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

Optimal. Leaf size=53 \[ \frac{(a-b)^2 \tanh (c+d x)}{d}+\frac{1}{2} b x (4 a-3 b)+\frac{b^2 \sinh (c+d x) \cosh (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0949116, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3191, 390, 385, 206} \[ \frac{(a-b)^2 \tanh (c+d x)}{d}+\frac{1}{2} b x (4 a-3 b)+\frac{b^2 \sinh (c+d x) \cosh (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a - 3*b)*b*x)/2 + (b^2*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + ((a - b)^2*Tanh[c + d*x])/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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 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 \text{sech}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-(a-b) x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left ((a-b)^2+\frac{(2 a-b) b-2 (a-b) b x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a-b)^2 \tanh (c+d x)}{d}+\frac{\operatorname{Subst}\left (\int \frac{(2 a-b) b-2 (a-b) b x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{(a-b)^2 \tanh (c+d x)}{d}+\frac{((4 a-3 b) b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{1}{2} (4 a-3 b) b x+\frac{b^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{(a-b)^2 \tanh (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.291894, size = 50, normalized size = 0.94 \[ \frac{2 b (4 a-3 b) (c+d x)+4 (a-b)^2 \tanh (c+d x)+b^2 \sinh (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.037, size = 71, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{2}\tanh \left ( dx+c \right ) +2\,ab \left ( dx+c-\tanh \left ( dx+c \right ) \right ) +{b}^{2} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{2\,\cosh \left ( dx+c \right ) }}-{\frac{3\,dx}{2}}-{\frac{3\,c}{2}}+{\frac{3\,\tanh \left ( dx+c \right ) }{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.21566, size = 161, normalized size = 3.04 \begin{align*} 2 \, a b{\left (x + \frac{c}{d} - \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} - \frac{1}{8} \, b^{2}{\left (\frac{12 \,{\left (d x + c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} + \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*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

2*a*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) - 1/8*b^2*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x
- 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c)))) + 2*a^2/(d*(e^(-2*d*x - 2*c) + 1))

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Fricas [A]  time = 1.53539, size = 235, normalized size = 4.43 \begin{align*} \frac{b^{2} \sinh \left (d x + c\right )^{3} + 4 \,{\left ({\left (4 \, a b - 3 \, b^{2}\right )} d x - 2 \, a^{2} + 4 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right ) +{\left (3 \, b^{2} \cosh \left (d x + c\right )^{2} + 8 \, a^{2} - 16 \, a b + 9 \, b^{2}\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(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/8*(b^2*sinh(d*x + c)^3 + 4*((4*a*b - 3*b^2)*d*x - 2*a^2 + 4*a*b - 2*b^2)*cosh(d*x + c) + (3*b^2*cosh(d*x + c
)^2 + 8*a^2 - 16*a*b + 9*b^2)*sinh(d*x + c))/(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(sech(d*x+c)**2*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.18936, size = 184, normalized size = 3.47 \begin{align*} \frac{b^{2} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{{\left (4 \, a b - 3 \, b^{2}\right )}{\left (d x + c\right )}}{2 \, d} - \frac{4 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 16 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 28 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 14 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}}{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(sech(d*x+c)^2*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

1/8*b^2*e^(2*d*x + 2*c)/d + 1/2*(4*a*b - 3*b^2)*(d*x + c)/d - 1/8*(4*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*x + 4*
c) + 16*a^2*e^(2*d*x + 2*c) - 28*a*b*e^(2*d*x + 2*c) + 14*b^2*e^(2*d*x + 2*c) + b^2)/(d*(e^(4*d*x + 4*c) + e^(
2*d*x + 2*c)))