3.484 \(\int \frac{\coth ^5(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx\)
Optimal. Leaf size=126 \[ -\frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f}-\frac{(8 a-3 b) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{8 a^2 f}-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f} \]
[Out]
-((8*a^2 - 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]])/(8*a^(5/2)*f) - ((8*a - 3*b)*Csch[e +
f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2])/(8*a^2*f) - (Csch[e + f*x]^4*Sqrt[a + b*Sinh[e + f*x]^2])/(4*a*f)
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Rubi [A] time = 0.147265, antiderivative size = 126, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3194, 89, 78, 63, 208} \[ -\frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f}-\frac{(8 a-3 b) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{8 a^2 f}-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f} \]
Antiderivative was successfully verified.
[In]
Int[Coth[e + f*x]^5/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
-((8*a^2 - 8*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]])/(8*a^(5/2)*f) - ((8*a - 3*b)*Csch[e +
f*x]^2*Sqrt[a + b*Sinh[e + f*x]^2])/(8*a^2*f) - (Csch[e + f*x]^4*Sqrt[a + b*Sinh[e + f*x]^2])/(4*a*f)
Rule 3194
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rule 89
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Rule 78
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] || LtQ
[p, n]))))
Rule 63
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 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{\coth ^5(e+f x)}{\sqrt{a+b \sinh ^2(e+f x)}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(1+x)^2}{x^3 \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{2 f}\\ &=-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f}+\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (8 a-3 b)+2 a x}{x^2 \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{4 a f}\\ &=-\frac{(8 a-3 b) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{8 a^2 f}-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f}+\frac{\left (8 a^2-8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sinh ^2(e+f x)\right )}{16 a^2 f}\\ &=-\frac{(8 a-3 b) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{8 a^2 f}-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f}+\frac{\left (8 a^2-8 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^2(e+f x)}\right )}{8 a^2 b f}\\ &=-\frac{\left (8 a^2-8 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )}{8 a^{5/2} f}-\frac{(8 a-3 b) \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{8 a^2 f}-\frac{\text{csch}^4(e+f x) \sqrt{a+b \sinh ^2(e+f x)}}{4 a f}\\ \end{align*}
Mathematica [A] time = 0.362039, size = 100, normalized size = 0.79 \[ \frac{\left (-8 a^2+8 a b-3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sinh ^2(e+f x)}}{\sqrt{a}}\right )+\sqrt{a} \text{csch}^2(e+f x) \sqrt{a+b \sinh ^2(e+f x)} \left (-2 a \text{csch}^2(e+f x)-8 a+3 b\right )}{8 a^{5/2} f} \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[e + f*x]^5/Sqrt[a + b*Sinh[e + f*x]^2],x]
[Out]
((-8*a^2 + 8*a*b - 3*b^2)*ArcTanh[Sqrt[a + b*Sinh[e + f*x]^2]/Sqrt[a]] + Sqrt[a]*Csch[e + f*x]^2*(-8*a + 3*b -
2*a*Csch[e + f*x]^2)*Sqrt[a + b*Sinh[e + f*x]^2])/(8*a^(5/2)*f)
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Maple [C] time = 0.102, size = 54, normalized size = 0.4 \begin{align*}{\frac{1}{f}\mbox{{\tt ` int/indef0`}} \left ({ \left ( \left ( \sinh \left ( fx+e \right ) \right ) ^{-1}+2\, \left ( \sinh \left ( fx+e \right ) \right ) ^{-3}+ \left ( \sinh \left ( fx+e \right ) \right ) ^{-5} \right ){\frac{1}{\sqrt{a+b \left ( \sinh \left ( fx+e \right ) \right ) ^{2}}}}},\sinh \left ( fx+e \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2),x)
[Out]
`int/indef0`((1/sinh(f*x+e)+2/sinh(f*x+e)^3+1/sinh(f*x+e)^5)/(a+b*sinh(f*x+e)^2)^(1/2),sinh(f*x+e))/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{5}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
integrate(coth(f*x + e)^5/sqrt(b*sinh(f*x + e)^2 + a), x)
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Fricas [B] time = 2.84606, size = 7682, normalized size = 60.97 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/16*(((8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^8 + 8*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)*sinh(f*x + e)^7 + (8
*a^2 - 8*a*b + 3*b^2)*sinh(f*x + e)^8 - 4*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^6 + 4*(7*(8*a^2 - 8*a*b + 3*b^
2)*cosh(f*x + e)^2 - 8*a^2 + 8*a*b - 3*b^2)*sinh(f*x + e)^6 + 8*(7*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 - 3
*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e))*sinh(f*x + e)^5 + 6*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 2*(35*(8
*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 - 30*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 + 24*a^2 - 24*a*b + 9*b^2)*
sinh(f*x + e)^4 + 8*(7*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^5 - 10*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 +
3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e))*sinh(f*x + e)^3 - 4*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 + 4*(7*(8
*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^6 - 15*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 9*(8*a^2 - 8*a*b + 3*b^2)
*cosh(f*x + e)^2 - 8*a^2 + 8*a*b - 3*b^2)*sinh(f*x + e)^2 + 8*a^2 - 8*a*b + 3*b^2 + 8*((8*a^2 - 8*a*b + 3*b^2)
*cosh(f*x + e)^7 - 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^5 + 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 - (8*
a^2 - 8*a*b + 3*b^2)*cosh(f*x + e))*sinh(f*x + e))*sqrt(a)*log((b*cosh(f*x + e)^4 + 4*b*cosh(f*x + e)*sinh(f*x
+ e)^3 + b*sinh(f*x + e)^4 + 2*(4*a - b)*cosh(f*x + e)^2 + 2*(3*b*cosh(f*x + e)^2 + 4*a - b)*sinh(f*x + e)^2
- 4*sqrt(2)*sqrt(a)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*
sinh(f*x + e) + sinh(f*x + e)^2))*(cosh(f*x + e) + sinh(f*x + e)) + 4*(b*cosh(f*x + e)^3 + (4*a - b)*cosh(f*x
+ e))*sinh(f*x + e) + b)/(cosh(f*x + e)^4 + 4*cosh(f*x + e)*sinh(f*x + e)^3 + sinh(f*x + e)^4 + 2*(3*cosh(f*x
+ e)^2 - 1)*sinh(f*x + e)^2 - 2*cosh(f*x + e)^2 + 4*(cosh(f*x + e)^3 - cosh(f*x + e))*sinh(f*x + e) + 1)) - 4*
sqrt(2)*((8*a^2 - 3*a*b)*cosh(f*x + e)^5 + 5*(8*a^2 - 3*a*b)*cosh(f*x + e)*sinh(f*x + e)^4 + (8*a^2 - 3*a*b)*s
inh(f*x + e)^5 - 2*(4*a^2 - 3*a*b)*cosh(f*x + e)^3 + 2*(5*(8*a^2 - 3*a*b)*cosh(f*x + e)^2 - 4*a^2 + 3*a*b)*sin
h(f*x + e)^3 + 2*(5*(8*a^2 - 3*a*b)*cosh(f*x + e)^3 - 3*(4*a^2 - 3*a*b)*cosh(f*x + e))*sinh(f*x + e)^2 + (8*a^
2 - 3*a*b)*cosh(f*x + e) + (5*(8*a^2 - 3*a*b)*cosh(f*x + e)^4 - 6*(4*a^2 - 3*a*b)*cosh(f*x + e)^2 + 8*a^2 - 3*
a*b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*
sinh(f*x + e) + sinh(f*x + e)^2)))/(a^3*f*cosh(f*x + e)^8 + 8*a^3*f*cosh(f*x + e)*sinh(f*x + e)^7 + a^3*f*sinh
(f*x + e)^8 - 4*a^3*f*cosh(f*x + e)^6 + 6*a^3*f*cosh(f*x + e)^4 + 4*(7*a^3*f*cosh(f*x + e)^2 - a^3*f)*sinh(f*x
+ e)^6 - 4*a^3*f*cosh(f*x + e)^2 + 8*(7*a^3*f*cosh(f*x + e)^3 - 3*a^3*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(3
5*a^3*f*cosh(f*x + e)^4 - 30*a^3*f*cosh(f*x + e)^2 + 3*a^3*f)*sinh(f*x + e)^4 + a^3*f + 8*(7*a^3*f*cosh(f*x +
e)^5 - 10*a^3*f*cosh(f*x + e)^3 + 3*a^3*f*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(7*a^3*f*cosh(f*x + e)^6 - 15*a^3
*f*cosh(f*x + e)^4 + 9*a^3*f*cosh(f*x + e)^2 - a^3*f)*sinh(f*x + e)^2 + 8*(a^3*f*cosh(f*x + e)^7 - 3*a^3*f*cos
h(f*x + e)^5 + 3*a^3*f*cosh(f*x + e)^3 - a^3*f*cosh(f*x + e))*sinh(f*x + e)), 1/8*(((8*a^2 - 8*a*b + 3*b^2)*co
sh(f*x + e)^8 + 8*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)*sinh(f*x + e)^7 + (8*a^2 - 8*a*b + 3*b^2)*sinh(f*x + e
)^8 - 4*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^6 + 4*(7*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 - 8*a^2 + 8*a*b
- 3*b^2)*sinh(f*x + e)^6 + 8*(7*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 - 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x
+ e))*sinh(f*x + e)^5 + 6*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 2*(35*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e
)^4 - 30*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 + 24*a^2 - 24*a*b + 9*b^2)*sinh(f*x + e)^4 + 8*(7*(8*a^2 - 8*
a*b + 3*b^2)*cosh(f*x + e)^5 - 10*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 + 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x
+ e))*sinh(f*x + e)^3 - 4*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 + 4*(7*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e
)^6 - 15*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^4 + 9*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^2 - 8*a^2 + 8*a*b -
3*b^2)*sinh(f*x + e)^2 + 8*a^2 - 8*a*b + 3*b^2 + 8*((8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^7 - 3*(8*a^2 - 8*a*
b + 3*b^2)*cosh(f*x + e)^5 + 3*(8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)^3 - (8*a^2 - 8*a*b + 3*b^2)*cosh(f*x + e)
)*sinh(f*x + e))*sqrt(-a)*arctan(1/2*sqrt(2)*sqrt(-a)*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(
cosh(f*x + e)^2 - 2*cosh(f*x + e)*sinh(f*x + e) + sinh(f*x + e)^2))/(a*cosh(f*x + e) + a*sinh(f*x + e))) - 2*s
qrt(2)*((8*a^2 - 3*a*b)*cosh(f*x + e)^5 + 5*(8*a^2 - 3*a*b)*cosh(f*x + e)*sinh(f*x + e)^4 + (8*a^2 - 3*a*b)*si
nh(f*x + e)^5 - 2*(4*a^2 - 3*a*b)*cosh(f*x + e)^3 + 2*(5*(8*a^2 - 3*a*b)*cosh(f*x + e)^2 - 4*a^2 + 3*a*b)*sinh
(f*x + e)^3 + 2*(5*(8*a^2 - 3*a*b)*cosh(f*x + e)^3 - 3*(4*a^2 - 3*a*b)*cosh(f*x + e))*sinh(f*x + e)^2 + (8*a^2
- 3*a*b)*cosh(f*x + e) + (5*(8*a^2 - 3*a*b)*cosh(f*x + e)^4 - 6*(4*a^2 - 3*a*b)*cosh(f*x + e)^2 + 8*a^2 - 3*a
*b)*sinh(f*x + e))*sqrt((b*cosh(f*x + e)^2 + b*sinh(f*x + e)^2 + 2*a - b)/(cosh(f*x + e)^2 - 2*cosh(f*x + e)*s
inh(f*x + e) + sinh(f*x + e)^2)))/(a^3*f*cosh(f*x + e)^8 + 8*a^3*f*cosh(f*x + e)*sinh(f*x + e)^7 + a^3*f*sinh(
f*x + e)^8 - 4*a^3*f*cosh(f*x + e)^6 + 6*a^3*f*cosh(f*x + e)^4 + 4*(7*a^3*f*cosh(f*x + e)^2 - a^3*f)*sinh(f*x
+ e)^6 - 4*a^3*f*cosh(f*x + e)^2 + 8*(7*a^3*f*cosh(f*x + e)^3 - 3*a^3*f*cosh(f*x + e))*sinh(f*x + e)^5 + 2*(35
*a^3*f*cosh(f*x + e)^4 - 30*a^3*f*cosh(f*x + e)^2 + 3*a^3*f)*sinh(f*x + e)^4 + a^3*f + 8*(7*a^3*f*cosh(f*x + e
)^5 - 10*a^3*f*cosh(f*x + e)^3 + 3*a^3*f*cosh(f*x + e))*sinh(f*x + e)^3 + 4*(7*a^3*f*cosh(f*x + e)^6 - 15*a^3*
f*cosh(f*x + e)^4 + 9*a^3*f*cosh(f*x + e)^2 - a^3*f)*sinh(f*x + e)^2 + 8*(a^3*f*cosh(f*x + e)^7 - 3*a^3*f*cosh
(f*x + e)^5 + 3*a^3*f*cosh(f*x + e)^3 - a^3*f*cosh(f*x + e))*sinh(f*x + e))]
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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(coth(f*x+e)**5/(a+b*sinh(f*x+e)**2)**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (f x + e\right )^{5}}{\sqrt{b \sinh \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(f*x+e)^5/(a+b*sinh(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
integrate(coth(f*x + e)^5/sqrt(b*sinh(f*x + e)^2 + a), x)