3.300 \(\int \text{sech}^4(c+d x) (a+b \sinh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=47 \[ \frac{\left (a^2-b^2\right ) \tanh (c+d x)}{d}-\frac{(a-b)^2 \tanh ^3(c+d x)}{3 d}+b^2 x \]

[Out]

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

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Rubi [A]  time = 0.062746, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3191, 390, 206} \[ \frac{\left (a^2-b^2\right ) \tanh (c+d x)}{d}-\frac{(a-b)^2 \tanh ^3(c+d x)}{3 d}+b^2 x \]

Antiderivative was successfully verified.

[In]

Int[Sech[c + d*x]^4*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

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

Mathematica [A]  time = 0.331525, size = 57, normalized size = 1.21 \[ \frac{(a-b) \tanh (c+d x) \text{sech}^2(c+d x) ((a+2 b) \cosh (2 (c+d x))+2 a+b)+3 b^2 (c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[c + d*x]^4*(a + b*Sinh[c + d*x]^2)^2,x]

[Out]

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

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Maple [B]  time = 0.046, size = 96, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{2}{3}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{3}} \right ) \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 ( dx+c-\tanh \left ( dx+c \right ) -{\frac{ \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{3}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*(2/3+1/3*sech(d*x+c)^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*(d*x+c-tanh(d*x+c)-1/3*tanh(d*x+c)^3))

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Maxima [B]  time = 1.11575, size = 360, normalized size = 7.66 \begin{align*} \frac{1}{3} \, b^{2}{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\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 )}}\right )} + \frac{4}{3} \, a^{2}{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, 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 )} + \frac{4}{3} \, a b{\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(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

1/3*b^2*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x -
4*c) + e^(-6*d*x - 6*c) + 1))) + 4/3*a^2*(3*e^(-2*d*x - 2*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))) + 4/3*a*b*(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 = 1.48203, size = 495, normalized size = 10.53 \begin{align*} \frac{{\left (3 \, b^{2} d x - 2 \, a^{2} - 2 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (3 \, b^{2} d x - 2 \, a^{2} - 2 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \,{\left (a^{2} + a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 3 \,{\left (3 \, b^{2} d x - 2 \, a^{2} - 2 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right ) + 6 \,{\left ({\left (a^{2} + a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} - a b\right )} \sinh \left (d x + c\right )}{3 \,{\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(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

1/3*((3*b^2*d*x - 2*a^2 - 2*a*b + 4*b^2)*cosh(d*x + c)^3 + 3*(3*b^2*d*x - 2*a^2 - 2*a*b + 4*b^2)*cosh(d*x + c)
*sinh(d*x + c)^2 + 2*(a^2 + a*b - 2*b^2)*sinh(d*x + c)^3 + 3*(3*b^2*d*x - 2*a^2 - 2*a*b + 4*b^2)*cosh(d*x + c)
+ 6*((a^2 + a*b - 2*b^2)*cosh(d*x + c)^2 + a^2 - a*b)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*s
inh(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(sech(d*x+c)**4*(a+b*sinh(d*x+c)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.1663, size = 132, normalized size = 2.81 \begin{align*} \frac{{\left (d x + c\right )} b^{2}}{d} - \frac{4 \,{\left (3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + a b - 2 \, 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(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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