3.18.72 \(\int \frac {a+b x}{(c+d x) (e+f x)^{7/2}} \, dx\) [1772]

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

[Out]

-2/5*(-a*f+b*e)/f/(-c*f+d*e)/(f*x+e)^(5/2)-2/3*(-a*d+b*c)/(-c*f+d*e)^2/(f*x+e)^(3/2)+2*d^(3/2)*(-a*d+b*c)*arct
anh(d^(1/2)*(f*x+e)^(1/2)/(-c*f+d*e)^(1/2))/(-c*f+d*e)^(7/2)-2*d*(-a*d+b*c)/(-c*f+d*e)^3/(f*x+e)^(1/2)

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Rubi [A]
time = 0.08, 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 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}}-\frac {2 d (b c-a d)}{\sqrt {e+f x} (d e-c f)^3}-\frac {2 (b c-a d)}{3 (e+f x)^{3/2} (d e-c f)^2}-\frac {2 (b e-a f)}{5 f (e+f x)^{5/2} (d e-c f)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*e - a*f))/(5*f*(d*e - c*f)*(e + f*x)^(5/2)) - (2*(b*c - a*d))/(3*(d*e - c*f)^2*(e + f*x)^(3/2)) - (2*d*
(b*c - a*d))/((d*e - c*f)^3*Sqrt[e + f*x]) + (2*d^(3/2)*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[d*e -
c*f]])/(d*e - c*f)^(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}{(c+d x) (e+f x)^{7/2}} \, dx &=-\frac {2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac {(b c-a d) \int \frac {1}{(c+d x) (e+f x)^{5/2}} \, dx}{d e-c f}\\ &=-\frac {2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac {2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {(d (b c-a d)) \int \frac {1}{(c+d x) (e+f x)^{3/2}} \, dx}{(d e-c f)^2}\\ &=-\frac {2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac {2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 d (b c-a d)}{(d e-c f)^3 \sqrt {e+f x}}-\frac {\left (d^2 (b c-a d)\right ) \int \frac {1}{(c+d x) \sqrt {e+f x}} \, dx}{(d e-c f)^3}\\ &=-\frac {2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac {2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 d (b c-a d)}{(d e-c f)^3 \sqrt {e+f x}}-\frac {\left (2 d^2 (b c-a d)\right ) \text {Subst}\left (\int \frac {1}{c-\frac {d e}{f}+\frac {d x^2}{f}} \, dx,x,\sqrt {e+f x}\right )}{f (d e-c f)^3}\\ &=-\frac {2 (b e-a f)}{5 f (d e-c f) (e+f x)^{5/2}}-\frac {2 (b c-a d)}{3 (d e-c f)^2 (e+f x)^{3/2}}-\frac {2 d (b c-a d)}{(d e-c f)^3 \sqrt {e+f x}}+\frac {2 d^{3/2} (b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {d e-c f}}\right )}{(d e-c f)^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 183, normalized size = 1.21 \begin {gather*} \frac {2 b \left (3 d^2 e^3-c^2 f^2 (2 e+5 f x)+c d f \left (14 e^2+35 e f x+15 f^2 x^2\right )\right )-2 a f \left (3 c^2 f^2-c d f (11 e+5 f x)+d^2 \left (23 e^2+35 e f x+15 f^2 x^2\right )\right )}{15 f (-d e+c f)^3 (e+f x)^{5/2}}-\frac {2 d^{3/2} (-b c+a d) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {e+f x}}{\sqrt {-d e+c f}}\right )}{(-d e+c f)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*(3*d^2*e^3 - c^2*f^2*(2*e + 5*f*x) + c*d*f*(14*e^2 + 35*e*f*x + 15*f^2*x^2)) - 2*a*f*(3*c^2*f^2 - c*d*f*(
11*e + 5*f*x) + d^2*(23*e^2 + 35*e*f*x + 15*f^2*x^2)))/(15*f*(-(d*e) + c*f)^3*(e + f*x)^(5/2)) - (2*d^(3/2)*(-
(b*c) + a*d)*ArcTan[(Sqrt[d]*Sqrt[e + f*x])/Sqrt[-(d*e) + c*f]])/(-(d*e) + c*f)^(7/2)

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Maple [A]
time = 0.11, size = 149, normalized size = 0.99

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*(-d^2*f*(a*d-b*c)/(c*f-d*e)^3/((c*f-d*e)*d)^(1/2)*arctan(d*(f*x+e)^(1/2)/((c*f-d*e)*d)^(1/2))-1/5*(a*f-b*e
)/(c*f-d*e)/(f*x+e)^(5/2)-f*(a*d-b*c)/(c*f-d*e)^3*d/(f*x+e)^(1/2)+1/3*f*(a*d-b*c)/(c*f-d*e)^2/(f*x+e)^(3/2))

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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)/(d*x+c)/(f*x+e)^(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(c*f-%e*d>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (141) = 282\).
time = 1.51, size = 902, normalized size = 5.97 \begin {gather*} \left [\frac {15 \, {\left ({\left (b c d - a d^{2}\right )} f^{4} x^{3} + 3 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} e + 3 \, {\left (b c d - a d^{2}\right )} f^{2} x e^{2} + {\left (b c d - a d^{2}\right )} f e^{3}\right )} \sqrt {-\frac {d}{c f - d e}} \log \left (\frac {d f x - c f + 2 \, {\left (c f - d e\right )} \sqrt {f x + e} \sqrt {-\frac {d}{c f - d e}} + 2 \, d e}{d x + c}\right ) - 2 \, {\left (3 \, a c^{2} f^{3} - 15 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} + 5 \, {\left (b c^{2} - a c d\right )} f^{3} x - 3 \, b d^{2} e^{3} - {\left (14 \, b c d - 23 \, a d^{2}\right )} f e^{2} - {\left (35 \, {\left (b c d - a d^{2}\right )} f^{2} x - {\left (2 \, b c^{2} - 11 \, a c d\right )} f^{2}\right )} e\right )} \sqrt {f x + e}}{15 \, {\left (c^{3} f^{7} x^{3} - d^{3} f e^{6} - 3 \, {\left (d^{3} f^{2} x - c d^{2} f^{2}\right )} e^{5} - 3 \, {\left (d^{3} f^{3} x^{2} - 3 \, c d^{2} f^{3} x + c^{2} d f^{3}\right )} e^{4} - {\left (d^{3} f^{4} x^{3} - 9 \, c d^{2} f^{4} x^{2} + 9 \, c^{2} d f^{4} x - c^{3} f^{4}\right )} e^{3} + 3 \, {\left (c d^{2} f^{5} x^{3} - 3 \, c^{2} d f^{5} x^{2} + c^{3} f^{5} x\right )} e^{2} - 3 \, {\left (c^{2} d f^{6} x^{3} - c^{3} f^{6} x^{2}\right )} e\right )}}, \frac {2 \, {\left (15 \, {\left ({\left (b c d - a d^{2}\right )} f^{4} x^{3} + 3 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} e + 3 \, {\left (b c d - a d^{2}\right )} f^{2} x e^{2} + {\left (b c d - a d^{2}\right )} f e^{3}\right )} \sqrt {\frac {d}{c f - d e}} \arctan \left (-\frac {{\left (c f - d e\right )} \sqrt {f x + e} \sqrt {\frac {d}{c f - d e}}}{d f x + d e}\right ) - {\left (3 \, a c^{2} f^{3} - 15 \, {\left (b c d - a d^{2}\right )} f^{3} x^{2} + 5 \, {\left (b c^{2} - a c d\right )} f^{3} x - 3 \, b d^{2} e^{3} - {\left (14 \, b c d - 23 \, a d^{2}\right )} f e^{2} - {\left (35 \, {\left (b c d - a d^{2}\right )} f^{2} x - {\left (2 \, b c^{2} - 11 \, a c d\right )} f^{2}\right )} e\right )} \sqrt {f x + e}\right )}}{15 \, {\left (c^{3} f^{7} x^{3} - d^{3} f e^{6} - 3 \, {\left (d^{3} f^{2} x - c d^{2} f^{2}\right )} e^{5} - 3 \, {\left (d^{3} f^{3} x^{2} - 3 \, c d^{2} f^{3} x + c^{2} d f^{3}\right )} e^{4} - {\left (d^{3} f^{4} x^{3} - 9 \, c d^{2} f^{4} x^{2} + 9 \, c^{2} d f^{4} x - c^{3} f^{4}\right )} e^{3} + 3 \, {\left (c d^{2} f^{5} x^{3} - 3 \, c^{2} d f^{5} x^{2} + c^{3} f^{5} x\right )} e^{2} - 3 \, {\left (c^{2} d f^{6} x^{3} - c^{3} f^{6} x^{2}\right )} e\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 32.86, size = 136, normalized size = 0.90 \begin {gather*} - \frac {2 d \left (a d - b c\right )}{\sqrt {e + f x} \left (c f - d e\right )^{3}} - \frac {2 d \left (a d - b c\right ) \operatorname {atan}{\left (\frac {\sqrt {e + f x}}{\sqrt {\frac {c f - d e}{d}}} \right )}}{\sqrt {\frac {c f - d e}{d}} \left (c f - d e\right )^{3}} + \frac {2 \left (a d - b c\right )}{3 \left (e + f x\right )^{\frac {3}{2}} \left (c f - d e\right )^{2}} - \frac {2 \left (a f - b e\right )}{5 f \left (e + f x\right )^{\frac {5}{2}} \left (c f - d e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(d*x+c)/(f*x+e)**(7/2),x)

[Out]

-2*d*(a*d - b*c)/(sqrt(e + f*x)*(c*f - d*e)**3) - 2*d*(a*d - b*c)*atan(sqrt(e + f*x)/sqrt((c*f - d*e)/d))/(sqr
t((c*f - d*e)/d)*(c*f - d*e)**3) + 2*(a*d - b*c)/(3*(e + f*x)**(3/2)*(c*f - d*e)**2) - 2*(a*f - b*e)/(5*f*(e +
f*x)**(5/2)*(c*f - d*e))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 285 vs. \(2 (141) = 282\).
time = 0.73, size = 285, normalized size = 1.89 \begin {gather*} \frac {2 \, {\left (b c d^{2} - a d^{3}\right )} \arctan \left (\frac {\sqrt {f x + e} d}{\sqrt {c d f - d^{2} e}}\right )}{{\left (c^{3} f^{3} - 3 \, c^{2} d f^{2} e + 3 \, c d^{2} f e^{2} - d^{3} e^{3}\right )} \sqrt {c d f - d^{2} e}} + \frac {2 \, {\left (15 \, {\left (f x + e\right )}^{2} b c d f - 15 \, {\left (f x + e\right )}^{2} a d^{2} f - 5 \, {\left (f x + e\right )} b c^{2} f^{2} + 5 \, {\left (f x + e\right )} a c d f^{2} - 3 \, a c^{2} f^{3} + 5 \, {\left (f x + e\right )} b c d f e - 5 \, {\left (f x + e\right )} a d^{2} f e + 3 \, b c^{2} f^{2} e + 6 \, a c d f^{2} e - 6 \, b c d f e^{2} - 3 \, a d^{2} f e^{2} + 3 \, b d^{2} e^{3}\right )}}{15 \, {\left (c^{3} f^{4} - 3 \, c^{2} d f^{3} e + 3 \, c d^{2} f^{2} e^{2} - d^{3} f e^{3}\right )} {\left (f x + e\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*c*d^2 - a*d^3)*arctan(sqrt(f*x + e)*d/sqrt(c*d*f - d^2*e))/((c^3*f^3 - 3*c^2*d*f^2*e + 3*c*d^2*f*e^2 - d^
3*e^3)*sqrt(c*d*f - d^2*e)) + 2/15*(15*(f*x + e)^2*b*c*d*f - 15*(f*x + e)^2*a*d^2*f - 5*(f*x + e)*b*c^2*f^2 +
5*(f*x + e)*a*c*d*f^2 - 3*a*c^2*f^3 + 5*(f*x + e)*b*c*d*f*e - 5*(f*x + e)*a*d^2*f*e + 3*b*c^2*f^2*e + 6*a*c*d*
f^2*e - 6*b*c*d*f*e^2 - 3*a*d^2*f*e^2 + 3*b*d^2*e^3)/((c^3*f^4 - 3*c^2*d*f^3*e + 3*c*d^2*f^2*e^2 - d^3*f*e^3)*
(f*x + e)^(5/2))

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Mupad [B]
time = 1.36, size = 173, normalized size = 1.15 \begin {gather*} -\frac {\frac {2\,\left (a\,f-b\,e\right )}{5\,\left (c\,f-d\,e\right )}-\frac {2\,\left (e+f\,x\right )\,\left (a\,d\,f-b\,c\,f\right )}{3\,{\left (c\,f-d\,e\right )}^2}+\frac {2\,d\,{\left (e+f\,x\right )}^2\,\left (a\,d\,f-b\,c\,f\right )}{{\left (c\,f-d\,e\right )}^3}}{f\,{\left (e+f\,x\right )}^{5/2}}-\frac {2\,d^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {d}\,\sqrt {e+f\,x}\,\left (c^3\,f^3-3\,c^2\,d\,e\,f^2+3\,c\,d^2\,e^2\,f-d^3\,e^3\right )}{{\left (c\,f-d\,e\right )}^{7/2}}\right )\,\left (a\,d-b\,c\right )}{{\left (c\,f-d\,e\right )}^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)/((e + f*x)^(7/2)*(c + d*x)),x)

[Out]

- ((2*(a*f - b*e))/(5*(c*f - d*e)) - (2*(e + f*x)*(a*d*f - b*c*f))/(3*(c*f - d*e)^2) + (2*d*(e + f*x)^2*(a*d*f
- b*c*f))/(c*f - d*e)^3)/(f*(e + f*x)^(5/2)) - (2*d^(3/2)*atan((d^(1/2)*(e + f*x)^(1/2)*(c^3*f^3 - d^3*e^3 +
3*c*d^2*e^2*f - 3*c^2*d*e*f^2))/(c*f - d*e)^(7/2))*(a*d - b*c))/(c*f - d*e)^(7/2)

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