3*b^4 + 146*A*a 
^2*b^5)*d^3*e^2 + 2*(4*B*a^4*b^3 - 129*A*a^3*b^4)*d^2*e^3 - 5*(7*B*a^5*b^2 
 + A*a^4*b^3)*d*e^4 + 15*(B*a^6*b + A*a^5*b^2)*e^5 - 15*(2*B*b^7*d^2*e^3 - 
 (3*B*a*b^6 + A*b^7)*d*e^4 + (B*a^2*b^5 + A*a*b^6)*e^5)*x^4 + 10*(2*B*b^7* 
d^3*e^2 - (17*B*a*b^6 + A*b^7)*d^2*e^3 + 2*(11*B*a^2*b^5 + 4*A*a*b^6)*d*e^ 
4 - 7*(B*a^3*b^4 + A*a^2*b^5)*e^5)*x^3 + 2*(120*B*b^7*d^4*e - 2*(227*B*a*b 
^6 - 2*A*b^7)*d^3*e^2 + 27*(21*B*a^2*b^5 - A*a*b^6)*d^2*e^3 - 3*(99*B*a^3* 
b^4 - 29*A*a^2*b^5)*d*e^4 + 64*(B*a^4*b^3 - A*a^3*b^4)*e^5)*x^2 + 2*(80*B* 
b^7*d^5 - 8*(31*B*a*b^6 - 11*A*b^7)*d^4*e + 2*(107*B*a^2*b^5 - 172*A*a*b^6 
)*d^3*e^2 + (B*a^3*b^4 + 489*A*a^2*b^5)*d^2*e^3 - 2*(41*B*a^4*b^3 + 134*A* 
a^3*b^4)*d*e^4 + 35*(B*a^5*b^2 + A*a^4*b^3)*e^5)*x)*sqrt(e*x + d))/(a^5*b^ 
8*d^4 - 4*a^6*b^7*d^3*e + 6*a^7*b^6*d^2*e^2 - 4*a^8*b^5*d*e^3 + a^9*b^4*e^ 
4 + (b^13*d^4 - 4*a*b^12*d^3*e + 6*a^2*b^11*d^2*e^2 - 4*a^3*b^10*d*e^3 + a 
^4*b^9*e^4)*x^5 + 5*(a*b^12*d^4 - 4*a^2*b^11*d^3*e + 6*a^3*b^10*d^2*e^2...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxi 
ma")
 

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. 814 vs. \(2 (260) = 520\).

Time = 0.19 (sec) , antiderivative size = 814, normalized size of antiderivative = 2.79 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

3/128*(2*B*b*d*e^4 - B*a*e^5 - A*b*e^5)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d 
 + a*b*e))/((b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a^3*b^3*e^3)*sqrt 
(-b^2*d + a*b*e)) + 1/640*(30*(e*x + d)^(9/2)*B*b^5*d*e^4 - 140*(e*x + d)^ 
(7/2)*B*b^5*d^2*e^4 + 140*(e*x + d)^(3/2)*B*b^5*d^4*e^4 - 30*sqrt(e*x + d) 
*B*b^5*d^5*e^4 - 15*(e*x + d)^(9/2)*B*a*b^4*e^5 - 15*(e*x + d)^(9/2)*A*b^5 
*e^5 + 210*(e*x + d)^(7/2)*B*a*b^4*d*e^5 + 70*(e*x + d)^(7/2)*A*b^5*d*e^5 
+ 128*(e*x + d)^(5/2)*B*a*b^4*d^2*e^5 - 128*(e*x + d)^(5/2)*A*b^5*d^2*e^5 
- 490*(e*x + d)^(3/2)*B*a*b^4*d^3*e^5 - 70*(e*x + d)^(3/2)*A*b^5*d^3*e^5 + 
 135*sqrt(e*x + d)*B*a*b^4*d^4*e^5 + 15*sqrt(e*x + d)*A*b^5*d^4*e^5 - 70*( 
e*x + d)^(7/2)*B*a^2*b^3*e^6 - 70*(e*x + d)^(7/2)*A*a*b^4*e^6 - 256*(e*x + 
 d)^(5/2)*B*a^2*b^3*d*e^6 + 256*(e*x + d)^(5/2)*A*a*b^4*d*e^6 + 630*(e*x + 
 d)^(3/2)*B*a^2*b^3*d^2*e^6 + 210*(e*x + d)^(3/2)*A*a*b^4*d^2*e^6 - 240*sq 
rt(e*x + d)*B*a^2*b^3*d^3*e^6 - 60*sqrt(e*x + d)*A*a*b^4*d^3*e^6 + 128*(e* 
x + d)^(5/2)*B*a^3*b^2*e^7 - 128*(e*x + d)^(5/2)*A*a^2*b^3*e^7 - 350*(e*x 
+ d)^(3/2)*B*a^3*b^2*d*e^7 - 210*(e*x + d)^(3/2)*A*a^2*b^3*d*e^7 + 210*sqr 
t(e*x + d)*B*a^3*b^2*d^2*e^7 + 90*sqrt(e*x + d)*A*a^2*b^3*d^2*e^7 + 70*(e* 
x + d)^(3/2)*B*a^4*b*e^8 + 70*(e*x + d)^(3/2)*A*a^3*b^2*e^8 - 90*sqrt(e*x 
+ d)*B*a^4*b*d*e^8 - 60*sqrt(e*x + d)*A*a^3*b^2*d*e^8 + 15*sqrt(e*x + d)*B 
*a^5*e^9 + 15*sqrt(e*x + d)*A*a^4*b*e^9)/((b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2 
*b^4*d*e^2 - a^3*b^3*e^3)*((e*x + d)*b - b*d + a*e)^5)
 

Mupad [B] (verification not implemented)

Time = 11.27 (sec) , antiderivative size = 535, normalized size of antiderivative = 1.83 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {7\,{\left (d+e\,x\right )}^{7/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{64\,{\left (a\,e-b\,d\right )}^2}-\frac {7\,{\left (d+e\,x\right )}^{3/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{64\,b^2}-\frac {3\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{128\,b^3}+\frac {3\,b\,{\left (d+e\,x\right )}^{9/2}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}{128\,{\left (a\,e-b\,d\right )}^3}+\frac {\left (A\,b\,e^5-B\,a\,e^5\right )\,{\left (d+e\,x\right )}^{5/2}}{5\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )-{\left (d+e\,x\right )}^2\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )+b^5\,{\left (d+e\,x\right )}^5-\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,{\left (d+e\,x\right )}^4+a^5\,e^5-b^5\,d^5+{\left (d+e\,x\right )}^3\,\left (10\,a^2\,b^3\,e^2-20\,a\,b^4\,d\,e+10\,b^5\,d^2\right )-10\,a^2\,b^3\,d^3\,e^2+10\,a^3\,b^2\,d^2\,e^3+5\,a\,b^4\,d^4\,e-5\,a^4\,b\,d\,e^4}+\frac {3\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^4\,\sqrt {d+e\,x}\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^5+B\,a\,e^5-2\,B\,b\,d\,e^4\right )}\right )\,\left (A\,b\,e+B\,a\,e-2\,B\,b\,d\right )}{128\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \] Input:

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

Output:

((7*(d + e*x)^(7/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(64*(a*e - b*d)^2) 
- (7*(d + e*x)^(3/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(64*b^2) - (3*(a*e 
 - b*d)*(d + e*x)^(1/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(128*b^3) + (3* 
b*(d + e*x)^(9/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4))/(128*(a*e - b*d)^3) + 
 ((A*b*e^5 - B*a*e^5)*(d + e*x)^(5/2))/(5*b*(a*e - b*d)))/((d + e*x)*(5*b^ 
5*d^4 + 5*a^4*b*e^4 - 20*a^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 - 20*a*b^4*d^3 
*e) - (d + e*x)^2*(10*b^5*d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b 
^4*d^2*e) + b^5*(d + e*x)^5 - (5*b^5*d - 5*a*b^4*e)*(d + e*x)^4 + a^5*e^5 
- b^5*d^5 + (d + e*x)^3*(10*b^5*d^2 + 10*a^2*b^3*e^2 - 20*a*b^4*d*e) - 10* 
a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4) + (3 
*e^4*atan((b^(1/2)*e^4*(d + e*x)^(1/2)*(A*b*e + B*a*e - 2*B*b*d))/((a*e - 
b*d)^(1/2)*(A*b*e^5 + B*a*e^5 - 2*B*b*d*e^4)))*(A*b*e + B*a*e - 2*B*b*d))/ 
(128*b^(7/2)*(a*e - b*d)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 742, normalized size of antiderivative = 2.54 \[ \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{4} e^{4}+12 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{3} b \,e^{4} x +18 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} b^{2} e^{4} x^{2}+12 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a \,b^{3} e^{4} x^{3}+3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{4} e^{4} x^{4}-3 \sqrt {e x +d}\, a^{4} b \,e^{4}+\sqrt {e x +d}\, a^{3} b^{2} d \,e^{3}-11 \sqrt {e x +d}\, a^{3} b^{2} e^{4} x +26 \sqrt {e x +d}\, a^{2} b^{3} d^{2} e^{2}+55 \sqrt {e x +d}\, a^{2} b^{3} d \,e^{3} x +11 \sqrt {e x +d}\, a^{2} b^{3} e^{4} x^{2}-40 \sqrt {e x +d}\, a \,b^{4} d^{3} e -68 \sqrt {e x +d}\, a \,b^{4} d^{2} e^{2} x -13 \sqrt {e x +d}\, a \,b^{4} d \,e^{3} x^{2}+3 \sqrt {e x +d}\, a \,b^{4} e^{4} x^{3}+16 \sqrt {e x +d}\, b^{5} d^{4}+24 \sqrt {e x +d}\, b^{5} d^{3} e x +2 \sqrt {e x +d}\, b^{5} d^{2} e^{2} x^{2}-3 \sqrt {e x +d}\, b^{5} d \,e^{3} x^{3}}{64 b^{3} \left (a^{3} b^{4} e^{3} x^{4}-3 a^{2} b^{5} d \,e^{2} x^{4}+3 a \,b^{6} d^{2} e \,x^{4}-b^{7} d^{3} x^{4}+4 a^{4} b^{3} e^{3} x^{3}-12 a^{3} b^{4} d \,e^{2} x^{3}+12 a^{2} b^{5} d^{2} e \,x^{3}-4 a \,b^{6} d^{3} x^{3}+6 a^{5} b^{2} e^{3} x^{2}-18 a^{4} b^{3} d \,e^{2} x^{2}+18 a^{3} b^{4} d^{2} e \,x^{2}-6 a^{2} b^{5} d^{3} x^{2}+4 a^{6} b \,e^{3} x -12 a^{5} b^{2} d \,e^{2} x +12 a^{4} b^{3} d^{2} e x -4 a^{3} b^{4} d^{3} x +a^{7} e^{3}-3 a^{6} b d \,e^{2}+3 a^{5} b^{2} d^{2} e -a^{4} b^{3} d^{3}\right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d) 
))*a**4*e**4 + 12*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)* 
sqrt(a*e - b*d)))*a**3*b*e**4*x + 18*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d 
+ e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**2*b**2*e**4*x**2 + 12*sqrt(b)*sqrt 
(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**3*e**4* 
x**3 + 3*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e 
- b*d)))*b**4*e**4*x**4 - 3*sqrt(d + e*x)*a**4*b*e**4 + sqrt(d + e*x)*a**3 
*b**2*d*e**3 - 11*sqrt(d + e*x)*a**3*b**2*e**4*x + 26*sqrt(d + e*x)*a**2*b 
**3*d**2*e**2 + 55*sqrt(d + e*x)*a**2*b**3*d*e**3*x + 11*sqrt(d + e*x)*a** 
2*b**3*e**4*x**2 - 40*sqrt(d + e*x)*a*b**4*d**3*e - 68*sqrt(d + e*x)*a*b** 
4*d**2*e**2*x - 13*sqrt(d + e*x)*a*b**4*d*e**3*x**2 + 3*sqrt(d + e*x)*a*b* 
*4*e**4*x**3 + 16*sqrt(d + e*x)*b**5*d**4 + 24*sqrt(d + e*x)*b**5*d**3*e*x 
 + 2*sqrt(d + e*x)*b**5*d**2*e**2*x**2 - 3*sqrt(d + e*x)*b**5*d*e**3*x**3) 
/(64*b**3*(a**7*e**3 - 3*a**6*b*d*e**2 + 4*a**6*b*e**3*x + 3*a**5*b**2*d** 
2*e - 12*a**5*b**2*d*e**2*x + 6*a**5*b**2*e**3*x**2 - a**4*b**3*d**3 + 12* 
a**4*b**3*d**2*e*x - 18*a**4*b**3*d*e**2*x**2 + 4*a**4*b**3*e**3*x**3 - 4* 
a**3*b**4*d**3*x + 18*a**3*b**4*d**2*e*x**2 - 12*a**3*b**4*d*e**2*x**3 + a 
**3*b**4*e**3*x**4 - 6*a**2*b**5*d**3*x**2 + 12*a**2*b**5*d**2*e*x**3 - 3* 
a**2*b**5*d*e**2*x**4 - 4*a*b**6*d**3*x**3 + 3*a*b**6*d**2*e*x**4 - b**7*d 
**3*x**4))