\(\int e^{a+b x} \coth ^4(a+b x) \, dx\) [213]

Optimal result

Integrand size = 16, antiderivative size = 113 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {arctanh}\left (e^{a+b x}\right )}{b} \]

[Out]

exp(b*x+a)/b+8/3*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^3-14/3*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^2+5*exp(b*x+a)/b/(1-ex
p(2*b*x+2*a))-3*arctanh(exp(b*x+a))/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2320, 398, 1272, 1171, 393, 212} \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=-\frac {3 \text {arctanh}\left (e^{a+b x}\right )}{b}+\frac {e^{a+b x}}{b}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3} \]

[In]

Int[E^(a + b*x)*Coth[a + b*x]^4,x]

[Out]

E^(a + b*x)/b + (8*E^(a + b*x))/(3*b*(1 - E^(2*a + 2*b*x))^3) - (14*E^(a + b*x))/(3*b*(1 - E^(2*a + 2*b*x))^2)
+ (5*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (3*ArcTanh[E^(a + b*x)])/b

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1272

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2
+ a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Dist[1/(2*e^(2*p + m/2)*(q + 1)), Int[(d +
e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e*x^2))*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + c*x^4)^p - (-d)^(m/2 -
1)*(c*d^2 + a*e^2)^p*(d + e*(2*q + 3)*x^2))], x], x], x] /; FreeQ[{a, c, d, e}, x] && IGtQ[p, 0] && ILtQ[q, -1
] && IGtQ[m/2, 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]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^4}{\left (1-x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {8 x^2 \left (1+x^4\right )}{\left (1-x^2\right )^4}\right ) \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {8 \text {Subst}\left (\int \frac {x^2 \left (1+x^4\right )}{\left (1-x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}+\frac {4 \text {Subst}\left (\int \frac {-2-6 x^2-6 x^4}{\left (1-x^2\right )^3} \, dx,x,e^{a+b x}\right )}{3 b} \\ & = \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}-\frac {\text {Subst}\left (\int \frac {-6-24 x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{3 b} \\ & = \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {3 \text {arctanh}\left (e^{a+b x}\right )}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.13 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.02 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=\frac {-24 e^{a+b x}+50 e^{3 (a+b x)}-48 e^{5 (a+b x)}+6 e^{7 (a+b x)}+9 \left (-1+e^{2 (a+b x)}\right )^3 \log \left (1-e^{a+b x}\right )-9 \left (-1+e^{2 (a+b x)}\right )^3 \log \left (1+e^{a+b x}\right )}{6 b \left (-1+e^{2 (a+b x)}\right )^3} \]

[In]

Integrate[E^(a + b*x)*Coth[a + b*x]^4,x]

[Out]

(-24*E^(a + b*x) + 50*E^(3*(a + b*x)) - 48*E^(5*(a + b*x)) + 6*E^(7*(a + b*x)) + 9*(-1 + E^(2*(a + b*x)))^3*Lo
g[1 - E^(a + b*x)] - 9*(-1 + E^(2*(a + b*x)))^3*Log[1 + E^(a + b*x)])/(6*b*(-1 + E^(2*(a + b*x)))^3)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {{\mathrm e}^{b x +a}}{b}-\frac {{\mathrm e}^{b x +a} \left (15 \,{\mathrm e}^{4 b x +4 a}-16 \,{\mathrm e}^{2 b x +2 a}+9\right )}{3 b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{3}}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}\)	\(88\)
derivativedivides	\(\frac {\frac {\cosh \left (b x +a \right )^{3}}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\operatorname {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\cosh \left (b x +a \right )^{4}}{\sinh \left (b x +a \right )^{3}}-\frac {4 \cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )^{3}}+\frac {8}{3 \sinh \left (b x +a \right )^{3}}}{b}\)	\(107\)
default	\(\frac {\frac {\cosh \left (b x +a \right )^{3}}{\sinh \left (b x +a \right )^{2}}-\frac {3 \cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {3 \,\operatorname {csch}\left (b x +a \right ) \coth \left (b x +a \right )}{2}-3 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\cosh \left (b x +a \right )^{4}}{\sinh \left (b x +a \right )^{3}}-\frac {4 \cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )^{3}}+\frac {8}{3 \sinh \left (b x +a \right )^{3}}}{b}\)	\(107\)

[In]

int(exp(b*x+a)*coth(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

exp(b*x+a)/b-1/3*exp(b*x+a)*(15*exp(4*b*x+4*a)-16*exp(2*b*x+2*a)+9)/b/(exp(2*b*x+2*a)-1)^3-3/2/b*ln(exp(b*x+a)
+1)+3/2/b*ln(exp(b*x+a)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 796 vs. \(2 (95) = 190\).

Time = 0.26 (sec) , antiderivative size = 796, normalized size of antiderivative = 7.04 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=\text {Too large to display} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^4,x, algorithm="fricas")

[Out]

1/6*(6*cosh(b*x + a)^7 + 42*cosh(b*x + a)*sinh(b*x + a)^6 + 6*sinh(b*x + a)^7 + 6*(21*cosh(b*x + a)^2 - 8)*sin
h(b*x + a)^5 - 48*cosh(b*x + a)^5 + 30*(7*cosh(b*x + a)^3 - 8*cosh(b*x + a))*sinh(b*x + a)^4 + 10*(21*cosh(b*x
+ a)^4 - 48*cosh(b*x + a)^2 + 5)*sinh(b*x + a)^3 + 50*cosh(b*x + a)^3 + 6*(21*cosh(b*x + a)^5 - 80*cosh(b*x +
a)^3 + 25*cosh(b*x + a))*sinh(b*x + a)^2 - 9*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x +
a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))
*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh
(b*x + a)^5 - 2*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) - 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 9
*(cosh(b*x + a)^6 + 6*cosh(b*x + a)*sinh(b*x + a)^5 + sinh(b*x + a)^6 + 3*(5*cosh(b*x + a)^2 - 1)*sinh(b*x + a
)^4 - 3*cosh(b*x + a)^4 + 4*(5*cosh(b*x + a)^3 - 3*cosh(b*x + a))*sinh(b*x + a)^3 + 3*(5*cosh(b*x + a)^4 - 6*c
osh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 3*cosh(b*x + a)^2 + 6*(cosh(b*x + a)^5 - 2*cosh(b*x + a)^3 + cosh(b*x +
a))*sinh(b*x + a) - 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 6*(7*cosh(b*x + a)^6 - 40*cosh(b*x + a)^4 + 25
*cosh(b*x + a)^2 - 4)*sinh(b*x + a) - 24*cosh(b*x + a))/(b*cosh(b*x + a)^6 + 6*b*cosh(b*x + a)*sinh(b*x + a)^5
+ b*sinh(b*x + a)^6 - 3*b*cosh(b*x + a)^4 + 3*(5*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^4 + 4*(5*b*cosh(b*x + a
)^3 - 3*b*cosh(b*x + a))*sinh(b*x + a)^3 + 3*b*cosh(b*x + a)^2 + 3*(5*b*cosh(b*x + a)^4 - 6*b*cosh(b*x + a)^2
+ b)*sinh(b*x + a)^2 + 6*(b*cosh(b*x + a)^5 - 2*b*cosh(b*x + a)^3 + b*cosh(b*x + a))*sinh(b*x + a) - b)

Sympy [F]

\[ \int e^{a+b x} \coth ^4(a+b x) \, dx=e^{a} \int e^{b x} \coth ^{4}{\left (a + b x \right )}\, dx \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)**4,x)

[Out]

exp(a)*Integral(exp(b*x)*coth(a + b*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.97 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=\frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, \log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {3 \, \log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^4,x, algorithm="maxima")

[Out]

e^(b*x + a)/b - 3/2*log(e^(b*x + a) + 1)/b + 3/2*log(e^(b*x + a) - 1)/b - 1/3*(15*e^(5*b*x + 5*a) - 16*e^(3*b*
x + 3*a) + 9*e^(b*x + a))/(b*(e^(6*b*x + 6*a) - 3*e^(4*b*x + 4*a) + 3*e^(2*b*x + 2*a) - 1))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, e^{\left (5 \, b x + 5 \, a\right )} - 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{3}} - 6 \, e^{\left (b x + a\right )} + 9 \, \log \left (e^{\left (b x + a\right )} + 1\right ) - 9 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{6 \, b} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^4,x, algorithm="giac")

[Out]

-1/6*(2*(15*e^(5*b*x + 5*a) - 16*e^(3*b*x + 3*a) + 9*e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^3 - 6*e^(b*x + a) + 9*
log(e^(b*x + a) + 1) - 9*log(abs(e^(b*x + a) - 1)))/b

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.42 \[ \int e^{a+b x} \coth ^4(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {\frac {4\,{\mathrm {e}}^{a+b\,x}}{3\,b}+\frac {4\,{\mathrm {e}}^{5\,a+5\,b\,x}}{3\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {11\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

[In]

int(coth(a + b*x)^4*exp(a + b*x),x)

[Out]

exp(a + b*x)/b - (3*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - ((4*exp(a + b*x))/(3*b) + (4*exp(5*
a + 5*b*x))/(3*b))/(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) - 1) - (2*exp(a + b*x))/(b*(exp
(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1)) - (11*exp(a + b*x))/(3*b*(exp(2*a + 2*b*x) - 1))