\(\int \frac {(d+e x)^5}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\) [266]

Optimal result

Integrand size = 37, antiderivative size = 214 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)^4}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {10 e (d+e x)^2}{3 c^2 d^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {5 e^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c^3 d^3}+\frac {5 e^{3/2} \left (c d^2-a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} (d+e x)}{\sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{c^{7/2} d^{7/2}} \] Output:

-2/3*(e*x+d)^4/c/d/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)-10/3*e*(e*x+d)^ 
2/c^2/d^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+5*e^2*(a*d*e+(a*e^2+c*d^ 
2)*x+c*d*e*x^2)^(1/2)/c^3/d^3+5*e^(3/2)*(-a*e^2+c*d^2)*arctanh(c^(1/2)*d^( 
1/2)*(e*x+d)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/c^(7/2)/d^(7 
/2)
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {((a e+c d x) (d+e x))^{5/2} \left (-\frac {\sqrt {c} \sqrt {d} \left (-15 a^2 e^4+10 a c d e^2 (d-2 e x)+c^2 d^2 \left (2 d^2+14 d e x-3 e^2 x^2\right )\right )}{(a e+c d x)^4 (d+e x)^2}+\frac {15 e^{3/2} \left (c d^2-a e^2\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{(a e+c d x)^{5/2} (d+e x)^{5/2}}\right )}{3 c^{7/2} d^{7/2}} \] Input:

Integrate[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]
 

Output:

(((a*e + c*d*x)*(d + e*x))^(5/2)*(-((Sqrt[c]*Sqrt[d]*(-15*a^2*e^4 + 10*a*c 
*d*e^2*(d - 2*e*x) + c^2*d^2*(2*d^2 + 14*d*e*x - 3*e^2*x^2)))/((a*e + c*d* 
x)^4*(d + e*x)^2)) + (15*e^(3/2)*(c*d^2 - a*e^2)*ArcTanh[(Sqrt[c]*Sqrt[d]* 
Sqrt[d + e*x])/(Sqrt[e]*Sqrt[a*e + c*d*x])])/((a*e + c*d*x)^(5/2)*(d + e*x 
)^(5/2))))/(3*c^(7/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1133, 1124, 27, 1160, 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.

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

\(\Big \downarrow \) 1133

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

\(\Big \downarrow \) 1124

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1160

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

\(\Big \downarrow \) 1092

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

\(\Big \downarrow \) 219

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

Input:

Int[(d + e*x)^5/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]
 

Output:

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

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 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]
 

rule 1133

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)/(c*(p + 1))), 
 x] - Simp[e^2*((m + p)/(c*(p + 1)))   Int[(d + e*x)^(m - 2)*(a + b*x + c*x 
^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e 
^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
 

rule 1160

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b 
*e)/(2*c)   Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] 
 && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4446\) vs. \(2(190)=380\).

Time = 1.75 (sec) , antiderivative size = 4447, normalized size of antiderivative = 20.78

method	result	size
default	\(\text {Expression too large to display}\)	\(4447\)

Input:

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/2),x,method=_RETURNVERB 
OSE)
 

Output:

d^5*(2/3*(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*x^2*e)^(3/2)+16/3*d*e*c/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)^ 
2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2))+e^5*(x^ 
4/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)-5/2*(a*e^2+c*d^2)/d/e/c*(- 
1/3*x^3/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)-1/2*(a*e^2+c*d^2)/d/ 
e/c*(-x^2/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)+1/2*(a*e^2+c*d^2)/ 
d/e/c*(-1/2*x/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)-1/4*(a*e^2+c*d 
^2)/d/e/c*(-1/3/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)-1/2*(a*e^2+c 
*d^2)/d/e/c*(2/3*(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*x^2*e)^(3/2)+16/3*d*e*c/(4*a*c*d^2*e^2-(a*e^2+c* 
d^2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2)) 
)+1/2*a/c*(2/3*(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*x^2*e)^(3/2)+16/3*d*e*c/(4*a*c*d^2*e^2-(a*e^2+c*d^ 
2)^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2)))+ 
2*a/c*(-1/3/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/2)-1/2*(a*e^2+c*d^2 
)/d/e/c*(2/3*(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*x^2*e)^(3/2)+16/3*d*e*c/(4*a*c*d^2*e^2-(a*e^2+c*d^2) 
^2)^2*(2*c*d*e*x+a*e^2+c*d^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2))))+1 
/d/e/c*(-x/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2)-1/2*(a*e^2+c*d^2) 
/d/e/c*(-1/d/e/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(1/2)-(a*e^2+c*d^2)/...
 

Fricas [A] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.00 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\left [-\frac {15 \, {\left (a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{2} - a^{2} c d e^{4}\right )} x\right )} \sqrt {\frac {e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x - 4 \, {\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {e}{c d}}\right ) - 4 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - 2 \, {\left (7 \, c^{2} d^{3} e - 10 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{12 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (a^{2} c d^{2} e^{3} - a^{3} e^{5} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x^{2} + 2 \, {\left (a c^{2} d^{3} e^{2} - a^{2} c d e^{4}\right )} x\right )} \sqrt {-\frac {e}{c d}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {e}{c d}}}{2 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) - 2 \, {\left (3 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 10 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - 2 \, {\left (7 \, c^{2} d^{3} e - 10 \, a c d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{6 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm=" 
fricas")
 

Output:

[-1/12*(15*(a^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4*e - a*c^2*d^2*e^3)*x^2 + 2* 
(a*c^2*d^3*e^2 - a^2*c*d*e^4)*x)*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)*s 
qrt(e/(c*d))) - 4*(3*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 10*a*c*d^2*e^2 + 15*a^2 
*e^4 - 2*(7*c^2*d^3*e - 10*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + 
 a*e^2)*x))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2), -1/6*(15*(a 
^2*c*d^2*e^3 - a^3*e^5 + (c^3*d^4*e - a*c^2*d^2*e^3)*x^2 + 2*(a*c^2*d^3*e^ 
2 - a^2*c*d*e^4)*x)*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*(3*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 10*a* 
c*d^2*e^2 + 15*a^2*e^4 - 2*(7*c^2*d^3*e - 10*a*c*d*e^3)*x)*sqrt(c*d*e*x^2 
+ a*d*e + (c*d^2 + a*e^2)*x))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3 
*e^2)]
 

Sympy [F]

\[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{5}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((e*x+d)**5/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)
 

Output:

Integral((d + e*x)**5/((d + e*x)*(a*e + c*d*x))**(5/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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 [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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%%%{%%{[%%%{2,[5,5,8]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[4,4] 
%%%}+%%%{
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^5}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}} \,d x \] Input:

int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)
 

Output:

int((d + e*x)^5/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.29 \[ \int \frac {(d+e x)^5}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx=\frac {-30 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}}{\sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} e^{4}+30 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}}{\sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c \,d^{2} e^{2}-30 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}}{\sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c d \,e^{3} x +30 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, \mathrm {log}\left (\frac {\sqrt {e}\, \sqrt {c d x +a e}+\sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}}{\sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{3} e x -5 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, a^{2} e^{4}+5 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, a c \,d^{2} e^{2}-5 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, a c d \,e^{3} x +5 \sqrt {e}\, \sqrt {d}\, \sqrt {c}\, \sqrt {c d x +a e}\, c^{2} d^{3} e x +30 \sqrt {e x +d}\, a^{2} c d \,e^{4}-20 \sqrt {e x +d}\, a \,c^{2} d^{3} e^{2}+40 \sqrt {e x +d}\, a \,c^{2} d^{2} e^{3} x -4 \sqrt {e x +d}\, c^{3} d^{5}-28 \sqrt {e x +d}\, c^{3} d^{4} e x +6 \sqrt {e x +d}\, c^{3} d^{3} e^{2} x^{2}}{6 \sqrt {c d x +a e}\, c^{4} d^{4} \left (c d x +a e \right )} \] Input:

int((e*x+d)^5/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x)
 

Output:

( - 30*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*log((sqrt(e)*sqrt(a*e + c 
*d*x) + sqrt(d)*sqrt(c)*sqrt(d + e*x))/sqrt(a*e**2 - c*d**2))*a**2*e**4 + 
30*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*log((sqrt(e)*sqrt(a*e + c*d*x 
) + sqrt(d)*sqrt(c)*sqrt(d + e*x))/sqrt(a*e**2 - c*d**2))*a*c*d**2*e**2 - 
30*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*log((sqrt(e)*sqrt(a*e + c*d*x 
) + sqrt(d)*sqrt(c)*sqrt(d + e*x))/sqrt(a*e**2 - c*d**2))*a*c*d*e**3*x + 3 
0*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*log((sqrt(e)*sqrt(a*e + c*d*x) 
 + sqrt(d)*sqrt(c)*sqrt(d + e*x))/sqrt(a*e**2 - c*d**2))*c**2*d**3*e*x - 5 
*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*a**2*e**4 + 5*sqrt(e)*sqrt(d)*s 
qrt(c)*sqrt(a*e + c*d*x)*a*c*d**2*e**2 - 5*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a* 
e + c*d*x)*a*c*d*e**3*x + 5*sqrt(e)*sqrt(d)*sqrt(c)*sqrt(a*e + c*d*x)*c**2 
*d**3*e*x + 30*sqrt(d + e*x)*a**2*c*d*e**4 - 20*sqrt(d + e*x)*a*c**2*d**3* 
e**2 + 40*sqrt(d + e*x)*a*c**2*d**2*e**3*x - 4*sqrt(d + e*x)*c**3*d**5 - 2 
8*sqrt(d + e*x)*c**3*d**4*e*x + 6*sqrt(d + e*x)*c**3*d**3*e**2*x**2)/(6*sq 
rt(a*e + c*d*x)*c**4*d**4*(a*e + c*d*x))