3.3 \(\int (c+d x) \tanh (e+f x) \, dx\)

Optimal. Leaf size=57 \[ \frac {(c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^2}{2 d}+\frac {d \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3718, 2190, 2279, 2391} \[ \frac {d \text {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac {(c+d x) \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Tanh[e + f*x],x]

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [C]  time = 4.42, size = 210, normalized size = 3.68 \[ \frac {c \log (\cosh (e+f x))}{f}+\frac {d \text {csch}(e) \text {sech}(e) \left (f^2 x^2 e^{-\tanh ^{-1}(\coth (e))}-\frac {i \coth (e) \left (i \text {Li}_2\left (e^{2 i \left (i f x+i \tanh ^{-1}(\coth (e))\right )}\right )-f x \left (-\pi +2 i \tanh ^{-1}(\coth (e))\right )-2 \left (i \tanh ^{-1}(\coth (e))+i f x\right ) \log \left (1-e^{2 i \left (i \tanh ^{-1}(\coth (e))+i f x\right )}\right )+2 i \tanh ^{-1}(\coth (e)) \log \left (i \sinh \left (\tanh ^{-1}(\coth (e))+f x\right )\right )-\pi \log \left (e^{2 f x}+1\right )+\pi \log (\cosh (f x))\right )}{\sqrt {1-\coth ^2(e)}}\right )}{2 f^2 \sqrt {\text {csch}^2(e) \left (\sinh ^2(e)-\cosh ^2(e)\right )}}+\frac {1}{2} d x^2 \tanh (e) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Tanh[e + f*x],x]

[Out]

(c*Log[Cosh[e + f*x]])/f + (d*Csch[e]*((f^2*x^2)/E^ArcTanh[Coth[e]] - (I*Coth[e]*(-(f*x*(-Pi + (2*I)*ArcTanh[C
oth[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)]
) + (d*x^2*Tanh[e])/2

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fricas [C]  time = 0.46, size = 171, normalized size = 3.00 \[ -\frac {d f^{2} x^{2} + 2 \, c f^{2} x - 2 \, d {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 2 \, d {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 2 \, {\left (d e - c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 2 \, {\left (d e - c f\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 2 \, {\left (d f x + d e\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 2 \, {\left (d f x + d e\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right )}{2 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*tanh(f*x+e),x, algorithm="fricas")

[Out]

-1/2*(d*f^2*x^2 + 2*c*f^2*x - 2*d*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) - 2*d*dilog(-I*cosh(f*x + e) - I*si
nh(f*x + e)) + 2*(d*e - c*f)*log(cosh(f*x + e) + sinh(f*x + e) + I) + 2*(d*e - c*f)*log(cosh(f*x + e) + sinh(f
*x + e) - I) - 2*(d*f*x + d*e)*log(I*cosh(f*x + e) + I*sinh(f*x + e) + 1) - 2*(d*f*x + d*e)*log(-I*cosh(f*x +
e) - I*sinh(f*x + e) + 1))/f^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \tanh \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)*tanh(f*x + e), x)

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maple [B]  time = 0.17, size = 109, normalized size = 1.91 \[ -\frac {d \,x^{2}}{2}+c x +\frac {c \ln \left ({\mathrm e}^{2 f x +2 e}+1\right )}{f}-\frac {2 c \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {2 d e x}{f}-\frac {d \,e^{2}}{f^{2}}+\frac {d \ln \left ({\mathrm e}^{2 f x +2 e}+1\right ) x}{f}+\frac {d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}+\frac {2 d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.48, size = 78, normalized size = 1.37 \[ -\frac {1}{2} \, d x^{2} + \frac {c \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac {c \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac {{\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} d}{2 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*d*x^2 + 1/2*c*log(e^(2*f*x + 2*e) + 1)/f + 1/2*c*log(e^(-2*f*x - 2*e) + 1)/f + 1/2*(2*f*x*log(e^(2*f*x +
2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*d/f^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {tanh}\left (e+f\,x\right )\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(e + f*x)*(c + d*x),x)

[Out]

int(tanh(e + f*x)*(c + d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \tanh {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c + d*x)*tanh(e + f*x), x)

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