3.55 \(\int (c+d x) (a+b \tanh (e+f x)) \, dx\)

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

[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 = 0.127623, 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, 3718, 2190, 2279, 2391} \[ \frac{b d \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac{a (c+d x)^2}{2 d}+\frac{b (c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{b (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*Tanh[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 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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 \tanh (e+f x)) \, dx &=\int (a (c+d x)+b (c+d x) \tanh (e+f x)) \, dx\\ &=\frac{a (c+d x)^2}{2 d}+b \int (c+d x) \tanh (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 [C]  time = 4.05068, size = 227, normalized size = 3.03 \[ -\frac{b d \text{csch}(e) \text{sech}(e) \left (-f^2 x^2 e^{-\tanh ^{-1}(\coth (e))}+\frac{i \coth (e) \left (i \text{PolyLog}\left (2,e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )-f x \left (-\pi +2 i \tanh ^{-1}(\coth (e))\right )-2 \left (i \tanh ^{-1}(\coth (e))+i f x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )+2 i \tanh ^{-1}(\coth (e)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (e))+f x\right )\right )-\pi \log \left (e^{2 f x}+1\right )+\pi \log (\cosh (f x))\right )}{\sqrt{1-\coth ^2(e)}}\right )}{2 f^2 \sqrt{\text{csch}^2(e) \left (\sinh ^2(e)-\cosh ^2(e)\right )}}+a c x+\frac{1}{2} a d x^2+\frac{b c \log (\cosh (e+f x))}{f}+\frac{1}{2} b d x^2 \tanh (e) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*c*x + (a*d*x^2)/2 + (b*c*Log[Cosh[e + f*x]])/f - (b*d*Csch[e]*(-((f^2*x^2)/E^ArcTanh[Coth[e]]) + (I*Coth[e]*
(-(f*x*(-Pi + (2*I)*ArcTanh[Coth[e]])) - Pi*Log[1 + E^(2*f*x)] - 2*(I*f*x + I*ArcTanh[Coth[e]])*Log[1 - E^((2*
I)*(I*f*x + I*ArcTanh[Coth[e]]))] + Pi*Log[Cosh[f*x]] + (2*I)*ArcTanh[Coth[e]]*Log[I*Sinh[f*x + ArcTanh[Coth[e
]]]] + I*PolyLog[2, E^((2*I)*(I*f*x + I*ArcTanh[Coth[e]]))]))/Sqrt[1 - Coth[e]^2])*Sech[e])/(2*f^2*Sqrt[Csch[e
]^2*(-Cosh[e]^2 + Sinh[e]^2)]) + (b*d*x^2*Tanh[e])/2

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Maple [A]  time = 0.044, size = 129, normalized size = 1.7 \begin{align*}{\frac{ad{x}^{2}}{2}}-{\frac{bd{x}^{2}}{2}}+acx+bcx+{\frac{cb\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}}-2\,{\frac{cb\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}-2\,{\frac{bdex}{f}}-{\frac{bd{e}^{2}}{{f}^{2}}}+{\frac{bd\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{f}}+{\frac{bd{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{2}}}+2\,{\frac{bde\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}} \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*d*x^2-1/2*b*d*x^2+a*c*x+b*c*x+b/f*c*ln(exp(2*f*x+2*e)+1)-2*b/f*c*ln(exp(f*x+e))-2*b/f*d*e*x-b/f^2*d*e^2+
b/f*d*ln(exp(2*f*x+2*e)+1)*x+1/2*b*d*polylog(2,-exp(2*f*x+2*e))/f^2+2*b/f^2*d*e*ln(exp(f*x+e))

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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} - 4 \, \int \frac{x}{e^{\left (2 \, f x + 2 \, e\right )} + 1}\,{d x}\right )} b d + a c x + \frac{b c \log \left (\cosh \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*tanh(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.44589, size = 536, normalized size = 7.15 \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 (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 2 \, b d{\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 2 \,{\left (b d e - b c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 2 \,{\left (b d e - b c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 2 \,{\left (b d f x + b d e\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) + 2 \,{\left (b d f x + b d e\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \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*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(I*cosh(f*x + e) + I*sinh(f*x + e)) + 2*b*d*dilog(-I*c
osh(f*x + e) - I*sinh(f*x + e)) - 2*(b*d*e - b*c*f)*log(cosh(f*x + e) + sinh(f*x + e) + I) - 2*(b*d*e - b*c*f)
*log(cosh(f*x + e) + sinh(f*x + e) - I) + 2*(b*d*f*x + b*d*e)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) + 2*(
b*d*f*x + b*d*e)*log(-I*cosh(f*x + e) - I*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 \tanh{\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*tanh(f*x+e)),x)

[Out]

Integral((a + b*tanh(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 \tanh \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*tanh(f*x+e)),x, algorithm="giac")

[Out]

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