3.90 \(\int \text {sech}^4(a+b x) \tanh ^2(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b} \]

[Out]

1/3*tanh(b*x+a)^3/b-1/5*tanh(b*x+a)^5/b

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Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2607, 14} \[ \frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Sech[a + b*x]^4*Tanh[a + b*x]^2,x]

[Out]

Tanh[a + b*x]^3/(3*b) - Tanh[a + b*x]^5/(5*b)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

\begin {align*} \int \text {sech}^4(a+b x) \tanh ^2(a+b x) \, dx &=\frac {i \operatorname {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {i \operatorname {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,i \tanh (a+b x)\right )}{b}\\ &=\frac {\tanh ^3(a+b x)}{3 b}-\frac {\tanh ^5(a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 56, normalized size = 1.81 \[ \frac {2 \tanh (a+b x)}{15 b}-\frac {\tanh (a+b x) \text {sech}^4(a+b x)}{5 b}+\frac {\tanh (a+b x) \text {sech}^2(a+b x)}{15 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[a + b*x]^4*Tanh[a + b*x]^2,x]

[Out]

(2*Tanh[a + b*x])/(15*b) + (Sech[a + b*x]^2*Tanh[a + b*x])/(15*b) - (Sech[a + b*x]^4*Tanh[a + b*x])/(5*b)

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fricas [B]  time = 0.44, size = 304, normalized size = 9.81 \[ -\frac {8 \, {\left (8 \, \cosh \left (b x + a\right )^{3} + 24 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 7 \, \sinh \left (b x + a\right )^{3} + {\left (21 \, \cosh \left (b x + a\right )^{2} - 5\right )} \sinh \left (b x + a\right )\right )}}{15 \, {\left (b \cosh \left (b x + a\right )^{7} + 7 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + b \sinh \left (b x + a\right )^{7} + 5 \, b \cosh \left (b x + a\right )^{5} + {\left (21 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )^{5} + 5 \, {\left (7 \, b \cosh \left (b x + a\right )^{3} + 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 11 \, b \cosh \left (b x + a\right )^{3} + {\left (35 \, b \cosh \left (b x + a\right )^{4} + 50 \, b \cosh \left (b x + a\right )^{2} + 9 \, b\right )} \sinh \left (b x + a\right )^{3} + {\left (21 \, b \cosh \left (b x + a\right )^{5} + 50 \, b \cosh \left (b x + a\right )^{3} + 33 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 15 \, b \cosh \left (b x + a\right ) + {\left (7 \, b \cosh \left (b x + a\right )^{6} + 25 \, b \cosh \left (b x + a\right )^{4} + 27 \, b \cosh \left (b x + a\right )^{2} + 5 \, b\right )} \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="fricas")

[Out]

-8/15*(8*cosh(b*x + a)^3 + 24*cosh(b*x + a)*sinh(b*x + a)^2 + 7*sinh(b*x + a)^3 + (21*cosh(b*x + a)^2 - 5)*sin
h(b*x + a))/(b*cosh(b*x + a)^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 + 5*b*cosh(b*x + a)^5 +
(21*b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a)^3 + 5*b*cosh(b*x + a))*sinh(b*x + a)^4 +
11*b*cosh(b*x + a)^3 + (35*b*cosh(b*x + a)^4 + 50*b*cosh(b*x + a)^2 + 9*b)*sinh(b*x + a)^3 + (21*b*cosh(b*x +
a)^5 + 50*b*cosh(b*x + a)^3 + 33*b*cosh(b*x + a))*sinh(b*x + a)^2 + 15*b*cosh(b*x + a) + (7*b*cosh(b*x + a)^6
+ 25*b*cosh(b*x + a)^4 + 27*b*cosh(b*x + a)^2 + 5*b)*sinh(b*x + a))

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giac [A]  time = 0.14, size = 53, normalized size = 1.71 \[ -\frac {4 \, {\left (15 \, e^{\left (6 \, b x + 6 \, a\right )} - 5 \, e^{\left (4 \, b x + 4 \, a\right )} + 5 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}}{15 \, b {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="giac")

[Out]

-4/15*(15*e^(6*b*x + 6*a) - 5*e^(4*b*x + 4*a) + 5*e^(2*b*x + 2*a) + 1)/(b*(e^(2*b*x + 2*a) + 1)^5)

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maple [A]  time = 0.32, size = 52, normalized size = 1.68 \[ \frac {-\frac {\sinh \left (b x +a \right )}{4 \cosh \left (b x +a \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\mathrm {sech}\left (b x +a \right )^{4}}{5}+\frac {4 \mathrm {sech}\left (b x +a \right )^{2}}{15}\right ) \tanh \left (b x +a \right )}{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(b*x+a)^4*tanh(b*x+a)^2,x)

[Out]

1/b*(-1/4*sinh(b*x+a)/cosh(b*x+a)^5+1/4*(8/15+1/5*sech(b*x+a)^4+4/15*sech(b*x+a)^2)*tanh(b*x+a))

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maxima [B]  time = 0.39, size = 276, normalized size = 8.90 \[ \frac {4 \, e^{\left (-2 \, b x - 2 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} - \frac {4 \, e^{\left (-4 \, b x - 4 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac {4 \, e^{\left (-6 \, b x - 6 \, a\right )}}{b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} + \frac {4}{15 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} + 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)^4*tanh(b*x+a)^2,x, algorithm="maxima")

[Out]

4/3*e^(-2*b*x - 2*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) +
e^(-10*b*x - 10*a) + 1)) - 4/3*e^(-4*b*x - 4*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x -
6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1)) + 4*e^(-6*b*x - 6*a)/(b*(5*e^(-2*b*x - 2*a) + 10*e^(-4*b
*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1)) + 4/15/(b*(5*e^(-2*b*x - 2*a)
+ 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) + 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) + 1))

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mupad [B]  time = 0.13, size = 270, normalized size = 8.71 \[ \frac {\frac {8}{15\,b}-\frac {4\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}}{3\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}+1}-\frac {\frac {2}{5\,b}-\frac {8\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}+\frac {6\,{\mathrm {e}}^{4\,a+4\,b\,x}}{5\,b}}{4\,{\mathrm {e}}^{2\,a+2\,b\,x}+6\,{\mathrm {e}}^{4\,a+4\,b\,x}+4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1}-\frac {\frac {8\,{\mathrm {e}}^{2\,a+2\,b\,x}}{5\,b}-\frac {16\,{\mathrm {e}}^{4\,a+4\,b\,x}}{5\,b}+\frac {8\,{\mathrm {e}}^{6\,a+6\,b\,x}}{5\,b}}{5\,{\mathrm {e}}^{2\,a+2\,b\,x}+10\,{\mathrm {e}}^{4\,a+4\,b\,x}+10\,{\mathrm {e}}^{6\,a+6\,b\,x}+5\,{\mathrm {e}}^{8\,a+8\,b\,x}+{\mathrm {e}}^{10\,a+10\,b\,x}+1}-\frac {2}{5\,b\,\left (2\,{\mathrm {e}}^{2\,a+2\,b\,x}+{\mathrm {e}}^{4\,a+4\,b\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(a + b*x)^2/cosh(a + b*x)^4,x)

[Out]

(8/(15*b) - (4*exp(2*a + 2*b*x))/(5*b))/(3*exp(2*a + 2*b*x) + 3*exp(4*a + 4*b*x) + exp(6*a + 6*b*x) + 1) - (2/
(5*b) - (8*exp(2*a + 2*b*x))/(5*b) + (6*exp(4*a + 4*b*x))/(5*b))/(4*exp(2*a + 2*b*x) + 6*exp(4*a + 4*b*x) + 4*
exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1) - ((8*exp(2*a + 2*b*x))/(5*b) - (16*exp(4*a + 4*b*x))/(5*b) + (8*exp(
6*a + 6*b*x))/(5*b))/(5*exp(2*a + 2*b*x) + 10*exp(4*a + 4*b*x) + 10*exp(6*a + 6*b*x) + 5*exp(8*a + 8*b*x) + ex
p(10*a + 10*b*x) + 1) - 2/(5*b*(2*exp(2*a + 2*b*x) + exp(4*a + 4*b*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \tanh ^{2}{\left (a + b x \right )} \operatorname {sech}^{4}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(b*x+a)**4*tanh(b*x+a)**2,x)

[Out]

Integral(tanh(a + b*x)**2*sech(a + b*x)**4, x)

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