3.38 \(\int \frac{\text{csch}^2(c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)

Optimal. Leaf size=92 \[ \frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 d (a+b)^{5/2}}-\frac{3 \coth (c+d x)}{2 d (a+b)^2}+\frac{\coth (c+d x)}{2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

(3*Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*(a + b)^(5/2)*d) - (3*Coth[c + d*x])/(2*(a + b)^2*
d) + Coth[c + d*x]/(2*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

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Rubi [A]  time = 0.0844725, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4132, 290, 325, 208} \[ \frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 d (a+b)^{5/2}}-\frac{3 \coth (c+d x)}{2 d (a+b)^2}+\frac{\coth (c+d x)}{2 d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^2/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

(3*Sqrt[b]*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*(a + b)^(5/2)*d) - (3*Coth[c + d*x])/(2*(a + b)^2*
d) + Coth[c + d*x]/(2*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\text{csch}^2(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\coth (c+d x)}{2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 (a+b) d}\\ &=-\frac{3 \coth (c+d x)}{2 (a+b)^2 d}+\frac{\coth (c+d x)}{2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 (a+b)^2 d}\\ &=\frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tanh (c+d x)}{\sqrt{a+b}}\right )}{2 (a+b)^{5/2} d}-\frac{3 \coth (c+d x)}{2 (a+b)^2 d}+\frac{\coth (c+d x)}{2 (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 2.46124, size = 220, normalized size = 2.39 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (2 \text{csch}(c) \sinh (d x) \text{csch}(c+d x) (a \cosh (2 (c+d x))+a+2 b)+\frac{3 b (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac{(\cosh (2 c)-\sinh (2 c)) \text{sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cosh (c)-\sinh (c))^4}}-\frac{b (a+2 b) \tanh (2 c)}{a}+b \text{sech}(2 c) \sinh (2 d x)\right )}{8 d (a+b)^2 \left (a+b \text{sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^2/(a + b*Sech[c + d*x]^2)^2,x]

[Out]

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^4*((3*b*ArcTanh[(Sech[d*x]*(Cosh[2*c] - Sinh[2*c])*((a + 2*b)*S
inh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])*
(Cosh[2*c] - Sinh[2*c]))/(Sqrt[a + b]*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + 2*(a + 2*b + a*Cosh[2*(c + d*x)])*Csch[
c]*Csch[c + d*x]*Sinh[d*x] + b*Sech[2*c]*Sinh[2*d*x] - (b*(a + 2*b)*Tanh[2*c])/a))/(8*(a + b)^2*d*(a + b*Sech[
c + d*x]^2)^2)

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Maple [B]  time = 0.085, size = 310, normalized size = 3.4 \begin{align*} -{\frac{1}{2\,d \left ({a}^{2}+2\,ab+{b}^{2} \right ) }\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{b}{d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}+{\frac{b}{d \left ( a+b \right ) ^{2}}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) ^{-1}}+{\frac{3}{4\,d}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}-{\frac{3}{4\,d}\sqrt{b}\ln \left ( \sqrt{a+b} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( 1/2\,dx+c/2 \right ) \sqrt{b}+\sqrt{a+b} \right ) \left ( a+b \right ) ^{-{\frac{5}{2}}}}-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x)

[Out]

-1/2/d/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c)+1/d*b/(a+b)^2/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*ta
nh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/2*c)^3+1/d*b/(a+b)^2/(tanh(1/2*d*x+1/2*c)^
4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)*tanh(1/2*d*x+1/2*c)+3/4/d
*b^(1/2)/(a+b)^(5/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))-3/4/d*b^(
1/2)/(a+b)^(5/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2))-1/2/d/(a+b)^2
/tanh(1/2*d*x+1/2*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.46985, size = 5806, normalized size = 63.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

[-1/4*(4*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^4 + 16*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 4*(2
*a^2 + a*b + 2*b^2)*sinh(d*x + c)^4 + 8*(2*a^2 + 4*a*b - b^2)*cosh(d*x + c)^2 + 8*(3*(2*a^2 + a*b + 2*b^2)*cos
h(d*x + c)^2 + 2*a^2 + 4*a*b - b^2)*sinh(d*x + c)^2 - 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x +
c)^5 + a^2*sinh(d*x + c)^6 + (a^2 + 4*a*b)*cosh(d*x + c)^4 + (15*a^2*cosh(d*x + c)^2 + a^2 + 4*a*b)*sinh(d*x +
c)^4 + 4*(5*a^2*cosh(d*x + c)^3 + (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c)^
2 + (15*a^2*cosh(d*x + c)^4 + 6*(a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - 4*a*b)*sinh(d*x + c)^2 - a^2 + 2*(3*a^2*
cosh(d*x + c)^5 + 2*(a^2 + 4*a*b)*cosh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c))*sqrt(b/(a + b)
)*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(
d*x + c)^2 + 2*(3*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*cosh(d*x +
c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) - 4*((a^2 + a*b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x +
c)*sinh(d*x + c) + (a^2 + a*b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c)^4 + 4
*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 +
a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 8*a^2 - 4*a*b
+ 16*((2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^3 + (2*a^2 + 4*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + 2*a
^3*b + a^2*b^2)*d*cosh(d*x + c)^6 + 6*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5 + (a^4 + 2*a^3
*b + a^2*b^2)*d*sinh(d*x + c)^6 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^4 + (15*(a^4 + 2*a^3*b
+ a^2*b^2)*d*cosh(d*x + c)^2 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^4 - (a^4 + 6*a^3*b + 9*
a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^3 + (a^4 + 6*a^3*b + 9*a
^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^4 + 6*(a^4
+ 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^2
- (a^4 + 2*a^3*b + a^2*b^2)*d + 2*(3*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^5 + 2*(a^4 + 6*a^3*b + 9*a^2*b^
2 + 4*a*b^3)*d*cosh(d*x + c)^3 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)), -1/2*(
2*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^4 + 8*(2*a^2 + a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(2*a^2 + a
*b + 2*b^2)*sinh(d*x + c)^4 + 4*(2*a^2 + 4*a*b - b^2)*cosh(d*x + c)^2 + 4*(3*(2*a^2 + a*b + 2*b^2)*cosh(d*x +
c)^2 + 2*a^2 + 4*a*b - b^2)*sinh(d*x + c)^2 - 3*(a^2*cosh(d*x + c)^6 + 6*a^2*cosh(d*x + c)*sinh(d*x + c)^5 + a
^2*sinh(d*x + c)^6 + (a^2 + 4*a*b)*cosh(d*x + c)^4 + (15*a^2*cosh(d*x + c)^2 + a^2 + 4*a*b)*sinh(d*x + c)^4 +
4*(5*a^2*cosh(d*x + c)^3 + (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c)^2 + (15*
a^2*cosh(d*x + c)^4 + 6*(a^2 + 4*a*b)*cosh(d*x + c)^2 - a^2 - 4*a*b)*sinh(d*x + c)^2 - a^2 + 2*(3*a^2*cosh(d*x
+ c)^5 + 2*(a^2 + 4*a*b)*cosh(d*x + c)^3 - (a^2 + 4*a*b)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/(a + b))*arcta
n(1/2*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b)
+ 4*a^2 - 2*a*b + 8*((2*a^2 + a*b + 2*b^2)*cosh(d*x + c)^3 + (2*a^2 + 4*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c
))/((a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^6 + 6*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^5
+ (a^4 + 2*a^3*b + a^2*b^2)*d*sinh(d*x + c)^6 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^4 + (15*
(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^2 + (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*sinh(d*x + c)^4 - (a^4
+ 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 + 4*(5*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^3 + (a^4 +
6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x + c)^3 + (15*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x +
c)^4 + 6*(a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^2 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d)*si
nh(d*x + c)^2 - (a^4 + 2*a^3*b + a^2*b^2)*d + 2*(3*(a^4 + 2*a^3*b + a^2*b^2)*d*cosh(d*x + c)^5 + 2*(a^4 + 6*a^
3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c)^3 - (a^4 + 6*a^3*b + 9*a^2*b^2 + 4*a*b^3)*d*cosh(d*x + c))*sinh(d*x
+ c))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}^{2}{\left (c + d x \right )}}{\left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**2/(a+b*sech(d*x+c)**2)**2,x)

[Out]

Integral(csch(c + d*x)**2/(a + b*sech(c + d*x)**2)**2, x)

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Giac [B]  time = 1.36499, size = 328, normalized size = 3.57 \begin{align*} \frac{3 \, b \arctan \left (\frac{a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt{-a b - b^{2}}}\right )}{2 \,{\left (a^{2} d + 2 \, a b d + b^{2} d\right )} \sqrt{-a b - b^{2}}} - \frac{2 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + a b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} - a b}{{\left (a^{3} d + 2 \, a^{2} b d + a b^{2} d\right )}{\left (a e^{\left (6 \, d x + 6 \, c\right )} + a e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (4 \, d x + 4 \, c\right )} - a e^{\left (2 \, d x + 2 \, c\right )} - 4 \, b e^{\left (2 \, d x + 2 \, c\right )} - a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^2/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")

[Out]

3/2*b*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2*b)/sqrt(-a*b - b^2))/((a^2*d + 2*a*b*d + b^2*d)*sqrt(-a*b - b^2))
- (2*a^2*e^(4*d*x + 4*c) + a*b*e^(4*d*x + 4*c) + 2*b^2*e^(4*d*x + 4*c) + 4*a^2*e^(2*d*x + 2*c) + 8*a*b*e^(2*d*
x + 2*c) - 2*b^2*e^(2*d*x + 2*c) + 2*a^2 - a*b)/((a^3*d + 2*a^2*b*d + a*b^2*d)*(a*e^(6*d*x + 6*c) + a*e^(4*d*x
+ 4*c) + 4*b*e^(4*d*x + 4*c) - a*e^(2*d*x + 2*c) - 4*b*e^(2*d*x + 2*c) - a))