\(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [188]

Optimal result

Integrand size = 37, antiderivative size = 128 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 e}{\left (c d^2-a e^2\right )^2 \sqrt {d+e x}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) \sqrt {d+e x}}+\frac {3 \sqrt {c} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {2 a e^2+c d (d+3 e x)}{\left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt {d+e x}}-\frac {3 \sqrt {c} \sqrt {d} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{5/2}} \] Input:

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

Output:

-((2*a*e^2 + c*d*(d + 3*e*x))/((c*d^2 - a*e^2)^2*(a*e + c*d*x)*Sqrt[d + e* 
x])) - (3*Sqrt[c]*Sqrt[d]*e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-( 
c*d^2) + a*e^2]])/(-(c*d^2) + a*e^2)^(5/2)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 52, 61, 73, 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.

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

\(\Big \downarrow \) 1121

\(\displaystyle \int \frac {1}{(d+e x)^{3/2} (a e+c d x)^2}dx\)

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 61

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {3 e \left (\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}\)

Input:

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

Output:

-(1/((c*d^2 - a*e^2)*(a*e + c*d*x)*Sqrt[d + e*x])) - (3*e*(2/((c*d^2 - a*e 
^2)*Sqrt[d + e*x]) - (2*Sqrt[c]*Sqrt[d]*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + 
e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(3/2)))/(2*(c*d^2 - a*e^2))
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 1121

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ 
Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] 
/; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int 
egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
 

Maple [A] (verified)

Time = 1.95 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {e \left (-\frac {2}{\sqrt {e x +d}}-\frac {c d \sqrt {e x +d}}{e \left (c d x +a e \right )}-\frac {3 c d \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{2}}\)	\(100\)
derivativedivides	\(2 e \left (-\frac {c d \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{2}}-\frac {1}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \sqrt {e x +d}}\right )\)	\(125\)
default	\(2 e \left (-\frac {c d \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{2}}-\frac {1}{\left (a \,e^{2}-c \,d^{2}\right )^{2} \sqrt {e x +d}}\right )\)	\(125\)

Input:

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

Output:

e/(a*e^2-c*d^2)^2*(-2/(e*x+d)^(1/2)-c*d*(e*x+d)^(1/2)/e/(c*d*x+a*e)-3*c*d/ 
(c*d*(a*e^2-c*d^2))^(1/2)*arctan(c*d*(e*x+d)^(1/2)/(c*d*(a*e^2-c*d^2))^(1/ 
2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 476, normalized size of antiderivative = 3.72 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [\frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x\right )}}, -\frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (\sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}\right ) + {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x}\right ] \] Input:

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

Output:

[1/2*(3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(c*d/(c*d^2 - a* 
e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2 + 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqr 
t(c*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) - 2*(3*c*d*e*x + c*d^2 + 2*a*e^2)*s 
qrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^3*d*e^5 + (c^3*d^5*e - 2* 
a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^ 
4 + a^3*e^6)*x), -(3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*sqrt(-c 
*d/(c*d^2 - a*e^2))*arctan(sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))) + (3* 
c*d*e*x + c*d^2 + 2*a*e^2)*sqrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + 
 a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - 
a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6)*x)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \left (a e + c d x\right )^{2}}\, dx \] Input:

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

Output:

Integral(1/((d + e*x)**(3/2)*(a*e + c*d*x)**2), x)
 

Maxima [F(-2)]

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

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^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 [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 \, c d e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {3 \, {\left (e x + d\right )} c d e - 2 \, c d^{2} e + 2 \, a e^{3}}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} c d - \sqrt {e x + d} c d^{2} + \sqrt {e x + d} a e^{2}\right )}} \] Input:

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

Output:

-3*c*d*e*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^2*d^4 - 
2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2)) - (3*(e*x + d)*c*d*e 
- 2*c*d^2*e + 2*a*e^3)/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*((e*x + d)^(3/ 
2)*c*d - sqrt(e*x + d)*c*d^2 + sqrt(e*x + d)*a*e^2))
 

Mupad [B] (verification not implemented)

Time = 5.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\frac {2\,e}{a\,e^2-c\,d^2}+\frac {3\,c\,d\,e\,\left (d+e\,x\right )}{{\left (a\,e^2-c\,d^2\right )}^2}}{\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}+c\,d\,{\left (d+e\,x\right )}^{3/2}}-\frac {3\,\sqrt {c}\,\sqrt {d}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}} \] Input:

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

Output:

- ((2*e)/(a*e^2 - c*d^2) + (3*c*d*e*(d + e*x))/(a*e^2 - c*d^2)^2)/((a*e^2 
- c*d^2)*(d + e*x)^(1/2) + c*d*(d + e*x)^(3/2)) - (3*c^(1/2)*d^(1/2)*e*ata 
n((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2))/(a 
*e^2 - c*d^2)^(5/2)))/(a*e^2 - c*d^2)^(5/2)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.11 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {-3 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,e^{2}-3 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c d e x -2 a^{2} e^{4}+a c \,d^{2} e^{2}-3 a c d \,e^{3} x +c^{2} d^{4}+3 c^{2} d^{3} e x}{\sqrt {e x +d}\, \left (a^{3} c d \,e^{6} x -3 a^{2} c^{2} d^{3} e^{4} x +3 a \,c^{3} d^{5} e^{2} x -c^{4} d^{7} x +a^{4} e^{7}-3 a^{3} c \,d^{2} e^{5}+3 a^{2} c^{2} d^{4} e^{3}-a \,c^{3} d^{6} e \right )} \] Input:

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

Output:

( - 3*sqrt(d)*sqrt(c)*sqrt(d + e*x)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e 
*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a*e**2 - 3*sqrt(d)*sqrt( 
c)*sqrt(d + e*x)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*s 
qrt(c)*sqrt(a*e**2 - c*d**2)))*c*d*e*x - 2*a**2*e**4 + a*c*d**2*e**2 - 3*a 
*c*d*e**3*x + c**2*d**4 + 3*c**2*d**3*e*x)/(sqrt(d + e*x)*(a**4*e**7 - 3*a 
**3*c*d**2*e**5 + a**3*c*d*e**6*x + 3*a**2*c**2*d**4*e**3 - 3*a**2*c**2*d* 
*3*e**4*x - a*c**3*d**6*e + 3*a*c**3*d**5*e**2*x - c**4*d**7*x))