\(\int \frac {(a+b x) \sqrt {d+e x}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [109]

Optimal result

Integrand size = 33, antiderivative size = 182 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\sqrt {d+e x}}{4 b (a+b x)^4}-\frac {e \sqrt {d+e x}}{24 b (b d-a e) (a+b x)^3}+\frac {5 e^2 \sqrt {d+e x}}{96 b (b d-a e)^2 (a+b x)^2}-\frac {5 e^3 \sqrt {d+e x}}{64 b (b d-a e)^3 (a+b x)}+\frac {5 e^4 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 b^{3/2} (b d-a e)^{7/2}} \] Output:

-1/4*(e*x+d)^(1/2)/b/(b*x+a)^4-1/24*e*(e*x+d)^(1/2)/b/(-a*e+b*d)/(b*x+a)^3 
+5/96*e^2*(e*x+d)^(1/2)/b/(-a*e+b*d)^2/(b*x+a)^2-5/64*e^3*(e*x+d)^(1/2)/b/ 
(-a*e+b*d)^3/(b*x+a)+5/64*e^4*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/ 
2))/b^(3/2)/(-a*e+b*d)^(7/2)
 

Mathematica [A] (verified)

Time = 1.67 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\sqrt {d+e x} \left (-15 a^3 e^3+a^2 b e^2 (118 d+73 e x)+a b^2 e \left (-136 d^2-36 d e x+55 e^2 x^2\right )+b^3 \left (48 d^3+8 d^2 e x-10 d e^2 x^2+15 e^3 x^3\right )\right )}{192 b (-b d+a e)^3 (a+b x)^4}+\frac {5 e^4 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{64 b^{3/2} (-b d+a e)^{7/2}} \] Input:

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

Output:

(Sqrt[d + e*x]*(-15*a^3*e^3 + a^2*b*e^2*(118*d + 73*e*x) + a*b^2*e*(-136*d 
^2 - 36*d*e*x + 55*e^2*x^2) + b^3*(48*d^3 + 8*d^2*e*x - 10*d*e^2*x^2 + 15* 
e^3*x^3)))/(192*b*(-(b*d) + a*e)^3*(a + b*x)^4) + (5*e^4*ArcTan[(Sqrt[b]*S 
qrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(64*b^(3/2)*(-(b*d) + a*e)^(7/2))
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {1184, 27, 51, 52, 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.

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

\(\Big \downarrow \) 1184

\(\displaystyle b^6 \int \frac {\sqrt {d+e x}}{b^6 (a+b x)^5}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {\sqrt {d+e x}}{(a+b x)^5}dx\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/4*Sqrt[d + e*x]/(b*(a + b*x)^4) + (e*(-1/3*Sqrt[d + e*x]/((b*d - a*e)*( 
a + b*x)^3) - (5*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]*Sqrt[d + e*x 
])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2))))/(4*(b*d - a*e))))/(6*(b 
*d - a*e))))/(8*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 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 = 1.49 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(-\frac {5 \left (\left (\left (-b^{3} x^{3}-\frac {11}{3} a \,b^{2} x^{2}-\frac {73}{15} a^{2} b x +a^{3}\right ) e^{3}-\frac {118 \left (-\frac {5}{59} b^{2} x^{2}-\frac {18}{59} a b x +a^{2}\right ) b d \,e^{2}}{15}+\frac {136 b^{2} d^{2} \left (-\frac {b x}{17}+a \right ) e}{15}-\frac {16 b^{3} d^{3}}{5}\right ) \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}-e^{4} \left (b x +a \right )^{4} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )\right )}{64 \sqrt {b \left (a e -b d \right )}\, \left (b x +a \right )^{4} b \left (a e -b d \right )^{3}}\)	\(170\)
derivativedivides	\(2 e^{4} \left (\frac {\frac {5 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{128 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {55 b \left (e x +d \right )^{\frac {5}{2}}}{384 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {73 \left (e x +d \right )^{\frac {3}{2}}}{384 \left (a e -b d \right )}-\frac {5 \sqrt {e x +d}}{128 b}}{\left (b \left (e x +d \right )+a e -b d \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{128 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(217\)
default	\(2 e^{4} \left (\frac {\frac {5 b^{2} \left (e x +d \right )^{\frac {7}{2}}}{128 \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {55 b \left (e x +d \right )^{\frac {5}{2}}}{384 \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {73 \left (e x +d \right )^{\frac {3}{2}}}{384 \left (a e -b d \right )}-\frac {5 \sqrt {e x +d}}{128 b}}{\left (b \left (e x +d \right )+a e -b d \right )^{4}}+\frac {5 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{128 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \sqrt {b \left (a e -b d \right )}}\right )\)	\(217\)

Input:

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

Output:

-5/64/(b*(a*e-b*d))^(1/2)*(((-b^3*x^3-11/3*a*b^2*x^2-73/15*a^2*b*x+a^3)*e^ 
3-118/15*(-5/59*b^2*x^2-18/59*a*b*x+a^2)*b*d*e^2+136/15*b^2*d^2*(-1/17*b*x 
+a)*e-16/5*b^3*d^3)*(b*(a*e-b*d))^(1/2)*(e*x+d)^(1/2)-e^4*(b*x+a)^4*arctan 
(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2)))/(b*x+a)^4/b/(a*e-b*d)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 581 vs. \(2 (154) = 308\).

Time = 0.11 (sec) , antiderivative size = 1176, normalized size of antiderivative = 6.46 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/384*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e 
^4*x + a^4*e^4)*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*(48*b^5*d^4 - 184*a*b^4*d^3*e + 2 
54*a^2*b^3*d^2*e^2 - 133*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a* 
b^4*e^4)*x^3 - 5*(2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 + 11*a^2*b^3*e^4)*x^2 + ( 
8*b^5*d^3*e - 44*a*b^4*d^2*e^2 + 109*a^2*b^3*d*e^3 - 73*a^3*b^2*e^4)*x)*sq 
rt(e*x + d))/(a^4*b^6*d^4 - 4*a^5*b^5*d^3*e + 6*a^6*b^4*d^2*e^2 - 4*a^7*b^ 
3*d*e^3 + a^8*b^2*e^4 + (b^10*d^4 - 4*a*b^9*d^3*e + 6*a^2*b^8*d^2*e^2 - 4* 
a^3*b^7*d*e^3 + a^4*b^6*e^4)*x^4 + 4*(a*b^9*d^4 - 4*a^2*b^8*d^3*e + 6*a^3* 
b^7*d^2*e^2 - 4*a^4*b^6*d*e^3 + a^5*b^5*e^4)*x^3 + 6*(a^2*b^8*d^4 - 4*a^3* 
b^7*d^3*e + 6*a^4*b^6*d^2*e^2 - 4*a^5*b^5*d*e^3 + a^6*b^4*e^4)*x^2 + 4*(a^ 
3*b^7*d^4 - 4*a^4*b^6*d^3*e + 6*a^5*b^5*d^2*e^2 - 4*a^6*b^4*d*e^3 + a^7*b^ 
3*e^4)*x), -1/192*(15*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 
 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e) 
*sqrt(e*x + d)/(b*e*x + b*d)) + (48*b^5*d^4 - 184*a*b^4*d^3*e + 254*a^2*b^ 
3*d^2*e^2 - 133*a^3*b^2*d*e^3 + 15*a^4*b*e^4 + 15*(b^5*d*e^3 - a*b^4*e^4)* 
x^3 - 5*(2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 + 11*a^2*b^3*e^4)*x^2 + (8*b^5*d^3 
*e - 44*a*b^4*d^2*e^2 + 109*a^2*b^3*d*e^3 - 73*a^3*b^2*e^4)*x)*sqrt(e*x + 
d))/(a^4*b^6*d^4 - 4*a^5*b^5*d^3*e + 6*a^6*b^4*d^2*e^2 - 4*a^7*b^3*d*e^3 + 
 a^8*b^2*e^4 + (b^10*d^4 - 4*a*b^9*d^3*e + 6*a^2*b^8*d^2*e^2 - 4*a^3*b^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((b*x+a)*(e*x+d)^(1/2)/(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. 311 vs. \(2 (154) = 308\).

Time = 0.22 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {5 \, e^{4} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {15 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{3} e^{4} - 55 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} d e^{4} + 73 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} + 15 \, \sqrt {e x + d} b^{3} d^{3} e^{4} + 55 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{2} e^{5} - 146 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} - 45 \, \sqrt {e x + d} a b^{2} d^{2} e^{5} + 73 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b e^{6} + 45 \, \sqrt {e x + d} a^{2} b d e^{6} - 15 \, \sqrt {e x + d} a^{3} e^{7}}{192 \, {\left (b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4}} \] Input:

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

Output:

-5/64*e^4*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^3 - 3*a*b^3 
*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*sqrt(-b^2*d + a*b*e)) - 1/192*(15*(e 
*x + d)^(7/2)*b^3*e^4 - 55*(e*x + d)^(5/2)*b^3*d*e^4 + 73*(e*x + d)^(3/2)* 
b^3*d^2*e^4 + 15*sqrt(e*x + d)*b^3*d^3*e^4 + 55*(e*x + d)^(5/2)*a*b^2*e^5 
- 146*(e*x + d)^(3/2)*a*b^2*d*e^5 - 45*sqrt(e*x + d)*a*b^2*d^2*e^5 + 73*(e 
*x + d)^(3/2)*a^2*b*e^6 + 45*sqrt(e*x + d)*a^2*b*d*e^6 - 15*sqrt(e*x + d)* 
a^3*e^7)/((b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^3*b*e^3)*((e*x + 
d)*b - b*d + a*e)^4)
 

Mupad [B] (verification not implemented)

Time = 11.00 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.63 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {73\,e^4\,{\left (d+e\,x\right )}^{3/2}}{192\,\left (a\,e-b\,d\right )}-\frac {5\,e^4\,\sqrt {d+e\,x}}{64\,b}+\frac {5\,b^2\,e^4\,{\left (d+e\,x\right )}^{7/2}}{64\,{\left (a\,e-b\,d\right )}^3}+\frac {55\,b\,e^4\,{\left (d+e\,x\right )}^{5/2}}{192\,{\left (a\,e-b\,d\right )}^2}}{b^4\,{\left (d+e\,x\right )}^4-\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,{\left (d+e\,x\right )}^3-\left (d+e\,x\right )\,\left (-4\,a^3\,b\,e^3+12\,a^2\,b^2\,d\,e^2-12\,a\,b^3\,d^2\,e+4\,b^4\,d^3\right )+a^4\,e^4+b^4\,d^4+{\left (d+e\,x\right )}^2\,\left (6\,a^2\,b^2\,e^2-12\,a\,b^3\,d\,e+6\,b^4\,d^2\right )+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e-4\,a^3\,b\,d\,e^3}+\frac {5\,e^4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{64\,b^{3/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \] Input:

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

Output:

((73*e^4*(d + e*x)^(3/2))/(192*(a*e - b*d)) - (5*e^4*(d + e*x)^(1/2))/(64* 
b) + (5*b^2*e^4*(d + e*x)^(7/2))/(64*(a*e - b*d)^3) + (55*b*e^4*(d + e*x)^ 
(5/2))/(192*(a*e - b*d)^2))/(b^4*(d + e*x)^4 - (4*b^4*d - 4*a*b^3*e)*(d + 
e*x)^3 - (d + e*x)*(4*b^4*d^3 - 4*a^3*b*e^3 + 12*a^2*b^2*d*e^2 - 12*a*b^3* 
d^2*e) + a^4*e^4 + b^4*d^4 + (d + e*x)^2*(6*b^4*d^2 + 6*a^2*b^2*e^2 - 12*a 
*b^3*d*e) + 6*a^2*b^2*d^2*e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3) + (5*e^4*at 
an((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(64*b^(3/2)*(a*e - b*d)^( 
7/2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 821, normalized size of antiderivative = 4.51 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {15 \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}+60 \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 +90 \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}+60 \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}+15 \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}-15 \sqrt {e x +d}\, a^{4} b \,e^{4}+133 \sqrt {e x +d}\, a^{3} b^{2} d \,e^{3}+73 \sqrt {e x +d}\, a^{3} b^{2} e^{4} x -254 \sqrt {e x +d}\, a^{2} b^{3} d^{2} e^{2}-109 \sqrt {e x +d}\, a^{2} b^{3} d \,e^{3} x +55 \sqrt {e x +d}\, a^{2} b^{3} e^{4} x^{2}+184 \sqrt {e x +d}\, a \,b^{4} d^{3} e +44 \sqrt {e x +d}\, a \,b^{4} d^{2} e^{2} x -65 \sqrt {e x +d}\, a \,b^{4} d \,e^{3} x^{2}+15 \sqrt {e x +d}\, a \,b^{4} e^{4} x^{3}-48 \sqrt {e x +d}\, b^{5} d^{4}-8 \sqrt {e x +d}\, b^{5} d^{3} e x +10 \sqrt {e x +d}\, b^{5} d^{2} e^{2} x^{2}-15 \sqrt {e x +d}\, b^{5} d \,e^{3} x^{3}}{192 b^{2} \left (a^{4} b^{4} e^{4} x^{4}-4 a^{3} b^{5} d \,e^{3} x^{4}+6 a^{2} b^{6} d^{2} e^{2} x^{4}-4 a \,b^{7} d^{3} e \,x^{4}+b^{8} d^{4} x^{4}+4 a^{5} b^{3} e^{4} x^{3}-16 a^{4} b^{4} d \,e^{3} x^{3}+24 a^{3} b^{5} d^{2} e^{2} x^{3}-16 a^{2} b^{6} d^{3} e \,x^{3}+4 a \,b^{7} d^{4} x^{3}+6 a^{6} b^{2} e^{4} x^{2}-24 a^{5} b^{3} d \,e^{3} x^{2}+36 a^{4} b^{4} d^{2} e^{2} x^{2}-24 a^{3} b^{5} d^{3} e \,x^{2}+6 a^{2} b^{6} d^{4} x^{2}+4 a^{7} b \,e^{4} x -16 a^{6} b^{2} d \,e^{3} x +24 a^{5} b^{3} d^{2} e^{2} x -16 a^{4} b^{4} d^{3} e x +4 a^{3} b^{5} d^{4} x +a^{8} e^{4}-4 a^{7} b d \,e^{3}+6 a^{6} b^{2} d^{2} e^{2}-4 a^{5} b^{3} d^{3} e +a^{4} b^{4} d^{4}\right )} \] Input:

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

Output:

(15*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d 
)))*a**4*e**4 + 60*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 + 90*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 + 60*sqrt(b)*sqr 
t(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**3*e**4 
*x**3 + 15*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 - 15*sqrt(d + e*x)*a**4*b*e**4 + 133*sqrt(d + e* 
x)*a**3*b**2*d*e**3 + 73*sqrt(d + e*x)*a**3*b**2*e**4*x - 254*sqrt(d + e*x 
)*a**2*b**3*d**2*e**2 - 109*sqrt(d + e*x)*a**2*b**3*d*e**3*x + 55*sqrt(d + 
 e*x)*a**2*b**3*e**4*x**2 + 184*sqrt(d + e*x)*a*b**4*d**3*e + 44*sqrt(d + 
e*x)*a*b**4*d**2*e**2*x - 65*sqrt(d + e*x)*a*b**4*d*e**3*x**2 + 15*sqrt(d 
+ e*x)*a*b**4*e**4*x**3 - 48*sqrt(d + e*x)*b**5*d**4 - 8*sqrt(d + e*x)*b** 
5*d**3*e*x + 10*sqrt(d + e*x)*b**5*d**2*e**2*x**2 - 15*sqrt(d + e*x)*b**5* 
d*e**3*x**3)/(192*b**2*(a**8*e**4 - 4*a**7*b*d*e**3 + 4*a**7*b*e**4*x + 6* 
a**6*b**2*d**2*e**2 - 16*a**6*b**2*d*e**3*x + 6*a**6*b**2*e**4*x**2 - 4*a* 
*5*b**3*d**3*e + 24*a**5*b**3*d**2*e**2*x - 24*a**5*b**3*d*e**3*x**2 + 4*a 
**5*b**3*e**4*x**3 + a**4*b**4*d**4 - 16*a**4*b**4*d**3*e*x + 36*a**4*b**4 
*d**2*e**2*x**2 - 16*a**4*b**4*d*e**3*x**3 + a**4*b**4*e**4*x**4 + 4*a**3* 
b**5*d**4*x - 24*a**3*b**5*d**3*e*x**2 + 24*a**3*b**5*d**2*e**2*x**3 - 4*a 
**3*b**5*d*e**3*x**4 + 6*a**2*b**6*d**4*x**2 - 16*a**2*b**6*d**3*e*x**3...