3.1 \(\int (c+d x)^3 \tanh (e+f x) \, dx\)

Optimal. Leaf size=117 \[ -\frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac{3 d^3 \text{PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}+\frac{(c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{(c+d x)^4}{4 d} \]

[Out]

-(c + d*x)^4/(4*d) + ((c + d*x)^3*Log[1 + E^(2*(e + f*x))])/f + (3*d*(c + d*x)^2*PolyLog[2, -E^(2*(e + f*x))])
/(2*f^2) - (3*d^2*(c + d*x)*PolyLog[3, -E^(2*(e + f*x))])/(2*f^3) + (3*d^3*PolyLog[4, -E^(2*(e + f*x))])/(4*f^
4)

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Rubi [A]  time = 0.186519, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3718, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac{3 d^3 \text{PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}+\frac{(c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{(c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Tanh[e + f*x],x]

[Out]

-(c + d*x)^4/(4*d) + ((c + d*x)^3*Log[1 + E^(2*(e + f*x))])/f + (3*d*(c + d*x)^2*PolyLog[2, -E^(2*(e + f*x))])
/(2*f^2) - (3*d^2*(c + d*x)*PolyLog[3, -E^(2*(e + f*x))])/(2*f^3) + (3*d^3*PolyLog[4, -E^(2*(e + f*x))])/(4*f^
4)

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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int (c+d x)^3 \tanh (e+f x) \, dx &=-\frac{(c+d x)^4}{4 d}+2 \int \frac{e^{2 (e+f x)} (c+d x)^3}{1+e^{2 (e+f x)}} \, dx\\ &=-\frac{(c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac{(3 d) \int (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f}\\ &=-\frac{(c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{\left (3 d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac{(c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{\left (3 d^3\right ) \int \text{Li}_3\left (-e^{2 (e+f x)}\right ) \, dx}{2 f^3}\\ &=-\frac{(c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{4 f^4}\\ &=-\frac{(c+d x)^4}{4 d}+\frac{(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 d^3 \text{Li}_4\left (-e^{2 (e+f x)}\right )}{4 f^4}\\ \end{align*}

Mathematica [A]  time = 1.67489, size = 156, normalized size = 1.33 \[ -\frac{3 d \left (2 f^2 (c+d x)^2 \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )+d \text{PolyLog}\left (4,-e^{-2 (e+f x)}\right )\right )\right )}{4 f^4}+\frac{1}{4} x \tanh (e) \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )+\frac{(c+d x)^3 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac{(c+d x)^4}{2 d \left (e^{2 e}+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^3*Tanh[e + f*x],x]

[Out]

(c + d*x)^4/(2*d*(1 + E^(2*e))) + ((c + d*x)^3*Log[1 + E^(-2*(e + f*x))])/f - (3*d*(2*f^2*(c + d*x)^2*PolyLog[
2, -E^(-2*(e + f*x))] + d*(2*f*(c + d*x)*PolyLog[3, -E^(-2*(e + f*x))] + d*PolyLog[4, -E^(-2*(e + f*x))])))/(4
*f^4) + (x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Tanh[e])/4

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Maple [B]  time = 0.112, size = 394, normalized size = 3.4 \begin{align*} -{\frac{{d}^{3}{x}^{4}}{4}}+{c}^{3}x+4\,{\frac{c{d}^{2}{e}^{3}}{{f}^{3}}}-2\,{\frac{{d}^{3}{e}^{3}x}{{f}^{3}}}-3\,{\frac{{c}^{2}d{e}^{2}}{{f}^{2}}}+{\frac{3\,{c}^{2}d{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{2}}}-{\frac{3\,c{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{3}}}+2\,{\frac{{d}^{3}{e}^{3}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{4}}}-{\frac{3\,{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{2\,{f}^{3}}}+{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{3}}{f}}+{\frac{3\,{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ){x}^{2}}{2\,{f}^{2}}}-c{d}^{2}{x}^{3}-{\frac{3\,{c}^{2}d{x}^{2}}{2}}-6\,{\frac{c{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+3\,{\frac{c{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{f}}+3\,{\frac{c{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{2}}}+3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{f}}+6\,{\frac{c{d}^{2}{e}^{2}x}{{f}^{2}}}-6\,{\frac{{c}^{2}dex}{f}}+6\,{\frac{{c}^{2}de\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}+{\frac{3\,{d}^{3}{\it polylog} \left ( 4,-{{\rm e}^{2\,fx+2\,e}} \right ) }{4\,{f}^{4}}}-{\frac{3\,{d}^{3}{e}^{4}}{2\,{f}^{4}}}-2\,{\frac{{c}^{3}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}+{\frac{{c}^{3}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^3*tanh(f*x+e),x)

[Out]

-1/4*d^3*x^4+c^3*x+4/f^3*c*d^2*e^3-2/f^3*d^3*e^3*x-3/f^2*c^2*d*e^2+3/2/f^2*c^2*d*polylog(2,-exp(2*f*x+2*e))-3/
2/f^3*c*d^2*polylog(3,-exp(2*f*x+2*e))+2/f^4*d^3*e^3*ln(exp(f*x+e))-3/2/f^3*d^3*polylog(3,-exp(2*f*x+2*e))*x+1
/f*d^3*ln(exp(2*f*x+2*e)+1)*x^3+3/2/f^2*d^3*polylog(2,-exp(2*f*x+2*e))*x^2-c*d^2*x^3-3/2*c^2*d*x^2-6/f^3*c*d^2
*e^2*ln(exp(f*x+e))+3/f*c*d^2*ln(exp(2*f*x+2*e)+1)*x^2+3/f^2*c*d^2*polylog(2,-exp(2*f*x+2*e))*x+3/f*c^2*d*ln(e
xp(2*f*x+2*e)+1)*x+6/f^2*c*d^2*e^2*x-6/f*c^2*d*e*x+6/f^2*c^2*d*e*ln(exp(f*x+e))+3/4*d^3*polylog(4,-exp(2*f*x+2
*e))/f^4-3/2/f^4*d^3*e^4-2/f*c^3*ln(exp(f*x+e))+1/f*c^3*ln(exp(2*f*x+2*e)+1)

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Maxima [B]  time = 1.74112, size = 386, normalized size = 3.3 \begin{align*} \frac{1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac{3}{2} \, c^{2} d x^{2} + \frac{c^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac{c^{3} \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac{3 \,{\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 )} c^{2} d}{2 \, f^{2}} + \frac{3 \,{\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x{\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) -{\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} c d^{2}}{2 \, f^{3}} + \frac{{\left (4 \, f^{3} x^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 6 \, f^{2} x^{2}{\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - 6 \, f x{\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )}) + 3 \,{\rm Li}_{4}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{3}}{3 \, f^{4}} - \frac{d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2}}{2 \, f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*tanh(f*x+e),x, algorithm="maxima")

[Out]

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + 1/2*c^3*log(e^(2*f*x + 2*e) + 1)/f + 1/2*c^3*log(e^(-2*f*x - 2*e) +
1)/f + 3/2*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*c^2*d/f^2 + 3/2*(2*f^2*x^2*log(e^(2*f*x
+ 2*e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - polylog(3, -e^(2*f*x + 2*e)))*c*d^2/f^3 + 1/3*(4*f^3*x^3*log(e^(
2*f*x + 2*e) + 1) + 6*f^2*x^2*dilog(-e^(2*f*x + 2*e)) - 6*f*x*polylog(3, -e^(2*f*x + 2*e)) + 3*polylog(4, -e^(
2*f*x + 2*e)))*d^3/f^4 - 1/2*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2)/f^4

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Fricas [C]  time = 1.80155, size = 1310, normalized size = 11.2 \begin{align*} -\frac{d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x - 24 \, d^{3}{\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 24 \, d^{3}{\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 12 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )}{\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 12 \,{\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )}{\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 4 \,{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 4 \,{\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 4 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 4 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right ) + 24 \,{\left (d^{3} f x + c d^{2} f\right )}{\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \,{\left (d^{3} f x + c d^{2} f\right )}{\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right )}{4 \, f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*tanh(f*x+e),x, algorithm="fricas")

[Out]

-1/4*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2 + 4*c^3*f^4*x - 24*d^3*polylog(4, I*cosh(f*x + e) + I*si
nh(f*x + e)) - 24*d^3*polylog(4, -I*cosh(f*x + e) - I*sinh(f*x + e)) - 12*(d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d
*f^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) - 12*(d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d*f^2)*dilog(-I*cosh(f*
x + e) - I*sinh(f*x + e)) + 4*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*log(cosh(f*x + e) + sinh(f*x
+ e) + I) + 4*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*log(cosh(f*x + e) + sinh(f*x + e) - I) - 4*
(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2)*log(I*cosh(f*x + e)
+ I*sinh(f*x + e) + 1) - 4*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*
e*f^2)*log(-I*cosh(f*x + e) - I*sinh(f*x + e) + 1) + 24*(d^3*f*x + c*d^2*f)*polylog(3, I*cosh(f*x + e) + I*sin
h(f*x + e)) + 24*(d^3*f*x + c*d^2*f)*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e)))/f^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \tanh{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*tanh(f*x+e),x)

[Out]

Integral((c + d*x)**3*tanh(e + f*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \tanh \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*tanh(f*x+e),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*tanh(f*x + e), x)