Integrand size = 16, antiderivative size = 120 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1-e^{2 d+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1-e^{2 d+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {3 e^{a-d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
exp(b*x+a)/b+8/3*exp(b*x+a)/b/(1-exp(2*b*x+2*d))^3-14/3*exp(b*x+a)/b/(1-ex p(2*b*x+2*d))^2+5*exp(b*x+a)/b/(1-exp(2*b*x+2*d))-3*exp(a-d)*arctanh(exp(b *x+d))/b
Time = 0.26 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.42 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=\frac {e^a \left (3 e^{b x}-9 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) \cosh (d)+9 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) \sinh (d)-\frac {8 e^{b x} (\cosh (d)-\sinh (d))^3}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^3}-\frac {14 e^{b x} (\cosh (d)-\sinh (d))^2}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}-\frac {15 e^{b x} (\cosh (d)-\sinh (d))}{\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)}\right )}{3 b} \] Input:
Integrate[E^(a + b*x)*Coth[d + b*x]^4,x]
Output:
(E^a*(3*E^(b*x) - 9*ArcTanh[E^(b*x)*(Cosh[d] + Sinh[d])]*Cosh[d] + 9*ArcTa nh[E^(b*x)*(Cosh[d] + Sinh[d])]*Sinh[d] - (8*E^(b*x)*(Cosh[d] - Sinh[d])^3 )/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^3 - (14*E^(b*x)*(Co sh[d] - Sinh[d])^2)/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^2 - (15*E^(b*x)*(Cosh[d] - Sinh[d]))/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2* b*x))*Sinh[d])))/(3*b)
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 27, 300, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{a+b x} \coth ^4(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {e^a \left (1+e^{2 b x}\right )^4}{\left (1-e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^a \int \frac {\left (1+e^{2 b x}\right )^4}{\left (1-e^{2 b x}\right )^4}de^{b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {e^a \int \left (\frac {8 e^{2 b x} \left (1+e^{4 b x}\right )}{\left (1-e^{2 b x}\right )^4}+1\right )de^{b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^a \left (-3 \text {arctanh}\left (e^{b x}\right )+e^{b x}+\frac {5 e^{b x}}{1-e^{2 b x}}-\frac {14 e^{b x}}{3 \left (1-e^{2 b x}\right )^2}+\frac {8 e^{b x}}{3 \left (1-e^{2 b x}\right )^3}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Coth[d + b*x]^4,x]
Output:
(E^a*(E^(b*x) + (8*E^(b*x))/(3*(1 - E^(2*b*x))^3) - (14*E^(b*x))/(3*(1 - E ^(2*b*x))^2) + (5*E^(b*x))/(1 - E^(2*b*x)) - 3*ArcTanh[E^(b*x)]))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 0.33 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {\left (15 \,{\mathrm e}^{4 b x +4 a +4 d}-16 \,{\mathrm e}^{2 b x +4 a +2 d}+9 \,{\mathrm e}^{4 a}\right ) {\mathrm e}^{b x +3 a}}{3 \left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{3} b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(133\) |
Input:
int(exp(b*x+a)*coth(b*x+d)^4,x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)/b+1/3/(-exp(2*b*x+2*a+2*d)+exp(2*a))^3/b*(15*exp(4*b*x+4*a+4*d) -16*exp(2*b*x+4*a+2*d)+9*exp(4*a))*exp(b*x+3*a)+3/2*ln(exp(b*x+a)-exp(a-d) )/b*exp(a-d)-3/2*ln(exp(b*x+a)+exp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 1693 vs. \(2 (101) = 202\).
Time = 0.09 (sec) , antiderivative size = 1693, normalized size of antiderivative = 14.11 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^4,x, algorithm="fricas")
Output:
1/6*(6*cosh(b*x + d)^7*cosh(-a + d) + 6*(cosh(-a + d) - sinh(-a + d))*sinh (b*x + d)^7 + 42*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d)) *sinh(b*x + d)^6 - 48*cosh(b*x + d)^5*cosh(-a + d) + 6*(21*cosh(b*x + d)^2 *cosh(-a + d) - (21*cosh(b*x + d)^2 - 8)*sinh(-a + d) - 8*cosh(-a + d))*si nh(b*x + d)^5 + 30*(7*cosh(b*x + d)^3*cosh(-a + d) - 8*cosh(b*x + d)*cosh( -a + d) - (7*cosh(b*x + d)^3 - 8*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d )^4 + 50*cosh(b*x + d)^3*cosh(-a + d) + 10*(21*cosh(b*x + d)^4*cosh(-a + d ) - 48*cosh(b*x + d)^2*cosh(-a + d) - (21*cosh(b*x + d)^4 - 48*cosh(b*x + d)^2 + 5)*sinh(-a + d) + 5*cosh(-a + d))*sinh(b*x + d)^3 + 6*(21*cosh(b*x + d)^5*cosh(-a + d) - 80*cosh(b*x + d)^3*cosh(-a + d) + 25*cosh(b*x + d)*c osh(-a + d) - (21*cosh(b*x + d)^5 - 80*cosh(b*x + d)^3 + 25*cosh(b*x + d)) *sinh(-a + d))*sinh(b*x + d)^2 - 24*cosh(b*x + d)*cosh(-a + d) - 9*(cosh(b *x + d)^6*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^6 + 6 *(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^5 - 3*cosh(b*x + d)^4*cosh(-a + d) + 3*(5*cosh(b*x + d)^2*cosh(-a + d) - (5 *cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh(b*x + d)^4 + 4*(5* cosh(b*x + d)^3*cosh(-a + d) - 3*cosh(b*x + d)*cosh(-a + d) - (5*cosh(b*x + d)^3 - 3*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^3 + 3*cosh(b*x + d)^ 2*cosh(-a + d) + 3*(5*cosh(b*x + d)^4*cosh(-a + d) - 6*cosh(b*x + d)^2*cos h(-a + d) - (5*cosh(b*x + d)^4 - 6*cosh(b*x + d)^2 + 1)*sinh(-a + d) + ...
\[ \int e^{a+b x} \coth ^4(d+b x) \, dx=e^{a} \int e^{b x} \coth ^{4}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*coth(b*x+d)**4,x)
Output:
exp(a)*Integral(exp(b*x)*coth(b*x + d)**4, x)
Time = 0.05 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.25 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=-\frac {3 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{2 \, b} + \frac {3 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{2 \, b} + \frac {e^{\left (b x + a\right )}}{b} - \frac {15 \, e^{\left (5 \, b x + 7 \, a + 4 \, d\right )} - 16 \, e^{\left (3 \, b x + 7 \, a + 2 \, d\right )} + 9 \, e^{\left (b x + 7 \, a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a + 6 \, d\right )} - 3 \, e^{\left (4 \, b x + 6 \, a + 4 \, d\right )} + 3 \, e^{\left (2 \, b x + 6 \, a + 2 \, d\right )} - e^{\left (6 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^4,x, algorithm="maxima")
Output:
-3/2*e^(a - d)*log(e^(b*x + a + d) + e^a)/b + 3/2*e^(a - d)*log(e^(b*x + a + d) - e^a)/b + e^(b*x + a)/b - 1/3*(15*e^(5*b*x + 7*a + 4*d) - 16*e^(3*b *x + 7*a + 2*d) + 9*e^(b*x + 7*a))/(b*(e^(6*b*x + 6*a + 6*d) - 3*e^(4*b*x + 6*a + 4*d) + 3*e^(2*b*x + 6*a + 2*d) - e^(6*a)))
Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.98 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=-\frac {9 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right ) - 9 \, e^{\left (a - d\right )} \log \left ({\left | e^{\left (b x + a + d\right )} - e^{a} \right |}\right ) + \frac {2 \, {\left (15 \, e^{\left (5 \, b x + 7 \, a + 4 \, d\right )} - 16 \, e^{\left (3 \, b x + 7 \, a + 2 \, d\right )} + 9 \, e^{\left (b x + 7 \, a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} - e^{\left (2 \, a\right )}\right )}^{3}} - 6 \, e^{\left (b x + a\right )}}{6 \, b} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^4,x, algorithm="giac")
Output:
-1/6*(9*e^(a - d)*log(e^(b*x + a + d) + e^a) - 9*e^(a - d)*log(abs(e^(b*x + a + d) - e^a)) + 2*(15*e^(5*b*x + 7*a + 4*d) - 16*e^(3*b*x + 7*a + 2*d) + 9*e^(b*x + 7*a))/(e^(2*b*x + 2*a + 2*d) - e^(2*a))^3 - 6*e^(b*x + a))/b
Timed out. \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=\int {\mathrm {coth}\left (d+b\,x\right )}^4\,{\mathrm {e}}^{a+b\,x} \,d x \] Input:
int(coth(d + b*x)^4*exp(a + b*x),x)
Output:
int(coth(d + b*x)^4*exp(a + b*x), x)
Time = 0.21 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.11 \[ \int e^{a+b x} \coth ^4(d+b x) \, dx=\frac {e^{a} \left (6 e^{7 b x +7 d}+9 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}-1\right )-9 e^{6 b x +6 d} \mathrm {log}\left (e^{b x +d}+1\right )-48 e^{5 b x +5 d}-27 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )+27 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )+50 e^{3 b x +3 d}+27 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )-27 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )-24 e^{b x +d}-9 \,\mathrm {log}\left (e^{b x +d}-1\right )+9 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{6 e^{d} b \left (e^{6 b x +6 d}-3 e^{4 b x +4 d}+3 e^{2 b x +2 d}-1\right )} \] Input:
int(exp(b*x+a)*coth(b*x+d)^4,x)
Output:
(e**a*(6*e**(7*b*x + 7*d) + 9*e**(6*b*x + 6*d)*log(e**(b*x + d) - 1) - 9*e **(6*b*x + 6*d)*log(e**(b*x + d) + 1) - 48*e**(5*b*x + 5*d) - 27*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) + 27*e**(4*b*x + 4*d)*log(e**(b*x + d) + 1) + 50*e**(3*b*x + 3*d) + 27*e**(2*b*x + 2*d)*log(e**(b*x + d) - 1) - 27*e** (2*b*x + 2*d)*log(e**(b*x + d) + 1) - 24*e**(b*x + d) - 9*log(e**(b*x + d) - 1) + 9*log(e**(b*x + d) + 1)))/(6*e**d*b*(e**(6*b*x + 6*d) - 3*e**(4*b* x + 4*d) + 3*e**(2*b*x + 2*d) - 1))