### 3.88 $$\int \text{sech}^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx$$

Optimal. Leaf size=48 $-\frac{(a-b) \tanh ^3(c+d x)}{3 d}+\frac{a \tanh (c+d x)}{d}-\frac{b \tanh ^5(c+d x)}{5 d}$

[Out]

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

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Rubi [A]  time = 0.0398919, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {3675, 373} $-\frac{(a-b) \tanh ^3(c+d x)}{3 d}+\frac{a \tanh (c+d x)}{d}-\frac{b \tanh ^5(c+d x)}{5 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[Sech[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 373

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

Rubi steps

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

Mathematica [A]  time = 0.0475146, size = 86, normalized size = 1.79 $-\frac{a \tanh ^3(c+d x)}{3 d}+\frac{a \tanh (c+d x)}{d}+\frac{2 b \tanh (c+d x)}{15 d}-\frac{b \tanh (c+d x) \text{sech}^4(c+d x)}{5 d}+\frac{b \tanh (c+d x) \text{sech}^2(c+d x)}{15 d}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sech[c + d*x]^4*(a + b*Tanh[c + d*x]^2),x]

[Out]

(a*Tanh[c + d*x])/d + (2*b*Tanh[c + d*x])/(15*d) + (b*Sech[c + d*x]^2*Tanh[c + d*x])/(15*d) - (b*Sech[c + d*x]
^4*Tanh[c + d*x])/(5*d) - (a*Tanh[c + d*x]^3)/(3*d)

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Maple [A]  time = 0.044, size = 75, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \right ) \tanh \left ( dx+c \right ) +b \left ( -{\frac{\sinh \left ( dx+c \right ) }{4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{\tanh \left ( dx+c \right ) }{4} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) } \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+b*(-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(8/15+1/5*sech(d*x+c)^4+4/15*
sech(d*x+c)^2)*tanh(d*x+c)))

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Maxima [B]  time = 1.10757, size = 501, normalized size = 10.44 \begin{align*} \frac{4}{15} \, b{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac{5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac{4}{3} \, a{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac{1}{d{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.91973, size = 934, normalized size = 19.46 \begin{align*} -\frac{8 \,{\left (2 \,{\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right )^{3} + 6 \,{\left (5 \, a + 4 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} +{\left (5 \, a + 7 \, b\right )} \sinh \left (d x + c\right )^{3} + 30 \, a \cosh \left (d x + c\right ) +{\left (3 \,{\left (5 \, a + 7 \, b\right )} \cosh \left (d x + c\right )^{2} + 5 \, a - 5 \, b\right )} \sinh \left (d x + c\right )\right )}}{15 \,{\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 5 \, d \cosh \left (d x + c\right )^{5} +{\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (7 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 11 \, d \cosh \left (d x + c\right )^{3} +{\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} +{\left (21 \, d \cosh \left (d x + c\right )^{5} + 50 \, d \cosh \left (d x + c\right )^{3} + 33 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) +{\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*(2*(5*a + 4*b)*cosh(d*x + c)^3 + 6*(5*a + 4*b)*cosh(d*x + c)*sinh(d*x + c)^2 + (5*a + 7*b)*sinh(d*x + c)
^3 + 30*a*cosh(d*x + c) + (3*(5*a + 7*b)*cosh(d*x + c)^2 + 5*a - 5*b)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*
cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 5*d*cosh(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*
x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c)^4 + 11*d*cosh(d*x + c)^3 + (35*d*cosh(d*x
+ c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + (21*d*cosh(d*x + c)^5 + 50*d*cosh(d*x + c)^3 + 33*d*co
sh(d*x + c))*sinh(d*x + c)^2 + 15*d*cosh(d*x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*
x + c)^2 + 5*d)*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname{sech}^{4}{\left (c + d x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.34274, size = 128, normalized size = 2.67 \begin{align*} -\frac{4 \,{\left (15 \, a e^{\left (6 \, d x + 6 \, c\right )} + 15 \, b e^{\left (6 \, d x + 6 \, c\right )} + 35 \, a e^{\left (4 \, d x + 4 \, c\right )} - 5 \, b e^{\left (4 \, d x + 4 \, c\right )} + 25 \, a e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b e^{\left (2 \, d x + 2 \, c\right )} + 5 \, a + b\right )}}{15 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*(15*a*e^(6*d*x + 6*c) + 15*b*e^(6*d*x + 6*c) + 35*a*e^(4*d*x + 4*c) - 5*b*e^(4*d*x + 4*c) + 25*a*e^(2*d*
x + 2*c) + 5*b*e^(2*d*x + 2*c) + 5*a + b)/(d*(e^(2*d*x + 2*c) + 1)^5)