3.39 \(\int (c+d x) (a+b \coth (e+f x)) \, dx\)

Optimal. Leaf size=75 \[ \frac{b d \text{PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}+\frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{b (c+d x)^2}{2 d} \]

[Out]

(a*(c + d*x)^2)/(2*d) - (b*(c + d*x)^2)/(2*d) + (b*(c + d*x)*Log[1 - E^(2*(e + f*x))])/f + (b*d*PolyLog[2, E^(
2*(e + f*x))])/(2*f^2)

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Rubi [A]  time = 0.126318, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {3722, 3716, 2190, 2279, 2391} \[ \frac{b d \text{PolyLog}\left (2,e^{2 (e+f x)}\right )}{2 f^2}+\frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{b (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*Coth[e + f*x]),x]

[Out]

(a*(c + d*x)^2)/(2*d) - (b*(c + d*x)^2)/(2*d) + (b*(c + d*x)*Log[1 - E^(2*(e + f*x))])/f + (b*d*PolyLog[2, E^(
2*(e + f*x))])/(2*f^2)

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && 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) (a+b \coth (e+f x)) \, dx &=\int (a (c+d x)+b (c+d x) \coth (e+f x)) \, dx\\ &=\frac{a (c+d x)^2}{2 d}+b \int (c+d x) \coth (e+f x) \, dx\\ &=\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x)^2}{2 d}-(2 b) \int \frac{e^{2 (e+f x)} (c+d x)}{1-e^{2 (e+f x)}} \, dx\\ &=\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{(b d) \int \log \left (1-e^{2 (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{2 f^2}\\ &=\frac{a (c+d x)^2}{2 d}-\frac{b (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (1-e^{2 (e+f x)}\right )}{f}+\frac{b d \text{Li}_2\left (e^{2 (e+f x)}\right )}{2 f^2}\\ \end{align*}

Mathematica [A]  time = 0.056351, size = 87, normalized size = 1.16 \[ \frac{b d \text{PolyLog}\left (2,e^{2 e+2 f x}\right )}{2 f^2}+a c x+\frac{1}{2} a d x^2+\frac{b c (\log (\tanh (e+f x))+\log (\cosh (e+f x)))}{f}+\frac{b d x \log \left (1-e^{2 e+2 f x}\right )}{f}-\frac{1}{2} b d x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*(a + b*Coth[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.043, size = 201, normalized size = 2.7 \begin{align*}{\frac{ad{x}^{2}}{2}}-{\frac{bd{x}^{2}}{2}}+acx+bcx-2\,{\frac{cb\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}+{\frac{cb\ln \left ({{\rm e}^{fx+e}}+1 \right ) }{f}}+{\frac{cb\ln \left ({{\rm e}^{fx+e}}-1 \right ) }{f}}-2\,{\frac{bdex}{f}}-{\frac{bd{e}^{2}}{{f}^{2}}}+{\frac{bd\ln \left ({{\rm e}^{fx+e}}+1 \right ) x}{f}}+{\frac{bd{\it polylog} \left ( 2,-{{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+{\frac{bd\ln \left ( 1-{{\rm e}^{fx+e}} \right ) x}{f}}+{\frac{bd\ln \left ( 1-{{\rm e}^{fx+e}} \right ) e}{{f}^{2}}}+{\frac{bd{\it polylog} \left ( 2,{{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+2\,{\frac{bde\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-{\frac{bde\ln \left ({{\rm e}^{fx+e}}-1 \right ) }{{f}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+b*coth(f*x+e)),x)

[Out]

1/2*a*d*x^2-1/2*b*d*x^2+a*c*x+b*c*x-2*b/f*c*ln(exp(f*x+e))+b/f*c*ln(exp(f*x+e)+1)+b/f*c*ln(exp(f*x+e)-1)-2*b/f
*d*e*x-b/f^2*d*e^2+b/f*d*ln(exp(f*x+e)+1)*x+b/f^2*d*polylog(2,-exp(f*x+e))+b/f*d*ln(1-exp(f*x+e))*x+b/f^2*d*ln
(1-exp(f*x+e))*e+b/f^2*d*polylog(2,exp(f*x+e))+2*b/f^2*d*e*ln(exp(f*x+e))-b/f^2*d*e*ln(exp(f*x+e)-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+b*coth(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.119, size = 435, normalized size = 5.8 \begin{align*} \frac{{\left (a - b\right )} d f^{2} x^{2} + 2 \,{\left (a - b\right )} c f^{2} x + 2 \, b d{\rm Li}_2\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right ) + 2 \, b d{\rm Li}_2\left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right )\right ) + 2 \,{\left (b d f x + b c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + 1\right ) - 2 \,{\left (b d e - b c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - 1\right ) + 2 \,{\left (b d f x + b d e\right )} \log \left (-\cosh \left (f x + e\right ) - \sinh \left (f x + e\right ) + 1\right )}{2 \, f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+b*coth(f*x+e)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+b*coth(f*x+e)),x)

[Out]

Integral((a + b*coth(e + f*x))*(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+b*coth(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*x + c)*(b*coth(f*x + e) + a), x)