3.270 \(\int F^{a+b (c+d x)^2} (c+d x)^6 \, dx\)

Optimal. Leaf size=145 \[ -\frac{15 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}-\frac{5 (c+d x)^3 F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac{15 (c+d x) F^{a+b (c+d x)^2}}{8 b^3 d \log ^3(F)}+\frac{(c+d x)^5 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

[Out]

(-15*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(16*b^(7/2)*d*Log[F]^(7/2)) + (15*F^(a + b*(c + d*x)^2
)*(c + d*x))/(8*b^3*d*Log[F]^3) - (5*F^(a + b*(c + d*x)^2)*(c + d*x)^3)/(4*b^2*d*Log[F]^2) + (F^(a + b*(c + d*
x)^2)*(c + d*x)^5)/(2*b*d*Log[F])

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Rubi [A]  time = 0.230178, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2212, 2204} \[ -\frac{15 \sqrt{\pi } F^a \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}-\frac{5 (c+d x)^3 F^{a+b (c+d x)^2}}{4 b^2 d \log ^2(F)}+\frac{15 (c+d x) F^{a+b (c+d x)^2}}{8 b^3 d \log ^3(F)}+\frac{(c+d x)^5 F^{a+b (c+d x)^2}}{2 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)*(c + d*x)^6,x]

[Out]

(-15*F^a*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(16*b^(7/2)*d*Log[F]^(7/2)) + (15*F^(a + b*(c + d*x)^2
)*(c + d*x))/(8*b^3*d*Log[F]^3) - (5*F^(a + b*(c + d*x)^2)*(c + d*x)^3)/(4*b^2*d*Log[F]^2) + (F^(a + b*(c + d*
x)^2)*(c + d*x)^5)/(2*b*d*Log[F])

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rubi steps

\begin{align*} \int F^{a+b (c+d x)^2} (c+d x)^6 \, dx &=\frac{F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}-\frac{5 \int F^{a+b (c+d x)^2} (c+d x)^4 \, dx}{2 b \log (F)}\\ &=-\frac{5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}+\frac{15 \int F^{a+b (c+d x)^2} (c+d x)^2 \, dx}{4 b^2 \log ^2(F)}\\ &=\frac{15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac{5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}-\frac{15 \int F^{a+b (c+d x)^2} \, dx}{8 b^3 \log ^3(F)}\\ &=-\frac{15 F^a \sqrt{\pi } \text{erfi}\left (\sqrt{b} (c+d x) \sqrt{\log (F)}\right )}{16 b^{7/2} d \log ^{\frac{7}{2}}(F)}+\frac{15 F^{a+b (c+d x)^2} (c+d x)}{8 b^3 d \log ^3(F)}-\frac{5 F^{a+b (c+d x)^2} (c+d x)^3}{4 b^2 d \log ^2(F)}+\frac{F^{a+b (c+d x)^2} (c+d x)^5}{2 b d \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.144107, size = 126, normalized size = 0.87 \[ \frac{F^a \left (-\frac{15 \sqrt{\pi } \text{Erfi}\left (\sqrt{b} \sqrt{\log (F)} (c+d x)\right )}{b^{5/2} \log ^{\frac{5}{2}}(F)}+\frac{30 (c+d x) F^{b (c+d x)^2}}{b^2 \log ^2(F)}+8 (c+d x)^5 F^{b (c+d x)^2}-\frac{20 (c+d x)^3 F^{b (c+d x)^2}}{b \log (F)}\right )}{16 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)*(c + d*x)^6,x]

[Out]

(F^a*(8*F^(b*(c + d*x)^2)*(c + d*x)^5 - (15*Sqrt[Pi]*Erfi[Sqrt[b]*(c + d*x)*Sqrt[Log[F]]])/(b^(5/2)*Log[F]^(5/
2)) + (30*F^(b*(c + d*x)^2)*(c + d*x))/(b^2*Log[F]^2) - (20*F^(b*(c + d*x)^2)*(c + d*x)^3)/(b*Log[F])))/(16*b*
d*Log[F])

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Maple [B]  time = 0.081, size = 561, normalized size = 3.9 \begin{align*}{\frac{5\,c{d}^{3}{x}^{4}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+5\,{\frac{{d}^{2}{c}^{2}{x}^{3}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{b\ln \left ( F \right ) }}+5\,{\frac{d{c}^{3}{x}^{2}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{b\ln \left ( F \right ) }}-{\frac{15\,cd{x}^{2}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}}+{\frac{15\,x{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{8\, \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}}}+{\frac{15\,\sqrt{\pi }{F}^{a}}{16\,d \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}}{\it Erf} \left ( -d\sqrt{-b\ln \left ( F \right ) }x+{bc\ln \left ( F \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}} \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}}+{\frac{{d}^{4}{x}^{5}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+{\frac{5\,{c}^{4}x{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,b\ln \left ( F \right ) }}+{\frac{{c}^{5}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{2\,d\ln \left ( F \right ) b}}-{\frac{5\,{c}^{3}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\,d \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}}-{\frac{15\,{c}^{2}x{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}}+{\frac{15\,c{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{8\,d \left ( \ln \left ( F \right ) \right ) ^{3}{b}^{3}}}-{\frac{5\,{d}^{2}{x}^{3}{F}^{b{d}^{2}{x}^{2}}{F}^{2\,bcdx}{F}^{{c}^{2}b}{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x)

[Out]

5/2*d^3*c/ln(F)/b*x^4*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+5*d^2*c^2/ln(F)/b*x^3*F^(b*d^2*x^2)*F^(2*b*c*d
*x)*F^(c^2*b)*F^a+5*d*c^3/ln(F)/b*x^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a-15/4*d*c/ln(F)^2/b^2*x^2*F^(b*
d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+15/8/ln(F)^3/b^3*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+15/16/d/ln(F
)^3/b^3*Pi^(1/2)*F^a/(-b*ln(F))^(1/2)*erf(-d*(-b*ln(F))^(1/2)*x+b*c*ln(F)/(-b*ln(F))^(1/2))+1/2*d^4/ln(F)/b*x^
5*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+5/2*c^4/ln(F)/b*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+1/2/d*
c^5/ln(F)/b*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a-5/4/d*c^3/ln(F)^2/b^2*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2
*b)*F^a-15/4*c^2/ln(F)^2/b^2*x*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a+15/8/d*c/ln(F)^3/b^3*F^(b*d^2*x^2)*F^
(2*b*c*d*x)*F^(c^2*b)*F^a-5/4*d^2/ln(F)^2/b^2*x^3*F^(b*d^2*x^2)*F^(2*b*c*d*x)*F^(c^2*b)*F^a

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Maxima [B]  time = 2.16137, size = 2712, normalized size = 18.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="maxima")

[Out]

-3*(sqrt(pi)*(b*d^2*x + b*c*d)*b*c*d*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^2/((b*d^2*log
(F))^(3/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - F^((b*d^2*x + b*c*d)^2/(b*d^2))*b*d^2*log(F)/(b*d^2*lo
g(F))^(3/2))*F^a*c^5*d/sqrt(b*d^2*log(F)) + 15/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^2*c^2*d^2*(erf(sqrt(-(b*d^2*x +
b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^3/((b*d^2*log(F))^(5/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 2*
F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^2*c*d^3*log(F)^2/(b*d^2*log(F))^(5/2) - (b*d^2*x + b*c*d)^3*gamma(3/2, -(b*d
^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/((b*d^2*log(F))^(5/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)))*F
^a*c^4*d^2/sqrt(b*d^2*log(F)) - 10*(sqrt(pi)*(b*d^2*x + b*c*d)*b^3*c^3*d^3*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(
F)/(b*d^2))) - 1)*log(F)^4/((b*d^2*log(F))^(7/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 3*F^((b*d^2*x +
b*c*d)^2/(b*d^2))*b^3*c^2*d^4*log(F)^3/(b*d^2*log(F))^(7/2) - 3*(b*d^2*x + b*c*d)^3*b*c*d*gamma(3/2, -(b*d^2*x
+ b*c*d)^2*log(F)/(b*d^2))*log(F)^4/((b*d^2*log(F))^(7/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + b^2*
d^4*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^2/(b*d^2*log(F))^(7/2))*F^a*c^3*d^3/sqrt(b*d^2*log(F)
) + 15/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^4*c^4*d^4*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 1)*log(F)^5
/((b*d^2*log(F))^(9/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 4*F^((b*d^2*x + b*c*d)^2/(b*d^2))*b^4*c^3*
d^5*log(F)^4/(b*d^2*log(F))^(9/2) - 6*(b*d^2*x + b*c*d)^3*b^2*c^2*d^2*gamma(3/2, -(b*d^2*x + b*c*d)^2*log(F)/(
b*d^2))*log(F)^5/((b*d^2*log(F))^(9/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 4*b^3*c*d^5*gamma(2, -(b
*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/(b*d^2*log(F))^(9/2) - (b*d^2*x + b*c*d)^5*gamma(5/2, -(b*d^2*x + b
*c*d)^2*log(F)/(b*d^2))*log(F)^5/((b*d^2*log(F))^(9/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2)))*F^a*c^2*d
^4/sqrt(b*d^2*log(F)) - 3*(sqrt(pi)*(b*d^2*x + b*c*d)*b^5*c^5*d^5*(erf(sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2
))) - 1)*log(F)^6/((b*d^2*log(F))^(11/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 5*F^((b*d^2*x + b*c*d)^2
/(b*d^2))*b^5*c^4*d^6*log(F)^5/(b*d^2*log(F))^(11/2) - 10*(b*d^2*x + b*c*d)^3*b^3*c^3*d^3*gamma(3/2, -(b*d^2*x
+ b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*d^2*log(F))^(11/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(3/2)) + 10*
b^4*c^2*d^6*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^4/(b*d^2*log(F))^(11/2) - b^3*d^6*gamma(3, -(
b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^3/(b*d^2*log(F))^(11/2) - 5*(b*d^2*x + b*c*d)^5*b*c*d*gamma(5/2, -(b
*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^6/((b*d^2*log(F))^(11/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(5/2))
)*F^a*c*d^5/sqrt(b*d^2*log(F)) + 1/2*(sqrt(pi)*(b*d^2*x + b*c*d)*b^6*c^6*d^6*(erf(sqrt(-(b*d^2*x + b*c*d)^2*lo
g(F)/(b*d^2))) - 1)*log(F)^7/((b*d^2*log(F))^(13/2)*sqrt(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))) - 6*F^((b*d^2*x
+ b*c*d)^2/(b*d^2))*b^6*c^5*d^7*log(F)^6/(b*d^2*log(F))^(13/2) - 15*(b*d^2*x + b*c*d)^3*b^4*c^4*d^4*gamma(3/2
, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^7/((b*d^2*log(F))^(13/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(
3/2)) + 20*b^5*c^3*d^7*gamma(2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^5/(b*d^2*log(F))^(13/2) - 6*b^4*c*
d^7*gamma(3, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^4/(b*d^2*log(F))^(13/2) - 15*(b*d^2*x + b*c*d)^5*b^2*
c^2*d^2*gamma(5/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^7/((b*d^2*log(F))^(13/2)*(-(b*d^2*x + b*c*d)^2*
log(F)/(b*d^2))^(5/2)) - (b*d^2*x + b*c*d)^7*gamma(7/2, -(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))*log(F)^7/((b*d^2*
log(F))^(13/2)*(-(b*d^2*x + b*c*d)^2*log(F)/(b*d^2))^(7/2)))*F^a*d^6/sqrt(b*d^2*log(F)) + 1/2*sqrt(pi)*F^(b*c^
2 + a)*c^6*erf(sqrt(-b*log(F))*d*x - b*c*log(F)/sqrt(-b*log(F)))/(sqrt(-b*log(F))*F^(b*c^2)*d)

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Fricas [A]  time = 1.52384, size = 493, normalized size = 3.4 \begin{align*} \frac{15 \, \sqrt{\pi } \sqrt{-b d^{2} \log \left (F\right )} F^{a} \operatorname{erf}\left (\frac{\sqrt{-b d^{2} \log \left (F\right )}{\left (d x + c\right )}}{d}\right ) + 2 \,{\left (4 \,{\left (b^{3} d^{6} x^{5} + 5 \, b^{3} c d^{5} x^{4} + 10 \, b^{3} c^{2} d^{4} x^{3} + 10 \, b^{3} c^{3} d^{3} x^{2} + 5 \, b^{3} c^{4} d^{2} x + b^{3} c^{5} d\right )} \log \left (F\right )^{3} - 10 \,{\left (b^{2} d^{4} x^{3} + 3 \, b^{2} c d^{3} x^{2} + 3 \, b^{2} c^{2} d^{2} x + b^{2} c^{3} d\right )} \log \left (F\right )^{2} + 15 \,{\left (b d^{2} x + b c d\right )} \log \left (F\right )\right )} F^{b d^{2} x^{2} + 2 \, b c d x + b c^{2} + a}}{16 \, b^{4} d^{2} \log \left (F\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="fricas")

[Out]

1/16*(15*sqrt(pi)*sqrt(-b*d^2*log(F))*F^a*erf(sqrt(-b*d^2*log(F))*(d*x + c)/d) + 2*(4*(b^3*d^6*x^5 + 5*b^3*c*d
^5*x^4 + 10*b^3*c^2*d^4*x^3 + 10*b^3*c^3*d^3*x^2 + 5*b^3*c^4*d^2*x + b^3*c^5*d)*log(F)^3 - 10*(b^2*d^4*x^3 + 3
*b^2*c*d^3*x^2 + 3*b^2*c^2*d^2*x + b^2*c^3*d)*log(F)^2 + 15*(b*d^2*x + b*c*d)*log(F))*F^(b*d^2*x^2 + 2*b*c*d*x
+ b*c^2 + a))/(b^4*d^2*log(F)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**2)*(d*x+c)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.27098, size = 178, normalized size = 1.23 \begin{align*} \frac{{\left (4 \, b^{2} d^{4}{\left (x + \frac{c}{d}\right )}^{5} \log \left (F\right )^{2} - 10 \, b d^{2}{\left (x + \frac{c}{d}\right )}^{3} \log \left (F\right ) + 15 \, x + \frac{15 \, c}{d}\right )} e^{\left (b d^{2} x^{2} \log \left (F\right ) + 2 \, b c d x \log \left (F\right ) + b c^{2} \log \left (F\right ) + a \log \left (F\right )\right )}}{8 \, b^{3} \log \left (F\right )^{3}} + \frac{15 \, \sqrt{\pi } F^{a} \operatorname{erf}\left (-\sqrt{-b \log \left (F\right )} d{\left (x + \frac{c}{d}\right )}\right )}{16 \, \sqrt{-b \log \left (F\right )} b^{3} d \log \left (F\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)*(d*x+c)^6,x, algorithm="giac")

[Out]

1/8*(4*b^2*d^4*(x + c/d)^5*log(F)^2 - 10*b*d^2*(x + c/d)^3*log(F) + 15*x + 15*c/d)*e^(b*d^2*x^2*log(F) + 2*b*c
*d*x*log(F) + b*c^2*log(F) + a*log(F))/(b^3*log(F)^3) + 15/16*sqrt(pi)*F^a*erf(-sqrt(-b*log(F))*d*(x + c/d))/(
sqrt(-b*log(F))*b^3*d*log(F)^3)