3.207 \(\int e^{a+b x} \tanh ^3(a+b x) \, dx\)

Optimal. Leaf size=77 \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac {3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2282, 390, 1158, 12, 288, 203} \[ \frac {e^{a+b x}}{b}+\frac {3 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac {2 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )^2}-\frac {3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Tanh[a + b*x]^3,x]

[Out]

E^(a + b*x)/b - (2*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))^2) + (3*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))) - (3*A
rcTan[E^(a + b*x)])/b

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*x*(d + e*x
^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*
(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int e^{a+b x} \tanh ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \left (1-\frac {2 \left (1+3 x^4\right )}{\left (1+x^2\right )^3}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 \operatorname {Subst}\left (\int \frac {1+3 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}+\frac {\operatorname {Subst}\left (\int -\frac {12 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{2 b}\\ &=\frac {e^{a+b x}}{b}-\frac {2 e^{a+b x}}{b \left (1+e^{2 a+2 b x}\right )^2}-\frac {6 \operatorname {Subst}\left (\int \frac {x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\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 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\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 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 60, normalized size = 0.78 \[ \frac {e^{a+b x} \left (5 e^{2 (a+b x)}+e^{4 (a+b x)}+2\right )}{b \left (e^{2 (a+b x)}+1\right )^2}-\frac {3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Tanh[a + b*x]^3,x]

[Out]

(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

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fricas [B]  time = 0.51, size = 339, normalized size = 4.40 \[ \frac {\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} + \sinh \left (b x + a\right )^{5} + 5 \, {\left (2 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{3} + 5 \, \cosh \left (b x + a\right )^{3} + 5 \, {\left (2 \, \cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + {\left (5 \, \cosh \left (b x + a\right )^{4} + 15 \, \cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right ) + 2 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} + 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} + b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} + b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="fricas")

[Out]

(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)

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giac [A]  time = 0.65, size = 52, normalized size = 0.68 \[ \frac {\frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{2}} - 3 \, \arctan \left (e^{\left (b x + a\right )}\right ) + e^{\left (b x + a\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="giac")

[Out]

((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

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maple [A]  time = 0.22, size = 103, normalized size = 1.34 \[ \frac {\sinh ^{3}\left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}+\frac {3 \sinh \left (b x +a \right )}{b \cosh \left (b x +a \right )^{2}}-\frac {3 \,\mathrm {sech}\left (b x +a \right ) \tanh \left (b x +a \right )}{2 b}-\frac {3 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}+\frac {\sinh ^{2}\left (b x +a \right )}{b \cosh \left (b x +a \right )}+\frac {2}{b \cosh \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*tanh(b*x+a)^3,x)

[Out]

1/b*sinh(b*x+a)^3/cosh(b*x+a)^2+3/b*sinh(b*x+a)/cosh(b*x+a)^2-3/2/b*sech(b*x+a)*tanh(b*x+a)-3*arctan(exp(b*x+a
))/b+1/b*sinh(b*x+a)^2/cosh(b*x+a)+2/b/cosh(b*x+a)

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maxima [A]  time = 0.43, size = 69, normalized size = 0.90 \[ -\frac {3 \, \arctan \left (e^{\left (b x + a\right )}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} + \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} + e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} + 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^3,x, algorithm="maxima")

[Out]

-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))

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mupad [B]  time = 0.08, size = 93, normalized size = 1.21 \[ \frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {3\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )}+\frac {3\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)*tanh(a + b*x)^3,x)

[Out]

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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a} \int e^{b x} \tanh ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)**3,x)

[Out]

exp(a)*Integral(exp(b*x)*tanh(a + b*x)**3, x)

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