Integrand size = 16, antiderivative size = 84 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1+e^{2 d+2 b x}\right )}-\frac {3 e^{a-d} \arctan \left (e^{d+b x}\right )}{b} \] 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)*arctan(exp(b*x+d))/b
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.49 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=\frac {e^a \left (e^{b x}-3 \arctan \left (e^{b x} (\cosh (d)+\sinh (d))\right ) \cosh (d)+3 \arctan \left (e^{b x} (\cosh (d)+\sinh (d))\right ) \sinh (d)-\frac {2 e^{b x} (\cosh (d)-\sinh (d))^2}{\left (\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)\right )^2}+\frac {3 e^{b x} (\cosh (d)-\sinh (d))}{\left (1+e^{2 b x}\right ) \cosh (d)+\left (-1+e^{2 b x}\right ) \sinh (d)}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Tanh[d + b*x]^3,x]
Output:
(E^a*(E^(b*x) - 3*ArcTan[E^(b*x)*(Cosh[d] + Sinh[d])]*Cosh[d] + 3*ArcTan[E ^(b*x)*(Cosh[d] + Sinh[d])]*Sinh[d] - (2*E^(b*x)*(Cosh[d] - Sinh[d])^2)/(( 1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2*b*x))*Sinh[d])^2 + (3*E^(b*x)*(Cosh[d] - Sinh[d]))/((1 + E^(2*b*x))*Cosh[d] + (-1 + E^(2*b*x))*Sinh[d])))/b
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69, 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.
\(\displaystyle \int e^{a+b x} \tanh ^3(b x+d) \, dx\) |
\(\Big \downarrow \) 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 |
\(\displaystyle -\frac {e^a \left (3 \arctan \left (e^{b x}\right )-e^{b x}-\frac {3 e^{b x}}{e^{2 b x}+1}+\frac {2 e^{b x}}{\left (e^{2 b x}+1\right )^2}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Tanh[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*ArcTan[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]]]
Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.43
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {\left (3 \,{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) {\mathrm e}^{b x +3 a}}{\left ({\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right )^{2} b}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i {\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{2 b}\) | \(120\) |
Input:
int(exp(b*x+a)*tanh(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)+exp (2*a))*exp(b*x+3*a)+3/2*I*ln(exp(b*x+a)-I*exp(a-d))/b*exp(a-d)-3/2*I*ln(ex p(b*x+a)+I*exp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 702 vs. \(2 (77) = 154\).
Time = 0.10 (sec) , antiderivative size = 702, normalized size of antiderivative = 8.36 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*tanh(b*x+d)^3,x, algorithm="fricas")
Output:
(cosh(b*x + d)^5*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d )^5 + 5*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^4 + 5*cosh(b*x + d)^3*cosh(-a + d) + 5*(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 + 5*(2*cosh(b*x + d)^3*cosh(-a + d) + 3*cosh(b*x + d)*cosh(-a + d) - (2*co sh(b*x + d)^3 + 3*cosh(b*x + d))*sinh(-a + d))*sinh(b*x + d)^2 - 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*(co sh(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)^2 + 1)*sinh(-a + d) + cosh(-a + d))*arctan(cosh(b*x + d) + sinh(b *x + d)) + 2*cosh(b*x + d)*cosh(-a + d) + (5*cosh(b*x + d)^4*cosh(-a + d) + 15*cosh(b*x + d)^2*cosh(-a + d) - (5*cosh(b*x + d)^4 + 15*cosh(b*x + d)^ 2 + 2)*sinh(-a + d) + 2*cosh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^5 + 5 *cosh(b*x + d)^3 + 2*cosh(b*x + d))*sinh(-a + d))/(b*cosh(b*x + d)^4 + 4*b *cosh(b*x + d)*sinh(b*x + d)^3 + b*sinh(b*x + d)^4 + 2*b*cosh(b*x + d)^2 + 2*(3*b*cosh(b*x + d)^2 + b)*sinh(b*x + d)^2 + 4*(b*cosh(b*x + d)^3 + b*co sh(b*x + d))*sinh(b*x + d) + b)
\[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=e^{a} \int e^{b x} \tanh ^{3}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*tanh(b*x+d)**3,x)
Output:
exp(a)*Integral(exp(b*x)*tanh(b*x + d)**3, x)
Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.06 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=-\frac {3 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )}}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {3 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )} + e^{\left (b x + 5 \, a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a + 4 \, d\right )} + 2 \, e^{\left (2 \, b x + 4 \, a + 2 \, d\right )} + e^{\left (4 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*tanh(b*x+d)^3,x, algorithm="maxima")
Output:
-3*arctan(e^(b*x + d))*e^(a - d)/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 = 73, normalized size of antiderivative = 0.87 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=-\frac {3 \, \arctan \left (e^{\left (b x + d\right )}\right ) e^{\left (a - d\right )} - \frac {3 \, e^{\left (3 \, b x + 5 \, a + 2 \, d\right )} + e^{\left (b x + 5 \, a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} + e^{\left (2 \, a\right )}\right )}^{2}} - e^{\left (b x + a\right )}}{b} \] Input:
integrate(exp(b*x+a)*tanh(b*x+d)^3,x, algorithm="giac")
Output:
-(3*arctan(e^(b*x + d))*e^(a - d) - (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 - e^(b*x + a))/b
Timed out. \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=\int {\mathrm {e}}^{a+b\,x}\,{\mathrm {tanh}\left (d+b\,x\right )}^3 \,d x \] Input:
int(exp(a + b*x)*tanh(d + b*x)^3,x)
Output:
int(exp(a + b*x)*tanh(d + b*x)^3, x)
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.43 \[ \int e^{a+b x} \tanh ^3(d+b x) \, dx=\frac {e^{a} \left (-3 e^{4 b x +4 d} \mathit {atan} \left (e^{b x +d}\right )-6 e^{2 b x +2 d} \mathit {atan} \left (e^{b x +d}\right )-3 \mathit {atan} \left (e^{b x +d}\right )+e^{5 b x +5 d}+5 e^{3 b x +3 d}+2 e^{b x +d}\right )}{e^{d} b \left (e^{4 b x +4 d}+2 e^{2 b x +2 d}+1\right )} \] Input:
int(exp(b*x+a)*tanh(b*x+d)^3,x)
Output:
(e**a*( - 3*e**(4*b*x + 4*d)*atan(e**(b*x + d)) - 6*e**(2*b*x + 2*d)*atan( e**(b*x + d)) - 3*atan(e**(b*x + d)) + e**(5*b*x + 5*d) + 5*e**(3*b*x + 3* d) + 2*e**(b*x + d)))/(e**d*b*(e**(4*b*x + 4*d) + 2*e**(2*b*x + 2*d) + 1))