Integrand size = 16, antiderivative size = 77 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {3 \arctan \left (e^{a+b x}\right )}{b} \] Output:
exp(b*x+a)/b-2*exp(b*x+a)/b/(1+exp(2*b*x+2*a))^2+3*exp(b*x+a)/b/(1+exp(2*b *x+2*a))-3*arctan(exp(b*x+a))/b
Time = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.78 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {e^{a+b x} \left (2+5 e^{2 (a+b x)}+e^{4 (a+b x)}\right )}{b \left (1+e^{2 (a+b x)}\right )^2}-\frac {3 \arctan \left (e^{a+b x}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Tanh[a + b*x]^3,x]
Output:
(E^(a + b*x)*(2 + 5*E^(2*(a + b*x)) + E^(4*(a + b*x))))/(b*(1 + E^(2*(a + b*x)))^2) - (3*ArcTan[E^(a + b*x)])/b
Time = 0.36 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2720, 25, 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.
\(\displaystyle \int e^{a+b x} \tanh ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {\left (1-e^{2 a+2 b x}\right )^3}{\left (1+e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\left (1-e^{2 a+2 b x}\right )^3}{\left (1+e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {\int \left (\frac {2 \left (1+3 e^{4 a+4 b x}\right )}{\left (1+e^{2 a+2 b x}\right )^3}-1\right )de^{a+b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-3 \arctan \left (e^{a+b x}\right )+e^{a+b x}+\frac {3 e^{a+b x}}{e^{2 a+2 b x}+1}-\frac {2 e^{a+b x}}{\left (e^{2 a+2 b x}+1\right )^2}}{b}\) |
Input:
Int[E^(a + b*x)*Tanh[a + b*x]^3,x]
Output:
(E^(a + b*x) - (2*E^(a + b*x))/(1 + E^(2*a + 2*b*x))^2 + (3*E^(a + b*x))/( 1 + E^(2*a + 2*b*x)) - 3*ArcTan[E^(a + b*x)])/b
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 = 0.71 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {{\mathrm e}^{b x +a} \left (3 \,{\mathrm e}^{2 b x +2 a}+1\right )}{b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{2}}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}\) | \(80\) |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )+\frac {\sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )}+\frac {2}{\cosh \left (b x +a \right )}}{b}\) | \(89\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )+\frac {\sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )}+\frac {2}{\cosh \left (b x +a \right )}}{b}\) | \(89\) |
Input:
int(exp(b*x+a)*tanh(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)/b+exp(b*x+a)*(3*exp(2*b*x+2*a)+1)/b/(1+exp(2*b*x+2*a))^2+3/2*I/ b*ln(exp(b*x+a)-I)-3/2*I/b*ln(exp(b*x+a)+I)
Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (71) = 142\).
Time = 0.09 (sec) , antiderivative size = 339, normalized size of antiderivative = 4.40 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 5 \, {\left (2 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )^{3} + 5 \, {\left (2 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (5 \, \cosh \left (b x + a\right )^{4} + 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 2 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="fricas")
Output:
(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a)^4 + sinh(b*x + a)^5 + 5*( 2*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^3 + 5*cosh(b*x + a)^3 + 5*(2*cosh(b*x + a)^3 + 3*cosh(b*x + a))*sinh(b*x + a)^2 - 3*(cosh(b*x + a)^4 + 4*cosh(b *x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 2*(3*cosh(b*x + a)^2 + 1)*sinh (b*x + a)^2 + 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 + cosh(b*x + a))*sinh (b*x + a) + 1)*arctan(cosh(b*x + a) + sinh(b*x + a)) + (5*cosh(b*x + a)^4 + 15*cosh(b*x + a)^2 + 2)*sinh(b*x + a) + 2*cosh(b*x + a))/(b*cosh(b*x + a )^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh(b*x + a)^4 + 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 + b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a) ^3 + b*cosh(b*x + a))*sinh(b*x + a) + b)
\[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=e^{a} \int e^{b x} \tanh ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)**3,x)
Output:
exp(a)*Integral(exp(b*x)*tanh(a + b*x)**3, x)
Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=-\frac {3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="maxima")
Output:
-3*arctan(e^(b*x + a))/b + e^(b*x + a)/b + (3*e^(3*b*x + 3*a) + e^(b*x + a ))/(b*(e^(4*b*x + 4*a) + 2*e^(2*b*x + 2*a) + 1))
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.68 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {\frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} - 3 \, \arctan \left (e^{\left (b x + a\right )}\right ) + e^{\left (b x + a\right )}}{b} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="giac")
Output:
((3*e^(3*b*x + 3*a) + e^(b*x + a))/(e^(2*b*x + 2*a) + 1)^2 - 3*arctan(e^(b *x + a)) + e^(b*x + a))/b
Time = 2.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.21 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \] Input:
int(exp(a + b*x)*tanh(a + b*x)^3,x)
Output:
exp(a + b*x)/b - (3*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - ( 2*exp(a + b*x))/(b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1)) + (3*exp(a + b*x))/(b*(exp(2*a + 2*b*x) + 1))
Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.45 \[ \int e^{a+b x} \tanh ^3(a+b x) \, dx=\frac {-3 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right )-6 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right )-3 \mathit {atan} \left (e^{b x +a}\right )+e^{5 b x +5 a}+5 e^{3 b x +3 a}+2 e^{b x +a}}{b \left (e^{4 b x +4 a}+2 e^{2 b x +2 a}+1\right )} \] Input:
int(exp(b*x+a)*tanh(b*x+a)^3,x)
Output:
( - 3*e**(4*a + 4*b*x)*atan(e**(a + b*x)) - 6*e**(2*a + 2*b*x)*atan(e**(a + b*x)) - 3*atan(e**(a + b*x)) + e**(5*a + 5*b*x) + 5*e**(3*a + 3*b*x) + 2 *e**(a + b*x))/(b*(e**(4*a + 4*b*x) + 2*e**(2*a + 2*b*x) + 1))