\(\int (a+b \text {sech}^2(c+d x)) \tanh ^4(c+d x) \, dx\) [102]

Optimal result

Integrand size = 21, antiderivative size = 48 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=a x-\frac {a \tanh (c+d x)}{d}-\frac {a \tanh ^3(c+d x)}{3 d}+\frac {b \tanh ^5(c+d x)}{5 d} \] Output:

a*x-a*tanh(d*x+c)/d-1/3*a*tanh(d*x+c)^3/d+1/5*b*tanh(d*x+c)^5/d
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\frac {a \text {arctanh}(\tanh (c+d x))}{d}-\frac {a \tanh (c+d x)}{d}-\frac {a \tanh ^3(c+d x)}{3 d}+\frac {b \tanh ^5(c+d x)}{5 d} \] Input:

Integrate[(a + b*Sech[c + d*x]^2)*Tanh[c + d*x]^4,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4629, 2075, 363, 254, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \tanh ^4(c+d x) \left (a+b \text {sech}^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (i c+i d x)^4 \left (a+b \sec (i c+i d x)^2\right )dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (a+b \left (1-\tanh ^2(c+d x)\right )\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\tanh ^4(c+d x) \left (-b \tanh ^2(c+d x)+a+b\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {a \int \frac {\tanh ^4(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {1}{5} b \tanh ^5(c+d x)}{d}\)

\(\Big \downarrow \) 254

\(\displaystyle \frac {a \int \left (-\tanh ^2(c+d x)+\frac {1}{1-\tanh ^2(c+d x)}-1\right )d\tanh (c+d x)+\frac {1}{5} b \tanh ^5(c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a \left (\text {arctanh}(\tanh (c+d x))-\frac {1}{3} \tanh ^3(c+d x)-\tanh (c+d x)\right )+\frac {1}{5} b \tanh ^5(c+d x)}{d}\)

Input:

Int[(a + b*Sech[c + d*x]^2)*Tanh[c + d*x]^4,x]
 

Output:

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

Defintions of rubi rules used

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2075

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*Expa 
ndToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{e, m, p, q}, x] && Binomi 
alQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  ! 
BinomialMatchQ[{u, v}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4629

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f 
_.)*(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim 
p[ff/f   Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2*x^2 
)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [A] (verified)

Time = 4.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.29

method	result	size
parts	\(\frac {a \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (-1+\tanh \left (d x +c \right )\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {b \tanh \left (d x +c \right )^{5}}{5 d}\)	\(62\)
derivativedivides	\(\frac {a \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )+b \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(98\)
default	\(\frac {a \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )+b \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(98\)
risch	\(x a +\frac {4 \,{\mathrm e}^{8 d x +8 c} a -2 \,{\mathrm e}^{8 d x +8 c} b +12 \,{\mathrm e}^{6 d x +6 c} a +\frac {44 \,{\mathrm e}^{4 d x +4 c} a}{3}-4 \,{\mathrm e}^{4 d x +4 c} b +\frac {28 a \,{\mathrm e}^{2 d x +2 c}}{3}+\frac {8 a}{3}-\frac {2 b}{5}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\)	\(102\)

Input:

int((a+b*sech(d*x+c)^2)*tanh(d*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (44) = 88\).

Time = 0.19 (sec) , antiderivative size = 327, normalized size of antiderivative = 6.81 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\frac {{\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (20 \, a - 3 \, b\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (2 \, {\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, a d x + 20 \, a - 3 \, b\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (20 \, a - 3 \, b\right )} \cosh \left (d x + c\right )^{4} + 3 \, {\left (8 \, a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a - 6 \, b\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \] Input:

integrate((a+b*sech(d*x+c)^2)*tanh(d*x+c)^4,x, algorithm="fricas")
 

Output:

1/15*((15*a*d*x + 20*a - 3*b)*cosh(d*x + c)^5 + 5*(15*a*d*x + 20*a - 3*b)* 
cosh(d*x + c)*sinh(d*x + c)^4 - (20*a - 3*b)*sinh(d*x + c)^5 + 5*(15*a*d*x 
 + 20*a - 3*b)*cosh(d*x + c)^3 - 5*(2*(20*a - 3*b)*cosh(d*x + c)^2 + 8*a + 
 3*b)*sinh(d*x + c)^3 + 5*(2*(15*a*d*x + 20*a - 3*b)*cosh(d*x + c)^3 + 3*( 
15*a*d*x + 20*a - 3*b)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*a*d*x + 20* 
a - 3*b)*cosh(d*x + c) - 5*((20*a - 3*b)*cosh(d*x + c)^4 + 3*(8*a + 3*b)*c 
osh(d*x + c)^2 + 4*a - 6*b)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5*d*cosh(d 
*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3 
*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))
 

Sympy [F]

\[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\int \left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right ) \tanh ^{4}{\left (c + d x \right )}\, dx \] Input:

integrate((a+b*sech(d*x+c)**2)*tanh(d*x+c)**4,x)
 

Output:

Integral((a + b*sech(c + d*x)**2)*tanh(c + d*x)**4, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (44) = 88\).

Time = 0.04 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.92 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\frac {b \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac {1}{3} \, a {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \] Input:

integrate((a+b*sech(d*x+c)^2)*tanh(d*x+c)^4,x, algorithm="maxima")
                                                                                    
                                                                                    
 

Output:

1/5*b*tanh(d*x + c)^5/d + 1/3*a*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e 
^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d 
*x - 6*c) + 1)))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 108 vs. \(2 (44) = 88\).

Time = 0.15 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.25 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\frac {15 \, {\left (d x + c\right )} a + \frac {2 \, {\left (30 \, a e^{\left (8 \, d x + 8 \, c\right )} - 15 \, b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a e^{\left (6 \, d x + 6 \, c\right )} + 110 \, a e^{\left (4 \, d x + 4 \, c\right )} - 30 \, b e^{\left (4 \, d x + 4 \, c\right )} + 70 \, a e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a - 3 \, b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \] Input:

integrate((a+b*sech(d*x+c)^2)*tanh(d*x+c)^4,x, algorithm="giac")
 

Output:

1/15*(15*(d*x + c)*a + 2*(30*a*e^(8*d*x + 8*c) - 15*b*e^(8*d*x + 8*c) + 90 
*a*e^(6*d*x + 6*c) + 110*a*e^(4*d*x + 4*c) - 30*b*e^(4*d*x + 4*c) + 70*a*e 
^(2*d*x + 2*c) + 20*a - 3*b)/(e^(2*d*x + 2*c) + 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 2.32 (sec) , antiderivative size = 433, normalized size of antiderivative = 9.02 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=a\,x+\frac {\frac {2\,\left (2\,a-3\,b\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {2\,\left (2\,a-b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-3\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}+\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {2\,\left (a+b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-3\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {2\,\left (2\,a-b\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:

int(tanh(c + d*x)^4*(a + b/cosh(c + d*x)^2),x)
 

Output:

a*x + ((2*(2*a - 3*b))/(15*d) + (4*exp(2*c + 2*d*x)*(a + b))/(5*d) + (2*ex 
p(4*c + 4*d*x)*(2*a - b))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) 
+ exp(6*c + 6*d*x) + 1) + ((2*(2*a - b))/(5*d) + (8*exp(2*c + 2*d*x)*(a + 
b))/(5*d) + (8*exp(6*c + 6*d*x)*(a + b))/(5*d) + (4*exp(4*c + 4*d*x)*(2*a 
- 3*b))/(5*d) + (2*exp(8*c + 8*d*x)*(2*a - b))/(5*d))/(5*exp(2*c + 2*d*x) 
+ 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10* 
c + 10*d*x) + 1) + ((2*(a + b))/(5*d) + (2*exp(2*c + 2*d*x)*(2*a - b))/(5* 
d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) + ((2*(a + b))/(5*d) + (6* 
exp(4*c + 4*d*x)*(a + b))/(5*d) + (2*exp(2*c + 2*d*x)*(2*a - 3*b))/(5*d) + 
 (2*exp(6*c + 6*d*x)*(2*a - b))/(5*d))/(4*exp(2*c + 2*d*x) + 6*exp(4*c + 4 
*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) + (2*(2*a - b))/(5*d*(e 
xp(2*c + 2*d*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 407, normalized size of antiderivative = 8.48 \[ \int \left (a+b \text {sech}^2(c+d x)\right ) \tanh ^4(c+d x) \, dx=\frac {-5 e^{10 d x +10 c} \tanh \left (d x +c \right )^{3} a -15 e^{10 d x +10 c} \tanh \left (d x +c \right ) a +15 e^{10 d x +10 c} a d x +6 e^{10 d x +10 c} b -25 e^{8 d x +8 c} \tanh \left (d x +c \right )^{3} a -75 e^{8 d x +8 c} \tanh \left (d x +c \right ) a +75 e^{8 d x +8 c} a d x -50 e^{6 d x +6 c} \tanh \left (d x +c \right )^{3} a -150 e^{6 d x +6 c} \tanh \left (d x +c \right ) a +150 e^{6 d x +6 c} a d x +60 e^{6 d x +6 c} b -50 e^{4 d x +4 c} \tanh \left (d x +c \right )^{3} a -150 e^{4 d x +4 c} \tanh \left (d x +c \right ) a +150 e^{4 d x +4 c} a d x -25 e^{2 d x +2 c} \tanh \left (d x +c \right )^{3} a -75 e^{2 d x +2 c} \tanh \left (d x +c \right ) a +75 e^{2 d x +2 c} a d x +30 e^{2 d x +2 c} b -5 \tanh \left (d x +c \right )^{3} a -15 \tanh \left (d x +c \right ) a +15 a d x}{15 d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:

int((a+b*sech(d*x+c)^2)*tanh(d*x+c)^4,x)
 

Output:

( - 5*e**(10*c + 10*d*x)*tanh(c + d*x)**3*a - 15*e**(10*c + 10*d*x)*tanh(c 
 + d*x)*a + 15*e**(10*c + 10*d*x)*a*d*x + 6*e**(10*c + 10*d*x)*b - 25*e**( 
8*c + 8*d*x)*tanh(c + d*x)**3*a - 75*e**(8*c + 8*d*x)*tanh(c + d*x)*a + 75 
*e**(8*c + 8*d*x)*a*d*x - 50*e**(6*c + 6*d*x)*tanh(c + d*x)**3*a - 150*e** 
(6*c + 6*d*x)*tanh(c + d*x)*a + 150*e**(6*c + 6*d*x)*a*d*x + 60*e**(6*c + 
6*d*x)*b - 50*e**(4*c + 4*d*x)*tanh(c + d*x)**3*a - 150*e**(4*c + 4*d*x)*t 
anh(c + d*x)*a + 150*e**(4*c + 4*d*x)*a*d*x - 25*e**(2*c + 2*d*x)*tanh(c + 
 d*x)**3*a - 75*e**(2*c + 2*d*x)*tanh(c + d*x)*a + 75*e**(2*c + 2*d*x)*a*d 
*x + 30*e**(2*c + 2*d*x)*b - 5*tanh(c + d*x)**3*a - 15*tanh(c + d*x)*a + 1 
5*a*d*x)/(15*d*(e**(10*c + 10*d*x) + 5*e**(8*c + 8*d*x) + 10*e**(6*c + 6*d 
*x) + 10*e**(4*c + 4*d*x) + 5*e**(2*c + 2*d*x) + 1))