\(\int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx\) [175]

Optimal result

Integrand size = 24, antiderivative size = 140 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx=-\frac {(3 b B d-4 A b e+a B e) \sqrt {a+b x} \sqrt {d+e x}}{4 b e^2}+\frac {B (a+b x)^{3/2} \sqrt {d+e x}}{2 b e}+\frac {(b d-a e) (3 b B d-4 A b e+a B e) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a+b x}}{\sqrt {b} \sqrt {d+e x}}\right )}{4 b^{3/2} e^{5/2}} \] Output:

-1/4*(-4*A*b*e+B*a*e+3*B*b*d)*(b*x+a)^(1/2)*(e*x+d)^(1/2)/b/e^2+1/2*B*(b*x 
+a)^(3/2)*(e*x+d)^(1/2)/b/e+1/4*(-a*e+b*d)*(-4*A*b*e+B*a*e+3*B*b*d)*arctan 
h(e^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(e*x+d)^(1/2))/b^(3/2)/e^(5/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx=\frac {b \sqrt {a+b x} \sqrt {d+e x} (a B e+b (-3 B d+4 A e+2 B e x))+\sqrt {\frac {b}{e}} (b d-a e) (-3 b B d+4 A b e-a B e) \log \left (\sqrt {a+b x}-\sqrt {\frac {b}{e}} \sqrt {d+e x}\right )}{4 b^2 e^2} \] Input:

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

Output:

(b*Sqrt[a + b*x]*Sqrt[d + e*x]*(a*B*e + b*(-3*B*d + 4*A*e + 2*B*e*x)) + Sq 
rt[b/e]*(b*d - a*e)*(-3*b*B*d + 4*A*b*e - a*B*e)*Log[Sqrt[a + b*x] - Sqrt[ 
b/e]*Sqrt[d + e*x]])/(4*b^2*e^2)
 

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {90, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {B (a+b x)^{3/2} \sqrt {d+e x}}{2 b e}-\frac {(a B e-4 A b e+3 b B d) \int \frac {\sqrt {a+b x}}{\sqrt {d+e x}}dx}{4 b e}\)

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(B*(a + b*x)^(3/2)*Sqrt[d + e*x])/(2*b*e) - ((3*b*B*d - 4*A*b*e + a*B*e)*( 
(Sqrt[a + b*x]*Sqrt[d + e*x])/e - ((b*d - a*e)*ArcTanh[(Sqrt[e]*Sqrt[a + b 
*x])/(Sqrt[b]*Sqrt[d + e*x])])/(Sqrt[b]*e^(3/2))))/(4*b*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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

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. \(375\) vs. \(2(114)=228\).

Time = 0.26 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.69

method	result	size
default	\(\frac {\sqrt {b x +a}\, \sqrt {e x +d}\, \left (4 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a b \,e^{2}-4 A \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{2} d e +4 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, b e x -B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a^{2} e^{2}-2 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) a b d e +3 B \ln \left (\frac {2 b e x +2 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}+a e +d b}{2 \sqrt {b e}}\right ) b^{2} d^{2}+8 A \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, b e +2 B \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, \sqrt {b e}\, a e -6 B \sqrt {b e}\, \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, b d \right )}{8 \sqrt {\left (e x +d \right ) \left (b x +a \right )}\, e^{2} b \sqrt {b e}}\)	\(376\)

Input:

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

Output:

1/8*(b*x+a)^(1/2)*(e*x+d)^(1/2)*(4*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))*a*b*e^2-4*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^2*d*e+4*B*((e*x+d)* 
(b*x+a))^(1/2)*(b*e)^(1/2)*b*e*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^2*e^2-2*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*b*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^2*d^2+8 
*A*(b*e)^(1/2)*((e*x+d)*(b*x+a))^(1/2)*b*e+2*B*((e*x+d)*(b*x+a))^(1/2)*(b* 
e)^(1/2)*a*e-6*B*(b*e)^(1/2)*((e*x+d)*(b*x+a))^(1/2)*b*d)/((e*x+d)*(b*x+a) 
)^(1/2)/e^2/b/(b*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.60 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx=\left [\frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (B a b + 2 \, A b^{2}\right )} d e - {\left (B a^{2} - 4 \, A a b\right )} e^{2}\right )} \sqrt {b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b e x + b d + a e\right )} \sqrt {b e} \sqrt {b x + a} \sqrt {e x + d} + 8 \, {\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \, {\left (2 \, B b^{2} e^{2} x - 3 \, B b^{2} d e + {\left (B a b + 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{16 \, b^{2} e^{3}}, -\frac {{\left (3 \, B b^{2} d^{2} - 2 \, {\left (B a b + 2 \, A b^{2}\right )} d e - {\left (B a^{2} - 4 \, A a b\right )} e^{2}\right )} \sqrt {-b e} \arctan \left (\frac {{\left (2 \, b e x + b d + a e\right )} \sqrt {-b e} \sqrt {b x + a} \sqrt {e x + d}}{2 \, {\left (b^{2} e^{2} x^{2} + a b d e + {\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \, {\left (2 \, B b^{2} e^{2} x - 3 \, B b^{2} d e + {\left (B a b + 4 \, A b^{2}\right )} e^{2}\right )} \sqrt {b x + a} \sqrt {e x + d}}{8 \, b^{2} e^{3}}\right ] \] Input:

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

Output:

[1/16*((3*B*b^2*d^2 - 2*(B*a*b + 2*A*b^2)*d*e - (B*a^2 - 4*A*a*b)*e^2)*sqr 
t(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*(2*B*b^2*e^2*x - 3*B*b^2*d*e + (B*a*b + 4*A*b^2)*e^2)*sqrt(b*x + a)*sq 
rt(e*x + d))/(b^2*e^3), -1/8*((3*B*b^2*d^2 - 2*(B*a*b + 2*A*b^2)*d*e - (B* 
a^2 - 4*A*a*b)*e^2)*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*(2*B*b^2*e^2*x - 3*B*b^2*d*e + (B*a*b + 4*A*b^2)*e^2)*sqrt(b*x + a 
)*sqrt(e*x + d))/(b^2*e^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((b*x+a)^(1/2)*(B*x+A)/(e*x+d)^(1/2),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>0)', see `assume?` for more de 
tails)Is e
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx=\frac {{\left (\sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \sqrt {b x + a} {\left (\frac {2 \, {\left (b x + a\right )} B}{b^{2} e} - \frac {3 \, B b^{3} d e + B a b^{2} e^{2} - 4 \, A b^{3} e^{2}}{b^{4} e^{3}}\right )} - \frac {{\left (3 \, B b^{2} d^{2} - 2 \, B a b d e - 4 \, A b^{2} d e - B a^{2} e^{2} + 4 \, A a b e^{2}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {b x + a} + \sqrt {b^{2} d + {\left (b x + a\right )} b e - a b e} \right |}\right )}{\sqrt {b e} b e^{2}}\right )} b}{4 \, {\left | b \right |}} \] Input:

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

Output:

1/4*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*B/(b^2 
*e) - (3*B*b^3*d*e + B*a*b^2*e^2 - 4*A*b^3*e^2)/(b^4*e^3)) - (3*B*b^2*d^2 
- 2*B*a*b*d*e - 4*A*b^2*d*e - B*a^2*e^2 + 4*A*a*b*e^2)*log(abs(-sqrt(b*e)* 
sqrt(b*x + a) + sqrt(b^2*d + (b*x + a)*b*e - a*b*e)))/(sqrt(b*e)*b*e^2))*b 
/abs(b)
 

Mupad [B] (verification not implemented)

Time = 14.50 (sec) , antiderivative size = 872, normalized size of antiderivative = 6.23 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx =\text {Too large to display} \] Input:

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

Output:

((((a + b*x)^(1/2) - a^(1/2))*((B*a^2*b^2*e^2)/2 - (3*B*b^4*d^2)/2 + B*a*b 
^3*d*e))/(e^6*((d + e*x)^(1/2) - d^(1/2))) + (((a + b*x)^(1/2) - a^(1/2))^ 
3*((11*B*b^3*d^2)/2 + (7*B*a^2*b*e^2)/2 + 23*B*a*b^2*d*e))/(e^5*((d + e*x) 
^(1/2) - d^(1/2))^3) + (((a + b*x)^(1/2) - a^(1/2))^5*((7*B*a^2*e^2)/2 + ( 
11*B*b^2*d^2)/2 + 23*B*a*b*d*e))/(e^4*((d + e*x)^(1/2) - d^(1/2))^5) + ((( 
a + b*x)^(1/2) - a^(1/2))^7*((B*a^2*e^2)/2 - (3*B*b^2*d^2)/2 + B*a*b*d*e)) 
/(b*e^3*((d + e*x)^(1/2) - d^(1/2))^7) - (a^(1/2)*d^(1/2)*((a + b*x)^(1/2) 
 - a^(1/2))^4*(32*B*b^2*d + 16*B*a*b*e))/(e^4*((d + e*x)^(1/2) - d^(1/2))^ 
4) - (8*B*a^(3/2)*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6)/(e^2*((d + e*x)^( 
1/2) - d^(1/2))^6) - (8*B*a^(3/2)*b^2*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^ 
2)/(e^4*((d + e*x)^(1/2) - d^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))^8/((d 
 + e*x)^(1/2) - d^(1/2))^8 + b^4/e^4 - (4*b^3*((a + b*x)^(1/2) - a^(1/2))^ 
2)/(e^3*((d + e*x)^(1/2) - d^(1/2))^2) + (6*b^2*((a + b*x)^(1/2) - a^(1/2) 
)^4)/(e^2*((d + e*x)^(1/2) - d^(1/2))^4) - (4*b*((a + b*x)^(1/2) - a^(1/2) 
)^6)/(e*((d + e*x)^(1/2) - d^(1/2))^6)) + ((((a + b*x)^(1/2) - a^(1/2))*(2 
*A*b^2*d + 2*A*a*b*e))/(e^3*((d + e*x)^(1/2) - d^(1/2))) + ((2*A*a*e + 2*A 
*b*d)*((a + b*x)^(1/2) - a^(1/2))^3)/(e^2*((d + e*x)^(1/2) - d^(1/2))^3) - 
 (8*A*a^(1/2)*b*d^(1/2)*((a + b*x)^(1/2) - a^(1/2))^2)/(e^2*((d + e*x)^(1/ 
2) - d^(1/2))^2))/(((a + b*x)^(1/2) - a^(1/2))^4/((d + e*x)^(1/2) - d^(1/2 
))^4 + b^2/e^2 - (2*b*((a + b*x)^(1/2) - a^(1/2))^2)/(e*((d + e*x)^(1/2...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.41 \[ \int \frac {\sqrt {a+b x} (A+B x)}{\sqrt {d+e x}} \, dx=\frac {5 \sqrt {e x +d}\, \sqrt {b x +a}\, a b \,e^{2}-3 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{2} d e +2 \sqrt {e x +d}\, \sqrt {b x +a}\, b^{2} e^{2} x +3 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a^{2} e^{2}-6 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) a b d e +3 \sqrt {e}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {e x +d}}{\sqrt {a e -b d}}\right ) b^{2} d^{2}}{4 b \,e^{3}} \] Input:

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

Output:

(5*sqrt(d + e*x)*sqrt(a + b*x)*a*b*e**2 - 3*sqrt(d + e*x)*sqrt(a + b*x)*b* 
*2*d*e + 2*sqrt(d + e*x)*sqrt(a + b*x)*b**2*e**2*x + 3*sqrt(e)*sqrt(b)*log 
((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*a**2*e** 
2 - 6*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + sqrt(b)*sqrt(d + e*x))/ 
sqrt(a*e - b*d))*a*b*d*e + 3*sqrt(e)*sqrt(b)*log((sqrt(e)*sqrt(a + b*x) + 
sqrt(b)*sqrt(d + e*x))/sqrt(a*e - b*d))*b**2*d**2)/(4*b*e**3)