3.1.1 \(\int x^3 \coth (a+b x) \, dx\) [1]

Optimal. Leaf size=87 \[ -\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {PolyLog}\left (2,e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {PolyLog}\left (3,e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {PolyLog}\left (4,e^{2 (a+b x)}\right )}{4 b^4} \]

[Out]

-1/4*x^4+x^3*ln(1-exp(2*b*x+2*a))/b+3/2*x^2*polylog(2,exp(2*b*x+2*a))/b^2-3/2*x*polylog(3,exp(2*b*x+2*a))/b^3+
3/4*polylog(4,exp(2*b*x+2*a))/b^4

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Rubi [A]
time = 0.11, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3797, 2221, 2611, 6744, 2320, 6724} \begin {gather*} \frac {3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {x^4}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Coth[a + b*x],x]

[Out]

-1/4*x^4 + (x^3*Log[1 - E^(2*(a + b*x))])/b + (3*x^2*PolyLog[2, E^(2*(a + b*x))])/(2*b^2) - (3*x*PolyLog[3, E^
(2*(a + b*x))])/(2*b^3) + (3*PolyLog[4, E^(2*(a + b*x))])/(4*b^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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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*} \int x^3 \coth (a+b x) \, dx &=-\frac {x^4}{4}-2 \int \frac {e^{2 (a+b x)} x^3}{1-e^{2 (a+b x)}} \, dx\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}-\frac {3 \int x^2 \log \left (1-e^{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 \int x \text {Li}_2\left (e^{2 (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \int \text {Li}_3\left (e^{2 (a+b x)}\right ) \, dx}{2 b^3}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 (a+b x)}\right )}{4 b^4}\\ &=-\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 (a+b x)}\right )}{b}+\frac {3 x^2 \text {Li}_2\left (e^{2 (a+b x)}\right )}{2 b^2}-\frac {3 x \text {Li}_3\left (e^{2 (a+b x)}\right )}{2 b^3}+\frac {3 \text {Li}_4\left (e^{2 (a+b x)}\right )}{4 b^4}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 91, normalized size = 1.05 \begin {gather*} -\frac {x^4}{4}+\frac {x^3 \log \left (1-e^{2 a+2 b x}\right )}{b}+\frac {3 x^2 \text {PolyLog}\left (2,e^{2 a+2 b x}\right )}{2 b^2}-\frac {3 x \text {PolyLog}\left (3,e^{2 a+2 b x}\right )}{2 b^3}+\frac {3 \text {PolyLog}\left (4,e^{2 a+2 b x}\right )}{4 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Coth[a + b*x],x]

[Out]

-1/4*x^4 + (x^3*Log[1 - E^(2*a + 2*b*x)])/b + (3*x^2*PolyLog[2, E^(2*a + 2*b*x)])/(2*b^2) - (3*x*PolyLog[3, E^
(2*a + 2*b*x)])/(2*b^3) + (3*PolyLog[4, E^(2*a + 2*b*x)])/(4*b^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(199\) vs. \(2(79)=158\).
time = 1.68, size = 200, normalized size = 2.30

method	result	size
risch	\(-\frac {x^{4}}{4}-\frac {3 a^{4}}{2 b^{4}}+\frac {6 \polylog \left (4, {\mathrm e}^{b x +a}\right )}{b^{4}}+\frac {6 \polylog \left (4, -{\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {2 a^{3} x}{b^{3}}-\frac {6 \polylog \left (3, -{\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) x^{3}}{b}+\frac {3 \polylog \left (2, {\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}-\frac {6 \polylog \left (3, {\mathrm e}^{b x +a}\right ) x}{b^{3}}+\frac {\ln \left ({\mathrm e}^{b x +a}+1\right ) x^{3}}{b}+\frac {3 \polylog \left (2, -{\mathrm e}^{b x +a}\right ) x^{2}}{b^{2}}+\frac {2 a^{3} \ln \left ({\mathrm e}^{b x +a}\right )}{b^{4}}-\frac {a^{3} \ln \left ({\mathrm e}^{b x +a}-1\right )}{b^{4}}+\frac {\ln \left (1-{\mathrm e}^{b x +a}\right ) a^{3}}{b^{4}}\)	\(200\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*coth(b*x+a),x,method=_RETURNVERBOSE)

[Out]

-1/4*x^4-3/2/b^4*a^4+6/b^4*polylog(4,exp(b*x+a))+6/b^4*polylog(4,-exp(b*x+a))-2/b^3*a^3*x-6/b^3*polylog(3,-exp
(b*x+a))*x+1/b*ln(1-exp(b*x+a))*x^3+3/b^2*polylog(2,exp(b*x+a))*x^2-6/b^3*polylog(3,exp(b*x+a))*x+1/b*ln(exp(b
*x+a)+1)*x^3+3/b^2*polylog(2,-exp(b*x+a))*x^2+2/b^4*a^3*ln(exp(b*x+a))-1/b^4*a^3*ln(exp(b*x+a)-1)+1/b^4*ln(1-e
xp(b*x+a))*a^3

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (78) = 156\).
time = 0.26, size = 170, normalized size = 1.95 \begin {gather*} \frac {1}{4} \, x^{4} \coth \left (b x + a\right ) - \frac {1}{2} \, {\left (\frac {x^{4}}{b e^{\left (2 \, b x + 2 \, a\right )} - b} + \frac {x^{4}}{b} - \frac {2 \, {\left (b^{3} x^{3} \log \left (e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (-e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(-e^{\left (b x + a\right )})\right )}}{b^{5}} - \frac {2 \, {\left (b^{3} x^{3} \log \left (-e^{\left (b x + a\right )} + 1\right ) + 3 \, b^{2} x^{2} {\rm Li}_2\left (e^{\left (b x + a\right )}\right ) - 6 \, b x {\rm Li}_{3}(e^{\left (b x + a\right )}) + 6 \, {\rm Li}_{4}(e^{\left (b x + a\right )})\right )}}{b^{5}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*coth(b*x+a),x, algorithm="maxima")

[Out]

1/4*x^4*coth(b*x + a) - 1/2*(x^4/(b*e^(2*b*x + 2*a) - b) + x^4/b - 2*(b^3*x^3*log(e^(b*x + a) + 1) + 3*b^2*x^2
*dilog(-e^(b*x + a)) - 6*b*x*polylog(3, -e^(b*x + a)) + 6*polylog(4, -e^(b*x + a)))/b^5 - 2*(b^3*x^3*log(-e^(b
*x + a) + 1) + 3*b^2*x^2*dilog(e^(b*x + a)) - 6*b*x*polylog(3, e^(b*x + a)) + 6*polylog(4, e^(b*x + a)))/b^5)*
b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (78) = 156\).
time = 0.35, size = 216, normalized size = 2.48 \begin {gather*} -\frac {b^{4} x^{4} - 4 \, b^{3} x^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - 12 \, b^{2} x^{2} {\rm Li}_2\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 12 \, b^{2} x^{2} {\rm Li}_2\left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) + 4 \, a^{3} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 24 \, b x {\rm polylog}\left (3, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + 24 \, b x {\rm polylog}\left (3, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right ) - 4 \, {\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\cosh \left (b x + a\right ) - \sinh \left (b x + a\right ) + 1\right ) - 24 \, {\rm polylog}\left (4, \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) - 24 \, {\rm polylog}\left (4, -\cosh \left (b x + a\right ) - \sinh \left (b x + a\right )\right )}{4 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*coth(b*x+a),x, algorithm="fricas")

[Out]

-1/4*(b^4*x^4 - 4*b^3*x^3*log(cosh(b*x + a) + sinh(b*x + a) + 1) - 12*b^2*x^2*dilog(cosh(b*x + a) + sinh(b*x +
a)) - 12*b^2*x^2*dilog(-cosh(b*x + a) - sinh(b*x + a)) + 4*a^3*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 24*b*
x*polylog(3, cosh(b*x + a) + sinh(b*x + a)) + 24*b*x*polylog(3, -cosh(b*x + a) - sinh(b*x + a)) - 4*(b^3*x^3 +
a^3)*log(-cosh(b*x + a) - sinh(b*x + a) + 1) - 24*polylog(4, cosh(b*x + a) + sinh(b*x + a)) - 24*polylog(4, -
cosh(b*x + a) - sinh(b*x + a)))/b^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \coth {\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*coth(b*x+a),x)

[Out]

Integral(x**3*coth(a + b*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(x^3*coth(b*x+a),x, algorithm="giac")

[Out]

integrate(x^3*coth(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {coth}\left (a+b\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*coth(a + b*x),x)

[Out]

int(x^3*coth(a + b*x), x)

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