3.18.49 \(\int \frac {A+B x}{(a+b x) (d+e x)^{7/2}} \, dx\) [1749]

Optimal. Leaf size=151 \[ -\frac {2 (B d-A e)}{5 e (b d-a e) (d+e x)^{5/2}}+\frac {2 (A b-a B)}{3 (b d-a e)^2 (d+e x)^{3/2}}+\frac {2 b (A b-a B)}{(b d-a e)^3 \sqrt {d+e x}}-\frac {2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}} \]

[Out]

-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)*arctan
h(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)

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {79, 53, 65, 214} \begin {gather*} -\frac {2 b^{3/2} (A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}+\frac {2 b (A b-a B)}{\sqrt {d+e x} (b d-a e)^3}+\frac {2 (A b-a B)}{3 (d+e x)^{3/2} (b d-a e)^2}-\frac {2 (B d-A e)}{5 e (d+e x)^{5/2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(-2*(B*d - A*e))/(5*e*(b*d - a*e)*(d + e*x)^(5/2)) + (2*(A*b - a*B))/(3*(b*d - a*e)^2*(d + e*x)^(3/2)) + (2*b*
(A*b - a*B))/((b*d - a*e)^3*Sqrt[d + e*x]) - (2*b^(3/2)*(A*b - a*B)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d -
a*e]])/(b*d - a*e)^(7/2)

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(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] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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]

Rubi steps

\begin {align*} \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 {(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 (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 {(b (A b-a B)) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{(b d-a e)^2}\\ &=-\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 {\left (b^2 (A b-a B)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{(b d-a e)^3}\\ &=-\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 {\left (2 b^2 (A b-a B)\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{e (b d-a e)^3}\\ &=-\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) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{(b d-a e)^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.36, size = 176, normalized size = 1.17 \begin {gather*} \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) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x]

[Out]

(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)

________________________________________________________________________________________

Maple [A]
time = 0.08, size = 149, normalized size = 0.99

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 e \left (A b -B a \right ) b}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 e \left (A b -B a \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 b^{2} e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}}{e}\)	\(149\)
default	\(\frac {-\frac {2 \left (A e -B d \right )}{5 \left (a e -b d \right ) \left (e x +d \right )^{\frac {5}{2}}}-\frac {2 e \left (A b -B a \right ) b}{\left (a e -b d \right )^{3} \sqrt {e x +d}}+\frac {2 e \left (A b -B a \right )}{3 \left (a e -b d \right )^{2} \left (e x +d \right )^{\frac {3}{2}}}-\frac {2 b^{2} e \left (A b -B a \right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{3} \sqrt {\left (a e -b d \right ) b}}}{e}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x,method=_RETURNVERBOSE)

[Out]

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)))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b*d-%e*a>0)', see `assume?` fo
r more detai

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 434 vs. \(2 (140) = 280\).
time = 0.87, size = 879, normalized size = 5.82 \begin {gather*} \left [\frac {15 \, {\left ({\left (B a b - A b^{2}\right )} x^{3} e^{4} + 3 \, {\left (B a b - A b^{2}\right )} d x^{2} e^{3} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} x e^{2} + {\left (B a b - A b^{2}\right )} d^{3} e\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {2 \, b d + 2 \, {\left (b d - a e\right )} \sqrt {x e + d} \sqrt {\frac {b}{b d - a e}} + {\left (b x - a\right )} e}{b x + a}\right ) - 2 \, {\left (3 \, B b^{2} d^{3} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} e^{3} + {\left (35 \, {\left (B a b - A b^{2}\right )} d x - {\left (2 \, B a^{2} - 11 \, A a b\right )} d\right )} e^{2}\right )} \sqrt {x e + d}}{15 \, {\left (b^{3} d^{6} e - a^{3} x^{3} e^{7} + 3 \, {\left (a^{2} b d x^{3} - a^{3} d x^{2}\right )} e^{6} - 3 \, {\left (a b^{2} d^{2} x^{3} - 3 \, a^{2} b d^{2} x^{2} + a^{3} d^{2} x\right )} e^{5} + {\left (b^{3} d^{3} x^{3} - 9 \, a b^{2} d^{3} x^{2} + 9 \, a^{2} b d^{3} x - a^{3} d^{3}\right )} e^{4} + 3 \, {\left (b^{3} d^{4} x^{2} - 3 \, a b^{2} d^{4} x + a^{2} b d^{4}\right )} e^{3} + 3 \, {\left (b^{3} d^{5} x - a b^{2} d^{5}\right )} e^{2}\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (B a b - A b^{2}\right )} x^{3} e^{4} + 3 \, {\left (B a b - A b^{2}\right )} d x^{2} e^{3} + 3 \, {\left (B a b - A b^{2}\right )} d^{2} x e^{2} + {\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 {x e + d} \sqrt {-\frac {b}{b d - a e}}}{b x e + b d}\right ) - {\left (3 \, B b^{2} d^{3} + {\left (14 \, B a b - 23 \, A b^{2}\right )} d^{2} e - {\left (3 \, A a^{2} - 15 \, {\left (B a b - A b^{2}\right )} x^{2} + 5 \, {\left (B a^{2} - A a b\right )} x\right )} e^{3} + {\left (35 \, {\left (B a b - A b^{2}\right )} d x - {\left (2 \, B a^{2} - 11 \, A a b\right )} d\right )} e^{2}\right )} \sqrt {x e + d}\right )}}{15 \, {\left (b^{3} d^{6} e - a^{3} x^{3} e^{7} + 3 \, {\left (a^{2} b d x^{3} - a^{3} d x^{2}\right )} e^{6} - 3 \, {\left (a b^{2} d^{2} x^{3} - 3 \, a^{2} b d^{2} x^{2} + a^{3} d^{2} x\right )} e^{5} + {\left (b^{3} d^{3} x^{3} - 9 \, a b^{2} d^{3} x^{2} + 9 \, a^{2} b d^{3} x - a^{3} d^{3}\right )} e^{4} + 3 \, {\left (b^{3} d^{4} x^{2} - 3 \, a b^{2} d^{4} x + a^{2} b d^{4}\right )} e^{3} + 3 \, {\left (b^{3} d^{5} x - a b^{2} d^{5}\right )} e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x, algorithm="fricas")

[Out]

[1/15*(15*((B*a*b - A*b^2)*x^3*e^4 + 3*(B*a*b - A*b^2)*d*x^2*e^3 + 3*(B*a*b - A*b^2)*d^2*x*e^2 + (B*a*b - A*b^
2)*d^3*e)*sqrt(b/(b*d - a*e))*log((2*b*d + 2*(b*d - a*e)*sqrt(x*e + d)*sqrt(b/(b*d - a*e)) + (b*x - a)*e)/(b*x
+ a)) - 2*(3*B*b^2*d^3 + (14*B*a*b - 23*A*b^2)*d^2*e - (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*
x)*e^3 + (35*(B*a*b - A*b^2)*d*x - (2*B*a^2 - 11*A*a*b)*d)*e^2)*sqrt(x*e + d))/(b^3*d^6*e - a^3*x^3*e^7 + 3*(a
^2*b*d*x^3 - a^3*d*x^2)*e^6 - 3*(a*b^2*d^2*x^3 - 3*a^2*b*d^2*x^2 + a^3*d^2*x)*e^5 + (b^3*d^3*x^3 - 9*a*b^2*d^3
*x^2 + 9*a^2*b*d^3*x - a^3*d^3)*e^4 + 3*(b^3*d^4*x^2 - 3*a*b^2*d^4*x + a^2*b*d^4)*e^3 + 3*(b^3*d^5*x - a*b^2*d
^5)*e^2), 2/15*(15*((B*a*b - A*b^2)*x^3*e^4 + 3*(B*a*b - A*b^2)*d*x^2*e^3 + 3*(B*a*b - A*b^2)*d^2*x*e^2 + (B*a
*b - A*b^2)*d^3*e)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(x*e + d)*sqrt(-b/(b*d - a*e))/(b*x*e + b*d))
- (3*B*b^2*d^3 + (14*B*a*b - 23*A*b^2)*d^2*e - (3*A*a^2 - 15*(B*a*b - A*b^2)*x^2 + 5*(B*a^2 - A*a*b)*x)*e^3 +
(35*(B*a*b - A*b^2)*d*x - (2*B*a^2 - 11*A*a*b)*d)*e^2)*sqrt(x*e + d))/(b^3*d^6*e - a^3*x^3*e^7 + 3*(a^2*b*d*x^
3 - a^3*d*x^2)*e^6 - 3*(a*b^2*d^2*x^3 - 3*a^2*b*d^2*x^2 + a^3*d^2*x)*e^5 + (b^3*d^3*x^3 - 9*a*b^2*d^3*x^2 + 9*
a^2*b*d^3*x - a^3*d^3)*e^4 + 3*(b^3*d^4*x^2 - 3*a*b^2*d^4*x + a^2*b*d^4)*e^3 + 3*(b^3*d^5*x - a*b^2*d^5)*e^2)]

________________________________________________________________________________________

Sympy [A]
time = 22.17, size = 136, normalized size = 0.90 \begin {gather*} \frac {2 b \left (- A b + B a\right )}{\sqrt {d + e x} \left (a e - b d\right )^{3}} + \frac {2 b \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 {2 \left (- A b + B a\right )}{3 \left (d + e x\right )^{\frac {3}{2}} \left (a e - b d\right )^{2}} + \frac {2 \left (- A e + B d\right )}{5 e \left (d + e x\right )^{\frac {5}{2}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)**(7/2),x)

[Out]

2*b*(-A*b + B*a)/(sqrt(d + e*x)*(a*e - b*d)**3) + 2*b*(-A*b + B*a)*atan(sqrt(d + e*x)/sqrt((a*e - b*d)/b))/(sq
rt((a*e - b*d)/b)*(a*e - b*d)**3) - 2*(-A*b + B*a)/(3*(d + e*x)**(3/2)*(a*e - b*d)**2) + 2*(-A*e + B*d)/(5*e*(
d + e*x)**(5/2)*(a*e - b*d))

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (140) = 280\).
time = 1.37, size = 284, normalized size = 1.88 \begin {gather*} -\frac {2 \, {\left (B a b^{2} - A b^{3}\right )} \arctan \left (\frac {\sqrt {x e + 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 (x e + d\right )}^{2} B a b e - 15 \, {\left (x e + d\right )}^{2} A b^{2} e + 5 \, {\left (x e + d\right )} B a b d e - 5 \, {\left (x e + d\right )} A b^{2} d e - 6 \, B a b d^{2} e - 3 \, A b^{2} d^{2} e - 5 \, {\left (x e + d\right )} B a^{2} e^{2} + 5 \, {\left (x e + 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 (x e + d\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(b*x+a)/(e*x+d)^(7/2),x, algorithm="giac")

[Out]

-2*(B*a*b^2 - A*b^3)*arctan(sqrt(x*e + 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*(x*e + d)^2*B*a*b*e - 15*(x*e + d)^2*A*b^2*e + 5*(x*e
+ d)*B*a*b*d*e - 5*(x*e + d)*A*b^2*d*e - 6*B*a*b*d^2*e - 3*A*b^2*d^2*e - 5*(x*e + d)*B*a^2*e^2 + 5*(x*e + 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)*(x*e + d)^(5/2))

________________________________________________________________________________________

Mupad [B]
time = 0.20, size = 173, normalized size = 1.15 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/((a + b*x)*(d + e*x)^(7/2)),x)

[Out]

- ((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)

________________________________________________________________________________________