3.114 \(\int (a+b \text{sech}^2(c+d x))^2 \tanh ^2(c+d x) \, dx\)

Optimal. Leaf size=59 \[ -\frac{a^2 \tanh (c+d x)}{d}+a^2 x+\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.0980225, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4141, 1802, 206} \[ -\frac{a^2 \tanh (c+d x)}{d}+a^2 x+\frac{b (2 a+b) \tanh ^3(c+d x)}{3 d}-\frac{b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4141

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

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 206

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

Rubi steps

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

Mathematica [B]  time = 0.89966, size = 281, normalized size = 4.76 \[ \frac{\text{sech}(c) \text{sech}^5(c+d x) \left (120 a^2 \sinh (2 c+d x)-120 a^2 \sinh (2 c+3 d x)+30 a^2 \sinh (4 c+3 d x)-30 a^2 \sinh (4 c+5 d x)+150 a^2 d x \cosh (2 c+d x)+75 a^2 d x \cosh (2 c+3 d x)+75 a^2 d x \cosh (4 c+3 d x)+15 a^2 d x \cosh (4 c+5 d x)+15 a^2 d x \cosh (6 c+5 d x)-180 a^2 \sinh (d x)+150 a^2 d x \cosh (d x)-120 a b \sinh (2 c+d x)+40 a b \sinh (2 c+3 d x)-60 a b \sinh (4 c+3 d x)+20 a b \sinh (4 c+5 d x)+80 a b \sinh (d x)-60 b^2 \sinh (2 c+d x)+20 b^2 \sinh (2 c+3 d x)+4 b^2 \sinh (4 c+5 d x)-20 b^2 \sinh (d x)\right )}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sech[c]*Sech[c + d*x]^5*(150*a^2*d*x*Cosh[d*x] + 150*a^2*d*x*Cosh[2*c + d*x] + 75*a^2*d*x*Cosh[2*c + 3*d*x] +
75*a^2*d*x*Cosh[4*c + 3*d*x] + 15*a^2*d*x*Cosh[4*c + 5*d*x] + 15*a^2*d*x*Cosh[6*c + 5*d*x] - 180*a^2*Sinh[d*x
] + 80*a*b*Sinh[d*x] - 20*b^2*Sinh[d*x] + 120*a^2*Sinh[2*c + d*x] - 120*a*b*Sinh[2*c + d*x] - 60*b^2*Sinh[2*c
+ d*x] - 120*a^2*Sinh[2*c + 3*d*x] + 40*a*b*Sinh[2*c + 3*d*x] + 20*b^2*Sinh[2*c + 3*d*x] + 30*a^2*Sinh[4*c + 3
*d*x] - 60*a*b*Sinh[4*c + 3*d*x] - 30*a^2*Sinh[4*c + 5*d*x] + 20*a*b*Sinh[4*c + 5*d*x] + 4*b^2*Sinh[4*c + 5*d*
x]))/(480*d)

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Maple [B]  time = 0.037, size = 115, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( dx+c-\tanh \left ( dx+c \right ) \right ) +2\,ab \left ( -1/2\,{\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+1/2\, \left ( 2/3+1/3\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) \tanh \left ( dx+c \right ) \right ) +{b}^{2} \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((a+b*sech(d*x+c)^2)^2*tanh(d*x+c)^2,x)

[Out]

1/d*(a^2*(d*x+c-tanh(d*x+c))+2*a*b*(-1/2*sinh(d*x+c)/cosh(d*x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c))+b^
2*(-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 = 1.18822, size = 439, normalized size = 7.44 \begin{align*} \frac{2 \, a b \tanh \left (d x + c\right )^{3}}{3 \, d} + a^{2}{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac{4}{15} \, b^{2}{\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*b*tanh(d*x + c)^3/d + a^2*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 4/15*b^2*(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)))

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Fricas [B]  time = 2.07574, size = 1099, normalized size = 18.63 \begin{align*} \frac{{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} -{\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} + 5 \,{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 5 \,{\left (2 \,{\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \,{\left (2 \,{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \,{\left (15 \, a^{2} d x + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \,{\left ({\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 3 \,{\left (9 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 6 \, a^{2} + 4 \, a b + 8 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^5 + 5*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d
*x + c)*sinh(d*x + c)^4 - (15*a^2 - 10*a*b - 2*b^2)*sinh(d*x + c)^5 + 5*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)
*cosh(d*x + c)^3 - 5*(2*(15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^2 + 9*a^2 + 2*a*b - 2*b^2)*sinh(d*x + c)^3 + 5
*(2*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c)^3 + 3*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x
+ c))*sinh(d*x + c)^2 + 10*(15*a^2*d*x + 15*a^2 - 10*a*b - 2*b^2)*cosh(d*x + c) - 5*((15*a^2 - 10*a*b - 2*b^2
)*cosh(d*x + c)^4 + 3*(9*a^2 + 2*a*b - 2*b^2)*cosh(d*x + c)^2 + 6*a^2 + 4*a*b + 8*b^2)*sinh(d*x + c))/(d*cosh(
d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x +
c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.23893, size = 261, normalized size = 4.42 \begin{align*} \frac{15 \, a^{2} d x + \frac{2 \,{\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} - 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 30 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 40 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^2*d*x + 2*(15*a^2*e^(8*d*x + 8*c) - 30*a*b*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) - 60*a*b*e^(6*d
*x + 6*c) - 30*b^2*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) - 40*a*b*e^(4*d*x + 4*c) + 10*b^2*e^(4*d*x + 4*c)
+ 60*a^2*e^(2*d*x + 2*c) - 20*a*b*e^(2*d*x + 2*c) - 10*b^2*e^(2*d*x + 2*c) + 15*a^2 - 10*a*b - 2*b^2)/(e^(2*d*
x + 2*c) + 1)^5)/d