Integrand size = 16, antiderivative size = 107 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {8 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^3}-\frac {14 e^{a+b x}}{3 b \left (1+e^{2 a+2 b x}\right )^2}+\frac {5 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )}-\frac {3 \arctan \left (e^{a+b x}\right )}{b} \] Output:
exp(b*x+a)/b+8/3*exp(b*x+a)/b/(1+exp(2*b*x+2*a))^3-14/3*exp(b*x+a)/b/(1+ex p(2*b*x+2*a))^2+5*exp(b*x+a)/b/(1+exp(2*b*x+2*a))-3*arctan(exp(b*x+a))/b
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.71 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=\frac {e^{a+b x} \left (12+25 e^{2 (a+b x)}+24 e^{4 (a+b x)}+3 e^{6 (a+b x)}\right )}{3 b \left (1+e^{2 (a+b x)}\right )^3}-\frac {3 \arctan \left (e^{a+b x}\right )}{b} \] Input:
Integrate[E^(a + b*x)*Tanh[a + b*x]^4,x]
Output:
(E^(a + b*x)*(12 + 25*E^(2*(a + b*x)) + 24*E^(4*(a + b*x)) + 3*E^(6*(a + b *x))))/(3*b*(1 + E^(2*(a + b*x)))^3) - (3*ArcTan[E^(a + b*x)])/b
Time = 0.41 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2720, 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.
\(\displaystyle \int e^{a+b x} \tanh ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {\left (1-e^{2 a+2 b x}\right )^4}{\left (1+e^{2 a+2 b x}\right )^4}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {\int \left (1-\frac {8 e^{2 a+2 b x} \left (1+e^{4 a+4 b x}\right )}{\left (1+e^{2 a+2 b x}\right )^4}\right )de^{a+b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-3 \arctan \left (e^{a+b x}\right )+e^{a+b x}+\frac {5 e^{a+b x}}{e^{2 a+2 b x}+1}-\frac {14 e^{a+b x}}{3 \left (e^{2 a+2 b x}+1\right )^2}+\frac {8 e^{a+b x}}{3 \left (e^{2 a+2 b x}+1\right )^3}}{b}\) |
Input:
Int[E^(a + b*x)*Tanh[a + b*x]^4,x]
Output:
(E^(a + b*x) + (8*E^(a + b*x))/(3*(1 + E^(2*a + 2*b*x))^3) - (14*E^(a + b* x))/(3*(1 + E^(2*a + 2*b*x))^2) + (5*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 = 0.80 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {{\mathrm e}^{b x +a} \left (15 \,{\mathrm e}^{4 b x +4 a}+16 \,{\mathrm e}^{2 b x +2 a}+9\right )}{3 b \left (1+{\mathrm e}^{2 b x +2 a}\right )^{3}}+\frac {3 i \ln \left ({\mathrm e}^{b x +a}-i\right )}{2 b}-\frac {3 i \ln \left ({\mathrm e}^{b x +a}+i\right )}{2 b}\) | \(92\) |
derivativedivides | \(\frac {\frac {\sinh \left (b x +a \right )^{4}}{\cosh \left (b x +a \right )^{3}}+\frac {4 \sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )^{3}}+\frac {8}{3 \cosh \left (b x +a \right )^{3}}+\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(107\) |
default | \(\frac {\frac {\sinh \left (b x +a \right )^{4}}{\cosh \left (b x +a \right )^{3}}+\frac {4 \sinh \left (b x +a \right )^{2}}{\cosh \left (b x +a \right )^{3}}+\frac {8}{3 \cosh \left (b x +a \right )^{3}}+\frac {\sinh \left (b x +a \right )^{3}}{\cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{\cosh \left (b x +a \right )^{2}}-\frac {3 \,\operatorname {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2}-3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\) | \(107\) |
Input:
int(exp(b*x+a)*tanh(b*x+a)^4,x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)/b+1/3*exp(b*x+a)*(15*exp(4*b*x+4*a)+16*exp(2*b*x+2*a)+9)/b/(1+e xp(2*b*x+2*a))^3+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. 604 vs. \(2 (95) = 190\).
Time = 0.10 (sec) , antiderivative size = 604, normalized size of antiderivative = 5.64 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx =\text {Too large to display} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="fricas")
Output:
1/3*(3*cosh(b*x + a)^7 + 21*cosh(b*x + a)*sinh(b*x + a)^6 + 3*sinh(b*x + a )^7 + 3*(21*cosh(b*x + a)^2 + 8)*sinh(b*x + a)^5 + 24*cosh(b*x + a)^5 + 15 *(7*cosh(b*x + a)^3 + 8*cosh(b*x + a))*sinh(b*x + a)^4 + 5*(21*cosh(b*x + a)^4 + 48*cosh(b*x + a)^2 + 5)*sinh(b*x + a)^3 + 25*cosh(b*x + a)^3 + 3*(2 1*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*cos h(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)*arctan(cosh(b *x + a) + sinh(b*x + a)) + 3*(7*cosh(b*x + a)^6 + 40*cosh(b*x + a)^4 + 25* cosh(b*x + a)^2 + 4)*sinh(b*x + a) + 12*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)
\[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=e^{a} \int e^{b x} \tanh ^{4}{\left (a + b x \right )}\, dx \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)**4,x)
Output:
exp(a)*Integral(exp(b*x)*tanh(a + b*x)**4, x)
Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.88 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=-\frac {3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} + 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{3 \, b {\left (e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="maxima")
Output:
-3*arctan(e^(b*x + a))/b + e^(b*x + a)/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))
Time = 0.27 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.64 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=\frac {\frac {15 \, e^{\left (5 \, b x + 5 \, a\right )} + 16 \, e^{\left (3 \, b x + 3 \, a\right )} + 9 \, e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{3}} - 9 \, \arctan \left (e^{\left (b x + a\right )}\right ) + 3 \, e^{\left (b x + a\right )}}{3 \, b} \] Input:
integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="giac")
Output:
1/3*((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 - 9*arctan(e^(b*x + a)) + 3*e^(b*x + a))/b
Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.45 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}+\frac {\frac {4\,{\mathrm {e}}^{a+b\,x}}{3\,b}+\frac {4\,{\mathrm {e}}^{5\,a+5\,b\,x}}{3\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {11\,{\mathrm {e}}^{a+b\,x}}{3\,b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \] Input:
int(exp(a + b*x)*tanh(a + b*x)^4,x)
Output:
exp(a + b*x)/b - (3*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) + ( (4*exp(a + b*x))/(3*b) + (4*exp(5*a + 5*b*x))/(3*b))/(3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1) - (2*exp(a + b*x))/(b*(2*exp(2 *a + 2*b*x) + exp(4*a + 4*b*x) + 1)) + (11*exp(a + b*x))/(3*b*(exp(2*a + 2 *b*x) + 1))
Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.49 \[ \int e^{a+b x} \tanh ^4(a+b x) \, dx=\frac {-9 e^{6 b x +6 a} \mathit {atan} \left (e^{b x +a}\right )-27 e^{4 b x +4 a} \mathit {atan} \left (e^{b x +a}\right )-27 e^{2 b x +2 a} \mathit {atan} \left (e^{b x +a}\right )-9 \mathit {atan} \left (e^{b x +a}\right )+3 e^{7 b x +7 a}+24 e^{5 b x +5 a}+25 e^{3 b x +3 a}+12 e^{b x +a}}{3 b \left (e^{6 b x +6 a}+3 e^{4 b x +4 a}+3 e^{2 b x +2 a}+1\right )} \] Input:
int(exp(b*x+a)*tanh(b*x+a)^4,x)
Output:
( - 9*e**(6*a + 6*b*x)*atan(e**(a + b*x)) - 27*e**(4*a + 4*b*x)*atan(e**(a + b*x)) - 27*e**(2*a + 2*b*x)*atan(e**(a + b*x)) - 9*atan(e**(a + b*x)) + 3*e**(7*a + 7*b*x) + 24*e**(5*a + 5*b*x) + 25*e**(3*a + 3*b*x) + 12*e**(a + b*x))/(3*b*(e**(6*a + 6*b*x) + 3*e**(4*a + 4*b*x) + 3*e**(2*a + 2*b*x) + 1))