\(\int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx\) [54]

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{d}-\frac {2 \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a x+b x^2}}\right )}{\sqrt {c} d} \] Output:

2*b^(1/2)*arctanh(b^(1/2)*x/(b*x^2+a*x)^(1/2))/d-2*(-a*d+b*c)^(1/2)*arctan 
h((-a*d+b*c)^(1/2)*x/c^(1/2)/(b*x^2+a*x)^(1/2))/c^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=-\frac {2 \sqrt {x} \sqrt {a+b x} \left (\sqrt {-b c+a d} \arctan \left (\frac {-d \sqrt {x} \sqrt {a+b x}+\sqrt {b} (c+d x)}{\sqrt {c} \sqrt {-b c+a d}}\right )+\sqrt {b} \sqrt {c} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {a+b x}\right )\right )}{\sqrt {c} d \sqrt {x (a+b x)}} \] Input:

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

Output:

(-2*Sqrt[x]*Sqrt[a + b*x]*(Sqrt[-(b*c) + a*d]*ArcTan[(-(d*Sqrt[x]*Sqrt[a + 
 b*x]) + Sqrt[b]*(c + d*x))/(Sqrt[c]*Sqrt[-(b*c) + a*d])] + Sqrt[b]*Sqrt[c 
]*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[a + b*x]]))/(Sqrt[c]*d*Sqrt[x*(a + b*x)])
 

Rubi [F]

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.

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

\(\Big \downarrow \) 1325

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 1325

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x 
_)^2)^(q_), x_Symbol] :> Unintegrable[(a + b*x + c*x^2)^p*(d + e*x + f*x^2) 
^q, x] /; FreeQ[{a, b, c, d, e, f, p, q}, x] &&  !IGtQ[p, 0] &&  !IGtQ[q, 0 
]
 

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.89

method	result	size
pseudoelliptic	\(-\frac {2 \left (-\sqrt {b}\, \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )+\frac {\left (a d -b c \right ) \arctan \left (\frac {\sqrt {x \left (b x +a \right )}\, c}{x \sqrt {c \left (a d -b c \right )}}\right )}{\sqrt {c \left (a d -b c \right )}}\right )}{d}\)	\(77\)
default	\(\frac {\sqrt {b \,x^{2}+a x}+\frac {a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 \sqrt {b}}}{c}-\frac {\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}+\frac {\left (a d -2 b c \right ) \ln \left (\frac {\frac {a d -2 b c}{2 d}+b \left (x +\frac {c}{d}\right )}{\sqrt {b}}+\sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}\right )}{2 d \sqrt {b}}+\frac {c \left (a d -b c \right ) \ln \left (\frac {-\frac {2 c \left (a d -b c \right )}{d^{2}}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}+\frac {\left (a d -2 b c \right ) \left (x +\frac {c}{d}\right )}{d}-\frac {c \left (a d -b c \right )}{d^{2}}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {c \left (a d -b c \right )}{d^{2}}}}}{c}\)	\(332\)

Input:

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

Output:

-2/d*(-b^(1/2)*arctanh((x*(b*x+a))^(1/2)/x/b^(1/2))+(a*d-b*c)/(c*(a*d-b*c) 
)^(1/2)*arctan((x*(b*x+a))^(1/2)/x*c/(c*(a*d-b*c))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 396, normalized size of antiderivative = 4.55 \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\left [\frac {\sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x - 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right )}{d}, -\frac {2 \, \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right ) - \sqrt {b} \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{d}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) - \sqrt {\frac {b c - a d}{c}} \log \left (\frac {a c + {\left (2 \, b c - a d\right )} x - 2 \, \sqrt {b x^{2} + a x} c \sqrt {\frac {b c - a d}{c}}}{d x + c}\right )}{d}, -\frac {2 \, {\left (\sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x + a}\right ) + \sqrt {-\frac {b c - a d}{c}} \arctan \left (-\frac {\sqrt {b x^{2} + a x} c \sqrt {-\frac {b c - a d}{c}}}{{\left (b c - a d\right )} x}\right )\right )}}{d}\right ] \] Input:

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

Output:

[(sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + a*x)*sqrt(b)) + sqrt((b*c - a*d)/ 
c)*log((a*c + (2*b*c - a*d)*x - 2*sqrt(b*x^2 + a*x)*c*sqrt((b*c - a*d)/c)) 
/(d*x + c)))/d, -(2*sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sqrt( 
-(b*c - a*d)/c)/((b*c - a*d)*x)) - sqrt(b)*log(2*b*x + a + 2*sqrt(b*x^2 + 
a*x)*sqrt(b)))/d, -(2*sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x + a) 
) - sqrt((b*c - a*d)/c)*log((a*c + (2*b*c - a*d)*x - 2*sqrt(b*x^2 + a*x)*c 
*sqrt((b*c - a*d)/c))/(d*x + c)))/d, -2*(sqrt(-b)*arctan(sqrt(b*x^2 + a*x) 
*sqrt(-b)/(b*x + a)) + sqrt(-(b*c - a*d)/c)*arctan(-sqrt(b*x^2 + a*x)*c*sq 
rt(-(b*c - a*d)/c)/((b*c - a*d)*x)))/d]
 

Sympy [F]

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

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

Output:

Integral(sqrt(x*(a + b*x))/(x*(c + d*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x^2+a*x)^(1/2)/(d*x^2+c*x),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*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\int \frac {\sqrt {b\,x^2+a\,x}}{d\,x^2+c\,x} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {a x+b x^2}}{c x+d x^2} \, dx=\frac {-2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}-\sqrt {d}\, \sqrt {b x +a}-\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right )-2 \sqrt {c}\, \sqrt {a d -b c}\, \mathit {atan} \left (\frac {\sqrt {a d -b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {x}\, \sqrt {d}\, \sqrt {b}}{\sqrt {c}\, \sqrt {b}}\right )+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b x +a}+\sqrt {x}\, \sqrt {b}}{\sqrt {a}}\right ) c}{c d} \] Input:

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

Output:

(2*( - sqrt(c)*sqrt(a*d - b*c)*atan((sqrt(a*d - b*c) - sqrt(d)*sqrt(a + b* 
x) - sqrt(x)*sqrt(d)*sqrt(b))/(sqrt(c)*sqrt(b))) - sqrt(c)*sqrt(a*d - b*c) 
*atan((sqrt(a*d - b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(x)*sqrt(d)*sqrt(b))/ 
(sqrt(c)*sqrt(b))) + sqrt(b)*log((sqrt(a + b*x) + sqrt(x)*sqrt(b))/sqrt(a) 
)*c))/(c*d)