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

Optimal. Leaf size=61 \[ -\frac{i d \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac{i d \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}+\frac{2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

(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

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Rubi [A]  time = 0.0380289, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4180, 2279, 2391} \[ -\frac{i d \text{PolyLog}\left (2,-i e^{a+b x}\right )}{b^2}+\frac{i d \text{PolyLog}\left (2,i e^{a+b x}\right )}{b^2}+\frac{2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Sech[a + b*x],x]

[Out]

(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

Rule 4180

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] + (-Dist[(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] + Dist[(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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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 2391

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]

Rubi steps

\begin{align*} \int (c+d x) \text{sech}(a+b x) \, dx &=\frac{2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}-\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) \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}+\frac{(i d) \operatorname{Subst}\left (\int \frac{\log (1+i x)}{x} \, dx,x,e^{a+b x}\right )}{b^2}\\ &=\frac{2 (c+d x) \tan ^{-1}\left (e^{a+b x}\right )}{b}-\frac{i d \text{Li}_2\left (-i e^{a+b x}\right )}{b^2}+\frac{i d \text{Li}_2\left (i e^{a+b x}\right )}{b^2}\\ \end{align*}

Mathematica [B]  time = 0.077762, size = 129, normalized size = 2.11 \[ \frac{-i d \left (\text{PolyLog}\left (2,-i e^{a+b x}\right )-\text{PolyLog}\left (2,i e^{a+b x}\right )\right )+b c \tan ^{-1}(\sinh (a+b x))-\frac{1}{2} d (-2 i a-2 i b x+\pi ) \left (\log \left (1-i e^{a+b x}\right )-\log \left (1+i e^{a+b x}\right )\right )+\frac{1}{2} (\pi -2 i a) d \log \left (\cot \left (\frac{1}{4} (2 i a+2 i b x+\pi )\right )\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Sech[a + b*x],x]

[Out]

(b*c*ArcTan[Sinh[a + b*x]] - (d*((-2*I)*a + Pi - (2*I)*b*x)*(Log[1 - I*E^(a + b*x)] - Log[1 + I*E^(a + b*x)]))
/2 + (d*((-2*I)*a + Pi)*Log[Cot[((2*I)*a + Pi + (2*I)*b*x)/4]])/2 - I*d*(PolyLog[2, (-I)*E^(a + b*x)] - PolyLo
g[2, I*E^(a + b*x)]))/b^2

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Maple [B]  time = 0.006, size = 449, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 2 \, d \int \frac{x e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} + 1}\,{d x} - \frac{2 \, c \arctan \left (e^{\left (-b x - a\right )}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.2176, size = 463, normalized size = 7.59 \begin{align*} \frac{i \, d{\rm Li}_2\left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right )\right ) - i \, d{\rm Li}_2\left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right )\right ) +{\left (i \, b c - i \, a d\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + i\right ) +{\left (-i \, b c + i \, a d\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - i\right ) +{\left (-i \, b d x - i \, a d\right )} \log \left (i \, \cosh \left (b x + a\right ) + i \, \sinh \left (b x + a\right ) + 1\right ) +{\left (i \, b d x + i \, a d\right )} \log \left (-i \, \cosh \left (b x + a\right ) - i \, \sinh \left (b x + a\right ) + 1\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(I*d*dilog(I*cosh(b*x + a) + I*sinh(b*x + a)) - I*d*dilog(-I*cosh(b*x + a) - I*sinh(b*x + a)) + (I*b*c - I*a*d
)*log(cosh(b*x + a) + sinh(b*x + a) + I) + (-I*b*c + I*a*d)*log(cosh(b*x + a) + sinh(b*x + a) - I) + (-I*b*d*x
- I*a*d)*log(I*cosh(b*x + a) + I*sinh(b*x + a) + 1) + (I*b*d*x + I*a*d)*log(-I*cosh(b*x + a) - I*sinh(b*x + a
) + 1))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*sech(a + b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*sech(b*x + a), x)