3.1.55 \(\int (c+d x) (a+b \tanh (e+f x)) \, dx\) [55]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, 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 = {3803, 3799, 2221, 2317, 2438} \begin {gather*} \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}+\frac {b d \text {Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3799

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 3803

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]

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) \text {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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 3.04, size = 226, normalized size = 3.01 \begin {gather*} a c x+\frac {1}{2} a d x^2+\frac {b c \log (\cosh (e+f x))}{f}+\frac {b d \text {csch}(e) \left (e^{-\tanh ^{-1}(\coth (e))} f^2 x^2-\frac {i \coth (e) \left (-f x \left (-\pi +2 i \tanh ^{-1}(\coth (e))\right )-\pi \log \left (1+e^{2 f x}\right )-2 \left (i f x+i \tanh ^{-1}(\coth (e))\right ) \log \left (1-e^{2 i \left (i f x+i \tanh ^{-1}(\coth (e))\right )}\right )+\pi \log (\cosh (f x))+2 i \tanh ^{-1}(\coth (e)) \log \left (i \sinh \left (f x+\tanh ^{-1}(\coth (e))\right )\right )+i \text {PolyLog}\left (2,e^{2 i \left (i f x+i \tanh ^{-1}(\coth (e))\right )}\right )\right )}{\sqrt {1-\coth ^2(e)}}\right ) \text {sech}(e)}{2 f^2 \sqrt {\text {csch}^2(e) \left (-\cosh ^2(e)+\sinh ^2(e)\right )}}+\frac {1}{2} b d x^2 \tanh (e) \end {gather*}

Antiderivative was successfully verified.

[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 = 1.90, size = 129, normalized size = 1.72

method	result	size
risch	\(\frac {a d \,x^{2}}{2}+a c x -\frac {b d \,x^{2}}{2}+b c x +\frac {b c \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 b c \ln \left ({\mathrm e}^{f x +e}\right )}{f}-\frac {2 b d e x}{f}-\frac {b d \,e^{2}}{f^{2}}+\frac {b d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {b d \polylog \left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}+\frac {2 b d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}\)	\(129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+b*tanh(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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] Result contains complex when optimal does not.
time = 0.38, size = 253, normalized size = 3.37 \begin {gather*} \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 + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) + 2 \, b d {\rm Li}_2\left (-i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right )\right ) + 2 \, {\left (b c f - b d \cosh \left (1\right ) - b d \sinh \left (1\right )\right )} \log \left (\cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i\right ) + 2 \, {\left (b c f - b d \cosh \left (1\right ) - b d \sinh \left (1\right )\right )} \log \left (\cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i\right ) + 2 \, {\left (b d f x + b d \cosh \left (1\right ) + b d \sinh \left (1\right )\right )} \log \left (i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 1\right ) + 2 \, {\left (b d f x + b d \cosh \left (1\right ) + b d \sinh \left (1\right )\right )} \log \left (-i \, \cosh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) - i \, \sinh \left (f x + \cosh \left (1\right ) + \sinh \left (1\right )\right ) + 1\right )}{2 \, f^{2}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \tanh {\left (e + f x \right )}\right ) \left (c + d x\right )\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )\,\left (c+d\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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