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

Optimal. Leaf size=113 \[ \frac{e^{a+b x}}{b}+\frac{5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac{8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[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

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Rubi [A]  time = 0.0716714, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2282, 390, 1258, 1157, 385, 206} \[ \frac{e^{a+b x}}{b}+\frac{5 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac{14 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^2}+\frac{8 e^{a+b x}}{3 b \left (1-e^{2 a+2 b x}\right )^3}-\frac{3 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[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 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 390

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 1258

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*(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)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, c, d, e}, x] && IGtQ[p, 0] && ILtQ[q, -1
] && IGtQ[m/2, 0]

Rule 1157

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] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

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 206

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

Rubi steps

\begin{align*} \int e^{a+b x} \coth ^4(a+b x) \, dx &=\frac{\operatorname{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{\operatorname{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 \operatorname{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 \operatorname{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{\operatorname{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 \operatorname{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 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 10.1054, size = 115, normalized size = 1.02 \[ \frac{-24 e^{a+b x}+50 e^{3 (a+b x)}-48 e^{5 (a+b x)}+6 e^{7 (a+b x)}+9 \left (e^{2 (a+b x)}-1\right )^3 \log \left (1-e^{a+b x}\right )-9 \left (e^{2 (a+b x)}-1\right )^3 \log \left (e^{a+b x}+1\right )}{6 b \left (e^{2 (a+b x)}-1\right )^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.014, size = 123, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}-3\,{\frac{\cosh \left ( bx+a \right ) }{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}}+{\frac{3\,{\rm csch} \left (bx+a\right ){\rm coth} \left (bx+a\right )}{2}}-3\,{\it Artanh} \left ({{\rm e}^{bx+a}} \right ) +{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}}-{\frac{4\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3\, \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}}-{\frac{8\, \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3\,\sinh \left ( bx+a \right ) }}+{\frac{8\,\sinh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(cosh(b*x+a)^3/sinh(b*x+a)^2-3/sinh(b*x+a)^2*cosh(b*x+a)+3/2*csch(b*x+a)*coth(b*x+a)-3*arctanh(exp(b*x+a))
+cosh(b*x+a)^4/sinh(b*x+a)^3-4/3/sinh(b*x+a)^3*cosh(b*x+a)^2-8/3*cosh(b*x+a)^2/sinh(b*x+a)+8/3*sinh(b*x+a))

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Maxima [A]  time = 1.1656, size = 149, normalized size = 1.32 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [B]  time = 2.11681, size = 2221, normalized size = 19.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.29063, size = 112, normalized size = 0.99 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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