Integrand size = 33, antiderivative size = 196 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {63 e^2}{20 (b d-a e)^3 (d+e x)^{5/2}}-\frac {1}{2 (b d-a e) (a+b x)^2 (d+e x)^{5/2}}+\frac {9 e}{4 (b d-a e)^2 (a+b x) (d+e x)^{5/2}}+\frac {21 b e^2}{4 (b d-a e)^4 (d+e x)^{3/2}}+\frac {63 b^2 e^2}{4 (b d-a e)^5 \sqrt {d+e x}}-\frac {63 b^{5/2} e^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{11/2}} \]
63/20*e^2/(-a*e+b*d)^3/(e*x+d)^(5/2)-1/2/(-a*e+b*d)/(b*x+a)^2/(e*x+d)^(5/2 )+9/4*e/(-a*e+b*d)^2/(b*x+a)/(e*x+d)^(5/2)+21/4*b*e^2/(-a*e+b*d)^4/(e*x+d) ^(3/2)-63/4*b^(5/2)*e^2*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/(- a*e+b*d)^(11/2)+63/4*b^2*e^2/(-a*e+b*d)^5/(e*x+d)^(1/2)
Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.12 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {8 a^4 e^4-8 a^3 b e^3 (7 d+3 e x)+24 a^2 b^2 e^2 \left (12 d^2+17 d e x+7 e^2 x^2\right )+a b^3 e \left (85 d^3+831 d^2 e x+1239 d e^2 x^2+525 e^3 x^3\right )+b^4 \left (-10 d^4+45 d^3 e x+483 d^2 e^2 x^2+735 d e^3 x^3+315 e^4 x^4\right )}{20 (b d-a e)^5 (a+b x)^2 (d+e x)^{5/2}}-\frac {63 b^{5/2} e^2 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 (-b d+a e)^{11/2}} \]
(8*a^4*e^4 - 8*a^3*b*e^3*(7*d + 3*e*x) + 24*a^2*b^2*e^2*(12*d^2 + 17*d*e*x + 7*e^2*x^2) + a*b^3*e*(85*d^3 + 831*d^2*e*x + 1239*d*e^2*x^2 + 525*e^3*x ^3) + b^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*(b*d - a*e)^5*(a + b*x)^2*(d + e*x)^(5/2)) - (63*b^(5/2)*e^2*A rcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(4*(-(b*d) + a*e)^(11/2 ))
Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1184, 27, 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.
| \(\displaystyle \int \frac {a+b x}{\left (a^2+2 a b x+b^2 x^2\right )^2 (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle b^4 \int \frac {1}{b^4 (a+b x)^3 (d+e x)^{7/2}}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {1}{(a+b x)^3 (d+e x)^{7/2}}dx\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {9 e \int \frac {1}{(a+b x)^2 (d+e x)^{7/2}}dx}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \int \frac {1}{(a+b x) (d+e x)^{7/2}}dx}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {9 e \left (-\frac {7 e \left (\frac {b \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 (b d-a e)}-\frac {1}{(a+b x) (d+e x)^{5/2} (b d-a e)}\right )}{4 (b d-a e)}-\frac {1}{2 (a+b x)^2 (d+e x)^{5/2} (b d-a e)}\) |
-1/2*1/((b*d - a*e)*(a + b*x)^2*(d + e*x)^(5/2)) - (9*e*(-(1/((b*d - a*e)* (a + b*x)*(d + e*x)^(5/2))) - (7*e*(2/(5*(b*d - a*e)*(d + e*x)^(5/2)) + (b *(2/(3*(b*d - a*e)*(d + e*x)^(3/2)) + (b*(2/((b*d - a*e)*Sqrt[d + e*x]) - (2*Sqrt[b]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^( 3/2)))/(b*d - a*e)))/(b*d - a*e)))/(2*(b*d - a*e))))/(4*(b*d - a*e))
3.21.82.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.36 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(2 e^{2} \left (-\frac {b^{3} \left (\frac {\frac {15 b \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 a e}{8}-\frac {17 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {63 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}-\frac {1}{5 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 b^{2}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b}{\left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}\right )\) | \(165\) |
| default | \(2 e^{2} \left (-\frac {b^{3} \left (\frac {\frac {15 b \left (e x +d \right )^{\frac {3}{2}}}{8}+\left (\frac {17 a e}{8}-\frac {17 b d}{8}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{2}}+\frac {63 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}-\frac {1}{5 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {5}{2}}}-\frac {6 b^{2}}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b}{\left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}\right )\) | \(165\) |
| pseudoelliptic | \(-\frac {2 \left (\frac {315 b^{3} e^{2} \left (e x +d \right )^{\frac {5}{2}} \left (b x +a \right )^{2} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{8}+\left (\left (\frac {315}{8} e^{4} x^{4}+\frac {735}{8} d \,e^{3} x^{3}+\frac {483}{8} d^{2} e^{2} x^{2}+\frac {45}{8} d^{3} e x -\frac {5}{4} d^{4}\right ) b^{4}+\frac {85 e \left (\frac {105}{17} e^{3} x^{3}+\frac {1239}{85} d \,e^{2} x^{2}+\frac {831}{85} d^{2} e x +d^{3}\right ) a \,b^{3}}{8}+36 e^{2} \left (\frac {7}{12} e^{2} x^{2}+\frac {17}{12} d e x +d^{2}\right ) a^{2} b^{2}-7 e^{3} a^{3} \left (\frac {3 e x}{7}+d \right ) b +e^{4} a^{4}\right ) \sqrt {\left (a e -b d \right ) b}\right )}{5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} \left (b x +a \right )^{2} \left (a e -b d \right )^{5}}\) | \(228\) |
2*e^2*(-1/(a*e-b*d)^5*b^3*((15/8*b*(e*x+d)^(3/2)+(17/8*a*e-17/8*b*d)*(e*x+ d)^(1/2))/(b*(e*x+d)+a*e-b*d)^2+63/8/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^ (1/2)/((a*e-b*d)*b)^(1/2)))-1/5/(a*e-b*d)^3/(e*x+d)^(5/2)-6/(a*e-b*d)^5*b^ 2/(e*x+d)^(1/2)+1/(a*e-b*d)^4*b/(e*x+d)^(3/2))
Leaf count of result is larger than twice the leaf count of optimal. 924 vs. \(2 (164) = 328\).
Time = 0.47 (sec) , antiderivative size = 1858, normalized size of antiderivative = 9.48 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \]
[-1/40*(315*(b^4*e^5*x^5 + a^2*b^2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x ^4 + (3*b^4*d^2*e^3 + 6*a*b^3*d*e^4 + a^2*b^2*e^5)*x^3 + (b^4*d^3*e^2 + 6* a*b^3*d^2*e^3 + 3*a^2*b^2*d*e^4)*x^2 + (2*a*b^3*d^3*e^2 + 3*a^2*b^2*d^2*e^ 3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*sqrt(e* x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(315*b^4*e^4*x^4 - 10*b^4*d^4 + 85*a*b^3*d^3*e + 288*a^2*b^2*d^2*e^2 - 56*a^3*b*d*e^3 + 8*a^4*e^4 + 105*( 7*b^4*d*e^3 + 5*a*b^3*e^4)*x^3 + 21*(23*b^4*d^2*e^2 + 59*a*b^3*d*e^3 + 8*a ^2*b^2*e^4)*x^2 + 3*(15*b^4*d^3*e + 277*a*b^3*d^2*e^2 + 136*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^2*b^5*d^8 - 5*a^3*b^4*d^7*e + 10*a^4*b ^3*d^6*e^2 - 10*a^5*b^2*d^5*e^3 + 5*a^6*b*d^4*e^4 - a^7*d^3*e^5 + (b^7*d^5 *e^3 - 5*a*b^6*d^4*e^4 + 10*a^2*b^5*d^3*e^5 - 10*a^3*b^4*d^2*e^6 + 5*a^4*b ^3*d*e^7 - a^5*b^2*e^8)*x^5 + (3*b^7*d^6*e^2 - 13*a*b^6*d^5*e^3 + 20*a^2*b ^5*d^4*e^4 - 10*a^3*b^4*d^3*e^5 - 5*a^4*b^3*d^2*e^6 + 7*a^5*b^2*d*e^7 - 2* a^6*b*e^8)*x^4 + (3*b^7*d^7*e - 9*a*b^6*d^6*e^2 + a^2*b^5*d^5*e^3 + 25*a^3 *b^4*d^4*e^4 - 35*a^4*b^3*d^3*e^5 + 17*a^5*b^2*d^2*e^6 - a^6*b*d*e^7 - a^7 *e^8)*x^3 + (b^7*d^8 + a*b^6*d^7*e - 17*a^2*b^5*d^6*e^2 + 35*a^3*b^4*d^5*e ^3 - 25*a^4*b^3*d^4*e^4 - a^5*b^2*d^3*e^5 + 9*a^6*b*d^2*e^6 - 3*a^7*d*e^7) *x^2 + (2*a*b^6*d^8 - 7*a^2*b^5*d^7*e + 5*a^3*b^4*d^6*e^2 + 10*a^4*b^3*d^5 *e^3 - 20*a^5*b^2*d^4*e^4 + 13*a^6*b*d^3*e^5 - 3*a^7*d^2*e^6)*x), -1/20*(3 15*(b^4*e^5*x^5 + a^2*b^2*d^3*e^2 + (3*b^4*d*e^4 + 2*a*b^3*e^5)*x^4 + (...
Timed out. \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
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-b*d>0)', see `assume?` for m ore detail
Leaf count of result is larger than twice the leaf count of optimal. 386 vs. \(2 (164) = 328\).
Time = 0.28 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.97 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {63 \, b^{3} e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} + \frac {15 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} e^{2} - 17 \, \sqrt {e x + d} b^{4} d e^{2} + 17 \, \sqrt {e x + d} a b^{3} e^{3}}{4 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} + \frac {2 \, {\left (30 \, {\left (e x + d\right )}^{2} b^{2} e^{2} + 5 \, {\left (e x + d\right )} b^{2} d e^{2} + b^{2} d^{2} e^{2} - 5 \, {\left (e x + d\right )} a b e^{3} - 2 \, a b d e^{3} + a^{2} e^{4}\right )}}{5 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]
63/4*b^3*e^2*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^5 - 5*a* b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5* e^5)*sqrt(-b^2*d + a*b*e)) + 1/4*(15*(e*x + d)^(3/2)*b^4*e^2 - 17*sqrt(e*x + d)*b^4*d*e^2 + 17*sqrt(e*x + d)*a*b^3*e^3)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*((e*x + d)*b - b*d + a*e)^2) + 2/5*(30*(e*x + d)^2*b^2*e^2 + 5*(e*x + d)*b^2*d*e^ 2 + b^2*d^2*e^2 - 5*(e*x + d)*a*b*e^3 - 2*a*b*d*e^3 + a^2*e^4)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*(e*x + d)^(5/2))
Time = 0.32 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.45 \[ \int \frac {a+b x}{(d+e x)^{7/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {\frac {2\,e^2}{5\,\left (a\,e-b\,d\right )}+\frac {42\,b^2\,e^2\,{\left (d+e\,x\right )}^2}{5\,{\left (a\,e-b\,d\right )}^3}+\frac {105\,b^3\,e^2\,{\left (d+e\,x\right )}^3}{4\,{\left (a\,e-b\,d\right )}^4}+\frac {63\,b^4\,e^2\,{\left (d+e\,x\right )}^4}{4\,{\left (a\,e-b\,d\right )}^5}-\frac {6\,b\,e^2\,\left (d+e\,x\right )}{5\,{\left (a\,e-b\,d\right )}^2}}{b^2\,{\left (d+e\,x\right )}^{9/2}-\left (2\,b^2\,d-2\,a\,b\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (a^2\,e^2-2\,a\,b\,d\,e+b^2\,d^2\right )}-\frac {63\,b^{5/2}\,e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}}\right )}{4\,{\left (a\,e-b\,d\right )}^{11/2}} \]
- ((2*e^2)/(5*(a*e - b*d)) + (42*b^2*e^2*(d + e*x)^2)/(5*(a*e - b*d)^3) + (105*b^3*e^2*(d + e*x)^3)/(4*(a*e - b*d)^4) + (63*b^4*e^2*(d + e*x)^4)/(4* (a*e - b*d)^5) - (6*b*e^2*(d + e*x))/(5*(a*e - b*d)^2))/(b^2*(d + e*x)^(9/ 2) - (2*b^2*d - 2*a*b*e)*(d + e*x)^(7/2) + (d + e*x)^(5/2)*(a^2*e^2 + b^2* d^2 - 2*a*b*d*e)) - (63*b^(5/2)*e^2*atan((b^(1/2)*(d + e*x)^(1/2)*(a^5*e^5 - b^5*d^5 - 10*a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a ^4*b*d*e^4))/(a*e - b*d)^(11/2)))/(4*(a*e - b*d)^(11/2))