\(\int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx\) [615]

Optimal result

Integrand size = 22, antiderivative size = 209 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx=-\frac {e \sqrt {a+b x+c x^2}}{2 \left (c d^2-b d e+a e^2\right ) (d+e x)^2}-\frac {3 e (2 c d-b e) \sqrt {a+b x+c x^2}}{4 \left (c d^2-b d e+a e^2\right )^2 (d+e x)}+\frac {\left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right ) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{8 \left (c d^2-b d e+a e^2\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 10.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx=-\frac {e \sqrt {a+x (b+c x)}}{2 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}-\frac {3 e (2 c d-b e) \sqrt {a+x (b+c x)}}{4 \left (c d^2+e (-b d+a e)\right )^2 (d+e x)}-\frac {\left (8 c^2 d^2+3 b^2 e^2-4 c e (2 b d+a e)\right ) \text {arctanh}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{8 \left (c d^2+e (-b d+a e)\right )^{5/2}} \] Input:

Integrate[1/((d + e*x)^3*Sqrt[a + b*x + c*x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.10, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1167, 27, 1228, 1154, 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.

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

\(\Big \downarrow \) 1167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1228

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

\(\Big \downarrow \) 1154

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

\(\Big \downarrow \) 219

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

Input:

Int[1/((d + e*x)^3*Sqrt[a + b*x + c*x^2]),x]
 

Output:

-1/2*(e*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) + ((- 
3*e*(2*c*d - b*e)*Sqrt[a + b*x + c*x^2])/((c*d^2 - b*d*e + a*e^2)*(d + e*x 
)) + ((8*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(2*b*d + a*e))*ArcTanh[(b*d - 2*a*e + 
 (2*c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/ 
(2*(c*d^2 - b*d*e + a*e^2)^(3/2)))/(4*(c*d^2 - b*d*e + a*e^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 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 1154

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym 
bol] :> Simp[-2   Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 
2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c 
, d, e}, x]
 

rule 1167

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d 
^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1)*(c*d^2 - b*d*e + a*e^2))   Int[ 
(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, 
 x]*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[m 
, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimp 
lerQ[m, 1] && IntegerQ[p]) || ILtQ[Simplify[m + 2*p + 3], 0])
 

rule 1228

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + 
 b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e 
*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2))   Int[(d + e*x)^ 
(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x 
] && EqQ[Simplify[m + 2*p + 3], 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(560\) vs. \(2(191)=382\).

Time = 1.24 (sec) , antiderivative size = 561, normalized size of antiderivative = 2.68

method	result	size
default	\(\frac {-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {3 \left (b e -2 c d \right ) e \left (-\frac {e^{2} \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {\left (b e -2 c d \right ) e \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\right )}{4 \left (a \,e^{2}-b d e +c \,d^{2}\right )}+\frac {c \,e^{2} \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x +\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x +\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{2 \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{e^{3}}\)	\(561\)

Input:

int(1/(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/e^3*(-1/2/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^2*(c*(x+d/e)^2+(b*e-2*c*d)/e*( 
x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)-3/4*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2 
)*(-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+( 
a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)/((a*e^ 
2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+d 
/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e)+( 
a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))+1/2*c/(a*e^2-b*d*e+c*d^2)*e^2/((a 
*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*( 
x+d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*(c*(x+d/e)^2+(b*e-2*c*d)/e*(x+d/e 
)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (191) = 382\).

Time = 0.95 (sec) , antiderivative size = 1362, normalized size of antiderivative = 6.52 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")
 

Output:

[-1/16*((8*c^2*d^4 - 8*b*c*d^3*e + (3*b^2 - 4*a*c)*d^2*e^2 + (8*c^2*d^2*e^ 
2 - 8*b*c*d*e^3 + (3*b^2 - 4*a*c)*e^4)*x^2 + 2*(8*c^2*d^3*e - 8*b*c*d^2*e^ 
2 + (3*b^2 - 4*a*c)*d*e^3)*x)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b*d*e - 
 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^ 
2)*x^2 + 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e 
+ (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x)/(e 
^2*x^2 + 2*d*e*x + d^2)) + 4*(8*c^2*d^4*e - 13*b*c*d^3*e^2 - 7*a*b*d*e^4 + 
 2*a^2*e^5 + 5*(b^2 + 2*a*c)*d^2*e^3 + 3*(2*c^2*d^3*e^2 - 3*b*c*d^2*e^3 - 
a*b*e^5 + (b^2 + 2*a*c)*d*e^4)*x)*sqrt(c*x^2 + b*x + a))/(c^3*d^8 - 3*b*c^ 
2*d^7*e - 3*a^2*b*d^3*e^5 + a^3*d^2*e^6 + 3*(b^2*c + a*c^2)*d^6*e^2 - (b^3 
 + 6*a*b*c)*d^5*e^3 + 3*(a*b^2 + a^2*c)*d^4*e^4 + (c^3*d^6*e^2 - 3*b*c^2*d 
^5*e^3 - 3*a^2*b*d*e^7 + a^3*e^8 + 3*(b^2*c + a*c^2)*d^4*e^4 - (b^3 + 6*a* 
b*c)*d^3*e^5 + 3*(a*b^2 + a^2*c)*d^2*e^6)*x^2 + 2*(c^3*d^7*e - 3*b*c^2*d^6 
*e^2 - 3*a^2*b*d^2*e^6 + a^3*d*e^7 + 3*(b^2*c + a*c^2)*d^5*e^3 - (b^3 + 6* 
a*b*c)*d^4*e^4 + 3*(a*b^2 + a^2*c)*d^3*e^5)*x), 1/8*((8*c^2*d^4 - 8*b*c*d^ 
3*e + (3*b^2 - 4*a*c)*d^2*e^2 + (8*c^2*d^2*e^2 - 8*b*c*d*e^3 + (3*b^2 - 4* 
a*c)*e^4)*x^2 + 2*(8*c^2*d^3*e - 8*b*c*d^2*e^2 + (3*b^2 - 4*a*c)*d*e^3)*x) 
*sqrt(-c*d^2 + b*d*e - a*e^2)*arctan(-1/2*sqrt(-c*d^2 + b*d*e - a*e^2)*sqr 
t(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x)/(a*c*d^2 - a*b*d*e + a^ 
2*e^2 + (c^2*d^2 - b*c*d*e + a*c*e^2)*x^2 + (b*c*d^2 - b^2*d*e + a*b*e^...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^(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-b*d*e>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (191) = 382\).

Time = 0.21 (sec) , antiderivative size = 795, normalized size of antiderivative = 3.80 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")
 

Output:

1/4*(8*c^2*d^2 - 8*b*c*d*e + 3*b^2*e^2 - 4*a*c*e^2)*arctan(-((sqrt(c)*x - 
sqrt(c*x^2 + b*x + a))*e + sqrt(c)*d)/sqrt(-c*d^2 + b*d*e - a*e^2))/((c^2* 
d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + a^2*e^4)*s 
qrt(-c*d^2 + b*d*e - a*e^2)) - 1/4*(8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^ 
3*c^2*d^2*e - 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*b*c*d*e^2 + 3*(sqrt( 
c)*x - sqrt(c*x^2 + b*x + a))^3*b^2*e^3 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x 
+ a))^3*a*c*e^3 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*c^(5/2)*d^3 - 2 
4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*b*c^(3/2)*d^2*e + 9*(sqrt(c)*x - s 
qrt(c*x^2 + b*x + a))^2*b^2*sqrt(c)*d*e^2 - 12*(sqrt(c)*x - sqrt(c*x^2 + b 
*x + a))^2*a*c^(3/2)*d*e^2 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b*c^2* 
d^3 - 20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*b^2*c*d^2*e - 40*(sqrt(c)*x - 
 sqrt(c*x^2 + b*x + a))*a*c^2*d^2*e + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x + a) 
)*b^3*d*e^2 + 28*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*a*b*c*d*e^2 - 5*(sqrt 
(c)*x - sqrt(c*x^2 + b*x + a))*a*b^2*e^3 - 4*(sqrt(c)*x - sqrt(c*x^2 + b*x 
 + a))*a^2*c*e^3 + 6*b^2*c^(3/2)*d^3 - 3*b^3*sqrt(c)*d^2*e - 20*a*b*c^(3/2 
)*d^2*e + 11*a*b^2*sqrt(c)*d*e^2 + 12*a^2*c^(3/2)*d*e^2 - 8*a^2*b*sqrt(c)* 
e^3)/((c^2*d^4 - 2*b*c*d^3*e + b^2*d^2*e^2 + 2*a*c*d^2*e^2 - 2*a*b*d*e^3 + 
 a^2*e^4)*((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*e + 2*(sqrt(c)*x - sqrt(c 
*x^2 + b*x + a))*sqrt(c)*d + b*d - a*e)^2)
 

Mupad [F(-1)]

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

int(1/((d + e*x)^3*(a + b*x + c*x^2)^(1/2)),x)
 

Output:

int(1/((d + e*x)^3*(a + b*x + c*x^2)^(1/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.71 (sec) , antiderivative size = 1920, normalized size of antiderivative = 9.19 \[ \int \frac {1}{(d+e x)^3 \sqrt {a+b x+c x^2}} \, dx =\text {Too large to display} \] Input:

int(1/(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x)
 

Output:

(4*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e* 
*2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a*c*d**2*e**2 + 8*sq 
rt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - 
b*d*e + c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a*c*d*e**3*x + 4*sqrt(a*e 
**2 - b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e 
+ c*d**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*a*c*e**4*x**2 - 3*sqrt(a*e**2 - 
 b*d*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d 
**2) - 2*a*e + b*d - b*e*x + 2*c*d*x)*b**2*d**2*e**2 - 6*sqrt(a*e**2 - b*d 
*e + c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) 
 - 2*a*e + b*d - b*e*x + 2*c*d*x)*b**2*d*e**3*x - 3*sqrt(a*e**2 - b*d*e + 
c*d**2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2* 
a*e + b*d - b*e*x + 2*c*d*x)*b**2*e**4*x**2 + 8*sqrt(a*e**2 - b*d*e + c*d* 
*2)*log( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e 
+ b*d - b*e*x + 2*c*d*x)*b*c*d**3*e + 16*sqrt(a*e**2 - b*d*e + c*d**2)*log 
( - 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e + b*d - 
 b*e*x + 2*c*d*x)*b*c*d**2*e**2*x + 8*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 
 2*sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e + b*d - b* 
e*x + 2*c*d*x)*b*c*d*e**3*x**2 - 8*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2* 
sqrt(a + b*x + c*x**2)*sqrt(a*e**2 - b*d*e + c*d**2) - 2*a*e + b*d - b*e*x 
 + 2*c*d*x)*c**2*d**4 - 16*sqrt(a*e**2 - b*d*e + c*d**2)*log( - 2*sqrt(...