\(\int (c+d x)^3 \tanh (e+f x) \, dx\) [1]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F]
   Maxima [B] (verification not implemented)
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=-\frac {(c+d x)^4}{4 d}+\frac {(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3799, 2221, 2611, 6744, 2320, 6724} \[ \int (c+d x)^3 \tanh (e+f x) \, dx=-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac {(c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {(c+d x)^4}{4 d}+\frac {3 d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4} \]

[In]

Int[(c + d*x)^3*Tanh[e + f*x],x]

[Out]

-1/4*(c + d*x)^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 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 2320

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 2611

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 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 6724

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]

Rule 6744

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]

Rubi steps \begin{align*} \text {integral}& = -\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 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {\left (3 d^2\right ) \int (c+d x) \operatorname {PolyLog}\left (2,-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 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {\left (3 d^3\right ) \int \operatorname {PolyLog}\left (3,-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 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {\left (3 d^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(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 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=\frac {1}{4} \left (\frac {2 (c+d x)^4}{d \left (1+e^{2 e}\right )}+\frac {4 (c+d x)^3 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 d \left (2 f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (4,-e^{-2 (e+f x)}\right )\right )\right )}{f^4}+x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \tanh (e)\right ) \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(109)=218\).

Time = 0.39 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.44

method result size
risch \(-d^{2} c \,x^{3}-\frac {3 d \,c^{2} x^{2}}{2}+c^{3} x -\frac {d^{3} x^{4}}{4}-\frac {3 d^{3} e^{4}}{2 f^{4}}+\frac {c^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {3 d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 f x +2 e}\right )}{4 f^{4}}-\frac {3 c^{2} d \,e^{2}}{f^{2}}+\frac {d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{3}}{f}-\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{2 f^{2}}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{2 f^{3}}+\frac {2 e^{3} d^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {3 c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}+\frac {4 c \,d^{2} e^{3}}{f^{3}}-\frac {2 d^{3} e^{3} x}{f^{3}}-\frac {6 c^{2} d e x}{f}+\frac {3 c \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}-\frac {6 c \,e^{2} d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {3 c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {6 c^{2} e d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}+\frac {6 c \,d^{2} e^{2} x}{f^{2}}+\frac {c^{4}}{4 d}\) \(402\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 531, normalized size of antiderivative = 4.54 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=-\frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x - 24 \, d^{3} {\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 24 \, d^{3} {\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 4 \, {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 4 \, {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right ) + 24 \, {\left (d^{3} f x + c d^{2} f\right )} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \, {\left (d^{3} f x + c d^{2} f\right )} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right )}{4 \, f^{4}} \]

[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

Sympy [F]

\[ \int (c+d x)^3 \tanh (e+f x) \, dx=\int \left (c + d x\right )^{3} \tanh {\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (108) = 216\).

Time = 0.32 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.44 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=\frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + \frac {c^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac {c^{3} \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac {3 \, {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} c^{2} d}{2 \, f^{2}} + \frac {3 \, {\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} c d^{2}}{2 \, f^{3}} + \frac {{\left (4 \, f^{3} x^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 6 \, f^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - 6 \, f x {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{3}}{3 \, f^{4}} - \frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2}}{2 \, f^{4}} \]

[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

Giac [F]

\[ \int (c+d x)^3 \tanh (e+f x) \, dx=\int { {\left (d x + c\right )}^{3} \tanh \left (f x + e\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \tanh (e+f x) \, dx=\int \mathrm {tanh}\left (e+f\,x\right )\,{\left (c+d\,x\right )}^3 \,d x \]

[In]

int(tanh(e + f*x)*(c + d*x)^3,x)

[Out]

int(tanh(e + f*x)*(c + d*x)^3, x)