3.19.11 \(\int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx\) [1811]

3.19.11.1 Optimal result

Integrand size = 33, antiderivative size = 140 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {(2 b B d+A b e-3 a B e) \sqrt {d+e x}}{b^2 (b d-a e)}-\frac {(A b-a B) (d+e x)^{3/2}}{b (b d-a e) (a+b x)}-\frac {(2 b B d+A b e-3 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{5/2} \sqrt {b d-a e}} \]

output

-(A*b-B*a)*(e*x+d)^(3/2)/b/(-a*e+b*d)/(b*x+a)-(A*b*e-3*B*a*e+2*B*b*d)*arct 
anh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/(-a*e+b*d)^(1/2)+(A*b* 
e-3*B*a*e+2*B*b*d)*(e*x+d)^(1/2)/b^2/(-a*e+b*d)
 

3.19.11.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.69 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {(-A b+3 a B+2 b B x) \sqrt {d+e x}}{b^2 (a+b x)}+\frac {(2 b B d+A b e-3 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{b^{5/2} \sqrt {-b d+a e}} \]

input

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

output

((-(A*b) + 3*a*B + 2*b*B*x)*Sqrt[d + e*x])/(b^2*(a + b*x)) + ((2*b*B*d + A 
*b*e - 3*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(b^(5/ 
2)*Sqrt[-(b*d) + a*e])
 

3.19.11.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1184, 27, 87, 60, 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.

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

\(\Big \downarrow \) 1184

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

-(((A*b - a*B)*(d + e*x)^(3/2))/(b*(b*d - a*e)*(a + b*x))) + ((2*b*B*d + A 
*b*e - 3*a*B*e)*((2*Sqrt[d + e*x])/b - (2*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b] 
*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/b^(3/2)))/(2*b*(b*d - a*e))
 

3.19.11.3.1 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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 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]
 

3.19.11.4 Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {-\left (\left (-2 B x +A \right ) b -3 B a \right ) \sqrt {e x +d}\, \sqrt {\left (a e -b d \right ) b}+\left (b x +a \right ) \left (b \left (A e +2 B d \right )-3 B a e \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}\, b^{2} \left (b x +a \right )}\)	\(103\)
risch	\(\frac {2 B \sqrt {e x +d}}{b^{2}}+\frac {\frac {2 \left (-\frac {1}{2} A b e +\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (A b e -3 B a e +2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{b^{2}}\)	\(107\)
derivativedivides	\(\frac {2 B \sqrt {e x +d}}{b^{2}}+\frac {\frac {2 \left (-\frac {1}{2} A b e +\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (A b e -3 B a e +2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{b^{2}}\)	\(108\)
default	\(\frac {2 B \sqrt {e x +d}}{b^{2}}+\frac {\frac {2 \left (-\frac {1}{2} A b e +\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (A b e -3 B a e +2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{b^{2}}\)	\(108\)

input

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

output

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

3.19.11.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\left [\frac {{\left (2 \, B a b d - {\left (3 \, B a^{2} - A a b\right )} e + {\left (2 \, B b^{2} d - {\left (3 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left ({\left (3 \, B a b^{2} - A b^{3}\right )} d - {\left (3 \, B a^{2} b - A a b^{2}\right )} e + 2 \, {\left (B b^{3} d - B a b^{2} e\right )} x\right )} \sqrt {e x + d}}{2 \, {\left (a b^{4} d - a^{2} b^{3} e + {\left (b^{5} d - a b^{4} e\right )} x\right )}}, \frac {{\left (2 \, B a b d - {\left (3 \, B a^{2} - A a b\right )} e + {\left (2 \, B b^{2} d - {\left (3 \, B a b - A b^{2}\right )} e\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left ({\left (3 \, B a b^{2} - A b^{3}\right )} d - {\left (3 \, B a^{2} b - A a b^{2}\right )} e + 2 \, {\left (B b^{3} d - B a b^{2} e\right )} x\right )} \sqrt {e x + d}}{a b^{4} d - a^{2} b^{3} e + {\left (b^{5} d - a b^{4} e\right )} x}\right ] \]

input

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

output

[1/2*((2*B*a*b*d - (3*B*a^2 - A*a*b)*e + (2*B*b^2*d - (3*B*a*b - A*b^2)*e) 
*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)*s 
qrt(e*x + d))/(b*x + a)) + 2*((3*B*a*b^2 - A*b^3)*d - (3*B*a^2*b - A*a*b^2 
)*e + 2*(B*b^3*d - B*a*b^2*e)*x)*sqrt(e*x + d))/(a*b^4*d - a^2*b^3*e + (b^ 
5*d - a*b^4*e)*x), ((2*B*a*b*d - (3*B*a^2 - A*a*b)*e + (2*B*b^2*d - (3*B*a 
*b - A*b^2)*e)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e* 
x + d)/(b*e*x + b*d)) + ((3*B*a*b^2 - A*b^3)*d - (3*B*a^2*b - A*a*b^2)*e + 
 2*(B*b^3*d - B*a*b^2*e)*x)*sqrt(e*x + d))/(a*b^4*d - a^2*b^3*e + (b^5*d - 
 a*b^4*e)*x)]
 

3.19.11.6 Sympy [F]

\[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\int \frac {\left (A + B x\right ) \sqrt {d + e x}}{\left (a + b x\right )^{2}}\, dx \]

input

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

output

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

3.19.11.7 Maxima [F(-2)]

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

input

integrate((B*x+A)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2),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
 

3.19.11.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2 \, \sqrt {e x + d} B}{b^{2}} + \frac {{\left (2 \, B b d - 3 \, B a e + A b e\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} b^{2}} + \frac {\sqrt {e x + d} B a e - \sqrt {e x + d} A b e}{{\left ({\left (e x + d\right )} b - b d + a e\right )} b^{2}} \]

input

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

output

2*sqrt(e*x + d)*B/b^2 + (2*B*b*d - 3*B*a*e + A*b*e)*arctan(sqrt(e*x + d)*b 
/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^2) + (sqrt(e*x + d)*B*a*e - 
 sqrt(e*x + d)*A*b*e)/(((e*x + d)*b - b*d + a*e)*b^2)
 

3.19.11.9 Mupad [B] (verification not implemented)

Time = 10.56 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {(A+B x) \sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx=\frac {2\,B\,\sqrt {d+e\,x}}{b^2}-\frac {\left (A\,b\,e-B\,a\,e\right )\,\sqrt {d+e\,x}}{b^3\,\left (d+e\,x\right )-b^3\,d+a\,b^2\,e}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )\,\left (A\,b\,e-3\,B\,a\,e+2\,B\,b\,d\right )}{b^{5/2}\,\sqrt {a\,e-b\,d}} \]

input

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

output

(2*B*(d + e*x)^(1/2))/b^2 - ((A*b*e - B*a*e)*(d + e*x)^(1/2))/(b^3*(d + e* 
x) - b^3*d + a*b^2*e) + (atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)) 
*(A*b*e - 3*B*a*e + 2*B*b*d))/(b^(5/2)*(a*e - b*d)^(1/2))