\(\int \frac {1}{(d+e x)^{3/2} (a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [190]

Optimal result

Integrand size = 37, antiderivative size = 192 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {7 e}{5 \left (c d^2-a e^2\right )^2 (d+e x)^{5/2}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) (d+e x)^{5/2}}-\frac {7 c d e}{3 \left (c d^2-a e^2\right )^3 (d+e x)^{3/2}}-\frac {7 c^2 d^2 e}{\left (c d^2-a e^2\right )^4 \sqrt {d+e x}}+\frac {7 c^{5/2} d^{5/2} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{9/2}} \] Output:

-7/5*e/(-a*e^2+c*d^2)^2/(e*x+d)^(5/2)-1/(-a*e^2+c*d^2)/(c*d*x+a*e)/(e*x+d) 
^(5/2)-7/3*c*d*e/(-a*e^2+c*d^2)^3/(e*x+d)^(3/2)-7*c^2*d^2*e/(-a*e^2+c*d^2) 
^4/(e*x+d)^(1/2)+7*c^(5/2)*d^(5/2)*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)^(9/2)
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {-6 a^3 e^6+2 a^2 c d e^4 (16 d+7 e x)-2 a c^2 d^2 e^2 \left (58 d^2+84 d e x+35 e^2 x^2\right )-c^3 d^3 \left (15 d^3+161 d^2 e x+245 d e^2 x^2+105 e^3 x^3\right )}{15 \left (c d^2-a e^2\right )^4 (a e+c d x) (d+e x)^{5/2}}-\frac {7 c^{5/2} d^{5/2} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{9/2}} \] Input:

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

Output:

(-6*a^3*e^6 + 2*a^2*c*d*e^4*(16*d + 7*e*x) - 2*a*c^2*d^2*e^2*(58*d^2 + 84* 
d*e*x + 35*e^2*x^2) - c^3*d^3*(15*d^3 + 161*d^2*e*x + 245*d*e^2*x^2 + 105* 
e^3*x^3))/(15*(c*d^2 - a*e^2)^4*(a*e + c*d*x)*(d + e*x)^(5/2)) - (7*c^(5/2 
)*d^(5/2)*e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]] 
)/(-(c*d^2) + a*e^2)^(9/2)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.23, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {1121, 52, 61, 61, 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.

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

\(\Big \downarrow \) 1121

\(\displaystyle \int \frac {1}{(d+e x)^{7/2} (a e+c d x)^2}dx\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {7 e \int \frac {1}{(a e+c d x) (d+e x)^{7/2}}dx}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 e \left (\frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{5/2}}dx}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 e \left (\frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{3/2}}dx}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{c d^2-a e^2}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {2 c d \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{e \left (c d^2-a e^2\right )}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\)

Input:

Int[1/((d + e*x)^(3/2)*(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)*(d + e*x)^(5/2))) - (7*e*(2/(5*(c*d^2 - 
 a*e^2)*(d + e*x)^(5/2)) + (c*d*(2/(3*(c*d^2 - a*e^2)*(d + e*x)^(3/2)) + ( 
c*d*(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)))/(c 
*d^2 - a*e^2)))/(c*d^2 - a*e^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 = 2.02 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(2 e \left (-\frac {c^{3} d^{3} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {7 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{4}}-\frac {1}{5 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {3 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {2 c d}{3 \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}\right )\)	\(183\)
default	\(2 e \left (-\frac {c^{3} d^{3} \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 a \,e^{2}-2 c \,d^{2}}+\frac {7 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{4}}-\frac {1}{5 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {3 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{4} \sqrt {e x +d}}+\frac {2 c d}{3 \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {3}{2}}}\right )\)	\(183\)
pseudoelliptic	\(-\frac {2 \left (\frac {35 c^{3} d^{3} e \left (e x +d \right )^{\frac {5}{2}} \left (c d x +a e \right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2}+\left (\frac {5 \left (7 e^{3} x^{3}+\frac {49}{3} d \,e^{2} x^{2}+\frac {161}{15} d^{2} e x +d^{3}\right ) d^{3} c^{3}}{2}+\frac {58 e^{2} a \,d^{2} \left (\frac {35}{58} e^{2} x^{2}+\frac {42}{29} d e x +d^{2}\right ) c^{2}}{3}-\frac {16 e^{4} a^{2} d \left (\frac {7 e x}{16}+d \right ) c}{3}+e^{6} a^{3}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (a \,e^{2}-c \,d^{2}\right )^{4} \left (c d x +a e \right )}\)	\(211\)

Input:

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

Output:

2*e*(-1/(a*e^2-c*d^2)^4*c^3*d^3*(1/2*(e*x+d)^(1/2)/(c*d*(e*x+d)+a*e^2-c*d^ 
2)+7/2/(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)))-1/5/(a*e^2-c*d^2)^2/(e*x+d)^(5/2)-3/(a*e^2-c*d^2)^4*c^2*d^2/(e 
*x+d)^(1/2)+2/3/(a*e^2-c*d^2)^3*c*d/(e*x+d)^(3/2))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 631 vs. \(2 (166) = 332\).

Time = 0.17 (sec) , antiderivative size = 1304, normalized size of antiderivative = 6.79 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/30*(105*(c^3*d^3*e^4*x^4 + a*c^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e 
^5)*x^3 + 3*(c^3*d^5*e^2 + a*c^2*d^3*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*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)*sqrt(c*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) - 2*(105 
*c^3*d^3*e^3*x^3 + 15*c^3*d^6 + 116*a*c^2*d^4*e^2 - 32*a^2*c*d^2*e^4 + 6*a 
^3*e^6 + 35*(7*c^3*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e + 24*a 
*c^2*d^3*e^3 - 2*a^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3* 
d^9*e^3 + 6*a^3*c^2*d^7*e^5 - 4*a^4*c*d^5*e^7 + a^5*d^3*e^9 + (c^5*d^9*e^3 
 - 4*a*c^4*d^7*e^5 + 6*a^2*c^3*d^5*e^7 - 4*a^3*c^2*d^3*e^9 + a^4*c*d*e^11) 
*x^4 + (3*c^5*d^10*e^2 - 11*a*c^4*d^8*e^4 + 14*a^2*c^3*d^6*e^6 - 6*a^3*c^2 
*d^4*e^8 - a^4*c*d^2*e^10 + a^5*e^12)*x^3 + 3*(c^5*d^11*e - 3*a*c^4*d^9*e^ 
3 + 2*a^2*c^3*d^7*e^5 + 2*a^3*c^2*d^5*e^7 - 3*a^4*c*d^3*e^9 + a^5*d*e^11)* 
x^2 + (c^5*d^12 - a*c^4*d^10*e^2 - 6*a^2*c^3*d^8*e^4 + 14*a^3*c^2*d^6*e^6 
- 11*a^4*c*d^4*e^8 + 3*a^5*d^2*e^10)*x), -1/15*(105*(c^3*d^3*e^4*x^4 + a*c 
^2*d^5*e^2 + (3*c^3*d^4*e^3 + a*c^2*d^2*e^5)*x^3 + 3*(c^3*d^5*e^2 + a*c^2* 
d^3*e^4)*x^2 + (c^3*d^6*e + 3*a*c^2*d^4*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))) + (105*c^3*d^3*e^3*x^3 + 
 15*c^3*d^6 + 116*a*c^2*d^4*e^2 - 32*a^2*c*d^2*e^4 + 6*a^3*e^6 + 35*(7*c^3 
*d^4*e^2 + 2*a*c^2*d^2*e^4)*x^2 + 7*(23*c^3*d^5*e + 24*a*c^2*d^3*e^3 - 2*a 
^2*c*d*e^5)*x)*sqrt(e*x + d))/(a*c^4*d^11*e - 4*a^2*c^3*d^9*e^3 + 6*a^3...
 

Sympy [F]

\[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {7}{2}} \left (a e + c d x\right )^{2}}\, dx \] Input:

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

Output:

Integral(1/((d + e*x)**(7/2)*(a*e + c*d*x)**2), x)
 

Maxima [F(-2)]

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

integrate(1/(e*x+d)^(3/2)/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (166) = 332\).

Time = 0.15 (sec) , antiderivative size = 340, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {7 \, c^{3} d^{3} e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {\sqrt {e x + d} c^{3} d^{3} e}{{\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}} - \frac {2 \, {\left (45 \, {\left (e x + d\right )}^{2} c^{2} d^{2} e + 10 \, {\left (e x + d\right )} c^{2} d^{3} e + 3 \, c^{2} d^{4} e - 10 \, {\left (e x + d\right )} a c d e^{3} - 6 \, a c d^{2} e^{3} + 3 \, a^{2} e^{5}\right )}}{15 \, {\left (c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \] Input:

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

Output:

-7*c^3*d^3*e*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^4*d^ 
8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*sqrt( 
-c^2*d^3 + a*c*d*e^2)) - sqrt(e*x + d)*c^3*d^3*e/((c^4*d^8 - 4*a*c^3*d^6*e 
^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*((e*x + d)*c*d - c*d^2 
 + a*e^2)) - 2/15*(45*(e*x + d)^2*c^2*d^2*e + 10*(e*x + d)*c^2*d^3*e + 3*c 
^2*d^4*e - 10*(e*x + d)*a*c*d*e^3 - 6*a*c*d^2*e^3 + 3*a^2*e^5)/((c^4*d^8 - 
 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8)*(e*x + d 
)^(5/2))
 

Mupad [B] (verification not implemented)

Time = 5.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\frac {2\,e}{5\,\left (a\,e^2-c\,d^2\right )}-\frac {14\,c\,d\,e\,\left (d+e\,x\right )}{15\,{\left (a\,e^2-c\,d^2\right )}^2}+\frac {14\,c^2\,d^2\,e\,{\left (d+e\,x\right )}^2}{3\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {7\,c^3\,d^3\,e\,{\left (d+e\,x\right )}^3}{{\left (a\,e^2-c\,d^2\right )}^4}}{\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^{5/2}+c\,d\,{\left (d+e\,x\right )}^{7/2}}-\frac {7\,c^{5/2}\,d^{5/2}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^4\,e^8-4\,a^3\,c\,d^2\,e^6+6\,a^2\,c^2\,d^4\,e^4-4\,a\,c^3\,d^6\,e^2+c^4\,d^8\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{9/2}} \] Input:

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

Output:

- ((2*e)/(5*(a*e^2 - c*d^2)) - (14*c*d*e*(d + e*x))/(15*(a*e^2 - c*d^2)^2) 
 + (14*c^2*d^2*e*(d + e*x)^2)/(3*(a*e^2 - c*d^2)^3) + (7*c^3*d^3*e*(d + e* 
x)^3)/(a*e^2 - c*d^2)^4)/((a*e^2 - c*d^2)*(d + e*x)^(5/2) + c*d*(d + e*x)^ 
(7/2)) - (7*c^(5/2)*d^(5/2)*e*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^4*e 
^8 + c^4*d^8 - 4*a*c^3*d^6*e^2 - 4*a^3*c*d^2*e^6 + 6*a^2*c^2*d^4*e^4))/(a* 
e^2 - c*d^2)^(9/2)))/(a*e^2 - c*d^2)^(9/2)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 933, normalized size of antiderivative = 4.86 \[ \int \frac {1}{(d+e x)^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 105*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*c**2*d**4*e**2 - 210 
*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*c**2*d**3*e**3*x - 105*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*c**2*d**2*e**4*x**2 - 105*sqrt(d 
)*sqrt(c)*sqrt(d + e*x)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sq 
rt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*c**3*d**5*e*x - 210*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)))*c**3*d**4*e**2*x**2 - 105*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)))*c**3*d**3*e**3*x**3 - 6*a**4*e**8 + 38*a**3*c*d**2 
*e**6 + 14*a**3*c*d*e**7*x - 148*a**2*c**2*d**4*e**4 - 182*a**2*c**2*d**3* 
e**5*x - 70*a**2*c**2*d**2*e**6*x**2 + 101*a*c**3*d**6*e**2 + 7*a*c**3*d** 
5*e**3*x - 175*a*c**3*d**4*e**4*x**2 - 105*a*c**3*d**3*e**5*x**3 + 15*c**4 
*d**8 + 161*c**4*d**7*e*x + 245*c**4*d**6*e**2*x**2 + 105*c**4*d**5*e**3*x 
**3)/(15*sqrt(d + e*x)*(a**6*d**2*e**11 + 2*a**6*d*e**12*x + a**6*e**13*x* 
*2 - 5*a**5*c*d**4*e**9 - 9*a**5*c*d**3*e**10*x - 3*a**5*c*d**2*e**11*x**2 
 + a**5*c*d*e**12*x**3 + 10*a**4*c**2*d**6*e**7 + 15*a**4*c**2*d**5*e**8*x 
 - 5*a**4*c**2*d**3*e**10*x**3 - 10*a**3*c**3*d**8*e**5 - 10*a**3*c**3*...