\(\int (c+d x) \text {sech}^3(a+b x) \, dx\) [38]

Optimal result

Integrand size = 14, antiderivative size = 102 \[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\frac {(c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{2 b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{2 b^2}+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \text {sech}(a+b x) \tanh (a+b x)}{2 b} \] Output:

(d*x+c)*arctan(exp(b*x+a))/b-1/2*I*d*polylog(2,-I*exp(b*x+a))/b^2+1/2*I*d* 
polylog(2,I*exp(b*x+a))/b^2+1/2*d*sech(b*x+a)/b^2+1/2*(d*x+c)*sech(b*x+a)* 
tanh(b*x+a)/b
 

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.53 \[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\frac {c \arctan (\sinh (a+b x))}{2 b}+\frac {i d \left (b x \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )-\operatorname {PolyLog}\left (2,-i e^{a+b x}\right )+\operatorname {PolyLog}\left (2,i e^{a+b x}\right )\right )}{2 b^2}+\frac {d \text {sech}(a) \text {sech}(a+b x) (\cosh (a)+b x \sinh (a))}{2 b^2}+\frac {d x \text {sech}(a) \text {sech}^2(a+b x) \sinh (b x)}{2 b}+\frac {c \text {sech}(a+b x) \tanh (a+b x)}{2 b} \] Input:

Integrate[(c + d*x)*Sech[a + b*x]^3,x]
 

Output:

(c*ArcTan[Sinh[a + b*x]])/(2*b) + ((I/2)*d*(b*x*(Log[1 - I*E^(a + b*x)] - 
Log[1 + I*E^(a + b*x)]) - PolyLog[2, (-I)*E^(a + b*x)] + PolyLog[2, I*E^(a 
 + b*x)]))/b^2 + (d*Sech[a]*Sech[a + b*x]*(Cosh[a] + b*x*Sinh[a]))/(2*b^2) 
 + (d*x*Sech[a]*Sech[a + b*x]^2*Sinh[b*x])/(2*b) + (c*Sech[a + b*x]*Tanh[a 
 + b*x])/(2*b)
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4673, 3042, 4668, 2715, 2838}

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.

\(\displaystyle \int (c+d x) \text {sech}^3(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x) \csc \left (i a+i b x+\frac {\pi }{2}\right )^3dx\)

\(\Big \downarrow \) 4673

\(\displaystyle \frac {1}{2} \int (c+d x) \text {sech}(a+b x)dx+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int (c+d x) \csc \left (i a+i b x+\frac {\pi }{2}\right )dx+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {1}{2} \left (-\frac {i d \int \log \left (1-i e^{a+b x}\right )dx}{b}+\frac {i d \int \log \left (1+i e^{a+b x}\right )dx}{b}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\right )+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b}\)

\(\Big \downarrow \) 2715

\(\displaystyle \frac {1}{2} \left (-\frac {i d \int e^{-a-b x} \log \left (1-i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {i d \int e^{-a-b x} \log \left (1+i e^{a+b x}\right )de^{a+b x}}{b^2}+\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}\right )+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b}\)

\(\Big \downarrow \) 2838

\(\displaystyle \frac {1}{2} \left (\frac {2 (c+d x) \arctan \left (e^{a+b x}\right )}{b}-\frac {i d \operatorname {PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac {i d \operatorname {PolyLog}\left (2,i e^{a+b x}\right )}{b^2}\right )+\frac {d \text {sech}(a+b x)}{2 b^2}+\frac {(c+d x) \tanh (a+b x) \text {sech}(a+b x)}{2 b}\)

Input:

Int[(c + d*x)*Sech[a + b*x]^3,x]
 

Output:

((2*(c + d*x)*ArcTan[E^(a + b*x)])/b - (I*d*PolyLog[2, (-I)*E^(a + b*x)])/ 
b^2 + (I*d*PolyLog[2, I*E^(a + b*x)])/b^2)/2 + (d*Sech[a + b*x])/(2*b^2) + 
 ((c + d*x)*Sech[a + b*x]*Tanh[a + b*x])/(2*b)
 

Defintions of rubi rules used

rule 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[1/(d*e*n*Log[F])   Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) 
))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
 

rule 2838

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 
, (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
 

rule 3042

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

rule 4668

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 4673

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> 
 Simp[(-b^2)*(c + d*x)*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), 
 x] + (-Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x] + S 
imp[b^2*((n - 2)/(n - 1))   Int[(c + d*x)*(b*Csc[e + f*x])^(n - 2), x], x]) 
 /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (87 ) = 174\).

Time = 0.56 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.12

method	result	size
risch	\(\frac {{\mathrm e}^{b x +a} \left (b d x \,{\mathrm e}^{2 b x +2 a}+b c \,{\mathrm e}^{2 b x +2 a}-d x b +{\mathrm e}^{2 b x +2 a} d -c b +d \right )}{b^{2} \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}+\frac {c \arctan \left ({\mathrm e}^{b x +a}\right )}{b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) x}{2 b}-\frac {i d \ln \left (1+i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) x}{2 b}+\frac {i d \ln \left (1-i {\mathrm e}^{b x +a}\right ) a}{2 b^{2}}-\frac {i d \operatorname {dilog}\left (1+i {\mathrm e}^{b x +a}\right )}{2 b^{2}}+\frac {i d \operatorname {dilog}\left (1-i {\mathrm e}^{b x +a}\right )}{2 b^{2}}-\frac {d a \arctan \left ({\mathrm e}^{b x +a}\right )}{b^{2}}\)	\(216\)

Input:

int((d*x+c)*sech(b*x+a)^3,x,method=_RETURNVERBOSE)
 

Output:

exp(b*x+a)*(b*d*x*exp(2*b*x+2*a)+b*c*exp(2*b*x+2*a)-d*x*b+exp(2*b*x+2*a)*d 
-c*b+d)/b^2/(1+exp(2*b*x+2*a))^2+1/b*c*arctan(exp(b*x+a))-1/2*I/b*d*ln(1+I 
*exp(b*x+a))*x-1/2*I/b^2*d*ln(1+I*exp(b*x+a))*a+1/2*I/b*d*ln(1-I*exp(b*x+a 
))*x+1/2*I/b^2*d*ln(1-I*exp(b*x+a))*a-1/2*I/b^2*d*dilog(1+I*exp(b*x+a))+1/ 
2*I/b^2*d*dilog(1-I*exp(b*x+a))-1/b^2*d*a*arctan(exp(b*x+a))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1267 vs. \(2 (81) = 162\).

Time = 0.13 (sec) , antiderivative size = 1267, normalized size of antiderivative = 12.42 \[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)*sech(b*x+a)^3,x, algorithm="fricas")
 

Output:

1/2*(2*(b*d*x + b*c + d)*cosh(b*x + a)^3 + 6*(b*d*x + b*c + d)*cosh(b*x + 
a)*sinh(b*x + a)^2 + 2*(b*d*x + b*c + d)*sinh(b*x + a)^3 - 2*(b*d*x + b*c 
- d)*cosh(b*x + a) + (I*d*cosh(b*x + a)^4 + 4*I*d*cosh(b*x + a)*sinh(b*x + 
 a)^3 + I*d*sinh(b*x + a)^4 + 2*I*d*cosh(b*x + a)^2 - 2*(-3*I*d*cosh(b*x + 
 a)^2 - I*d)*sinh(b*x + a)^2 - 4*(-I*d*cosh(b*x + a)^3 - I*d*cosh(b*x + a) 
)*sinh(b*x + a) + I*d)*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) + (-I*d*co 
sh(b*x + a)^4 - 4*I*d*cosh(b*x + a)*sinh(b*x + a)^3 - I*d*sinh(b*x + a)^4 
- 2*I*d*cosh(b*x + a)^2 - 2*(3*I*d*cosh(b*x + a)^2 + I*d)*sinh(b*x + a)^2 
- 4*(I*d*cosh(b*x + a)^3 + I*d*cosh(b*x + a))*sinh(b*x + a) - I*d)*dilog(- 
I*cosh(b*x + a) - I*sinh(b*x + a)) + ((I*b*c - I*a*d)*cosh(b*x + a)^4 - 4* 
(-I*b*c + I*a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (I*b*c - I*a*d)*sinh(b*x 
+ a)^4 - 2*(-I*b*c + I*a*d)*cosh(b*x + a)^2 - 2*(3*(-I*b*c + I*a*d)*cosh(b 
*x + a)^2 - I*b*c + I*a*d)*sinh(b*x + a)^2 + I*b*c - I*a*d - 4*((-I*b*c + 
I*a*d)*cosh(b*x + a)^3 + (-I*b*c + I*a*d)*cosh(b*x + a))*sinh(b*x + a))*lo 
g(cosh(b*x + a) + sinh(b*x + a) + I) + ((-I*b*c + I*a*d)*cosh(b*x + a)^4 - 
 4*(I*b*c - I*a*d)*cosh(b*x + a)*sinh(b*x + a)^3 + (-I*b*c + I*a*d)*sinh(b 
*x + a)^4 - 2*(I*b*c - I*a*d)*cosh(b*x + a)^2 - 2*(3*(I*b*c - I*a*d)*cosh( 
b*x + a)^2 + I*b*c - I*a*d)*sinh(b*x + a)^2 - I*b*c + I*a*d - 4*((I*b*c - 
I*a*d)*cosh(b*x + a)^3 + (I*b*c - I*a*d)*cosh(b*x + a))*sinh(b*x + a))*log 
(cosh(b*x + a) + sinh(b*x + a) - I) + ((-I*b*d*x - I*a*d)*cosh(b*x + a)...
 

Sympy [F]

\[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\int \left (c + d x\right ) \operatorname {sech}^{3}{\left (a + b x \right )}\, dx \] Input:

integrate((d*x+c)*sech(b*x+a)**3,x)
 

Output:

Integral((c + d*x)*sech(a + b*x)**3, x)
 

Maxima [F]

\[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {sech}\left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)*sech(b*x+a)^3,x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\int { {\left (d x + c\right )} \operatorname {sech}\left (b x + a\right )^{3} \,d x } \] Input:

integrate((d*x+c)*sech(b*x+a)^3,x, algorithm="giac")
 

Output:

integrate((d*x + c)*sech(b*x + a)^3, x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\int \frac {c+d\,x}{{\mathrm {cosh}\left (a+b\,x\right )}^3} \,d x \] Input:

int((c + d*x)/cosh(a + b*x)^3,x)
 

Output:

int((c + d*x)/cosh(a + b*x)^3, x)
 

Reduce [F]

\[ \int (c+d x) \text {sech}^3(a+b x) \, dx=\frac {3 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right ) b c +4 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right ) d +6 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right ) b c +8 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right ) d +3 \mathit {atan} \left (e^{b x +a}\right ) b c +4 \mathit {atan} \left (e^{b x +a}\right ) d +8 e^{4 b x +5 a} \left (\int \frac {e^{b x} x}{e^{6 b x +6 a}+3 e^{4 b x +4 a}+3 e^{2 b x +2 a}+1}d x \right ) b^{2} d +3 e^{3 b x +3 a} b c +4 e^{3 b x +3 a} d +16 e^{2 b x +3 a} \left (\int \frac {e^{b x} x}{e^{6 b x +6 a}+3 e^{4 b x +4 a}+3 e^{2 b x +2 a}+1}d x \right ) b^{2} d -3 e^{b x +a} b c -8 e^{b x +a} b d x +4 e^{b x +a} d +8 e^{a} \left (\int \frac {e^{b x} x}{e^{6 b x +6 a}+3 e^{4 b x +4 a}+3 e^{2 b x +2 a}+1}d x \right ) b^{2} d}{3 b^{2} \left (e^{4 b x +4 a}+2 e^{2 b x +2 a}+1\right )} \] Input:

int((d*x+c)*sech(b*x+a)^3,x)
 

Output:

(3*e**(4*a + 4*b*x)*atan(e**(a + b*x))*b*c + 4*e**(4*a + 4*b*x)*atan(e**(a 
 + b*x))*d + 6*e**(2*a + 2*b*x)*atan(e**(a + b*x))*b*c + 8*e**(2*a + 2*b*x 
)*atan(e**(a + b*x))*d + 3*atan(e**(a + b*x))*b*c + 4*atan(e**(a + b*x))*d 
 + 8*e**(5*a + 4*b*x)*int((e**(b*x)*x)/(e**(6*a + 6*b*x) + 3*e**(4*a + 4*b 
*x) + 3*e**(2*a + 2*b*x) + 1),x)*b**2*d + 3*e**(3*a + 3*b*x)*b*c + 4*e**(3 
*a + 3*b*x)*d + 16*e**(3*a + 2*b*x)*int((e**(b*x)*x)/(e**(6*a + 6*b*x) + 3 
*e**(4*a + 4*b*x) + 3*e**(2*a + 2*b*x) + 1),x)*b**2*d - 3*e**(a + b*x)*b*c 
 - 8*e**(a + b*x)*b*d*x + 4*e**(a + b*x)*d + 8*e**a*int((e**(b*x)*x)/(e**( 
6*a + 6*b*x) + 3*e**(4*a + 4*b*x) + 3*e**(2*a + 2*b*x) + 1),x)*b**2*d)/(3* 
b**2*(e**(4*a + 4*b*x) + 2*e**(2*a + 2*b*x) + 1))