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

Optimal. Leaf size=50 \[ -\frac {d \text {Li}_2\left (-e^{a+b x}\right )}{b^2}+\frac {d \text {Li}_2\left (e^{a+b x}\right )}{b^2}-\frac {2 (c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

-2*(d*x+c)*arctanh(exp(b*x+a))/b-d*polylog(2,-exp(b*x+a))/b^2+d*polylog(2,exp(b*x+a))/b^2

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4182, 2279, 2391} \[ -\frac {d \text {PolyLog}\left (2,-e^{a+b x}\right )}{b^2}+\frac {d \text {PolyLog}\left (2,e^{a+b x}\right )}{b^2}-\frac {2 (c+d x) \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Csch[a + b*x],x]

[Out]

(-2*(c + d*x)*ArcTanh[E^(a + b*x)])/b - (d*PolyLog[2, -E^(a + b*x)])/b^2 + (d*PolyLog[2, E^(a + b*x)])/b^2

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]

Rule 4182

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

Rubi steps

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

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Mathematica [C]  time = 0.07, size = 174, normalized size = 3.48 \[ \frac {d \left (-a \log \left (\tanh \left (\frac {1}{2} (a+b x)\right )\right )-i \left (i \left (\text {Li}_2\left (-e^{i (i a+i b x)}\right )-\text {Li}_2\left (e^{i (i a+i b x)}\right )\right )+(i a+i b x) \left (\log \left (1-e^{i (i a+i b x)}\right )-\log \left (1+e^{i (i a+i b x)}\right )\right )\right )\right )}{b^2}+\frac {c \log \left (\sinh \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}-\frac {c \log \left (\cosh \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Csch[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.72, size = 119, normalized size = 2.38 \[ \frac {d {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - d {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - {\left (b d x + b c\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (b d x + a d\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \operatorname {csch}\left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*csch(b*x + a), x)

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maple [A]  time = 0.03, size = 60, normalized size = 1.20 \[ \frac {\frac {d \left (2 \dilog \left ({\mathrm e}^{-b x -a}\right )-\frac {\dilog \left ({\mathrm e}^{-2 b x -2 a}\right )}{2}\right )}{b}+\frac {2 d a \arctanh \left ({\mathrm e}^{b x +a}\right )}{b}-2 c \arctanh \left ({\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/b*d*(2*dilog(exp(-b*x-a))-1/2*dilog(exp(-2*b*x-2*a)))+2/b*d*a*arctanh(exp(b*x+a))-2*c*arctanh(exp(b*x+a
)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -c {\left (\frac {\log \left (e^{\left (-b x - a\right )} + 1\right )}{b} - \frac {\log \left (e^{\left (-b x - a\right )} - 1\right )}{b}\right )} + 2 \, d {\left (\int \frac {x}{2 \, {\left (e^{\left (b x + a\right )} + 1\right )}}\,{d x} + \int \frac {x}{2 \, {\left (e^{\left (b x + a\right )} - 1\right )}}\,{d x}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {c+d\,x}{\mathrm {sinh}\left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/sinh(a + b*x),x)

[Out]

int((c + d*x)/sinh(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \operatorname {csch}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*csch(a + b*x), x)

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