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

Optimal. Leaf size=73 \[ \frac{a^2 \sinh ^2(c+d x) \tanh (c+d x)}{2 d}+\frac{a (a-4 b) \tanh (c+d x)}{2 d}-\frac{1}{2} a x (a-4 b)+\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.107428, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 463, 459, 321, 206} \[ \frac{a^2 \sinh ^2(c+d x) \tanh (c+d x)}{2 d}+\frac{a (a-4 b) \tanh (c+d x)}{2 d}-\frac{1}{2} a x (a-4 b)+\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4132

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = Fr
eeFactors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p)/(
1 + ff^2*x^2)^(m/2 + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && Integer
Q[n/2]

Rule 463

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> -Simp[((b*c - a*
d)^2*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b^2*e*n*(p + 1)), x] + Dist[1/(a*b^2*n*(p + 1)), Int[(e*x)^m*(a + b
*x^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 1)*x^n, x], x], x] /; FreeQ[{a,
b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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

Mathematica [A]  time = 0.93376, size = 126, normalized size = 1.73 \[ \frac{\text{sech}^3(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (3 a \cosh ^3(c+d x) (a \sinh (2 (c+d x))-2 d x (a-4 b))-4 b (6 a-b) \text{sech}(c) \sinh (d x) \cosh ^2(c+d x)-4 b^2 \tanh (c) \cosh (c+d x)-4 b^2 \text{sech}(c) \sinh (d x)\right )}{3 d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[c + d*x]^2)^2*Sech[c + d*x]^3*(-4*b^2*Sech[c]*Sinh[d*x] - 4*(6*a - b)*b*Cosh[c + d*x]^2*Sech[c]*S
inh[d*x] + 3*a*Cosh[c + d*x]^3*(-2*(a - 4*b)*d*x + a*Sinh[2*(c + d*x)]) - 4*b^2*Cosh[c + d*x]*Tanh[c]))/(3*d*(
a + 2*b + a*Cosh[2*(c + d*x)])^2)

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Maple [A]  time = 0.036, size = 90, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) +2\,ab \left ( dx+c-\tanh \left ( dx+c \right ) \right ) +{b}^{2} \left ( -{\frac{\sinh \left ( dx+c \right ) }{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+{\frac{\tanh \left ( dx+c \right ) }{2} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \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*sinh(d*x+c)^2,x)

[Out]

1/d*(a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+2*a*b*(d*x+c-tanh(d*x+c))+b^2*(-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)))

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Maxima [B]  time = 1.04187, size = 216, normalized size = 2.96 \begin{align*} -\frac{1}{8} \, a^{2}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 2 \, a b{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac{2}{3} \, b^{2}{\left (\frac{3 \, e^{\left (-4 \, d x - 4 \, 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((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^2,x, algorithm="maxima")

[Out]

-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 2*a*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + 2/3
*b^2*(3*e^(-4*d*x - 4*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 = 2.90141, size = 641, normalized size = 8.78 \begin{align*} \frac{3 \, a^{2} \sinh \left (d x + c\right )^{5} - 4 \,{\left (3 \,{\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 12 \,{\left (3 \,{\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} +{\left (30 \, a^{2} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} - 48 \, a b + 8 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \,{\left (3 \,{\left (a^{2} - 4 \, a b\right )} d x - 12 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \,{\left (5 \, a^{2} \cosh \left (d x + c\right )^{4} +{\left (9 \, a^{2} - 48 \, a b + 8 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} - 16 \, a b - 8 \, b^{2}\right )} \sinh \left (d x + c\right )}{24 \,{\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, 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*sinh(d*x+c)^2,x, algorithm="fricas")

[Out]

1/24*(3*a^2*sinh(d*x + c)^5 - 4*(3*(a^2 - 4*a*b)*d*x - 12*a*b + 2*b^2)*cosh(d*x + c)^3 - 12*(3*(a^2 - 4*a*b)*d
*x - 12*a*b + 2*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*a^2*cosh(d*x + c)^2 + 9*a^2 - 48*a*b + 8*b^2)*sinh(d*
x + c)^3 - 12*(3*(a^2 - 4*a*b)*d*x - 12*a*b + 2*b^2)*cosh(d*x + c) + 3*(5*a^2*cosh(d*x + c)^4 + (9*a^2 - 48*a*
b + 8*b^2)*cosh(d*x + c)^2 + 2*a^2 - 16*a*b - 8*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sin
h(d*x + c)^2 + 3*d*cosh(d*x + c))

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

[Out]

Timed out

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Giac [B]  time = 1.18011, size = 204, normalized size = 2.79 \begin{align*} \frac{a^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} - \frac{{\left (a^{2} - 4 \, a b\right )}{\left (d x + c\right )}}{2 \, d} + \frac{{\left (2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, a b e^{\left (2 \, d x + 2 \, c\right )} - a^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} + \frac{2 \,{\left (6 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b - b^{2}\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a^2*e^(2*d*x + 2*c)/d - 1/2*(a^2 - 4*a*b)*(d*x + c)/d + 1/8*(2*a^2*e^(2*d*x + 2*c) - 8*a*b*e^(2*d*x + 2*c)
- a^2)*e^(-2*d*x - 2*c)/d + 2/3*(6*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d*x + 2*c) + 6*a
*b - b^2)/(d*(e^(2*d*x + 2*c) + 1)^3)