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

Optimal. Leaf size=49 \[ \frac{a^2 \tanh (c+d x)}{d}+\frac{2 a b \tanh ^3(c+d x)}{3 d}+\frac{b^2 \tanh ^5(c+d x)}{5 d} \]

[Out]

(a^2*Tanh[c + d*x])/d + (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 = 0.0530052, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3675, 194} \[ \frac{a^2 \tanh (c+d x)}{d}+\frac{2 a b \tanh ^3(c+d x)}{3 d}+\frac{b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Tanh[c + d*x])/d + (2*a*b*Tanh[c + d*x]^3)/(3*d) + (b^2*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 194

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

Rubi steps

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

Mathematica [A]  time = 0.179594, size = 49, normalized size = 1. \[ \frac{a^2 \tanh (c+d x)}{d}+\frac{2 a b \tanh ^3(c+d x)}{3 d}+\frac{b^2 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.052, size = 126, normalized size = 2.6 \begin{align*}{\frac{1}{d} \left ({a}^{2}\tanh \left ( dx+c \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{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{3\,\tanh \left ( dx+c \right ) }{8} \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)^2*(a+b*tanh(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*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*(-1/2*
sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/8*(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 [A]  time = 1.16467, size = 72, normalized size = 1.47 \begin{align*} \frac{b^{2} \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac{2 \, a b \tanh \left (d x + c\right )^{3}}{3 \, d} + \frac{2 \, a^{2}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*tanh(d*x + c)^5/d + 2/3*a*b*tanh(d*x + c)^3/d + 2*a^2/(d*(e^(-2*d*x - 2*c) + 1))

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Fricas [B]  time = 1.86226, size = 1023, normalized size = 20.88 \begin{align*} -\frac{4 \,{\left ({\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 8 \,{\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \,{\left (3 \, a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 20 \, a b\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 20 \, a b + 15 \, b^{2} + 8 \,{\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \,{\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \,{\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \,{\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \,{\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \,{\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/15*((15*a^2 + 20*a*b + 9*b^2)*cosh(d*x + c)^4 + 8*(5*a*b + 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (15*a^2 +
20*a*b + 9*b^2)*sinh(d*x + c)^4 + 20*(3*a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(15*a^2 + 20*a*b + 9*b^2)*cosh(d*
x + c)^2 + 30*a^2 + 20*a*b)*sinh(d*x + c)^2 + 45*a^2 + 20*a*b + 15*b^2 + 8*((5*a*b + 3*b^2)*cosh(d*x + c)^3 +
5*a*b*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6
+ 6*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 + 4*d*cosh(d*x
+ c))*sinh(d*x + c)^3 + 15*d*cosh(d*x + c)^2 + 3*(5*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x
+ c)^2 + 2*(3*d*cosh(d*x + c)^5 + 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh(d*x + c) + 10*d)

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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 )^{2} \operatorname{sech}^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.41103, size = 228, normalized size = 4.65 \begin{align*} -\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 )} + 15 \, b^{2} 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 )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, 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 )} + 15 \, a^{2} + 10 \, a b + 3 \, b^{2}\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)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-2/15*(15*a^2*e^(8*d*x + 8*c) + 30*a*b*e^(8*d*x + 8*c) + 15*b^2*e^(8*d*x + 8*c) + 60*a^2*e^(6*d*x + 6*c) + 60*
a*b*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x + 4*c) + 40*a*b*e^(4*d*x + 4*c) + 30*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) + 15*a^2 + 10*a*b + 3*b^2)/(d*(e^(2*d*x + 2*c) + 1)^5)