Integrand size = 22, antiderivative size = 151 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=-\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 b (A b-a B)}{(b d-a e)^3 \sqrt {d+e x}}-\frac {2 b^{3/2} (A b-a B) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}} \]
-2/5*(-A*e+B*d)/e/(-a*e+b*d)/(e*x+d)^(5/2)+2/3*(A*b-B*a)/(-a*e+b*d)^2/(e*x +d)^(3/2)-2*b^(3/2)*(A*b-B*a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/ 2))/(-a*e+b*d)^(7/2)+2*b*(A*b-B*a)/(-a*e+b*d)^3/(e*x+d)^(1/2)
Time = 0.35 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.17 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=\frac {2 \left (-a^2 e^2 (2 B d+3 A e+5 B e x)+a b e \left (A e (11 d+5 e x)+B \left (14 d^2+35 d e x+15 e^2 x^2\right )\right )+b^2 \left (3 B d^3-A e \left (23 d^2+35 d e x+15 e^2 x^2\right )\right )\right )}{15 e (-b d+a e)^3 (d+e x)^{5/2}}-\frac {2 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}} \]
(2*(-(a^2*e^2*(2*B*d + 3*A*e + 5*B*e*x)) + a*b*e*(A*e*(11*d + 5*e*x) + B*( 14*d^2 + 35*d*e*x + 15*e^2*x^2)) + b^2*(3*B*d^3 - A*e*(23*d^2 + 35*d*e*x + 15*e^2*x^2))))/(15*e*(-(b*d) + a*e)^3*(d + e*x)^(5/2)) - (2*b^(3/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^ (7/2)
Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {87, 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.
| \(\displaystyle \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {(A b-a B) \int \frac {1}{(a+b x) (d+e x)^{5/2}}dx}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {b \int \frac {1}{(a+b x) (d+e x)^{3/2}}dx}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 61 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {b \left (\frac {b \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{b d-a e}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {b \left (\frac {2 b \int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{e (b d-a e)}+\frac {2}{\sqrt {d+e x} (b d-a e)}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {(A b-a B) \left (\frac {b \left (\frac {2}{\sqrt {d+e x} (b d-a e)}-\frac {2 \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{3/2}}\right )}{b d-a e}+\frac {2}{3 (d+e x)^{3/2} (b d-a e)}\right )}{b d-a e}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)}\) |
(-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + ((A*b - a*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)
3.18.49.3.1 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 = 2.94 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.99
| method | result | size |
| derivativedivides | \(\frac {-\frac {2 b^{2} e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 e \left (A b -B a \right ) b}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 e \left (A b -B a \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}}{e}\) | \(149\) |
| default | \(\frac {-\frac {2 b^{2} e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 e \left (A b -B a \right ) b}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 e \left (A b -B a \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}}{e}\) | \(149\) |
| pseudoelliptic | \(-\frac {2 \left (5 b^{2} e \left (e x +d \right )^{\frac {5}{2}} \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )+\left (\left (5 A \,b^{2} x^{2}-\frac {5 a x \left (3 B x +A \right ) b}{3}+a^{2} \left (\frac {5 B x}{3}+A \right )\right ) e^{3}-\frac {11 d \left (-\frac {35 A \,b^{2} x}{11}+a \left (\frac {35 B x}{11}+A \right ) b -\frac {2 a^{2} B}{11}\right ) e^{2}}{3}+\frac {23 d^{2} b \left (A b -\frac {14 B a}{23}\right ) e}{3}-b^{2} B \,d^{3}\right ) \sqrt {\left (a e -b d \right ) b}\right )}{5 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {5}{2}} e \left (a e -b d \right )^{3}}\) | \(181\) |
2/e*(-1/5*(A*e-B*d)/(a*e-b*d)/(e*x+d)^(5/2)-e*(A*b-B*a)/(a*e-b*d)^3*b/(e*x +d)^(1/2)+1/3*e*(A*b-B*a)/(a*e-b*d)^2/(e*x+d)^(3/2)-b^2*e*(A*b-B*a)/(a*e-b *d)^3/((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. 446 vs. \(2 (131) = 262\).
Time = 0.28 (sec) , antiderivative size = 902, normalized size of antiderivative = 5.97 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=\left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} e^{4} x^{3} + 3 \, {\left (B a b - A b^{2}\right )} d e^{3} x^{2} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} e^{2} x + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (3 \, B b^{2} d^{3} - 3 \, A a^{2} e^{3} + 15 \, {\left (B a b - A b^{2}\right )} e^{3} x^{2} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (2 \, B a^{2} - 11 \, A a b\right )} d e^{2} + 5 \, {\left (7 \, {\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x\right )} \sqrt {e x + d}\right )}}{15 \, {\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} + {\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \, {\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \, {\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\right )} x\right )}}\right ] \]
[1/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e^3*x^2 + 3*(B*a* b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^2)*d^3*e)*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*(3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a *b - 23*A*b^2)*d^2*e - (2*B*a^2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d *e^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2* b*d*e^6 - a^3*e^7)*x^3 + 3*(b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^ 5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*e^4 - a^3*d^2*e^5)*x), 2/15*(15*((B*a*b - A*b^2)*e^4*x^3 + 3*(B*a*b - A*b^2)*d*e ^3*x^2 + 3*(B*a*b - A*b^2)*d^2*e^2*x + (B*a*b - A*b^2)*d^3*e)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b *d)) - (3*B*b^2*d^3 - 3*A*a^2*e^3 + 15*(B*a*b - A*b^2)*e^3*x^2 + (14*B*a*b - 23*A*b^2)*d^2*e - (2*B*a^2 - 11*A*a*b)*d*e^2 + 5*(7*(B*a*b - A*b^2)*d*e ^2 - (B*a^2 - A*a*b)*e^3)*x)*sqrt(e*x + d))/(b^3*d^6*e - 3*a*b^2*d^5*e^2 + 3*a^2*b*d^4*e^3 - a^3*d^3*e^4 + (b^3*d^3*e^4 - 3*a*b^2*d^2*e^5 + 3*a^2*b* d*e^6 - a^3*e^7)*x^3 + 3*(b^3*d^4*e^3 - 3*a*b^2*d^3*e^4 + 3*a^2*b*d^2*e^5 - a^3*d*e^6)*x^2 + 3*(b^3*d^5*e^2 - 3*a*b^2*d^4*e^3 + 3*a^2*b*d^3*e^4 - a^ 3*d^2*e^5)*x)]
Time = 4.01 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.13 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b e \left (- A b + B a\right )}{\sqrt {d + e x} \left (a e - b d\right )^{3}} + \frac {b e \left (- A b + B a\right ) \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e - b d}{b}}} \right )}}{\sqrt {\frac {a e - b d}{b}} \left (a e - b d\right )^{3}} - \frac {e \left (- A b + B a\right )}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e - b d\right )^{2}} + \frac {- A e + B d}{5 \left (d + e x\right )^{\frac {5}{2}} \left (a e - b d\right )}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\frac {B x}{b} - \frac {\left (- A b + B a\right ) \left (\begin {cases} \frac {x}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x \right )}}{b} & \text {otherwise} \end {cases}\right )}{b}}{d^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]
Piecewise((2*(b*e*(-A*b + B*a)/(sqrt(d + e*x)*(a*e - b*d)**3) + b*e*(-A*b + B*a)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sqrt((a*e - b*d)/b)*(a*e - b*d)**3) - e*(-A*b + B*a)/(3*(d + e*x)**(3/2)*(a*e - b*d)**2) + (-A*e + B *d)/(5*(d + e*x)**(5/2)*(a*e - b*d)))/e, Ne(e, 0)), ((B*x/b - (-A*b + B*a) *Piecewise((x/a, Eq(b, 0)), (log(a + b*x)/b, True))/b)/d**(7/2), True))
Exception generated. \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]
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. 276 vs. \(2 (131) = 262\).
Time = 0.28 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.83 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=-\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, {\left (3 \, B b^{2} d^{3} + 15 \, {\left (e x + d\right )}^{2} B a b e - 15 \, {\left (e x + d\right )}^{2} A b^{2} e + 5 \, {\left (e x + d\right )} B a b d e - 5 \, {\left (e x + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (e x + d\right )} B a^{2} e^{2} + 5 \, {\left (e x + d\right )} A a b e^{2} + 3 \, B a^{2} d e^{2} + 6 \, A a b d e^{2} - 3 \, A a^{2} e^{3}\right )}}{15 \, {\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )} {\left (e x + d\right )}^{\frac {5}{2}}} \]
-2*(B*a*b^2 - A*b^3)*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^3*d^ 3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)*sqrt(-b^2*d + a*b*e)) - 2/15* (3*B*b^2*d^3 + 15*(e*x + d)^2*B*a*b*e - 15*(e*x + d)^2*A*b^2*e + 5*(e*x + d)*B*a*b*d*e - 5*(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 + 5*(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^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^2*b*d*e^3 - a^3*e^4)*( e*x + d)^(5/2))
Time = 0.23 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.15 \[ \int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx=-\frac {\frac {2\,\left (A\,e-B\,d\right )}{5\,\left (a\,e-b\,d\right )}-\frac {2\,\left (A\,b\,e-B\,a\,e\right )\,\left (d+e\,x\right )}{3\,{\left (a\,e-b\,d\right )}^2}+\frac {2\,b\,\left (A\,b\,e-B\,a\,e\right )\,{\left (d+e\,x\right )}^2}{{\left (a\,e-b\,d\right )}^3}}{e\,{\left (d+e\,x\right )}^{5/2}}-\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )}{{\left (a\,e-b\,d\right )}^{7/2}}\right )\,\left (A\,b-B\,a\right )}{{\left (a\,e-b\,d\right )}^{7/2}} \]
- ((2*(A*e - B*d))/(5*(a*e - b*d)) - (2*(A*b*e - B*a*e)*(d + e*x))/(3*(a*e - b*d)^2) + (2*b*(A*b*e - B*a*e)*(d + e*x)^2)/(a*e - b*d)^3)/(e*(d + e*x) ^(5/2)) - (2*b^(3/2)*atan((b^(1/2)*(d + e*x)^(1/2)*(a^3*e^3 - b^3*d^3 + 3* a*b^2*d^2*e - 3*a^2*b*d*e^2))/(a*e - b*d)^(7/2))*(A*b - B*a))/(a*e - b*d)^ (7/2)