\(\int \frac {1}{\sqrt {d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)^3} \, dx\) [199]

Optimal result

Integrand size = 37, antiderivative size = 244 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac {21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac {63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt {d+e x}}-\frac {63 c^{5/2} d^{5/2} e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}} \] Output:

63/20*e^2/(-a*e^2+c*d^2)^3/(e*x+d)^(5/2)-1/2/(-a*e^2+c*d^2)/(c*d*x+a*e)^2/ 
(e*x+d)^(5/2)+9/4*e/(-a*e^2+c*d^2)^2/(c*d*x+a*e)/(e*x+d)^(5/2)+21/4*c*d*e^ 
2/(-a*e^2+c*d^2)^4/(e*x+d)^(3/2)+63/4*c^2*d^2*e^2/(-a*e^2+c*d^2)^5/(e*x+d) 
^(1/2)-63/4*c^(5/2)*d^(5/2)*e^2*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)^(11/2)
 

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {8 a^4 e^8-8 a^3 c d e^6 (7 d+3 e x)+24 a^2 c^2 d^2 e^4 \left (12 d^2+17 d e x+7 e^2 x^2\right )+a c^3 d^3 e^2 \left (85 d^3+831 d^2 e x+1239 d e^2 x^2+525 e^3 x^3\right )+c^4 d^4 \left (-10 d^4+45 d^3 e x+483 d^2 e^2 x^2+735 d e^3 x^3+315 e^4 x^4\right )}{20 \left (c d^2-a e^2\right )^5 (a e+c d x)^2 (d+e x)^{5/2}}-\frac {63 c^{5/2} d^{5/2} e^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 \left (-c d^2+a e^2\right )^{11/2}} \] Input:

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

Output:

(8*a^4*e^8 - 8*a^3*c*d*e^6*(7*d + 3*e*x) + 24*a^2*c^2*d^2*e^4*(12*d^2 + 17 
*d*e*x + 7*e^2*x^2) + a*c^3*d^3*e^2*(85*d^3 + 831*d^2*e*x + 1239*d*e^2*x^2 
 + 525*e^3*x^3) + c^4*d^4*(-10*d^4 + 45*d^3*e*x + 483*d^2*e^2*x^2 + 735*d* 
e^3*x^3 + 315*e^4*x^4))/(20*(c*d^2 - a*e^2)^5*(a*e + c*d*x)^2*(d + e*x)^(5 
/2)) - (63*c^(5/2)*d^(5/2)*e^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt 
[-(c*d^2) + a*e^2]])/(4*(-(c*d^2) + a*e^2)^(11/2))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.216, Rules used = {1121, 52, 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.

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

\(\Big \downarrow \) 1121

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

\(\Big \downarrow \) 52

\(\displaystyle -\frac {9 e \int \frac {1}{(a e+c d x)^2 (d+e x)^{7/2}}dx}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {9 e \left (-\frac {7 e \int \frac {1}{(a e+c d x) (d+e x)^{7/2}}dx}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{5/2}}dx}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{3/2}}dx}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 61

\(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{c d^2-a e^2}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {2 c d \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{e \left (c d^2-a e^2\right )}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {c d \left (\frac {c d \left (\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{(d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}\)

Input:

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

Output:

-1/2*1/((c*d^2 - a*e^2)*(a*e + c*d*x)^2*(d + e*x)^(5/2)) - (9*e*(-(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))))/(4*(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 = 209, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(2 e^{2} \left (-\frac {1}{5 \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{5} \sqrt {e x +d}}+\frac {c d}{\left (a \,e^{2}-c \,d^{2}\right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {c^{3} d^{3} \left (\frac {\frac {15 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 a \,e^{2}}{8}-\frac {17 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {63 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{5}}\right )\)	\(209\)
default	\(2 e^{2} \left (-\frac {1}{5 \left (a \,e^{2}-c \,d^{2}\right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{5} \sqrt {e x +d}}+\frac {c d}{\left (a \,e^{2}-c \,d^{2}\right )^{4} \left (e x +d \right )^{\frac {3}{2}}}-\frac {c^{3} d^{3} \left (\frac {\frac {15 c d \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 a \,e^{2}}{8}-\frac {17 c \,d^{2}}{8}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {63 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{5}}\right )\)	\(209\)
pseudoelliptic	\(-\frac {2 \left (\frac {315 c^{3} d^{3} e^{2} \left (e x +d \right )^{\frac {5}{2}} \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8}+\left (-\frac {5 \left (-\frac {63}{2} e^{4} x^{4}-\frac {147}{2} d \,e^{3} x^{3}-\frac {483}{10} d^{2} e^{2} x^{2}-\frac {9}{2} d^{3} e x +d^{4}\right ) d^{4} c^{4}}{4}+\frac {85 \left (\frac {105}{17} e^{3} x^{3}+\frac {1239}{85} d \,e^{2} x^{2}+\frac {831}{85} d^{2} e x +d^{3}\right ) e^{2} a \,d^{3} c^{3}}{8}+36 e^{4} \left (\frac {7}{12} e^{2} x^{2}+\frac {17}{12} d e x +d^{2}\right ) a^{2} d^{2} c^{2}-7 e^{6} \left (\frac {3 e x}{7}+d \right ) a^{3} d c +a^{4} e^{8}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\right )}{5 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (a \,e^{2}-c \,d^{2}\right )^{5} \left (c d x +a e \right )^{2}}\)	\(268\)

Input:

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

Output:

2*e^2*(-1/5/(a*e^2-c*d^2)^3/(e*x+d)^(5/2)-6/(a*e^2-c*d^2)^5*c^2*d^2/(e*x+d 
)^(1/2)+1/(a*e^2-c*d^2)^4*c*d/(e*x+d)^(3/2)-1/(a*e^2-c*d^2)^5*c^3*d^3*((15 
/8*c*d*(e*x+d)^(3/2)+(17/8*a*e^2-17/8*c*d^2)*(e*x+d)^(1/2))/(c*d*(e*x+d)+a 
*e^2-c*d^2)^2+63/8/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 969 vs. \(2 (208) = 416\).

Time = 0.37 (sec) , antiderivative size = 1980, normalized size of antiderivative = 8.11 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/40*(315*(c^4*d^4*e^5*x^5 + a^2*c^2*d^5*e^4 + (3*c^4*d^5*e^4 + 2*a*c^3* 
d^3*e^6)*x^4 + (3*c^4*d^6*e^3 + 6*a*c^3*d^4*e^5 + a^2*c^2*d^2*e^7)*x^3 + ( 
c^4*d^7*e^2 + 6*a*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6)*x^2 + (2*a*c^3*d^6*e^3 
+ 3*a^2*c^2*d^4*e^5)*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*(315*c^4*d^4*e^4*x^4 - 10*c^4*d^8 + 85*a*c^3*d^6*e^2 + 288*a^ 
2*c^2*d^4*e^4 - 56*a^3*c*d^2*e^6 + 8*a^4*e^8 + 105*(7*c^4*d^5*e^3 + 5*a*c^ 
3*d^3*e^5)*x^3 + 21*(23*c^4*d^6*e^2 + 59*a*c^3*d^4*e^4 + 8*a^2*c^2*d^2*e^6 
)*x^2 + 3*(15*c^4*d^7*e + 277*a*c^3*d^5*e^3 + 136*a^2*c^2*d^3*e^5 - 8*a^3* 
c*d*e^7)*x)*sqrt(e*x + d))/(a^2*c^5*d^13*e^2 - 5*a^3*c^4*d^11*e^4 + 10*a^4 
*c^3*d^9*e^6 - 10*a^5*c^2*d^7*e^8 + 5*a^6*c*d^5*e^10 - a^7*d^3*e^12 + (c^7 
*d^12*e^3 - 5*a*c^6*d^10*e^5 + 10*a^2*c^5*d^8*e^7 - 10*a^3*c^4*d^6*e^9 + 5 
*a^4*c^3*d^4*e^11 - a^5*c^2*d^2*e^13)*x^5 + (3*c^7*d^13*e^2 - 13*a*c^6*d^1 
1*e^4 + 20*a^2*c^5*d^9*e^6 - 10*a^3*c^4*d^7*e^8 - 5*a^4*c^3*d^5*e^10 + 7*a 
^5*c^2*d^3*e^12 - 2*a^6*c*d*e^14)*x^4 + (3*c^7*d^14*e - 9*a*c^6*d^12*e^3 + 
 a^2*c^5*d^10*e^5 + 25*a^3*c^4*d^8*e^7 - 35*a^4*c^3*d^6*e^9 + 17*a^5*c^2*d 
^4*e^11 - a^6*c*d^2*e^13 - a^7*e^15)*x^3 + (c^7*d^15 + a*c^6*d^13*e^2 - 17 
*a^2*c^5*d^11*e^4 + 35*a^3*c^4*d^9*e^6 - 25*a^4*c^3*d^7*e^8 - a^5*c^2*d^5* 
e^10 + 9*a^6*c*d^3*e^12 - 3*a^7*d*e^14)*x^2 + (2*a*c^6*d^14*e - 7*a^2*c^5* 
d^12*e^3 + 5*a^3*c^4*d^10*e^5 + 10*a^4*c^3*d^8*e^7 - 20*a^5*c^2*d^6*e^9...
 

Sympy [F]

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

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

Output:

Integral(1/((d + e*x)**(7/2)*(a*e + c*d*x)**3), x)
 

Maxima [F(-2)]

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

integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,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. 433 vs. \(2 (208) = 416\).

Time = 0.19 (sec) , antiderivative size = 433, normalized size of antiderivative = 1.77 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {63 \, c^{3} d^{3} e^{2} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {15 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{4} d^{4} e^{2} - 17 \, \sqrt {e x + d} c^{4} d^{5} e^{2} + 17 \, \sqrt {e x + d} a c^{3} d^{3} e^{4}}{4 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} + \frac {2 \, {\left (30 \, {\left (e x + d\right )}^{2} c^{2} d^{2} e^{2} + 5 \, {\left (e x + d\right )} c^{2} d^{3} e^{2} + c^{2} d^{4} e^{2} - 5 \, {\left (e x + d\right )} a c d e^{4} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )}}{5 \, {\left (c^{5} d^{10} - 5 \, a c^{4} d^{8} e^{2} + 10 \, a^{2} c^{3} d^{6} e^{4} - 10 \, a^{3} c^{2} d^{4} e^{6} + 5 \, a^{4} c d^{2} e^{8} - a^{5} e^{10}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \] Input:

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

Output:

63/4*c^3*d^3*e^2*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^ 
5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4 
*c*d^2*e^8 - a^5*e^10)*sqrt(-c^2*d^3 + a*c*d*e^2)) + 1/4*(15*(e*x + d)^(3/ 
2)*c^4*d^4*e^2 - 17*sqrt(e*x + d)*c^4*d^5*e^2 + 17*sqrt(e*x + d)*a*c^3*d^3 
*e^4)/((c^5*d^10 - 5*a*c^4*d^8*e^2 + 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e 
^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*((e*x + d)*c*d - c*d^2 + a*e^2)^2) + 2/5* 
(30*(e*x + d)^2*c^2*d^2*e^2 + 5*(e*x + d)*c^2*d^3*e^2 + c^2*d^4*e^2 - 5*(e 
*x + d)*a*c*d*e^4 - 2*a*c*d^2*e^4 + a^2*e^6)/((c^5*d^10 - 5*a*c^4*d^8*e^2 
+ 10*a^2*c^3*d^6*e^4 - 10*a^3*c^2*d^4*e^6 + 5*a^4*c*d^2*e^8 - a^5*e^10)*(e 
*x + d)^(5/2))
 

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\frac {2\,e^2}{5\,\left (a\,e^2-c\,d^2\right )}-\frac {6\,c\,d\,e^2\,\left (d+e\,x\right )}{5\,{\left (a\,e^2-c\,d^2\right )}^2}+\frac {42\,c^2\,d^2\,e^2\,{\left (d+e\,x\right )}^2}{5\,{\left (a\,e^2-c\,d^2\right )}^3}+\frac {105\,c^3\,d^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e^2-c\,d^2\right )}^4}+\frac {63\,c^4\,d^4\,e^2\,{\left (d+e\,x\right )}^4}{4\,{\left (a\,e^2-c\,d^2\right )}^5}}{{\left (d+e\,x\right )}^{5/2}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,{\left (d+e\,x\right )}^{7/2}+c^2\,d^2\,{\left (d+e\,x\right )}^{9/2}}-\frac {63\,c^{5/2}\,d^{5/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^5\,e^{10}-5\,a^4\,c\,d^2\,e^8+10\,a^3\,c^2\,d^4\,e^6-10\,a^2\,c^3\,d^6\,e^4+5\,a\,c^4\,d^8\,e^2-c^5\,d^{10}\right )}{{\left (a\,e^2-c\,d^2\right )}^{11/2}}\right )}{4\,{\left (a\,e^2-c\,d^2\right )}^{11/2}} \] Input:

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

Output:

- ((2*e^2)/(5*(a*e^2 - c*d^2)) - (6*c*d*e^2*(d + e*x))/(5*(a*e^2 - c*d^2)^ 
2) + (42*c^2*d^2*e^2*(d + e*x)^2)/(5*(a*e^2 - c*d^2)^3) + (105*c^3*d^3*e^2 
*(d + e*x)^3)/(4*(a*e^2 - c*d^2)^4) + (63*c^4*d^4*e^2*(d + e*x)^4)/(4*(a*e 
^2 - c*d^2)^5))/((d + e*x)^(5/2)*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2) - (2* 
c^2*d^3 - 2*a*c*d*e^2)*(d + e*x)^(7/2) + c^2*d^2*(d + e*x)^(9/2)) - (63*c^ 
(5/2)*d^(5/2)*e^2*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^5*e^10 - c^5*d^ 
10 + 5*a*c^4*d^8*e^2 - 5*a^4*c*d^2*e^8 - 10*a^2*c^3*d^6*e^4 + 10*a^3*c^2*d 
^4*e^6))/(a*e^2 - c*d^2)^(11/2)))/(4*(a*e^2 - c*d^2)^(11/2))
 

Reduce [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1491, normalized size of antiderivative = 6.11 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 315*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**2*c**2*d**4*e**4 - 
630*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**2*c**2*d**3*e**5*x - 31 
5*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**2*c**2*d**2*e**6*x**2 - 6 
30*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**3*d**5*e**3*x - 1260*s 
qrt(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**3*d**4*e**4*x**2 - 630*sqr 
t(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**3*d**3*e**5*x**3 - 315*sqrt( 
d)*sqrt(c)*sqrt(d + e*x)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(s 
qrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*c**4*d**6*e**2*x**2 - 630*sqrt(d)*s 
qrt(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**4*d**5*e**3*x**3 - 315*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**4*d**4*e**4*x**4 - 8*a**5*e**10 + 64*a** 
4*c*d**2*e**8 + 24*a**4*c*d*e**9*x - 344*a**3*c**2*d**4*e**6 - 432*a**3*c* 
*2*d**3*e**7*x - 168*a**3*c**2*d**2*e**8*x**2 + 203*a**2*c**3*d**6*e**4...