Integrand size = 16, antiderivative size = 88 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {3 e^{a-d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
exp(b*x+a)/b-2*exp(b*x+a)/b/(1-exp(2*b*x+2*d))^2+3*exp(b*x+a)/b/(1-exp(2*b *x+2*d))-3*exp(a-d)*arctanh(exp(b*x+d))/b
Time = 0.31 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=\frac {e^a \left (-3 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) (\cosh (d)-\sinh (d))+\frac {e^{b x} \left (-5 e^{2 b x}+\left (2+e^{4 b x}\right ) \cosh (2 d)+\left (-2+e^{4 b x}\right ) \sinh (2 d)\right )}{\left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )^2}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Coth[d + b*x]^3,x]
Output:
(E^a*(-3*ArcTanh[E^(b*x)*(Cosh[d] + Sinh[d])]*(Cosh[d] - Sinh[d]) + (E^(b* x)*(-5*E^(2*b*x) + (2 + E^(4*b*x))*Cosh[2*d] + (-2 + E^(4*b*x))*Sinh[2*d]) )/((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d])^2))/b
Time = 0.22 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2720, 25, 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.
\(\displaystyle \int e^{a+b x} \coth ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^a \left (1+e^{2 b x}\right )^3}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^a \left (1+e^{2 b x}\right )^3}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^a \int \frac {\left (1+e^{2 b x}\right )^3}{\left (1-e^{2 b x}\right )^3}de^{b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {e^a \int \left (\frac {2 \left (1+3 e^{4 b x}\right )}{\left (1-e^{2 b x}\right )^3}-1\right )de^{b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^a \left (3 \text {arctanh}\left (e^{b x}\right )-e^{b x}-\frac {3 e^{b x}}{1-e^{2 b x}}+\frac {2 e^{b x}}{\left (1-e^{2 b x}\right )^2}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Coth[d + b*x]^3,x]
Output:
-((E^a*(-E^(b*x) + (2*E^(b*x))/(1 - E^(2*b*x))^2 - (3*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 = 0.29 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {\left (-3 \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +3 a}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}-\frac {3 \ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}+\frac {3 \ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(116\) |
Input:
int(exp(b*x+a)*coth(b*x+d)^3,x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)/b+1/(-exp(2*b*x+2*a+2*d)+exp(2*a))^2/b*(-3*exp(2*b*x+2*a+2*d)+e xp(2*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. 980 vs. \(2 (77) = 154\).
Time = 0.09 (sec) , antiderivative size = 980, normalized size of antiderivative = 11.14 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=\text {Too large to display} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^3,x, algorithm="fricas")
Output:
1/2*(2*cosh(b*x + d)^5*cosh(-a + d) + 2*(cosh(-a + d) - sinh(-a + d))*sinh (b*x + d)^5 + 10*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d)) *sinh(b*x + d)^4 - 10*cosh(b*x + d)^3*cosh(-a + d) + 10*(2*cosh(b*x + d)^2 *cosh(-a + d) - (2*cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh( b*x + d)^3 + 10*(2*cosh(b*x + d)^3*cosh(-a + d) - 3*cosh(b*x + d)*cosh(-a + d) - (2*cosh(b*x + d)^3 - 3*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^2 + 4*cosh(b*x + d)*cosh(-a + d) - 3*(cosh(b*x + d)^4*cosh(-a + d) + (cosh( -a + d) - sinh(-a + d))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^3 - 2*cosh(b*x + d)^2*cosh(-a + d) + 2*(3*cosh(b*x + d)^2*cosh(-a + d) - (3*cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d)^3*cosh(-a + d) - co sh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^3 - cosh(b*x + d))*sinh(-a + d)) *sinh(b*x + d) - (cosh(b*x + d)^4 - 2*cosh(b*x + d)^2 + 1)*sinh(-a + d) + cosh(-a + d))*log(cosh(b*x + d) + sinh(b*x + d) + 1) + 3*(cosh(b*x + d)^4* cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^4 + 4*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^3 - 2*cosh( b*x + d)^2*cosh(-a + d) + 2*(3*cosh(b*x + d)^2*cosh(-a + d) - (3*cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*sinh(b*x + d)^2 + 4*(cosh(b*x + d )^3*cosh(-a + d) - cosh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^3 - cosh(b* x + d))*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^4 - 2*cosh(b*x + d...
\[ \int e^{a+b x} \coth ^3(d+b x) \, dx=e^{a} \int e^{b x} \coth ^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*coth(b*x+d)**3,x)
Output:
exp(a)*Integral(exp(b*x)*coth(b*x + d)**3, x)
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.36 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=-\frac {3 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{2 \, b} + \frac {3 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{2 \, b} + \frac {e^{\left (b x + a\right )}}{b} - \frac {3 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )} - e^{\left (b x + 5 \, a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a + 4 \, d\right )} - 2 \, e^{\left (2 \, b x + 4 \, a + 2 \, d\right )} + e^{\left (4 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^3,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 - (3*e^(3*b*x + 5*a + 2*d) - e^(b*x + 5*a)) /(b*(e^(4*b*x + 4*a + 4*d) - 2*e^(2*b*x + 4*a + 2*d) + e^(4*a)))
Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.17 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=-\frac {3 \, e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right ) - 3 \, e^{\left (a - d\right )} \log \left ({\left | e^{\left (b x + a + d\right )} - e^{a} \right |}\right ) + \frac {2 \, {\left (3 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )} - e^{\left (b x + 5 \, a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} - e^{\left (2 \, a\right )}\right )}^{2}} - 2 \, e^{\left (b x + a\right )}}{2 \, b} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^3,x, algorithm="giac")
Output:
-1/2*(3*e^(a - d)*log(e^(b*x + a + d) + e^a) - 3*e^(a - d)*log(abs(e^(b*x + a + d) - e^a)) + 2*(3*e^(3*b*x + 5*a + 2*d) - e^(b*x + 5*a))/(e^(2*b*x + 2*a + 2*d) - e^(2*a))^2 - 2*e^(b*x + a))/b
Timed out. \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=\int {\mathrm {coth}\left (d+b\,x\right )}^3\,{\mathrm {e}}^{a+b\,x} \,d x \] Input:
int(coth(d + b*x)^3*exp(a + b*x),x)
Output:
int(coth(d + b*x)^3*exp(a + b*x), x)
Time = 0.23 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.10 \[ \int e^{a+b x} \coth ^3(d+b x) \, dx=\frac {e^{a} \left (2 e^{5 b x +5 d}+3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}-1\right )-3 e^{4 b x +4 d} \mathrm {log}\left (e^{b x +d}+1\right )-10 e^{3 b x +3 d}-6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )+6 e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )+4 e^{b x +d}+3 \,\mathrm {log}\left (e^{b x +d}-1\right )-3 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{2 e^{d} b \left (e^{4 b x +4 d}-2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*coth(b*x+d)^3,x)
Output:
(e**a*(2*e**(5*b*x + 5*d) + 3*e**(4*b*x + 4*d)*log(e**(b*x + d) - 1) - 3*e **(4*b*x + 4*d)*log(e**(b*x + d) + 1) - 10*e**(3*b*x + 3*d) - 6*e**(2*b*x + 2*d)*log(e**(b*x + d) - 1) + 6*e**(2*b*x + 2*d)*log(e**(b*x + d) + 1) + 4*e**(b*x + d) + 3*log(e**(b*x + d) - 1) - 3*log(e**(b*x + d) + 1)))/(2*e* *d*b*(e**(4*b*x + 4*d) - 2*e**(2*b*x + 2*d) + 1))