3.302 \(\int \text{sech}^6(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ \frac{a^2 \tanh (c+d x)}{d}+\frac{(a-b)^2 \tanh ^5(c+d x)}{5 d}-\frac{2 a (a-b) \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0601492, antiderivative size = 57, 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 = {3191, 194} \[ \frac{a^2 \tanh (c+d x)}{d}+\frac{(a-b)^2 \tanh ^5(c+d x)}{5 d}-\frac{2 a (a-b) \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

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

Rule 3191

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, T
an[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

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}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a-(a-b) x^2\right )^2 \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^2-2 a (a-b) x^2+(a-b)^2 x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{a^2 \tanh (c+d x)}{d}-\frac{2 a (a-b) \tanh ^3(c+d x)}{3 d}+\frac{(a-b)^2 \tanh ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.373003, size = 69, normalized size = 1.21 \[ \frac{\tanh (c+d x) \left (2 \left (2 a^2+a b-3 b^2\right ) \text{sech}^2(c+d x)+8 a^2+3 (a-b)^2 \text{sech}^4(c+d x)+4 a b+3 b^2\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

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

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Maple [B]  time = 0.048, size = 158, normalized size = 2.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \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 ) \tanh \left ( dx+c \right ) +2\,ab \left ( -1/4\,{\frac{\sinh \left ( dx+c \right ) }{ \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+1/4\, \left ({\frac{8}{15}}+1/5\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \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)^6*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)+2*a*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))+b^2*(-1/2*sinh(d*x+c)^3/cosh(d*x+c)^5-3/8*sinh(d*x+c)/co
sh(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 [B]  time = 1.12718, size = 942, normalized size = 16.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

16/15*a^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)) + 10*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)) + 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))) + 8/15*a*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))) + 2/5*b^2*(10*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)) + 5*e^(-8*d*x
- 8*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 = 1.45971, size = 1040, normalized size = 18.25 \begin{align*} -\frac{4 \,{\left ({\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - 8 \,{\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} +{\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \,{\left (a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \,{\left (3 \,{\left (4 \, a^{2} + 2 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 10 \, a^{2} + 20 \, a b\right )} \sinh \left (d x + c\right )^{2} + 40 \, a^{2} - 10 \, a b + 15 \, b^{2} - 8 \,{\left ({\left (2 \, a^{2} + a b - 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \,{\left (a^{2} - a b\right )} \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)^6*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-4/15*((4*a^2 + 2*a*b + 9*b^2)*cosh(d*x + c)^4 - 8*(2*a^2 + a*b - 3*b^2)*cosh(d*x + c)*sinh(d*x + c)^3 + (4*a^
2 + 2*a*b + 9*b^2)*sinh(d*x + c)^4 + 20*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(4*a^2 + 2*a*b + 9*b^2)*cosh(d*x
+ c)^2 + 10*a^2 + 20*a*b)*sinh(d*x + c)^2 + 40*a^2 - 10*a*b + 15*b^2 - 8*((2*a^2 + a*b - 3*b^2)*cosh(d*x + c)^
3 + 5*(a^2 - 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*sin
h(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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(d*x+c)**6*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.22235, size = 173, normalized size = 3.04 \begin{align*} -\frac{2 \,{\left (15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{2} + 4 \, 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)^6*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

-2/15*(15*b^2*e^(8*d*x + 8*c) + 60*a*b*e^(6*d*x + 6*c) + 80*a^2*e^(4*d*x + 4*c) - 20*a*b*e^(4*d*x + 4*c) + 30*
b^2*e^(4*d*x + 4*c) + 40*a^2*e^(2*d*x + 2*c) + 20*a*b*e^(2*d*x + 2*c) + 8*a^2 + 4*a*b + 3*b^2)/(d*(e^(2*d*x +
2*c) + 1)^5)