Integrand size = 22, antiderivative size = 221 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\frac {2 b B d-7 A b e+5 a B e}{5 b (b d-a e)^2 (d+e x)^{5/2}}-\frac {A b-a B}{b (b d-a e) (a+b x) (d+e x)^{5/2}}+\frac {2 b B d-7 A b e+5 a B e}{3 (b d-a e)^3 (d+e x)^{3/2}}+\frac {b (2 b B d-7 A b e+5 a B e)}{(b d-a e)^4 \sqrt {d+e x}}-\frac {b^{3/2} (2 b B d-7 A b e+5 a B e) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{9/2}} \]
1/5*(-7*A*b*e+5*B*a*e+2*B*b*d)/b/(-a*e+b*d)^2/(e*x+d)^(5/2)+(-A*b+B*a)/b/( -a*e+b*d)/(b*x+a)/(e*x+d)^(5/2)+1/3*(-7*A*b*e+5*B*a*e+2*B*b*d)/(-a*e+b*d)^ 3/(e*x+d)^(3/2)-b^(3/2)*(-7*A*b*e+5*B*a*e+2*B*b*d)*arctanh(b^(1/2)*(e*x+d) ^(1/2)/(-a*e+b*d)^(1/2))/(-a*e+b*d)^(9/2)+b*(-7*A*b*e+5*B*a*e+2*B*b*d)/(-a *e+b*d)^4/(e*x+d)^(1/2)
Time = 0.81 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\frac {B \left (-2 a^3 e^2 (2 d+5 e x)+2 b^3 d x \left (23 d^2+35 d e x+15 e^2 x^2\right )+2 a^2 b e \left (24 d^2+58 d e x+25 e^2 x^2\right )+a b^2 \left (61 d^3+163 d^2 e x+195 d e^2 x^2+75 e^3 x^3\right )\right )-A \left (6 a^3 e^3-2 a^2 b e^2 (16 d+7 e x)+2 a b^2 e \left (58 d^2+84 d e x+35 e^2 x^2\right )+b^3 \left (15 d^3+161 d^2 e x+245 d e^2 x^2+105 e^3 x^3\right )\right )}{15 (b d-a e)^4 (a+b x) (d+e x)^{5/2}}+\frac {b^{3/2} (2 b B d-7 A b e+5 a B e) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{9/2}} \]
(B*(-2*a^3*e^2*(2*d + 5*e*x) + 2*b^3*d*x*(23*d^2 + 35*d*e*x + 15*e^2*x^2) + 2*a^2*b*e*(24*d^2 + 58*d*e*x + 25*e^2*x^2) + a*b^2*(61*d^3 + 163*d^2*e*x + 195*d*e^2*x^2 + 75*e^3*x^3)) - A*(6*a^3*e^3 - 2*a^2*b*e^2*(16*d + 7*e*x ) + 2*a*b^2*e*(58*d^2 + 84*d*e*x + 35*e^2*x^2) + b^3*(15*d^3 + 161*d^2*e*x + 245*d*e^2*x^2 + 105*e^3*x^3)))/(15*(b*d - a*e)^4*(a + b*x)*(d + e*x)^(5 /2)) + (b^(3/2)*(2*b*B*d - 7*A*b*e + 5*a*B*e)*ArcTan[(Sqrt[b]*Sqrt[d + e*x ])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(9/2)
Time = 0.30 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {87, 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.
| \(\displaystyle \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \int \frac {1}{(a+b x) (d+e x)^{7/2}}dx}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \left (\frac {b \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \left (\frac {b \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(5 a B e-7 A b e+2 b B d) \left (\frac {b \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}+\frac {2}{5 (d+e x)^{5/2} (b d-a e)}\right )}{2 b (b d-a e)}-\frac {A b-a B}{b (a+b x) (d+e x)^{5/2} (b d-a e)}\) |
-((A*b - a*B)/(b*(b*d - a*e)*(a + b*x)*(d + e*x)^(5/2))) + ((2*b*B*d - 7*A *b*e + 5*a*B*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*(b*d - a*e))
3.18.57.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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]
Time = 1.04 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-2 A b e +B a e +B b d \right )}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 b \left (3 A b e -2 B a e -B b d \right )}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {2 b^{2} \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (7 A b e -5 B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4}}\) | \(202\) |
| default | \(-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 \left (-2 A b e +B a e +B b d \right )}{3 \left (a e -b d \right )^{3} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 b \left (3 A b e -2 B a e -B b d \right )}{\left (a e -b d \right )^{4} \sqrt {e x +d}}-\frac {2 b^{2} \left (\frac {\left (\frac {1}{2} A b e -\frac {1}{2} B a e \right ) \sqrt {e x +d}}{b \left (e x +d \right )+a e -b d}+\frac {\left (7 A b e -5 B a e -2 B b d \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{4}}\) | \(202\) |
| pseudoelliptic | \(-\frac {2 \left (\frac {35 \left (e x +d \right )^{\frac {5}{2}} \left (b x +a \right ) b^{2} \left (\left (A e -\frac {2 B d}{7}\right ) b -\frac {5 B a e}{7}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2}+\sqrt {\left (a e -b d \right ) b}\, \left (\frac {\left (35 A \,e^{3} x^{3}+\frac {245 d \,x^{2} \left (-\frac {6 B x}{49}+A \right ) e^{2}}{3}+\frac {161 d^{2} x \left (-\frac {10 B x}{23}+A \right ) e}{3}+5 d^{3} \left (-\frac {46 B x}{15}+A \right )\right ) b^{3}}{2}+\frac {58 \left (\frac {35 x^{2} \left (-\frac {15 B x}{14}+A \right ) e^{3}}{58}+\frac {42 d x \left (-\frac {65 B x}{56}+A \right ) e^{2}}{29}+d^{2} \left (-\frac {163 B x}{116}+A \right ) e -\frac {61 B \,d^{3}}{116}\right ) a \,b^{2}}{3}-\frac {16 \left (\frac {7 x \left (\frac {25 B x}{7}+A \right ) e^{2}}{16}+d \left (\frac {29 B x}{8}+A \right ) e +\frac {3 B \,d^{2}}{2}\right ) e \,a^{2} b}{3}+\left (\left (\frac {5 B x}{3}+A \right ) e +\frac {2 B d}{3}\right ) e^{2} a^{3}\right )\right )}{5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} \left (a e -b d \right )^{4} \left (b x +a \right )}\) | \(269\) |
-2/5*(A*e-B*d)/(a*e-b*d)^2/(e*x+d)^(5/2)-2/3*(-2*A*b*e+B*a*e+B*b*d)/(a*e-b *d)^3/(e*x+d)^(3/2)-2*b*(3*A*b*e-2*B*a*e-B*b*d)/(a*e-b*d)^4/(e*x+d)^(1/2)- 2/(a*e-b*d)^4*b^2*((1/2*A*b*e-1/2*B*a*e)*(e*x+d)^(1/2)/(b*(e*x+d)+a*e-b*d) +1/2*(7*A*b*e-5*B*a*e-2*B*b*d)/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/ ((a*e-b*d)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 869 vs. \(2 (198) = 396\).
Time = 0.34 (sec) , antiderivative size = 1749, normalized size of antiderivative = 7.91 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\text {Too large to display} \]
[-1/30*(15*(2*B*a*b^2*d^4 + (5*B*a^2*b - 7*A*a*b^2)*d^3*e + (2*B*b^3*d*e^3 + (5*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + (6*B*b^3*d^2*e^2 + (17*B*a*b^2 - 21*A* b^3)*d*e^3 + (5*B*a^2*b - 7*A*a*b^2)*e^4)*x^3 + 3*(2*B*b^3*d^3*e + 7*(B*a* b^2 - A*b^3)*d^2*e^2 + (5*B*a^2*b - 7*A*a*b^2)*d*e^3)*x^2 + (2*B*b^3*d^4 + (11*B*a*b^2 - 7*A*b^3)*d^3*e + 3*(5*B*a^2*b - 7*A*a*b^2)*d^2*e^2)*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)*sqr t(b/(b*d - a*e)))/(b*x + a)) + 2*(6*A*a^3*e^3 - (61*B*a*b^2 - 15*A*b^3)*d^ 3 - 4*(12*B*a^2*b - 29*A*a*b^2)*d^2*e + 4*(B*a^3 - 8*A*a^2*b)*d*e^2 - 15*( 2*B*b^3*d*e^2 + (5*B*a*b^2 - 7*A*b^3)*e^3)*x^3 - 5*(14*B*b^3*d^2*e + (39*B *a*b^2 - 49*A*b^3)*d*e^2 + 2*(5*B*a^2*b - 7*A*a*b^2)*e^3)*x^2 - (46*B*b^3* d^3 + (163*B*a*b^2 - 161*A*b^3)*d^2*e + 4*(29*B*a^2*b - 42*A*a*b^2)*d*e^2 - 2*(5*B*a^3 - 7*A*a^2*b)*e^3)*x)*sqrt(e*x + d))/(a*b^4*d^7 - 4*a^2*b^3*d^ 6*e + 6*a^3*b^2*d^5*e^2 - 4*a^4*b*d^4*e^3 + a^5*d^3*e^4 + (b^5*d^4*e^3 - 4 *a*b^4*d^3*e^4 + 6*a^2*b^3*d^2*e^5 - 4*a^3*b^2*d*e^6 + a^4*b*e^7)*x^4 + (3 *b^5*d^5*e^2 - 11*a*b^4*d^4*e^3 + 14*a^2*b^3*d^3*e^4 - 6*a^3*b^2*d^2*e^5 - a^4*b*d*e^6 + a^5*e^7)*x^3 + 3*(b^5*d^6*e - 3*a*b^4*d^5*e^2 + 2*a^2*b^3*d ^4*e^3 + 2*a^3*b^2*d^3*e^4 - 3*a^4*b*d^2*e^5 + a^5*d*e^6)*x^2 + (b^5*d^7 - a*b^4*d^6*e - 6*a^2*b^3*d^5*e^2 + 14*a^3*b^2*d^4*e^3 - 11*a^4*b*d^3*e^4 + 3*a^5*d^2*e^5)*x), -1/15*(15*(2*B*a*b^2*d^4 + (5*B*a^2*b - 7*A*a*b^2)*d^3 *e + (2*B*b^3*d*e^3 + (5*B*a*b^2 - 7*A*b^3)*e^4)*x^4 + (6*B*b^3*d^2*e^2...
Timed out. \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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. 422 vs. \(2 (198) = 396\).
Time = 0.28 (sec) , antiderivative size = 422, normalized size of antiderivative = 1.91 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\frac {{\left (2 \, B b^{3} d + 5 \, B a b^{2} e - 7 \, A b^{3} e\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} + \frac {\sqrt {e x + d} B a b^{2} e - \sqrt {e x + d} A b^{3} e}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}} + \frac {2 \, {\left (15 \, {\left (e x + d\right )}^{2} B b^{2} d + 5 \, {\left (e x + d\right )} B b^{2} d^{2} + 3 \, B b^{2} d^{3} + 30 \, {\left (e x + d\right )}^{2} B a b e - 45 \, {\left (e x + d\right )}^{2} A b^{2} e - 10 \, {\left (e x + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (e x + d\right )} B a^{2} e^{2} + 10 \, {\left (e x + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]
(2*B*b^3*d + 5*B*a*b^2*e - 7*A*b^3*e)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a ^4*e^4)*sqrt(-b^2*d + a*b*e)) + (sqrt(e*x + d)*B*a*b^2*e - sqrt(e*x + d)*A *b^3*e)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^ 4*e^4)*((e*x + d)*b - b*d + a*e)) + 2/15*(15*(e*x + d)^2*B*b^2*d + 5*(e*x + d)*B*b^2*d^2 + 3*B*b^2*d^3 + 30*(e*x + d)^2*B*a*b*e - 45*(e*x + d)^2*A*b ^2*e - 10*(e*x + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e - 5*(e*x + d )*B*a^2*e^2 + 10*(e*x + d)*A*a*b*e^2 + 3*B*a^2*d*e^2 + 6*A*a*b*d*e^2 - 3*A *a^2*e^3)/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*(e*x + d)^(5/2))
Time = 1.74 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.18 \[ \int \frac {A+B x}{(a+b x)^2 (d+e x)^{7/2}} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}}\right )\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^{9/2}}-\frac {\frac {2\,\left (A\,e-B\,d\right )}{5\,\left (a\,e-b\,d\right )}+\frac {2\,\left (d+e\,x\right )\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{15\,{\left (a\,e-b\,d\right )}^2}-\frac {b^2\,{\left (d+e\,x\right )}^3\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{{\left (a\,e-b\,d\right )}^4}-\frac {2\,b\,{\left (d+e\,x\right )}^2\,\left (5\,B\,a\,e-7\,A\,b\,e+2\,B\,b\,d\right )}{3\,{\left (a\,e-b\,d\right )}^3}}{b\,{\left (d+e\,x\right )}^{7/2}+\left (a\,e-b\,d\right )\,{\left (d+e\,x\right )}^{5/2}} \]
(b^(3/2)*atan((b^(1/2)*(d + e*x)^(1/2)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2* e^2 - 4*a*b^3*d^3*e - 4*a^3*b*d*e^3))/(a*e - b*d)^(9/2))*(5*B*a*e - 7*A*b* e + 2*B*b*d))/(a*e - b*d)^(9/2) - ((2*(A*e - B*d))/(5*(a*e - b*d)) + (2*(d + e*x)*(5*B*a*e - 7*A*b*e + 2*B*b*d))/(15*(a*e - b*d)^2) - (b^2*(d + e*x) ^3*(5*B*a*e - 7*A*b*e + 2*B*b*d))/(a*e - b*d)^4 - (2*b*(d + e*x)^2*(5*B*a* e - 7*A*b*e + 2*B*b*d))/(3*(a*e - b*d)^3))/(b*(d + e*x)^(7/2) + (a*e - b*d )*(d + e*x)^(5/2))