\(\int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx\) [121]

Optimal result

Integrand size = 24, antiderivative size = 116 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\frac {(4 B c d-3 b B e+4 A c e) \sqrt {b x+c x^2}}{4 c^2}+\frac {B e x \sqrt {b x+c x^2}}{2 c}+\frac {\left (8 A c^2 d+3 b^2 B e-4 b c (B d+A e)\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{5/2}} \] Output:

1/4*(4*A*c*e-3*B*b*e+4*B*c*d)*(c*x^2+b*x)^(1/2)/c^2+1/2*B*e*x*(c*x^2+b*x)^ 
(1/2)/c+1/4*(8*A*c^2*d+3*b^2*B*e-4*b*c*(A*e+B*d))*arctanh(c^(1/2)*x/(c*x^2 
+b*x)^(1/2))/c^(5/2)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.06 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {x} \left (\sqrt {c} \sqrt {x} (b+c x) (4 A c e+B (4 c d-3 b e+2 c e x))+\left (-8 A c^2 d-3 b^2 B e+4 b c (B d+A e)\right ) \sqrt {b+c x} \log \left (-\sqrt {c} \sqrt {x}+\sqrt {b+c x}\right )\right )}{4 c^{5/2} \sqrt {x (b+c x)}} \] Input:

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

Output:

(Sqrt[x]*(Sqrt[c]*Sqrt[x]*(b + c*x)*(4*A*c*e + B*(4*c*d - 3*b*e + 2*c*e*x) 
) + (-8*A*c^2*d - 3*b^2*B*e + 4*b*c*(B*d + A*e))*Sqrt[b + c*x]*Log[-(Sqrt[ 
c]*Sqrt[x]) + Sqrt[b + c*x]]))/(4*c^(5/2)*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1225, 1091, 219}

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.

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

\(\Big \downarrow \) 1225

\(\displaystyle \frac {\left (-4 b c (A e+B d)+8 A c^2 d+3 b^2 B e\right ) \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c^2}-\frac {\sqrt {b x+c x^2} (-4 c (A e+B d)+3 b B e-2 B c e x)}{4 c^2}\)

\(\Big \downarrow \) 1091

\(\displaystyle \frac {\left (-4 b c (A e+B d)+8 A c^2 d+3 b^2 B e\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c^2}-\frac {\sqrt {b x+c x^2} (-4 c (A e+B d)+3 b B e-2 B c e x)}{4 c^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (-4 b c (A e+B d)+8 A c^2 d+3 b^2 B e\right )}{4 c^{5/2}}-\frac {\sqrt {b x+c x^2} (-4 c (A e+B d)+3 b B e-2 B c e x)}{4 c^2}\)

Input:

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

Output:

-1/4*((3*b*B*e - 4*c*(B*d + A*e) - 2*B*c*e*x)*Sqrt[b*x + c*x^2])/c^2 + ((8 
*A*c^2*d + 3*b^2*B*e - 4*b*c*(B*d + A*e))*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c 
*x^2]])/(4*c^(5/2))
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1225

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( 
x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 
 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), 
 x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p 
+ 3))/(2*c^2*(2*p + 3))   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.74

method	result	size
pseudoelliptic	\(\frac {\left (\left (\frac {3}{4} e \,b^{2}-d b c \right ) B -c A \left (b e -2 c d \right )\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )+\sqrt {x \left (c x +b \right )}\, \left (\left (\left (\frac {e x}{2}+d \right ) B +A e \right ) c^{\frac {3}{2}}-\frac {3 B b e \sqrt {c}}{4}\right )}{c^{\frac {5}{2}}}\)	\(86\)
risch	\(\frac {\left (2 B c e x +4 A c e -3 B b e +4 B c d \right ) x \left (c x +b \right )}{4 c^{2} \sqrt {x \left (c x +b \right )}}-\frac {\left (4 A c e b -8 A \,c^{2} d -3 b^{2} B e +4 B b c d \right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {5}{2}}}\)	\(100\)
default	\(\frac {A d \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}+\left (A e +B d \right ) \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )+B e \left (\frac {x \sqrt {c \,x^{2}+b x}}{2 c}-\frac {3 b \left (\frac {\sqrt {c \,x^{2}+b x}}{c}-\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {3}{2}}}\right )}{4 c}\right )\)	\(159\)

Input:

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

Output:

1/c^(5/2)*(((3/4*e*b^2-d*b*c)*B-c*A*(b*e-2*c*d))*arctanh((x*(c*x+b))^(1/2) 
/x/c^(1/2))+(x*(c*x+b))^(1/2)*(((1/2*e*x+d)*B+A*e)*c^(3/2)-3/4*B*b*e*c^(1/ 
2)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\left [\frac {{\left (4 \, {\left (B b c - 2 \, A c^{2}\right )} d - {\left (3 \, B b^{2} - 4 \, A b c\right )} e\right )} \sqrt {c} \log \left (2 \, c x + b - 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, B c^{2} e x + 4 \, B c^{2} d - {\left (3 \, B b c - 4 \, A c^{2}\right )} e\right )} \sqrt {c x^{2} + b x}}{8 \, c^{3}}, \frac {{\left (4 \, {\left (B b c - 2 \, A c^{2}\right )} d - {\left (3 \, B b^{2} - 4 \, A b c\right )} e\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) + {\left (2 \, B c^{2} e x + 4 \, B c^{2} d - {\left (3 \, B b c - 4 \, A c^{2}\right )} e\right )} \sqrt {c x^{2} + b x}}{4 \, c^{3}}\right ] \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\begin {cases} \left (A d - \frac {b \left (A e - \frac {3 B b e}{4 c} + B d\right )}{2 c}\right ) \left (\begin {cases} \frac {\log {\left (b + 2 \sqrt {c} \sqrt {b x + c x^{2}} + 2 c x \right )}}{\sqrt {c}} & \text {for}\: \frac {b^{2}}{c} \neq 0 \\\frac {\left (\frac {b}{2 c} + x\right ) \log {\left (\frac {b}{2 c} + x \right )}}{\sqrt {c \left (\frac {b}{2 c} + x\right )^{2}}} & \text {otherwise} \end {cases}\right ) + \sqrt {b x + c x^{2}} \left (\frac {B e x}{2 c} + \frac {A e - \frac {3 B b e}{4 c} + B d}{c}\right ) & \text {for}\: c \neq 0 \\\frac {2 \left (A d \sqrt {b x} + \frac {B e \left (b x\right )^{\frac {5}{2}}}{5 b^{2}} + \frac {\left (b x\right )^{\frac {3}{2}} \left (A e + B d\right )}{3 b}\right )}{b} & \text {for}\: b \neq 0 \\\tilde {\infty } \left (A d x + \frac {B e x^{3}}{3} + \frac {x^{2} \left (A e + B d\right )}{2}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise(((A*d - b*(A*e - 3*B*b*e/(4*c) + B*d)/(2*c))*Piecewise((log(b + 
2*sqrt(c)*sqrt(b*x + c*x**2) + 2*c*x)/sqrt(c), Ne(b**2/c, 0)), ((b/(2*c) + 
 x)*log(b/(2*c) + x)/sqrt(c*(b/(2*c) + x)**2), True)) + sqrt(b*x + c*x**2) 
*(B*e*x/(2*c) + (A*e - 3*B*b*e/(4*c) + B*d)/c), Ne(c, 0)), (2*(A*d*sqrt(b* 
x) + B*e*(b*x)**(5/2)/(5*b**2) + (b*x)**(3/2)*(A*e + B*d)/(3*b))/b, Ne(b, 
0)), (zoo*(A*d*x + B*e*x**3/3 + x**2*(A*e + B*d)/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\frac {\sqrt {c x^{2} + b x} B e x}{2 \, c} + \frac {A d \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{\sqrt {c}} + \frac {3 \, B b^{2} e \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {5}{2}}} - \frac {3 \, \sqrt {c x^{2} + b x} B b e}{4 \, c^{2}} - \frac {{\left (B d + A e\right )} b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {3}{2}}} + \frac {\sqrt {c x^{2} + b x} {\left (B d + A e\right )}}{c} \] Input:

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

Output:

1/2*sqrt(c*x^2 + b*x)*B*e*x/c + A*d*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sq 
rt(c))/sqrt(c) + 3/8*B*b^2*e*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/ 
c^(5/2) - 3/4*sqrt(c*x^2 + b*x)*B*b*e/c^2 - 1/2*(B*d + A*e)*b*log(2*c*x + 
b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2) + sqrt(c*x^2 + b*x)*(B*d + A*e)/c
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\frac {1}{4} \, \sqrt {c x^{2} + b x} {\left (\frac {2 \, B e x}{c} + \frac {4 \, B c d - 3 \, B b e + 4 \, A c e}{c^{2}}\right )} + \frac {{\left (4 \, B b c d - 8 \, A c^{2} d - 3 \, B b^{2} e + 4 \, A b c e\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {5}{2}}} \] Input:

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

Output:

1/4*sqrt(c*x^2 + b*x)*(2*B*e*x/c + (4*B*c*d - 3*B*b*e + 4*A*c*e)/c^2) + 1/ 
8*(4*B*b*c*d - 8*A*c^2*d - 3*B*b^2*e + 4*A*b*c*e)*log(abs(2*(sqrt(c)*x - s 
qrt(c*x^2 + b*x))*sqrt(c) + b))/c^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.49 \[ \int \frac {(A+B x) (d+e x)}{\sqrt {b x+c x^2}} \, dx=\frac {4 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{2} e -3 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c e +4 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} d +2 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} e x -4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a b c e +8 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a \,c^{2} d +3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} e -4 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{2} c d}{4 c^{3}} \] Input:

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

Output:

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