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

Optimal. Leaf size=43 \[ -\frac{1}{2} x (a-2 b)+\frac{a \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b \tanh (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0561302, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {4132, 455, 388, 206} \[ -\frac{1}{2} x (a-2 b)+\frac{a \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a - 2*b)*x)/2 + (a*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) - (b*Tanh[c + d*x])/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 455

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

Rule 388

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

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

Mathematica [A]  time = 0.18568, size = 57, normalized size = 1.33 \[ \frac{a (-c-d x)}{2 d}+\frac{a \sinh (2 (c+d x))}{4 d}+\frac{b \tanh ^{-1}(\tanh (c+d x))}{d}-\frac{b \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.028, size = 45, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) +b \left ( dx+c-\tanh \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03543, size = 84, normalized size = 1.95 \begin{align*} -\frac{1}{8} \, a{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + b{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, 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)*sinh(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.44495, size = 177, normalized size = 4.12 \begin{align*} \frac{a \sinh \left (d x + c\right )^{3} - 4 \,{\left ({\left (a - 2 \, b\right )} d x - 2 \, b\right )} \cosh \left (d x + c\right ) +{\left (3 \, a \cosh \left (d x + c\right )^{2} + a - 8 \, b\right )} \sinh \left (d x + c\right )}{8 \, d \cosh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a*sinh(d*x + c)^3 - 4*((a - 2*b)*d*x - 2*b)*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a - 8*b)*sinh(d*x + c)
)/(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 ) \sinh ^{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)*sinh(d*x+c)**2,x)

[Out]

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

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Giac [B]  time = 1.1741, size = 130, normalized size = 3.02 \begin{align*} -\frac{{\left (d x + c\right )}{\left (a - 2 \, b\right )}}{2 \, d} + \frac{a e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{a e^{\left (4 \, d x + 4 \, c\right )} - 2 \, b e^{\left (4 \, d x + 4 \, c\right )} + 14 \, b e^{\left (2 \, d x + 2 \, c\right )} - a}{8 \, d{\left (e^{\left (4 \, d x + 4 \, c\right )} + e^{\left (2 \, d x + 2 \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(d*x + c)*(a - 2*b)/d + 1/8*a*e^(2*d*x + 2*c)/d + 1/8*(a*e^(4*d*x + 4*c) - 2*b*e^(4*d*x + 4*c) + 14*b*e^(
2*d*x + 2*c) - a)/(d*(e^(4*d*x + 4*c) + e^(2*d*x + 2*c)))