3.70 \(\int \frac{c+d x}{a+b \tanh (e+f x)} \, dx\)

Optimal. Leaf size=108 \[ \frac{b d \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^2 \left (a^2-b^2\right )}-\frac{b (c+d x) \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f \left (a^2-b^2\right )}+\frac{(c+d x)^2}{2 d (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.165925, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3732, 2190, 2279, 2391} \[ \frac{b d \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^2 \left (a^2-b^2\right )}-\frac{b (c+d x) \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f \left (a^2-b^2\right )}+\frac{(c+d x)^2}{2 d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && 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 \frac{c+d x}{a+b \tanh (e+f x)} \, dx &=\frac{(c+d x)^2}{2 (a+b) d}+(2 b) \int \frac{e^{-2 (e+f x)} (c+d x)}{(a+b)^2+\left (a^2-b^2\right ) e^{-2 (e+f x)}} \, dx\\ &=\frac{(c+d x)^2}{2 (a+b) d}-\frac{b (c+d x) \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{(b d) \int \log \left (1+\frac{\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac{(c+d x)^2}{2 (a+b) d}-\frac{b (c+d x) \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{\left (a^2-b^2\right ) x}{(a+b)^2}\right )}{x} \, dx,x,e^{-2 (e+f x)}\right )}{2 \left (a^2-b^2\right ) f^2}\\ &=\frac{(c+d x)^2}{2 (a+b) d}-\frac{b (c+d x) \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{b d \text{Li}_2\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2}\\ \end{align*}

Mathematica [A]  time = 3.01736, size = 144, normalized size = 1.33 \[ \frac{b \left (\frac{d \text{PolyLog}\left (2,\frac{(b-a) e^{-2 (e+f x)}}{a+b}\right )}{f^2 (a-b)}-\frac{2 (c+d x) \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f (a-b)}-\frac{2 (c+d x)^2}{d \left (a \left (e^{2 e}+1\right )+b \left (e^{2 e}-1\right )\right )}\right )}{2 (a+b)}+\frac{x \cosh (e) (2 c+d x)}{2 (a \cosh (e)+b \sinh (e))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.132, size = 357, normalized size = 3.3 \begin{align*}{\frac{d{x}^{2}}{2\,b+2\,a}}+{\frac{cx}{a+b}}+2\,{\frac{cb\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ( a+b \right ) f \left ( a-b \right ) }}-{\frac{cb\ln \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }{ \left ( a+b \right ) f \left ( a-b \right ) }}+{\frac{bdx}{ \left ( a+b \right ) f \left ( -a+b \right ) }\ln \left ( 1-{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) }+{\frac{bde}{ \left ( a+b \right ){f}^{2} \left ( -a+b \right ) }\ln \left ( 1-{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) }-{\frac{bd{x}^{2}}{ \left ( a+b \right ) \left ( -a+b \right ) }}-2\,{\frac{bdex}{ \left ( a+b \right ) f \left ( -a+b \right ) }}-{\frac{bd{e}^{2}}{ \left ( a+b \right ){f}^{2} \left ( -a+b \right ) }}+{\frac{bd}{ \left ( 2\,b+2\,a \right ){f}^{2} \left ( -a+b \right ) }{\it polylog} \left ( 2,{\frac{ \left ( a+b \right ){{\rm e}^{2\,fx+2\,e}}}{-a+b}} \right ) }-2\,{\frac{bde\ln \left ({{\rm e}^{fx+e}} \right ) }{ \left ( a+b \right ){f}^{2} \left ( a-b \right ) }}+{\frac{bde\ln \left ( a{{\rm e}^{2\,fx+2\,e}}+b{{\rm e}^{2\,fx+2\,e}}+a-b \right ) }{ \left ( a+b \right ){f}^{2} \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (4 \, b \int \frac{x}{a^{2} - b^{2} +{\left (a^{2} e^{\left (2 \, e\right )} + 2 \, a b e^{\left (2 \, e\right )} + b^{2} e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}\,{d x} + \frac{x^{2}}{a + b}\right )} d - c{\left (\frac{b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} - a - b\right )}{{\left (a^{2} - b^{2}\right )} f} - \frac{f x + e}{{\left (a + b\right )} f}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*b*integrate(x/(a^2 - b^2 + (a^2*e^(2*e) + 2*a*b*e^(2*e) + b^2*e^(2*e))*e^(2*f*x)), x) + x^2/(a + b))*d
- c*(b*log(-(a - b)*e^(-2*f*x - 2*e) - a - b)/((a^2 - b^2)*f) - (f*x + e)/((a + b)*f))

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Fricas [B]  time = 2.38339, size = 806, normalized size = 7.46 \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 (\sqrt{-\frac{a + b}{a - b}}{\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) - 2 \, b d{\rm Li}_2\left (-\sqrt{-\frac{a + b}{a - b}}{\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )}\right ) + 2 \,{\left (b d e - b c f\right )} \log \left (2 \,{\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \,{\left (a + b\right )} \sinh \left (f x + e\right ) + 2 \,{\left (a - b\right )} \sqrt{-\frac{a + b}{a - b}}\right ) + 2 \,{\left (b d e - b c f\right )} \log \left (2 \,{\left (a + b\right )} \cosh \left (f x + e\right ) + 2 \,{\left (a + b\right )} \sinh \left (f x + e\right ) - 2 \,{\left (a - b\right )} \sqrt{-\frac{a + b}{a - b}}\right ) - 2 \,{\left (b d f x + b d e\right )} \log \left (\sqrt{-\frac{a + b}{a - b}}{\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right ) - 2 \,{\left (b d f x + b d e\right )} \log \left (-\sqrt{-\frac{a + b}{a - b}}{\left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right )\right )} + 1\right )}{2 \,{\left (a^{2} - b^{2}\right )} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*tanh(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(sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)
)) - 2*b*d*dilog(-sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e))) + 2*(b*d*e - b*c*f)*log(2*(a + b)*co
sh(f*x + e) + 2*(a + b)*sinh(f*x + e) + 2*(a - b)*sqrt(-(a + b)/(a - b))) + 2*(b*d*e - b*c*f)*log(2*(a + b)*co
sh(f*x + e) + 2*(a + b)*sinh(f*x + e) - 2*(a - b)*sqrt(-(a + b)/(a - b))) - 2*(b*d*f*x + b*d*e)*log(sqrt(-(a +
b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e)) + 1) - 2*(b*d*f*x + b*d*e)*log(-sqrt(-(a + b)/(a - b))*(cosh(f*x
+ e) + sinh(f*x + e)) + 1))/((a^2 - b^2)*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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