\(\int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx\) [176]

Optimal result

Integrand size = 24, antiderivative size = 113 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {2 (B d-A e) \sqrt {a+b x}}{e^2 \sqrt {d+e x}}+\frac {B \sqrt {a+b x} \sqrt {d+e x}}{e^2}-\frac {(3 b B d-2 A b e-a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {a+b x} (3 B d-2 A e+B e x)}{e^2 \sqrt {d+e x}}+\frac {(-3 b B d+2 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{\sqrt {b} e^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.28, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {87, 60, 66, 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.

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(385\) vs. \(2(93)=186\).

Time = 0.26 (sec) , antiderivative size = 386, normalized size of antiderivative = 3.42

method	result	size
default	\(\frac {\sqrt {b x +a}\, \left (2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b \,e^{2} x +B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a \,e^{2} x -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b d e x +2 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b d e +B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a d e -3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b \,d^{2}+2 B e x \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}-4 A e \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}+6 B d \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}\right )}{2 \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, e^{2} \sqrt {e x +d}}\)	\(386\)

Input:

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

Output:

1/2*(b*x+a)^(1/2)*(2*A*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/ 
2)+a*e+d*b)/(b*e)^(1/2))*b*e^2*x+B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/ 
2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*a*e^2*x-3*B*ln(1/2*(2*b*e*x+2*((e*x+d 
)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b*d*e*x+2*A*ln(1/2*(2*b 
*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2))*b*d*e+B*l 
n(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/(b*e)^(1/2)) 
*a*d*e-3*B*ln(1/2*(2*b*e*x+2*((e*x+d)*(b*x+a))^(1/2)*(b*e)^(1/2)+a*e+d*b)/ 
(b*e)^(1/2))*b*d^2+2*B*e*x*(b*e)^(1/2)*((e*x+d)*(b*x+a))^(1/2)-4*A*e*(b*e) 
^(1/2)*((e*x+d)*(b*x+a))^(1/2)+6*B*d*(b*e)^(1/2)*((e*x+d)*(b*x+a))^(1/2))/ 
(b*e)^(1/2)/((e*x+d)*(b*x+a))^(1/2)/e^2/(e*x+d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 358, normalized size of antiderivative = 3.17 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\left [-\frac {{\left (3 \, B b d^{2} - {\left (B a + 2 \, A b\right )} d e + {\left (3 \, B b d e - {\left (B a + 2 \, A b\right )} e^{2}\right )} x\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) - 4 \, {\left (B b e^{2} x + 3 \, B b d e - 2 \, A b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{4 \, {\left (b e^{4} x + b d e^{3}\right )}}, \frac {{\left (3 \, B b d^{2} - {\left (B a + 2 \, A b\right )} d e + {\left (3 \, B b d e - {\left (B a + 2 \, A b\right )} e^{2}\right )} x\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) + 2 \, {\left (B b e^{2} x + 3 \, B b d e - 2 \, A b e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b e^{4} x + b d e^{3}\right )}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(3/2),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(e*(a*e-b*d)>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {\sqrt {b x + a} {\left (\frac {{\left (b x + a\right )} B {\left | b \right |}}{b e} + \frac {3 \, B b^{2} d e {\left | b \right |} - B a b e^{2} {\left | b \right |} - 2 \, A b^{2} e^{2} {\left | b \right |}}{b^{2} e^{3}}\right )}}{\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e}} + \frac {{\left (3 \, B b d {\left | b \right |} - B a e {\left | b \right |} - 2 \, A b e {\left | b \right |}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b e^{2}} \] Input:

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

Output:

sqrt(b*x + a)*((b*x + a)*B*abs(b)/(b*e) + (3*B*b^2*d*e*abs(b) - B*a*b*e^2* 
abs(b) - 2*A*b^2*e^2*abs(b))/(b^2*e^3))/sqrt(b^2*d + (b*x + a)*b*e - a*b*e 
) + (3*B*b*d*abs(b) - B*a*e*abs(b) - 2*A*b*e*abs(b))*log(abs(-sqrt(b*e)*sq 
rt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {a+b x} (A+B x)}{(d+e x)^{3/2}} \, dx=\frac {-8 \sqrt {e x +d}\, \sqrt {b x +a}\, a \,e^{2}+12 \sqrt {e x +d}\, \sqrt {b x +a}\, b d e +4 \sqrt {e x +d}\, \sqrt {b x +a}\, b \,e^{2} x +12 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a d e +12 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a \,e^{2} x -12 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b \,d^{2}-12 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b d e x -9 \sqrt {e}\, \sqrt {b}\, a d e -9 \sqrt {e}\, \sqrt {b}\, a \,e^{2} x +9 \sqrt {e}\, \sqrt {b}\, b \,d^{2}+9 \sqrt {e}\, \sqrt {b}\, b d e x}{4 e^{3} \left (e x +d \right )} \] Input:

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

Output:

( - 8*sqrt(d + e*x)*sqrt(a + b*x)*a*e**2 + 12*sqrt(d + e*x)*sqrt(a + b*x)* 
b*d*e + 4*sqrt(d + e*x)*sqrt(a + b*x)*b*e**2*x + 12*sqrt(e)*sqrt(b)*log((s 
qrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a*d*e + 12* 
sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a 
*e - b*d))*a*e**2*x - 12*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt 
(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b*d**2 - 12*sqrt(e)*sqrt(b)*log((sqrt( 
e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b*d*e*x - 9*sqr 
t(e)*sqrt(b)*a*d*e - 9*sqrt(e)*sqrt(b)*a*e**2*x + 9*sqrt(e)*sqrt(b)*b*d**2 
 + 9*sqrt(e)*sqrt(b)*b*d*e*x)/(4*e**3*(d + e*x))