Integrand size = 37, antiderivative size = 152 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {15 e^2 \sqrt {d+e x}}{4 c^3 d^3}-\frac {5 e (d+e x)^{3/2}}{4 c^2 d^2 (a e+c d x)}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}-\frac {15 e^2 \sqrt {c d^2-a e^2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{7/2} d^{7/2}} \] Output:
15/4*e^2*(e*x+d)^(1/2)/c^3/d^3-5/4*e*(e*x+d)^(3/2)/c^2/d^2/(c*d*x+a*e)-1/2 *(e*x+d)^(5/2)/c/d/(c*d*x+a*e)^2-15/4*e^2*(-a*e^2+c*d^2)^(1/2)*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)
Time = 0.58 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x} \left (-15 a^2 e^4+5 a c d e^2 (d-5 e x)+c^2 d^2 \left (2 d^2+9 d e x-8 e^2 x^2\right )\right )}{4 c^3 d^3 (a e+c d x)^2}-\frac {15 e^2 \sqrt {-c d^2+a e^2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{7/2} d^{7/2}} \] Input:
Integrate[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
Output:
-1/4*(Sqrt[d + e*x]*(-15*a^2*e^4 + 5*a*c*d*e^2*(d - 5*e*x) + c^2*d^2*(2*d^ 2 + 9*d*e*x - 8*e^2*x^2)))/(c^3*d^3*(a*e + c*d*x)^2) - (15*e^2*Sqrt[-(c*d^ 2) + a*e^2]*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]] )/(4*c^(7/2)*d^(7/2))
Time = 0.42 (sec) , antiderivative size = 176, 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 = {1121, 51, 51, 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.
| \(\displaystyle \int \frac {(d+e x)^{11/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1121 |
| \(\displaystyle \int \frac {(d+e x)^{5/2}}{(a e+c d x)^3}dx\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {5 e \int \frac {(d+e x)^{3/2}}{(a e+c d x)^2}dx}{4 c d}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {5 e \left (\frac {3 e \int \frac {\sqrt {d+e x}}{a e+c d x}dx}{2 c d}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {5 e \left (\frac {3 e \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{d}+\frac {2 \sqrt {d+e x}}{c d}\right )}{2 c d}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {5 e \left (\frac {3 e \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{d e}+\frac {2 \sqrt {d+e x}}{c d}\right )}{2 c d}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {5 e \left (\frac {3 e \left (\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} d^{3/2} \sqrt {c d^2-a e^2}}\right )}{2 c d}-\frac {(d+e x)^{3/2}}{c d (a e+c d x)}\right )}{4 c d}-\frac {(d+e x)^{5/2}}{2 c d (a e+c d x)^2}\) |
Input:
Int[(d + e*x)^(11/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^3,x]
Output:
-1/2*(d + e*x)^(5/2)/(c*d*(a*e + c*d*x)^2) + (5*e*(-((d + e*x)^(3/2)/(c*d* (a*e + c*d*x))) + (3*e*((2*Sqrt[d + e*x])/(c*d) - (2*(d^2 - (a*e^2)/c)*Arc Tanh[(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])))/(2*c*d)))/(4*c*d)
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]
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]
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]
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]
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]))
Time = 18.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.08
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {15 e^{2} \left (a \,e^{2}-c \,d^{2}\right ) \left (c d x +a e \right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{4}+\frac {15 \sqrt {e x +d}\, \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (-\frac {2 \left (-4 e^{2} x^{2}+\frac {9}{2} d e x +d^{2}\right ) d^{2} c^{2}}{15}-\frac {a d \,e^{2} \left (-5 e x +d \right ) c}{3}+a^{2} e^{4}\right )}{4}}{d^{3} c^{3} \left (c d x +a e \right )^{2} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\) | \(164\) |
| derivativedivides | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} a d \,e^{2} c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {15 \left (a \,e^{2}-c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{3} d^{3}}\right )\) | \(173\) |
| default | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{c^{3} d^{3}}-\frac {\frac {\left (-\frac {9}{8} a d \,e^{2} c +\frac {9}{8} c^{2} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}+\left (-\frac {7}{8} a^{2} e^{4}+\frac {7}{4} a c \,d^{2} e^{2}-\frac {7}{8} c^{2} d^{4}\right ) \sqrt {e x +d}}{\left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )^{2}}+\frac {15 \left (a \,e^{2}-c \,d^{2}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{8 \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c^{3} d^{3}}\right )\) | \(173\) |
| risch | \(\frac {2 e^{2} \textit {\_O1} \sqrt {e x +d}}{d^{3}}-\frac {\left (a \,e^{2}-c \,d^{2}\right ) e^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c^{2} d^{3} \textit {\_Z}^{6}+\left (3 a c \,d^{2} e^{2}-3 c^{2} d^{4}\right ) \textit {\_Z}^{4}+\left (3 a^{2} d \,e^{4}-6 a c \,d^{3} e^{2}+3 c^{2} d^{5}\right ) \textit {\_Z}^{2}+a^{3} c^{2} e^{6} \textit {\_O1} -3 a^{2} d^{2} e^{4}+3 a \,d^{4} e^{2} c -d^{6} c^{2}\right )}{\sum }\frac {\left (d^{2} \left (3 \textit {\_R}^{4}-3 \textit {\_R}^{2} d +d^{2}\right ) c^{2}-a d \,e^{2} c \left (-3 \textit {\_R}^{2}+2 d \right )+a^{2} e^{4}\right ) \ln \left (\sqrt {e x +d}-\textit {\_R} \right )}{\textit {\_R} \left (c d \,\textit {\_R}^{2}+a \,e^{2}-c \,d^{2}\right )^{2}}\right )}{3 c^{4} d^{4}}\) | \(234\) |
Input:
int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^3,x,method=_RETURNVER BOSE)
Output:
15/4*(-e^2*(a*e^2-c*d^2)*(c*d*x+a*e)^2*arctan(c*d*(e*x+d)^(1/2)/(c*d*(a*e^ 2-c*d^2))^(1/2))+(e*x+d)^(1/2)*(c*d*(a*e^2-c*d^2))^(1/2)*(-2/15*(-4*e^2*x^ 2+9/2*d*e*x+d^2)*d^2*c^2-1/3*a*d*e^2*(-5*e*x+d)*c+a^2*e^4))/(c*d*(a*e^2-c* d^2))^(1/2)/d^3/c^3/(c*d*x+a*e)^2
Time = 0.10 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.89 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}, -\frac {15 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (8 \, c^{2} d^{2} e^{2} x^{2} - 2 \, c^{2} d^{4} - 5 \, a c d^{2} e^{2} + 15 \, a^{2} e^{4} - {\left (9 \, c^{2} d^{3} e - 25 \, a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (c^{5} d^{5} x^{2} + 2 \, a c^{4} d^{4} e x + a^{2} c^{3} d^{3} e^{2}\right )}}\right ] \] Input:
integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= "fricas")
Output:
[1/8*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*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*(8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c*d ^2*e^2 + 15*a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)*sqrt(e*x + d))/(c^5* d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2), -1/4*(15*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*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)) - (8*c^2*d^2*e^2*x^2 - 2*c^2*d^4 - 5*a*c*d^2*e^2 + 15*a^2*e^4 - (9*c^2*d^3*e - 25*a*c*d*e^3)*x)* sqrt(e*x + d))/(c^5*d^5*x^2 + 2*a*c^4*d^4*e*x + a^2*c^3*d^3*e^2)]
Timed out. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**(11/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**3,x)
Output:
Timed out
Exception generated. \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^(11/2)/(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
Time = 0.17 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.34 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2 \, \sqrt {e x + d} e^{2}}{c^{3} d^{3}} + \frac {15 \, {\left (c d^{2} e^{2} - a e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} c^{3} d^{3}} - \frac {9 \, {\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{3} e^{2} - 7 \, \sqrt {e x + d} c^{2} d^{4} e^{2} - 9 \, {\left (e x + d\right )}^{\frac {3}{2}} a c d e^{4} + 14 \, \sqrt {e x + d} a c d^{2} e^{4} - 7 \, \sqrt {e x + d} a^{2} e^{6}}{4 \, {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2} c^{3} d^{3}} \] Input:
integrate((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x, algorithm= "giac")
Output:
2*sqrt(e*x + d)*e^2/(c^3*d^3) + 15/4*(c*d^2*e^2 - a*e^4)*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) - 1/4*(9*(e*x + d)^(3/2)*c^2*d^3*e^2 - 7*sqrt(e*x + d)*c^2*d^4*e^2 - 9*(e*x + d)^(3/2)*a*c*d*e^4 + 14*sqrt(e*x + d)*a*c*d^2*e^4 - 7*sqrt(e*x + d)*a^2 *e^6)/(((e*x + d)*c*d - c*d^2 + a*e^2)^2*c^3*d^3)
Time = 5.34 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.58 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {2\,e^2\,\sqrt {d+e\,x}}{c^3\,d^3}-\frac {\left (\frac {9\,c^2\,d^3\,e^2}{4}-\frac {9\,a\,c\,d\,e^4}{4}\right )\,{\left (d+e\,x\right )}^{3/2}-\sqrt {d+e\,x}\,\left (\frac {7\,a^2\,e^6}{4}-\frac {7\,a\,c\,d^2\,e^4}{2}+\frac {7\,c^2\,d^4\,e^2}{4}\right )}{c^5\,d^7-\left (2\,c^5\,d^6-2\,a\,c^4\,d^4\,e^2\right )\,\left (d+e\,x\right )+c^5\,d^5\,{\left (d+e\,x\right )}^2-2\,a\,c^4\,d^5\,e^2+a^2\,c^3\,d^3\,e^4}-\frac {15\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,e^2\,\sqrt {a\,e^2-c\,d^2}\,\sqrt {d+e\,x}}{a\,e^4-c\,d^2\,e^2}\right )\,\sqrt {a\,e^2-c\,d^2}}{4\,c^{7/2}\,d^{7/2}} \] Input:
int((d + e*x)^(11/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^3,x)
Output:
(2*e^2*(d + e*x)^(1/2))/(c^3*d^3) - (((9*c^2*d^3*e^2)/4 - (9*a*c*d*e^4)/4) *(d + e*x)^(3/2) - (d + e*x)^(1/2)*((7*a^2*e^6)/4 + (7*c^2*d^4*e^2)/4 - (7 *a*c*d^2*e^4)/2))/(c^5*d^7 - (2*c^5*d^6 - 2*a*c^4*d^4*e^2)*(d + e*x) + c^5 *d^5*(d + e*x)^2 - 2*a*c^4*d^5*e^2 + a^2*c^3*d^3*e^4) - (15*e^2*atan((c^(1 /2)*d^(1/2)*e^2*(a*e^2 - c*d^2)^(1/2)*(d + e*x)^(1/2))/(a*e^4 - c*d^2*e^2) )*(a*e^2 - c*d^2)^(1/2))/(4*c^(7/2)*d^(7/2))
Time = 0.22 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.11 \[ \int \frac {(d+e x)^{11/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {-15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a^{2} e^{4}-30 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a c d \,e^{3} x -15 \sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}+15 \sqrt {e x +d}\, a^{2} c d \,e^{4}-5 \sqrt {e x +d}\, a \,c^{2} d^{3} e^{2}+25 \sqrt {e x +d}\, a \,c^{2} d^{2} e^{3} x -2 \sqrt {e x +d}\, c^{3} d^{5}-9 \sqrt {e x +d}\, c^{3} d^{4} e x +8 \sqrt {e x +d}\, c^{3} d^{3} e^{2} x^{2}}{4 c^{4} d^{4} \left (c^{2} d^{2} x^{2}+2 a c d e x +a^{2} e^{2}\right )} \] Input:
int((e*x+d)^(11/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^3,x)
Output:
( - 15*sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqr t(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*a**2*e**4 - 30*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*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)))*c**2*d**2*e**2*x** 2 + 15*sqrt(d + e*x)*a**2*c*d*e**4 - 5*sqrt(d + e*x)*a*c**2*d**3*e**2 + 25 *sqrt(d + e*x)*a*c**2*d**2*e**3*x - 2*sqrt(d + e*x)*c**3*d**5 - 9*sqrt(d + e*x)*c**3*d**4*e*x + 8*sqrt(d + e*x)*c**3*d**3*e**2*x**2)/(4*c**4*d**4*(a **2*e**2 + 2*a*c*d*e*x + c**2*d**2*x**2))