\(\int \frac {d+e x}{x \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [76]

Optimal result

Integrand size = 38, antiderivative size = 137 \[ \int \frac {d+e x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} (d+e x)}{\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{\sqrt {c} \sqrt {d}}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} (d+e x)}{\sqrt {d} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{\sqrt {a} \sqrt {e}} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(485\) vs. \(2(137)=274\).

Time = 2.27 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.54 \[ \int \frac {d+e x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=-\frac {2 \sqrt {a e+c d x} \sqrt {d+e x} \left (\sqrt {c} d \left (-\left (\left (-\sqrt {c} d+\sqrt {c d^2-a e^2}\right ) \sqrt {-2 c d^2+a e^2-2 \sqrt {c} d \sqrt {c d^2-a e^2}} \arctan \left (\frac {\sqrt {-2 c d^2+a e^2-2 \sqrt {c} d \sqrt {c d^2-a e^2}} \sqrt {a e+c d x}}{\sqrt {a} \sqrt {c} \sqrt {d} \sqrt {e} \left (\sqrt {d-\frac {a e^2}{c d}}-\sqrt {d+e x}\right )}\right )\right )+\left (\sqrt {c} d+\sqrt {c d^2-a e^2}\right ) \sqrt {-2 c d^2+a e^2+2 \sqrt {c} d \sqrt {c d^2-a e^2}} \arctan \left (\frac {\sqrt {-2 c d^2+a e^2+2 \sqrt {c} d \sqrt {c d^2-a e^2}} \sqrt {a e+c d x}}{\sqrt {a} \sqrt {c} \sqrt {d} \sqrt {e} \left (\sqrt {d-\frac {a e^2}{c d}}-\sqrt {d+e x}\right )}\right )\right )+2 a^{3/2} e^3 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \left (\sqrt {d-\frac {a e^2}{c d}}-\sqrt {d+e x}\right )}\right )\right )}{a^{3/2} \sqrt {c} \sqrt {d} e^{5/2} \sqrt {(a e+c d x) (d+e x)}} \] Input:

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

Output:

(-2*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*(Sqrt[c]*d*(-((-(Sqrt[c]*d) + Sqrt[c*d 
^2 - a*e^2])*Sqrt[-2*c*d^2 + a*e^2 - 2*Sqrt[c]*d*Sqrt[c*d^2 - a*e^2]]*ArcT 
an[(Sqrt[-2*c*d^2 + a*e^2 - 2*Sqrt[c]*d*Sqrt[c*d^2 - a*e^2]]*Sqrt[a*e + c* 
d*x])/(Sqrt[a]*Sqrt[c]*Sqrt[d]*Sqrt[e]*(Sqrt[d - (a*e^2)/(c*d)] - Sqrt[d + 
 e*x]))]) + (Sqrt[c]*d + Sqrt[c*d^2 - a*e^2])*Sqrt[-2*c*d^2 + a*e^2 + 2*Sq 
rt[c]*d*Sqrt[c*d^2 - a*e^2]]*ArcTan[(Sqrt[-2*c*d^2 + a*e^2 + 2*Sqrt[c]*d*S 
qrt[c*d^2 - a*e^2]]*Sqrt[a*e + c*d*x])/(Sqrt[a]*Sqrt[c]*Sqrt[d]*Sqrt[e]*(S 
qrt[d - (a*e^2)/(c*d)] - Sqrt[d + e*x]))]) + 2*a^(3/2)*e^3*ArcTanh[(Sqrt[e 
]*Sqrt[a*e + c*d*x])/(Sqrt[c]*Sqrt[d]*(Sqrt[d - (a*e^2)/(c*d)] - Sqrt[d + 
e*x]))]))/(a^(3/2)*Sqrt[c]*Sqrt[d]*e^(5/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1268, 140, 27, 66, 104, 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.

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

\(\Big \downarrow \) 1268

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

\(\Big \downarrow \) 140

\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (e \int \frac {1}{\sqrt {a e+c d x} \sqrt {d+e x}}dx+\int \frac {d}{x \sqrt {a e+c d x} \sqrt {d+e x}}dx\right )}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (e \int \frac {1}{\sqrt {a e+c d x} \sqrt {d+e x}}dx+d \int \frac {1}{x \sqrt {a e+c d x} \sqrt {d+e x}}dx\right )}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (d \int \frac {1}{x \sqrt {a e+c d x} \sqrt {d+e x}}dx+2 e \int \frac {1}{c d-\frac {e (a e+c d x)}{d+e x}}d\frac {\sqrt {a e+c d x}}{\sqrt {d+e x}}\right )}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (2 e \int \frac {1}{c d-\frac {e (a e+c d x)}{d+e x}}d\frac {\sqrt {a e+c d x}}{\sqrt {d+e x}}+2 d \int \frac {1}{\frac {a e (d+e x)}{a e+c d x}-d}d\frac {\sqrt {d+e x}}{\sqrt {a e+c d x}}\right )}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {d+e x} \sqrt {a e+c d x} \left (\frac {2 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {c} \sqrt {d}}-\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {a} \sqrt {e} \sqrt {d+e x}}{\sqrt {d} \sqrt {a e+c d x}}\right )}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

Input:

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

Output:

(Sqrt[a*e + c*d*x]*Sqrt[d + e*x]*((2*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[a*e + c 
*d*x])/(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(Sqrt[c]*Sqrt[d]) - (2*Sqrt[d]*Ar 
cTanh[(Sqrt[a]*Sqrt[e]*Sqrt[d + e*x])/(Sqrt[d]*Sqrt[a*e + c*d*x])])/(Sqrt[ 
a]*Sqrt[e])))/Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]
 

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 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 140

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*d^(m + n)*f^p   Int[(a + b*x)^(m - 1)/(c + d*x)^m, x] 
, x] + Int[(a + b*x)^(m - 1)*((e + f*x)^p/(c + d*x)^m)*ExpandToSum[(a + b*x 
)*(c + d*x)^(-p - 1) - (b*d^(-p - 1)*f^p)/(e + f*x)^p, x], x] /; FreeQ[{a, 
b, c, d, e, f, m, n}, x] && EqQ[m + n + p + 1, 0] && ILtQ[p, 0] && (GtQ[m, 
0] || SumSimplerQ[m, -1] ||  !(GtQ[n, 0] || SumSimplerQ[n, -1]))
 

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 1268

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x_ 
) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/((d 
 + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p])   Int[(d + e*x)^(m + p)*(f 
 + g*x)^n*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x 
] && EqQ[c*d^2 - b*d*e + a*e^2, 0]
 

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {e \ln \left (\frac {\frac {1}{2} a \,e^{2}+\frac {1}{2} c \,d^{2}+c d x e}{\sqrt {d e c}}+\sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e}\right )}{\sqrt {d e c}}-\frac {d \ln \left (\frac {2 a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +2 \sqrt {a d e}\, \sqrt {a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e}}{x}\right )}{\sqrt {a d e}}\)	\(132\)

Input:

int((e*x+d)/x/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

e*ln((1/2*a*e^2+1/2*c*d^2+c*d*x*e)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c* 
d*x^2*e)^(1/2))/(d*e*c)^(1/2)-d/(a*d*e)^(1/2)*ln((2*a*d*e+(a*e^2+c*d^2)*x+ 
2*(a*d*e)^(1/2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2))/x)
 

Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 975, normalized size of antiderivative = 7.12 \[ \int \frac {d+e x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/2*sqrt(e/(c*d))*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e 
^4 + 8*(c^2*d^3*e + a*c*d*e^3)*x + 4*(2*c^2*d^2*e*x + c^2*d^3 + a*c*d*e^2) 
*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d))) + 1/2*sqrt(d/( 
a*e))*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)*x^2 + 8*(a* 
c*d^3*e + a^2*d*e^3)*x - 4*(2*a^2*d*e^2 + (a*c*d^2*e + a^2*e^3)*x)*sqrt(c* 
d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(d/(a*e)))/x^2), -sqrt(-e/(c*d))* 
arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 
+ a*e^2)*sqrt(-e/(c*d))/(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)) + 1 
/2*sqrt(d/(a*e))*log((8*a^2*d^2*e^2 + (c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4)* 
x^2 + 8*(a*c*d^3*e + a^2*d*e^3)*x - 4*(2*a^2*d*e^2 + (a*c*d^2*e + a^2*e^3) 
*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(d/(a*e)))/x^2), sqrt( 
-d/(a*e))*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d*e 
+ (c*d^2 + a*e^2)*x)*sqrt(-d/(a*e))/(c*d^2*e*x^2 + a*d^2*e + (c*d^3 + a*d* 
e^2)*x)) + 1/2*sqrt(e/(c*d))*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e 
^2 + a^2*e^4 + 8*(c^2*d^3*e + a*c*d*e^3)*x + 4*(2*c^2*d^2*e*x + c^2*d^3 + 
a*c*d*e^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d))), sqr 
t(-d/(a*e))*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*a*d* 
e + (c*d^2 + a*e^2)*x)*sqrt(-d/(a*e))/(c*d^2*e*x^2 + a*d^2*e + (c*d^3 + a* 
d*e^2)*x)) - sqrt(-e/(c*d))*arctan(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a 
*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-e/(c*d))/(c*d*e^2*x^2 + a*d*...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m operator + Error: 
Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 6.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.86 \[ \int \frac {d+e x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {e\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )}{\sqrt {c\,d\,e}}-\frac {d\,\ln \left (\frac {a\,e^2}{2}+\frac {c\,d^2}{2}+\frac {\sqrt {a\,d\,e}\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{x}+\frac {a\,d\,e}{x}\right )}{\sqrt {a\,d\,e}} \] Input:

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

Output:

(e*log(2*((a*e + c*d*x)*(d + e*x))^(1/2)*(c*d*e)^(1/2) + a*e^2 + c*d^2 + 2 
*c*d*e*x))/(c*d*e)^(1/2) - (d*log((a*e^2)/2 + (c*d^2)/2 + ((a*d*e)^(1/2)*( 
x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(1/2))/x + (a*d*e)/x))/(a*d*e)^(1/2 
)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.55 \[ \int \frac {d+e x}{x \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx=\frac {\sqrt {e}\, \sqrt {d}\, \left (\sqrt {a}\, \mathrm {log}\left (\sqrt {e}\, \sqrt {c d x +a e}-\sqrt {2 \sqrt {c}\, \sqrt {a}\, d e +a \,e^{2}+c \,d^{2}}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\right ) c d +\sqrt {a}\, \mathrm {log}\left (\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {2 \sqrt {c}\, \sqrt {a}\, d e +a \,e^{2}+c \,d^{2}}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\right ) c d -\sqrt {a}\, \mathrm {log}\left (2 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {c d x +a e}+2 \sqrt {c}\, \sqrt {a}\, d e +2 c d e x \right ) c d +2 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}}{\sqrt {a \,e^{2}-c \,d^{2}}}\right ) a e \right )}{a c d e} \] Input:

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

Output:

(sqrt(e)*sqrt(d)*(sqrt(a)*log(sqrt(e)*sqrt(a*e + c*d*x) - sqrt(2*sqrt(c)*s 
qrt(a)*d*e + a*e**2 + c*d**2) + sqrt(d)*sqrt(c)*sqrt(d + e*x))*c*d + sqrt( 
a)*log(sqrt(e)*sqrt(a*e + c*d*x) + sqrt(2*sqrt(c)*sqrt(a)*d*e + a*e**2 + c 
*d**2) + sqrt(d)*sqrt(c)*sqrt(d + e*x))*c*d - sqrt(a)*log(2*sqrt(e)*sqrt(d 
)*sqrt(c)*sqrt(d + e*x)*sqrt(a*e + c*d*x) + 2*sqrt(c)*sqrt(a)*d*e + 2*c*d* 
e*x)*c*d + 2*sqrt(c)*log((sqrt(e)*sqrt(a*e + c*d*x) + sqrt(d)*sqrt(c)*sqrt 
(d + e*x))/sqrt(a*e**2 - c*d**2))*a*e))/(a*c*d*e)