\(\int \frac {\sqrt {d+e x}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [383]

Optimal result

Integrand size = 30, antiderivative size = 168 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {e \sqrt {d+e x}}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {\sqrt {b} \sqrt {-b d+a e} \sqrt {d+e x} (a e-b (2 d+e x))-e^2 (a+b x)^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{3/2} (-b d+a e)^{3/2} (a+b x) \sqrt {(a+b x)^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1102, 27, 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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((a + b*x)*(-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]*Sqrt[d + e*x])/Sqrt[b*d - a*e]]) 
/(Sqrt[b]*(b*d - a*e)^(3/2))))/(4*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 1102

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

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\left (\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{2} e^{2} x^{2}+2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a b \,e^{2} x +\left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b +\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) e^{2} a^{2}-\sqrt {e x +d}\, a e \sqrt {b \left (a e -b d \right )}+\sqrt {e x +d}\, d b \sqrt {b \left (a e -b d \right )}\right ) \left (b x +a \right )}{4 \sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right ) b \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(200\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 456, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} + {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} + {\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \, {\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}, -\frac {{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} + {\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} + {\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \, {\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {{\left (e x + d\right )}^{\frac {3}{2}} b e^{2} + \sqrt {e x + d} b d e^{2} - \sqrt {e x + d} a e^{3}}{4 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.79 \[ \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a^{2} e^{2}+2 \sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) a b \,e^{2} x +\sqrt {b}\, \sqrt {a e -b d}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, b}{\sqrt {b}\, \sqrt {a e -b d}}\right ) b^{2} e^{2} x^{2}-\sqrt {e x +d}\, a^{2} b \,e^{2}+3 \sqrt {e x +d}\, a \,b^{2} d e +\sqrt {e x +d}\, a \,b^{2} e^{2} x -2 \sqrt {e x +d}\, b^{3} d^{2}-\sqrt {e x +d}\, b^{3} d e x}{4 b^{2} \left (a^{2} b^{2} e^{2} x^{2}-2 a \,b^{3} d e \,x^{2}+b^{4} d^{2} x^{2}+2 a^{3} b \,e^{2} x -4 a^{2} b^{2} d e x +2 a \,b^{3} d^{2} x +a^{4} e^{2}-2 a^{3} b d e +a^{2} b^{2} d^{2}\right )} \] Input:

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

Output:

(sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d))) 
*a**2*e**2 + 2*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqr 
t(a*e - b*d)))*a*b*e**2*x + sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b) 
/(sqrt(b)*sqrt(a*e - b*d)))*b**2*e**2*x**2 - sqrt(d + e*x)*a**2*b*e**2 + 3 
*sqrt(d + e*x)*a*b**2*d*e + sqrt(d + e*x)*a*b**2*e**2*x - 2*sqrt(d + e*x)* 
b**3*d**2 - sqrt(d + e*x)*b**3*d*e*x)/(4*b**2*(a**4*e**2 - 2*a**3*b*d*e + 
2*a**3*b*e**2*x + a**2*b**2*d**2 - 4*a**2*b**2*d*e*x + a**2*b**2*e**2*x**2 
 + 2*a*b**3*d**2*x - 2*a*b**3*d*e*x**2 + b**4*d**2*x**2))