3.206 \(\int e^{a+b x} \tanh ^4(a+b x) \, dx\)

Optimal. Leaf size=107 \[ \frac{e^{a+b x}}{b}+\frac{5 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{14 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^2}+\frac{8 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

E^(a + b*x)/b + (8*E^(a + b*x))/(3*b*(1 + E^(2*a + 2*b*x))^3) - (14*E^(a + b*x))/(3*b*(1 + E^(2*a + 2*b*x))^2)
+ (5*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))) - (3*ArcTan[E^(a + b*x)])/b

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Rubi [A]  time = 0.0685635, antiderivative size = 107, normalized size of antiderivative = 1., 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, 1258, 1157, 385, 203} \[ \frac{e^{a+b x}}{b}+\frac{5 e^{a+b x}}{b \left (e^{2 a+2 b x}+1\right )}-\frac{14 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^2}+\frac{8 e^{a+b x}}{3 b \left (e^{2 a+2 b x}+1\right )^3}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Tanh[a + b*x]^4,x]

[Out]

E^(a + b*x)/b + (8*E^(a + b*x))/(3*b*(1 + E^(2*a + 2*b*x))^3) - (14*E^(a + b*x))/(3*b*(1 + E^(2*a + 2*b*x))^2)
+ (5*E^(a + b*x))/(b*(1 + E^(2*a + 2*b*x))) - (3*ArcTan[E^(a + b*x)])/b

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]]]

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 1258

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(m/2 - 1)*(c*d^
2 + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[1/(2*e^(2*p + m/2)*(q + 1)), Int[(d +
e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*e^(2*p + m/2)*(q + 1)*x^m*(a + c*x^4)^p - (-d)^(m/2 - 1)*(c*d^2 + a
*e^2)^p*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x], x] /; FreeQ[{a, c, d, e}, x] && IGtQ[p, 0] && ILtQ[q, -1
] && IGtQ[m/2, 0]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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])

Rubi steps

\begin{align*} \int e^{a+b x} \tanh ^4(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^4}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1-\frac{8 x^2 \left (1+x^4\right )}{\left (1+x^2\right )^4}\right ) \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{e^{a+b x}}{b}-\frac{8 \operatorname{Subst}\left (\int \frac{x^2 \left (1+x^4\right )}{\left (1+x^2\right )^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\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{4 \operatorname{Subst}\left (\int \frac{-2+6 x^2-6 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\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{\operatorname{Subst}\left (\int \frac{-6+24 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{a+b x}\right )}{3 b}\\ &=\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 \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\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 \tan ^{-1}\left (e^{a+b x}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 0.131919, size = 76, normalized size = 0.71 \[ \frac{e^{a+b x} \left (25 e^{2 (a+b x)}+24 e^{4 (a+b x)}+3 e^{6 (a+b x)}+12\right )}{3 b \left (e^{2 (a+b x)}+1\right )^3}-\frac{3 \tan ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Tanh[a + b*x]^4,x]

[Out]

(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

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Maple [A]  time = 0.013, size = 143, normalized size = 1.3 \begin{align*}{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{4}}{b \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}+{\frac{4\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}}-{\frac{8\, \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3\,b\cosh \left ( bx+a \right ) }}+{\frac{8\,\cosh \left ( bx+a \right ) }{3\,b}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{3}}{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}+3\,{\frac{\sinh \left ( bx+a \right ) }{b \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}}-{\frac{3\,{\rm sech} \left (bx+a\right )\tanh \left ( bx+a \right ) }{2\,b}}-3\,{\frac{\arctan \left ({{\rm e}^{bx+a}} \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*tanh(b*x+a)^4,x)

[Out]

1/b*sinh(b*x+a)^4/cosh(b*x+a)^3+4/3/b*sinh(b*x+a)^2/cosh(b*x+a)^3-8/3/b*sinh(b*x+a)^2/cosh(b*x+a)+8/3/b*cosh(b
*x+a)+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

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Maxima [A]  time = 1.57505, size = 127, normalized size = 1.19 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="maxima")

[Out]

-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))

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Fricas [B]  time = 2.24032, size = 1685, normalized size = 15.75 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="fricas")

[Out]

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)*sin
h(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*(21*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*cosh(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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \tanh ^{4}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)**4,x)

[Out]

exp(a)*Integral(exp(b*x)*tanh(a + b*x)**4, x)

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Giac [A]  time = 1.80442, size = 92, normalized size = 0.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*tanh(b*x+a)^4,x, algorithm="giac")

[Out]

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