\(\int \frac {(a+b x) \sqrt {d+e x}}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [250]

Optimal result

Integrand size = 35, antiderivative size = 224 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^2 \sqrt {d+e x}}{8 b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{3 b (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e \sqrt {d+e x}}{12 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{3/2} (b d-a e)^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

1/8*e^2*(e*x+d)^(1/2)/b/(-a*e+b*d)^2/((b*x+a)^2)^(1/2)-1/3*(e*x+d)^(1/2)/b 
/(b*x+a)^2/((b*x+a)^2)^(1/2)-1/12*e*(e*x+d)^(1/2)/b/(-a*e+b*d)/(b*x+a)/((b 
*x+a)^2)^(1/2)-1/8*e^3*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1 
/2))/b^(3/2)/(-a*e+b*d)^(5/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.02 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.67 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) \left (\frac {\sqrt {b} \sqrt {d+e x} \left (-3 a^2 e^2+2 a b e (7 d+4 e x)+b^2 \left (-8 d^2-2 d e x+3 e^2 x^2\right )\right )}{e^3 (b d-a e)^2 (a+b x)^3}+\frac {3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{5/2}}\right )}{24 b^{3/2} \sqrt {(a+b x)^2}} \] Input:

Integrate[((a + b*x)*Sqrt[d + e*x])/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(e^3*(a + b*x)*((Sqrt[b]*Sqrt[d + e*x]*(-3*a^2*e^2 + 2*a*b*e*(7*d + 4*e*x) 
 + b^2*(-8*d^2 - 2*d*e*x + 3*e^2*x^2)))/(e^3*(b*d - a*e)^2*(a + b*x)^3) + 
(3*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(5/2 
)))/(24*b^(3/2)*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.80, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1187, 27, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[((a + b*x)*Sqrt[d + e*x])/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

((a + b*x)*(-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]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e) 
^(3/2))))/(4*(b*d - a*e))))/(6*b)))/Sqrt[a^2 + 2*a*b*x + b^2*x^2]
 

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 1187

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^ 
IntPart[p]*(b/2 + c*x)^(2*FracPart[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, p}, x] && EqQ[b^2 
 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(325\) vs. \(2(156)=312\).

Time = 1.39 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.46

method	result	size
default	\(\frac {\left (b x +a \right )^{2} \left (3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{3} e^{3} x^{3}+9 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{2} e^{3} x^{2}+3 \left (e x +d \right )^{\frac {5}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2}+9 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b \,e^{3} x +8 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, a b e -8 \left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b^{2} d +3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} e^{3}-3 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{2} e^{2}+6 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a b d e -3 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{2} d^{2}\right )}{24 \sqrt {b \left (a e -b d \right )}\, b \left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(326\)

Input:

int((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/24*(b*x+a)^2*(3*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*b^3*e^3*x^3+ 
9*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^2*e^3*x^2+3*(e*x+d)^(5/2 
)*(b*(a*e-b*d))^(1/2)*b^2+9*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^ 
2*b*e^3*x+8*(e*x+d)^(3/2)*(b*(a*e-b*d))^(1/2)*a*b*e-8*(e*x+d)^(3/2)*(b*(a* 
e-b*d))^(1/2)*b^2*d+3*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a^3*e^3- 
3*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a^2*e^2+6*(e*x+d)^(1/2)*(b*(a*e-b*d))^ 
(1/2)*a*b*d*e-3*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^2*d^2)/(b*(a*e-b*d))^( 
1/2)/b/(a*e-b*d)^2/((b*x+a)^2)^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (156) = 312\).

Time = 0.10 (sec) , antiderivative size = 785, normalized size of antiderivative = 3.50 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, 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} - 22 \, a b^{3} d^{2} e + 17 \, 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 (b^{4} d^{2} e - 5 \, 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^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\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} - 22 \, a b^{3} d^{2} e + 17 \, 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 (b^{4} d^{2} e - 5 \, 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^{3} - 3 \, a^{4} b^{4} d^{2} e + 3 \, a^{5} b^{3} d e^{2} - a^{6} b^{2} e^{3} + {\left (b^{8} d^{3} - 3 \, a b^{7} d^{2} e + 3 \, a^{2} b^{6} d e^{2} - a^{3} b^{5} e^{3}\right )} x^{3} + 3 \, {\left (a b^{7} d^{3} - 3 \, a^{2} b^{6} d^{2} e + 3 \, a^{3} b^{5} d e^{2} - a^{4} b^{4} e^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{6} d^{3} - 3 \, a^{3} b^{5} d^{2} e + 3 \, a^{4} b^{4} d e^{2} - a^{5} b^{3} e^{3}\right )} x\right )}}\right ] \] Input:

integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),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 - 22*a*b^3*d^2*e + 17*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*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^ 
2*b^2*e^3)*x)*sqrt(e*x + d))/(a^3*b^5*d^3 - 3*a^4*b^4*d^2*e + 3*a^5*b^3*d* 
e^2 - a^6*b^2*e^3 + (b^8*d^3 - 3*a*b^7*d^2*e + 3*a^2*b^6*d*e^2 - a^3*b^5*e 
^3)*x^3 + 3*(a*b^7*d^3 - 3*a^2*b^6*d^2*e + 3*a^3*b^5*d*e^2 - a^4*b^4*e^3)* 
x^2 + 3*(a^2*b^6*d^3 - 3*a^3*b^5*d^2*e + 3*a^4*b^4*d*e^2 - a^5*b^3*e^3)*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 - 22*a*b^3*d^2*e + 17*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*(b^4*d^2*e - 5*a*b^3*d*e^2 + 4*a^2*b^2*e^3)*x)*sqrt 
(e*x + d))/(a^3*b^5*d^3 - 3*a^4*b^4*d^2*e + 3*a^5*b^3*d*e^2 - a^6*b^2*e^3 
+ (b^8*d^3 - 3*a*b^7*d^2*e + 3*a^2*b^6*d*e^2 - a^3*b^5*e^3)*x^3 + 3*(a*b^7 
*d^3 - 3*a^2*b^6*d^2*e + 3*a^3*b^5*d*e^2 - a^4*b^4*e^3)*x^2 + 3*(a^2*b^6*d 
^3 - 3*a^3*b^5*d^2*e + 3*a^4*b^4*d*e^2 - a^5*b^3*e^3)*x)]
 

Sympy [F]

\[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x\right ) \sqrt {d + e x}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((b*x+a)*(e*x+d)**(1/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Integral((a + b*x)*sqrt(d + e*x)/((a + b*x)**2)**(5/2), x)
 

Maxima [F]

\[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )} \sqrt {e x + d}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm=" 
maxima")
 

Output:

integrate((b*x + a)*sqrt(e*x + d)/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{2} e^{3} - 8 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{2} d e^{3} - 3 \, \sqrt {e x + d} b^{2} d^{2} e^{3} + 8 \, {\left (e x + d\right )}^{\frac {3}{2}} a b e^{4} + 6 \, \sqrt {e x + d} a b d e^{4} - 3 \, \sqrt {e x + d} a^{2} e^{5}}{24 \, {\left (b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{3}} \] Input:

integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm=" 
giac")
 

Output:

1/8*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^2*sgn(b*x + a 
) - 2*a*b^2*d*e*sgn(b*x + a) + a^2*b*e^2*sgn(b*x + a))*sqrt(-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*sqr 
t(e*x + d)*b^2*d^2*e^3 + 8*(e*x + d)^(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)/((b^3*d^2*sgn(b*x + a) - 2*a*b^2*d*e*sgn(b 
*x + a) + a^2*b*e^2*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,\sqrt {d+e\,x}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((a + b*x)*(d + e*x)^(1/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int(((a + b*x)*(d + e*x)^(1/2))/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 536, normalized size of antiderivative = 2.39 \[ \int \frac {(a+b x) \sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{3} e^{3}+9 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} b \,e^{3} x +9 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a \,b^{2} e^{3} x^{2}+3 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{3} e^{3} x^{3}-3 \sqrt {e x +d}\, a^{3} b \,e^{3}+17 \sqrt {e x +d}\, a^{2} b^{2} d \,e^{2}+8 \sqrt {e x +d}\, a^{2} b^{2} e^{3} x -22 \sqrt {e x +d}\, a \,b^{3} d^{2} e -10 \sqrt {e x +d}\, a \,b^{3} d \,e^{2} x +3 \sqrt {e x +d}\, a \,b^{3} e^{3} x^{2}+8 \sqrt {e x +d}\, b^{4} d^{3}+2 \sqrt {e x +d}\, b^{4} d^{2} e x -3 \sqrt {e x +d}\, b^{4} d \,e^{2} x^{2}}{24 b^{2} \left (a^{3} b^{3} e^{3} x^{3}-3 a^{2} b^{4} d \,e^{2} x^{3}+3 a \,b^{5} d^{2} e \,x^{3}-b^{6} d^{3} x^{3}+3 a^{4} b^{2} e^{3} x^{2}-9 a^{3} b^{3} d \,e^{2} x^{2}+9 a^{2} b^{4} d^{2} e \,x^{2}-3 a \,b^{5} d^{3} x^{2}+3 a^{5} b \,e^{3} x -9 a^{4} b^{2} d \,e^{2} x +9 a^{3} b^{3} d^{2} e x -3 a^{2} b^{4} d^{3} x +a^{6} e^{3}-3 a^{5} b d \,e^{2}+3 a^{4} b^{2} d^{2} e -a^{3} b^{3} d^{3}\right )} \] Input:

int((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),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 + 17*sqrt(d + e*x)*a**2*b**2*d*e**2 + 8*sqrt(d 
+ e*x)*a**2*b**2*e**3*x - 22*sqrt(d + e*x)*a*b**3*d**2*e - 10*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 + 2*sqrt(d + e*x)*b**4*d**2*e*x - 3*sqrt(d + e*x)*b**4*d*e**2*x**2) 
/(24*b**2*(a**6*e**3 - 3*a**5*b*d*e**2 + 3*a**5*b*e**3*x + 3*a**4*b**2*d** 
2*e - 9*a**4*b**2*d*e**2*x + 3*a**4*b**2*e**3*x**2 - a**3*b**3*d**3 + 9*a* 
*3*b**3*d**2*e*x - 9*a**3*b**3*d*e**2*x**2 + a**3*b**3*e**3*x**3 - 3*a**2* 
b**4*d**3*x + 9*a**2*b**4*d**2*e*x**2 - 3*a**2*b**4*d*e**2*x**3 - 3*a*b**5 
*d**3*x**2 + 3*a*b**5*d**2*e*x**3 - b**6*d**3*x**3))