Integrand size = 37, antiderivative size = 153 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2}{5 \left (c d^2-a e^2\right ) (d+e x)^{5/2}}+\frac {2 c d}{3 \left (c d^2-a e^2\right )^2 (d+e x)^{3/2}}+\frac {2 c^2 d^2}{\left (c d^2-a e^2\right )^3 \sqrt {d+e x}}-\frac {2 c^{5/2} d^{5/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{7/2}} \] Output:
2/5/(-a*e^2+c*d^2)/(e*x+d)^(5/2)+2/3*c*d/(-a*e^2+c*d^2)^2/(e*x+d)^(3/2)+2* c^2*d^2/(-a*e^2+c*d^2)^3/(e*x+d)^(1/2)-2*c^(5/2)*d^(5/2)*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)^(7/2)
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {6 a^2 e^4-2 a c d e^2 (11 d+5 e x)+2 c^2 d^2 \left (23 d^2+35 d e x+15 e^2 x^2\right )}{15 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac {2 c^{5/2} d^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{7/2}} \] Input:
Integrate[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]
Output:
(6*a^2*e^4 - 2*a*c*d*e^2*(11*d + 5*e*x) + 2*c^2*d^2*(23*d^2 + 35*d*e*x + 1 5*e^2*x^2))/(15*(c*d^2 - a*e^2)^3*(d + e*x)^(5/2)) - (2*c^(5/2)*d^(5/2)*Ar cTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(-(c*d^2) + a*e^2)^(7/2)
Time = 0.48 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1121, 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.
\(\displaystyle \int \frac {1}{(d+e x)^{5/2} \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {1}{(d+e x)^{7/2} (a e+c d x)}dx\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{5/2}}dx}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) (d+e x)^{3/2}}dx}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {c d \left (\frac {c d \left (\frac {c d \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{c d^2-a e^2}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {c d \left (\frac {c d \left (\frac {2 c d \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{e \left (c d^2-a e^2\right )}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {c d \left (\frac {c d \left (\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}\right )}{c d^2-a e^2}+\frac {2}{3 (d+e x)^{3/2} \left (c d^2-a e^2\right )}\right )}{c d^2-a e^2}+\frac {2}{5 (d+e x)^{5/2} \left (c d^2-a e^2\right )}\) |
Input:
Int[1/((d + e*x)^(5/2)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]
Output:
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]*A rcTanh[(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)
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]
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 = 1.92 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(-\frac {2}{5 \left (a \,e^{2}-c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {2 c d}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 c^{3} d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\) | \(146\) |
default | \(-\frac {2}{5 \left (a \,e^{2}-c \,d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 c^{2} d^{2}}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {e x +d}}+\frac {2 c d}{3 \left (a \,e^{2}-c \,d^{2}\right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 c^{3} d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\left (a \,e^{2}-c \,d^{2}\right )^{3} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\) | \(146\) |
pseudoelliptic | \(-\frac {2 \left (5 c^{3} d^{3} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right ) \left (e x +d \right )^{\frac {5}{2}}+\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (\left (5 d^{2} e^{2} x^{2}+\frac {35}{3} d^{3} e x +\frac {23}{3} d^{4}\right ) c^{2}-\frac {11 e^{2} \left (\frac {5 e x}{11}+d \right ) a d c}{3}+a^{2} e^{4}\right )\right )}{5 \left (e x +d \right )^{\frac {5}{2}} \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}\, \left (a \,e^{2}-c \,d^{2}\right )^{3}}\) | \(153\) |
Input:
int(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e),x,method=_RETURNVERB OSE)
Output:
-2/5/(a*e^2-c*d^2)/(e*x+d)^(5/2)-2/(a*e^2-c*d^2)^3*c^2*d^2/(e*x+d)^(1/2)+2 /3/(a*e^2-c*d^2)^2*c*d/(e*x+d)^(3/2)-2*c^3*d^3/(a*e^2-c*d^2)^3/(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))
Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (129) = 258\).
Time = 0.10 (sec) , antiderivative size = 756, normalized size of antiderivative = 4.94 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\left [-\frac {15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, {\left (15 \, c^{2} d^{2} e^{2} x^{2} + 23 \, c^{2} d^{4} - 11 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 5 \, {\left (7 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (c^{3} d^{9} - 3 \, a c^{2} d^{7} e^{2} + 3 \, a^{2} c d^{5} e^{4} - a^{3} d^{3} e^{6} + {\left (c^{3} d^{6} e^{3} - 3 \, a c^{2} d^{4} e^{5} + 3 \, a^{2} c d^{2} e^{7} - a^{3} e^{9}\right )} x^{3} + 3 \, {\left (c^{3} d^{7} e^{2} - 3 \, a c^{2} d^{5} e^{4} + 3 \, a^{2} c d^{3} e^{6} - a^{3} d e^{8}\right )} x^{2} + 3 \, {\left (c^{3} d^{8} e - 3 \, a c^{2} d^{6} e^{3} + 3 \, a^{2} c d^{4} e^{5} - a^{3} d^{2} e^{7}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left (c^{2} d^{2} e^{3} x^{3} + 3 \, c^{2} d^{3} e^{2} x^{2} + 3 \, c^{2} d^{4} e x + c^{2} d^{5}\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (\sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}\right ) + {\left (15 \, c^{2} d^{2} e^{2} x^{2} + 23 \, c^{2} d^{4} - 11 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4} + 5 \, {\left (7 \, c^{2} d^{3} e - a c d e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (c^{3} d^{9} - 3 \, a c^{2} d^{7} e^{2} + 3 \, a^{2} c d^{5} e^{4} - a^{3} d^{3} e^{6} + {\left (c^{3} d^{6} e^{3} - 3 \, a c^{2} d^{4} e^{5} + 3 \, a^{2} c d^{2} e^{7} - a^{3} e^{9}\right )} x^{3} + 3 \, {\left (c^{3} d^{7} e^{2} - 3 \, a c^{2} d^{5} e^{4} + 3 \, a^{2} c d^{3} e^{6} - a^{3} d e^{8}\right )} x^{2} + 3 \, {\left (c^{3} d^{8} e - 3 \, a c^{2} d^{6} e^{3} + 3 \, a^{2} c d^{4} e^{5} - a^{3} d^{2} e^{7}\right )} x\right )}}\right ] \] Input:
integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm=" fricas")
Output:
[-1/15*(15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5) *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*(15*c^2*d^ 2*e^2*x^2 + 23*c^2*d^4 - 11*a*c*d^2*e^2 + 3*a^2*e^4 + 5*(7*c^2*d^3*e - a*c *d*e^3)*x)*sqrt(e*x + d))/(c^3*d^9 - 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5*e^4 - a ^3*d^3*e^6 + (c^3*d^6*e^3 - 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*e^7 - a^3*e^9)*x ^3 + 3*(c^3*d^7*e^2 - 3*a*c^2*d^5*e^4 + 3*a^2*c*d^3*e^6 - a^3*d*e^8)*x^2 + 3*(c^3*d^8*e - 3*a*c^2*d^6*e^3 + 3*a^2*c*d^4*e^5 - a^3*d^2*e^7)*x), 2/15* (15*(c^2*d^2*e^3*x^3 + 3*c^2*d^3*e^2*x^2 + 3*c^2*d^4*e*x + c^2*d^5)*sqrt(- c*d/(c*d^2 - a*e^2))*arctan(sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))) + (1 5*c^2*d^2*e^2*x^2 + 23*c^2*d^4 - 11*a*c*d^2*e^2 + 3*a^2*e^4 + 5*(7*c^2*d^3 *e - a*c*d*e^3)*x)*sqrt(e*x + d))/(c^3*d^9 - 3*a*c^2*d^7*e^2 + 3*a^2*c*d^5 *e^4 - a^3*d^3*e^6 + (c^3*d^6*e^3 - 3*a*c^2*d^4*e^5 + 3*a^2*c*d^2*e^7 - a^ 3*e^9)*x^3 + 3*(c^3*d^7*e^2 - 3*a*c^2*d^5*e^4 + 3*a^2*c*d^3*e^6 - a^3*d*e^ 8)*x^2 + 3*(c^3*d^8*e - 3*a*c^2*d^6*e^3 + 3*a^2*c*d^4*e^5 - a^3*d^2*e^7)*x )]
Time = 13.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {c^{2} d^{2} e}{\sqrt {d + e x} \left (a e^{2} - c d^{2}\right )^{3}} - \frac {c^{2} d^{2} e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}} \left (a e^{2} - c d^{2}\right )^{3}} + \frac {c d e}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e^{2} - c d^{2}\right )^{2}} - \frac {e}{5 \left (d + e x\right )^{\frac {5}{2}} \left (a e^{2} - c d^{2}\right )}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\log {\left (x \right )}}{c d^{\frac {9}{2}}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2),x)
Output:
Piecewise((2*(-c**2*d**2*e/(sqrt(d + e*x)*(a*e**2 - c*d**2)**3) - c**2*d** 2*e*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d)))/(sqrt((a*e**2 - c*d* *2)/(c*d))*(a*e**2 - c*d**2)**3) + c*d*e/(3*(d + e*x)**(3/2)*(a*e**2 - c*d **2)**2) - e/(5*(d + e*x)**(5/2)*(a*e**2 - c*d**2)))/e, Ne(e, 0)), (log(x) /(c*d**(9/2)), True))
Exception generated. \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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
Time = 0.14 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.42 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2 \, c^{3} d^{3} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} c^{2} d^{2} + 5 \, {\left (e x + d\right )} c^{2} d^{3} + 3 \, c^{2} d^{4} - 5 \, {\left (e x + d\right )} a c d e^{2} - 6 \, a c d^{2} e^{2} + 3 \, a^{2} e^{4}\right )}}{15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \] Input:
integrate(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x, algorithm=" giac")
Output:
2*c^3*d^3*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c^2*d^3 + a*c*d*e^2)) + 2/15*(15*(e*x + d)^2*c^2*d^2 + 5*(e*x + d)*c^2*d^3 + 3*c^2*d^4 - 5*(e*x + d)*a*c*d*e^2 - 6*a*c*d^2*e^2 + 3*a^2*e^4)/((c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*(e*x + d)^(5/2))
Time = 5.33 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=-\frac {\frac {2}{5\,\left (a\,e^2-c\,d^2\right )}+\frac {2\,c^2\,d^2\,{\left (d+e\,x\right )}^2}{{\left (a\,e^2-c\,d^2\right )}^3}-\frac {2\,c\,d\,\left (d+e\,x\right )}{3\,{\left (a\,e^2-c\,d^2\right )}^2}}{{\left (d+e\,x\right )}^{5/2}}-\frac {2\,c^{5/2}\,d^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^3\,e^6-3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2-c^3\,d^6\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{7/2}} \] Input:
int(1/((d + e*x)^(5/2)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)),x)
Output:
- (2/(5*(a*e^2 - c*d^2)) + (2*c^2*d^2*(d + e*x)^2)/(a*e^2 - c*d^2)^3 - (2* c*d*(d + e*x))/(3*(a*e^2 - c*d^2)^2))/(d + e*x)^(5/2) - (2*c^(5/2)*d^(5/2) *atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^3*e^6 - c^3*d^6 + 3*a*c^2*d^4*e^ 2 - 3*a^2*c*d^2*e^4))/(a*e^2 - c*d^2)^(7/2)))/(a*e^2 - c*d^2)^(7/2)
Time = 0.23 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.29 \[ \int \frac {1}{(d+e x)^{5/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {-2 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{4}-4 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{3} e x -2 \sqrt {d}\, \sqrt {c}\, \sqrt {e x +d}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c^{2} d^{2} e^{2} x^{2}-\frac {2 a^{3} e^{6}}{5}+\frac {28 a^{2} c \,d^{2} e^{4}}{15}+\frac {2 a^{2} c d \,e^{5} x}{3}-\frac {68 a \,c^{2} d^{4} e^{2}}{15}-\frac {16 a \,c^{2} d^{3} e^{3} x}{3}-2 a \,c^{2} d^{2} e^{4} x^{2}+\frac {46 c^{3} d^{6}}{15}+\frac {14 c^{3} d^{5} e x}{3}+2 c^{3} d^{4} e^{2} x^{2}}{\sqrt {e x +d}\, \left (a^{4} e^{10} x^{2}-4 a^{3} c \,d^{2} e^{8} x^{2}+6 a^{2} c^{2} d^{4} e^{6} x^{2}-4 a \,c^{3} d^{6} e^{4} x^{2}+c^{4} d^{8} e^{2} x^{2}+2 a^{4} d \,e^{9} x -8 a^{3} c \,d^{3} e^{7} x +12 a^{2} c^{2} d^{5} e^{5} x -8 a \,c^{3} d^{7} e^{3} x +2 c^{4} d^{9} e x +a^{4} d^{2} e^{8}-4 a^{3} c \,d^{4} e^{6}+6 a^{2} c^{2} d^{6} e^{4}-4 a \,c^{3} d^{8} e^{2}+c^{4} d^{10}\right )} \] Input:
int(1/(e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2),x)
Output:
(2*( - 15*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**2*d**4 - 30*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**2*d**3*e*x - 15*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**2*d**2*e**2*x**2 - 3*a**3*e**6 + 14*a**2*c* d**2*e**4 + 5*a**2*c*d*e**5*x - 34*a*c**2*d**4*e**2 - 40*a*c**2*d**3*e**3* x - 15*a*c**2*d**2*e**4*x**2 + 23*c**3*d**6 + 35*c**3*d**5*e*x + 15*c**3*d **4*e**2*x**2))/(15*sqrt(d + e*x)*(a**4*d**2*e**8 + 2*a**4*d*e**9*x + a**4 *e**10*x**2 - 4*a**3*c*d**4*e**6 - 8*a**3*c*d**3*e**7*x - 4*a**3*c*d**2*e* *8*x**2 + 6*a**2*c**2*d**6*e**4 + 12*a**2*c**2*d**5*e**5*x + 6*a**2*c**2*d **4*e**6*x**2 - 4*a*c**3*d**8*e**2 - 8*a*c**3*d**7*e**3*x - 4*a*c**3*d**6* e**4*x**2 + c**4*d**10 + 2*c**4*d**9*e*x + c**4*d**8*e**2*x**2))