\(\int \frac {(d+e x)^{9/2}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [184]

Optimal result

Integrand size = 37, antiderivative size = 144 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {5 e \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^3 d^3}+\frac {5 e (d+e x)^{3/2}}{3 c^2 d^2}-\frac {(d+e x)^{5/2}}{c d (a e+c d x)}-\frac {5 e \left (c d^2-a e^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{7/2} d^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\sqrt {d+e x} \left (15 a^2 e^4+10 a c d e^2 (-2 d+e x)+c^2 d^2 \left (3 d^2-14 d e x-2 e^2 x^2\right )\right )}{3 c^3 d^3 (a e+c d x)}+\frac {5 e \left (-c d^2+a e^2\right )^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{7/2} d^{7/2}} \] Input:

Integrate[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
 

Output:

-1/3*(Sqrt[d + e*x]*(15*a^2*e^4 + 10*a*c*d*e^2*(-2*d + e*x) + c^2*d^2*(3*d 
^2 - 14*d*e*x - 2*e^2*x^2)))/(c^3*d^3*(a*e + c*d*x)) + (5*e*(-(c*d^2) + a* 
e^2)^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]]) 
/(c^(7/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1121, 51, 60, 60, 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.

\(\displaystyle \int \frac {(d+e x)^{9/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2} \, dx\)

\(\Big \downarrow \) 1121

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[(d + e*x)^(9/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 = 4.16 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05

method	result	size
risch	\(-\frac {2 e \left (-c d x e +6 a \,e^{2}-7 c \,d^{2}\right ) \sqrt {e x +d}}{3 d^{3} c^{3}}+\frac {\left (2 a^{2} e^{4}-4 a c \,d^{2} e^{2}+2 c^{2} d^{4}\right ) e \left (-\frac {\sqrt {e x +d}}{2 \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {5 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{c^{3} d^{3}}\)	\(151\)
pseudoelliptic	\(-\frac {5 \left (-e \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (c d x +a e \right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )+\sqrt {e x +d}\, \left (\frac {\left (-\frac {2}{3} e^{2} x^{2}-\frac {14}{3} d e x +d^{2}\right ) d^{2} c^{2}}{5}-\frac {4 e^{2} \left (-\frac {e x}{2}+d \right ) a d c}{3}+a^{2} e^{4}\right ) \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\right )}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, d^{3} c^{3} \left (c d x +a e \right )}\)	\(162\)
derivativedivides	\(2 e \left (-\frac {-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} \sqrt {e x +d}-2 c \,d^{2} \sqrt {e x +d}}{c^{3} d^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} e^{4}+a c \,d^{2} e^{2}-\frac {1}{2} c^{2} d^{4}\right ) \sqrt {e x +d}}{c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}}+\frac {5 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{3} d^{3}}\right )\)	\(187\)
default	\(2 e \left (-\frac {-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+2 a \,e^{2} \sqrt {e x +d}-2 c \,d^{2} \sqrt {e x +d}}{c^{3} d^{3}}+\frac {\frac {\left (-\frac {1}{2} a^{2} e^{4}+a c \,d^{2} e^{2}-\frac {1}{2} c^{2} d^{4}\right ) \sqrt {e x +d}}{c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}}+\frac {5 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{3} d^{3}}\right )\)	\(187\)

Input:

int((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2,x,method=_RETURNVERB 
OSE)
 

Output:

-2/3*e*(-c*d*e*x+6*a*e^2-7*c*d^2)*(e*x+d)^(1/2)/d^3/c^3+1/c^3/d^3*(2*a^2*e 
^4-4*a*c*d^2*e^2+2*c^2*d^4)*e*(-1/2*(e*x+d)^(1/2)/(c*d*(e*x+d)+a*e^2-c*d^2 
)+5/2/(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)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 420, normalized size of antiderivative = 2.92 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [-\frac {15 \, {\left (a c d^{2} e^{2} - a^{2} e^{4} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 20 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} + 2 \, {\left (7 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{6 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}, -\frac {15 \, {\left (a c d^{2} e^{2} - a^{2} e^{4} + {\left (c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (2 \, c^{2} d^{2} e^{2} x^{2} - 3 \, c^{2} d^{4} + 20 \, a c d^{2} e^{2} - 15 \, a^{2} e^{4} + 2 \, {\left (7 \, c^{2} d^{3} e - 5 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (c^{4} d^{4} x + a c^{3} d^{3} e\right )}}\right ] \] Input:

integrate((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm=" 
fricas")
 

Output:

[-1/6*(15*(a*c*d^2*e^2 - a^2*e^4 + (c^2*d^3*e - a*c*d*e^3)*x)*sqrt((c*d^2 
- a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^2 + 2*sqrt(e*x + d)*c*d*sqrt( 
(c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) - 2*(2*c^2*d^2*e^2*x^2 - 3*c^2*d^4 
+ 20*a*c*d^2*e^2 - 15*a^2*e^4 + 2*(7*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqrt(e*x 
+ d))/(c^4*d^4*x + a*c^3*d^3*e), -1/3*(15*(a*c*d^2*e^2 - a^2*e^4 + (c^2*d^ 
3*e - a*c*d*e^3)*x)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d 
*sqrt(-(c*d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (2*c^2*d^2*e^2*x^2 - 3*c^ 
2*d^4 + 20*a*c*d^2*e^2 - 15*a^2*e^4 + 2*(7*c^2*d^3*e - 5*a*c*d*e^3)*x)*sqr 
t(e*x + d))/(c^4*d^4*x + a*c^3*d^3*e)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**(9/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((e*x+d)^(9/2)/(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 [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.51 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {5 \, {\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c^{3} d^{3}} - \frac {\sqrt {e x + d} c^{2} d^{4} e - 2 \, \sqrt {e x + d} a c d^{2} e^{3} + \sqrt {e x + d} a^{2} e^{5}}{{\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )} c^{3} d^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{4} d^{4} e + 6 \, \sqrt {e x + d} c^{4} d^{5} e - 6 \, \sqrt {e x + d} a c^{3} d^{3} e^{3}\right )}}{3 \, c^{6} d^{6}} \] Input:

integrate((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm=" 
giac")
 

Output:

5*(c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2 
*d^3 + a*c*d*e^2))/(sqrt(-c^2*d^3 + a*c*d*e^2)*c^3*d^3) - (sqrt(e*x + d)*c 
^2*d^4*e - 2*sqrt(e*x + d)*a*c*d^2*e^3 + sqrt(e*x + d)*a^2*e^5)/(((e*x + d 
)*c*d - c*d^2 + a*e^2)*c^3*d^3) + 2/3*((e*x + d)^(3/2)*c^4*d^4*e + 6*sqrt( 
e*x + d)*c^4*d^5*e - 6*sqrt(e*x + d)*a*c^3*d^3*e^3)/(c^6*d^6)
 

Mupad [B] (verification not implemented)

Time = 5.30 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {2\,e\,{\left (d+e\,x\right )}^{3/2}}{3\,c^2\,d^2}-\frac {\sqrt {d+e\,x}\,\left (a^2\,e^5-2\,a\,c\,d^2\,e^3+c^2\,d^4\,e\right )}{c^4\,d^4\,\left (d+e\,x\right )-c^4\,d^5+a\,c^3\,d^3\,e^2}+\frac {2\,e\,\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\sqrt {d+e\,x}}{c^4\,d^4}+\frac {5\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^5-2\,a\,c\,d^2\,e^3+c^2\,d^4\,e}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{c^{7/2}\,d^{7/2}} \] Input:

int((d + e*x)^(9/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.49 \[ \int \frac {(d+e x)^{9/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} e^{4}-15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c \,d^{2} e^{2}+15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c d \,e^{3} x -15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{3} e x -15 \sqrt {e x +d}\, a^{2} c d \,e^{4}+20 \sqrt {e x +d}\, a \,c^{2} d^{3} e^{2}-10 \sqrt {e x +d}\, a \,c^{2} d^{2} e^{3} x -3 \sqrt {e x +d}\, c^{3} d^{5}+14 \sqrt {e x +d}\, c^{3} d^{4} e x +2 \sqrt {e x +d}\, c^{3} d^{3} e^{2} x^{2}}{3 c^{4} d^{4} \left (c d x +a e \right )} \] Input:

int((e*x+d)^(9/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)
 

Output:

(15*sqrt(d)*sqrt(c)*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*e**4 - 15*sqrt(d)*sqrt(c)*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*d**2*e**2 + 15*sqrt(d)*sqrt(c)*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*d*e**3*x - 15*sq 
rt(d)*sqrt(c)*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**2*d**3*e*x - 15*sqrt(d + e*x)*a**2*c*d*e**4 
 + 20*sqrt(d + e*x)*a*c**2*d**3*e**2 - 10*sqrt(d + e*x)*a*c**2*d**2*e**3*x 
 - 3*sqrt(d + e*x)*c**3*d**5 + 14*sqrt(d + e*x)*c**3*d**4*e*x + 2*sqrt(d + 
 e*x)*c**3*d**3*e**2*x**2)/(3*c**4*d**4*(a*e + c*d*x))