3.20.57 \(\int \frac {(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx\) [1957]

3.20.57.1 Optimal result
3.20.57.2 Mathematica [A] (verified)
3.20.57.3 Rubi [A] (verified)
3.20.57.4 Maple [B] (verified)
3.20.57.5 Fricas [A] (verification not implemented)
3.20.57.6 Sympy [F]
3.20.57.7 Maxima [F(-2)]
3.20.57.8 Giac [F(-2)]
3.20.57.9 Mupad [F(-1)]

3.20.57.1 Optimal result

Integrand size = 37, antiderivative size = 125 \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=-\frac {2 (d+e x)}{c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {\sqrt {e} \text {arctanh}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{c^{3/2} d^{3/2}} \]

output
arctanh(1/2*(2*c*d*e*x+a*e^2+c*d^2)/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+ 
c*d^2)*x+c*d*e*x^2)^(1/2))*e^(1/2)/c^(3/2)/d^(3/2)-2*(e*x+d)/c/d/(a*d*e+(a 
*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)
 
3.20.57.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {c} \sqrt {d} (d+e x)+2 \sqrt {e} \sqrt {a e+c d x} \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{c^{3/2} d^{3/2} \sqrt {(a e+c d x) (d+e x)}} \]

input
Integrate[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]
 
output
(-2*Sqrt[c]*Sqrt[d]*(d + e*x) + 2*Sqrt[e]*Sqrt[a*e + c*d*x]*Sqrt[d + e*x]* 
ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c*d*x])])/(c^( 
3/2)*d^(3/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])
 
3.20.57.3 Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {1124, 1092, 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.

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

\(\Big \downarrow \) 1124

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

\(\Big \downarrow \) 1092

\(\displaystyle \frac {2 e \int \frac {1}{4 c d e-\frac {\left (c d^2+2 c e x d+a e^2\right )^2}{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}d\frac {c d^2+2 c e x d+a e^2}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}}{c d}-\frac {2 (d+e x)}{c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\)

\(\Big \downarrow \) 219

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

input
Int[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2),x]
 
output
(-2*(d + e*x))/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (Sqrt[e 
]*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d* 
e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(c^(3/2)*d^(3/2))
 

3.20.57.3.1 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 1092
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[I 
nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a 
, b, c}, x]
 

rule 1124
Int[((d_.) + (e_.)*(x_))^(m_.)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x 
_Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + 
b*x + c*x^2])), x] + Simp[e^2/c^(m - 1)   Int[(1/Sqrt[a + b*x + c*x^2])*Exp 
andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - 
 c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e 
^2, 0] && IGtQ[m, 0]
 
3.20.57.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(484\) vs. \(2(107)=214\).

Time = 2.94 (sec) , antiderivative size = 485, normalized size of antiderivative = 3.88

method result size
default \(\frac {2 d^{2} \left (2 x c d e +e^{2} a +c \,d^{2}\right )}{\left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}+e^{2} \left (-\frac {x}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 x c d e +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )}{2 c d e}+\frac {\ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+x c d e}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{c d e \sqrt {c d e}}\right )+2 d e \left (-\frac {1}{c d e \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}-\frac {\left (e^{2} a +c \,d^{2}\right ) \left (2 x c d e +e^{2} a +c \,d^{2}\right )}{c d e \left (4 a c \,d^{2} e^{2}-\left (e^{2} a +c \,d^{2}\right )^{2}\right ) \sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}\right )\) \(485\)

input
int((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x,method=_RETURNVERB 
OSE)
 
output
2*d^2*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^ 
2+c*d^2)*x+c*d*e*x^2)^(1/2)+e^2*(-x/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2 
)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2 
)^(1/2)-(a*e^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+ 
c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))+1/c/d/e*ln((1/2*e^2*a+1 
/2*c*d^2+x*c*d*e)/(c*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/( 
c*d*e)^(1/2))+2*d*e*(-1/c/d/e/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-(a*e 
^2+c*d^2)/c/d/e*(2*c*d*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a 
*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))
 
3.20.57.5 Fricas [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.79 \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\left [\frac {{\left (c d x + a e\right )} \sqrt {\frac {e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \, {\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {e}{c d}}\right ) - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{2 \, {\left (c^{2} d^{2} x + a c d e\right )}}, -\frac {{\left (c d x + a e\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{c^{2} d^{2} x + a c d e}\right ] \]

input
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm=" 
fricas")
 
output
[1/2*((c*d*x + a*e)*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))) 
 - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*x + a*c*d*e), - 
((c*d*x + a*e)*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)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x) 
)/(c^2*d^2*x + a*c*d*e)]
 
3.20.57.6 Sympy [F]

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

input
integrate((e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)
 
output
Integral((d + e*x)**2/((d + e*x)*(a*e + c*d*x))**(3/2), x)
 
3.20.57.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

input
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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
 
3.20.57.8 Giac [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \]

input
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm=" 
giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%{[%%%{1,[1,1,4]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[2,2] 
%%%}+%%%{
 
3.20.57.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/2}} \,d x \]

input
int((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2),x)
 
output
int((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/2), x)