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

Optimal. Leaf size=114 \[ -\frac{\left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)}{4 d}+\frac{1}{8} x \left (3 a^2-24 a b+8 b^2\right )+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{a (a-8 b) \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

[Out]

((3*a^2 - 24*a*b + 8*b^2)*x)/8 - (a*(a - 8*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) - ((a^2 - 8*a*b + 4*b^2)*Tanh
[c + d*x])/(4*d) + (a^2*Sinh[c + d*x]^4*Tanh[c + d*x])/(4*d) - (b^2*Tanh[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.138374, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4132, 463, 455, 1153, 206} \[ -\frac{\left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)}{4 d}+\frac{1}{8} x \left (3 a^2-24 a b+8 b^2\right )+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{a (a-8 b) \sinh (c+d x) \cosh (c+d x)}{8 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 24*a*b + 8*b^2)*x)/8 - (a*(a - 8*b)*Cosh[c + d*x]*Sinh[c + d*x])/(8*d) - ((a^2 - 8*a*b + 4*b^2)*Tanh
[c + d*x])/(4*d) + (a^2*Sinh[c + d*x]^4*Tanh[c + d*x])/(4*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 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 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 ^4(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b-b x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (5 a^2-4 (a+b)^2+4 b^2 x^2\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{a (a-8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-a (a-8 b)-2 a (a-8 b) x^2-8 b^2 x^4}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{a (a-8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \left (2 \left (a^2-8 a b+4 b^2\right )+8 b^2 x^2+\frac{-3 a^2+24 a b-8 b^2}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{a (a-8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac{\left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)}{4 d}+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}+\frac{\left (3 a^2-24 a b+8 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{1}{8} \left (3 a^2-24 a b+8 b^2\right ) x-\frac{a (a-8 b) \cosh (c+d x) \sinh (c+d x)}{8 d}-\frac{\left (a^2-8 a b+4 b^2\right ) \tanh (c+d x)}{4 d}+\frac{a^2 \sinh ^4(c+d x) \tanh (c+d x)}{4 d}-\frac{b^2 \tanh ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.6206, size = 153, normalized size = 1.34 \[ \frac{\text{sech}^3(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (3 \cosh ^3(c+d x) \left (4 d x \left (3 a^2-24 a b+8 b^2\right )+a^2 \sinh (4 (c+d x))-8 a (a-2 b) \sinh (2 (c+d x))\right )+64 b (3 a-2 b) \text{sech}(c) \sinh (d x) \cosh ^2(c+d x)+32 b^2 \tanh (c) \cosh (c+d x)+32 b^2 \text{sech}(c) \sinh (d x)\right )}{24 d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Cosh[c + d*x]^2)^2*Sech[c + d*x]^3*(32*b^2*Sech[c]*Sinh[d*x] + 64*(3*a - 2*b)*b*Cosh[c + d*x]^2*Sech[c
]*Sinh[d*x] + 3*Cosh[c + d*x]^3*(4*(3*a^2 - 24*a*b + 8*b^2)*d*x - 8*a*(a - 2*b)*Sinh[2*(c + d*x)] + a^2*Sinh[4
*(c + d*x)]) + 32*b^2*Cosh[c + d*x]*Tanh[c]))/(24*d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)

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Maple [A]  time = 0.04, size = 109, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +2\,ab \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}-3/2\,dx-3/2\,c+3/2\,\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((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^4,x)

[Out]

1/d*(a^2*((1/4*sinh(d*x+c)^3-3/8*sinh(d*x+c))*cosh(d*x+c)+3/8*d*x+3/8*c)+2*a*b*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-
3/2*d*x-3/2*c+3/2*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 = 1.05235, size = 285, normalized size = 2.5 \begin{align*} \frac{1}{64} \, a^{2}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \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{1}{4} \, a b{\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 )} \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)^4,x, algorithm="maxima")

[Out]

1/64*a^2*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 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))) - 1/4*a*b*(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))))

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Fricas [B]  time = 2.70513, size = 880, normalized size = 7.72 \begin{align*} \frac{3 \, a^{2} \sinh \left (d x + c\right )^{7} + 3 \,{\left (21 \, a^{2} \cosh \left (d x + c\right )^{2} - 5 \, a^{2} + 16 \, a b\right )} \sinh \left (d x + c\right )^{5} + 8 \,{\left (3 \,{\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 24 \,{\left (3 \,{\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} +{\left (105 \, a^{2} \cosh \left (d x + c\right )^{4} - 30 \,{\left (5 \, a^{2} - 16 \, a b\right )} \cosh \left (d x + c\right )^{2} - 63 \, a^{2} + 528 \, a b - 256 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 24 \,{\left (3 \,{\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )} d x - 48 \, a b + 32 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \,{\left (7 \, a^{2} \cosh \left (d x + c\right )^{6} - 5 \,{\left (5 \, a^{2} - 16 \, a b\right )} \cosh \left (d x + c\right )^{4} -{\left (63 \, a^{2} - 528 \, a b + 256 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 15 \, a^{2} + 160 \, a b\right )} \sinh \left (d x + c\right )}{192 \,{\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)^4,x, algorithm="fricas")

[Out]

1/192*(3*a^2*sinh(d*x + c)^7 + 3*(21*a^2*cosh(d*x + c)^2 - 5*a^2 + 16*a*b)*sinh(d*x + c)^5 + 8*(3*(3*a^2 - 24*
a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*cosh(d*x + c)^3 + 24*(3*(3*a^2 - 24*a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*co
sh(d*x + c)*sinh(d*x + c)^2 + (105*a^2*cosh(d*x + c)^4 - 30*(5*a^2 - 16*a*b)*cosh(d*x + c)^2 - 63*a^2 + 528*a*
b - 256*b^2)*sinh(d*x + c)^3 + 24*(3*(3*a^2 - 24*a*b + 8*b^2)*d*x - 48*a*b + 32*b^2)*cosh(d*x + c) + 3*(7*a^2*
cosh(d*x + c)^6 - 5*(5*a^2 - 16*a*b)*cosh(d*x + c)^4 - (63*a^2 - 528*a*b + 256*b^2)*cosh(d*x + c)^2 - 15*a^2 +
160*a*b)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(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)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.182, size = 328, normalized size = 2.88 \begin{align*} \frac{{\left (3 \, a^{2} - 24 \, a b + 8 \, b^{2}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{{\left (18 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} - 144 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 16 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{a^{2} d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, a^{2} d e^{\left (2 \, d x + 2 \, c\right )} + 16 \, a b d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} - \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 )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, 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((a+b*sech(d*x+c)^2)^2*sinh(d*x+c)^4,x, algorithm="giac")

[Out]

1/8*(3*a^2 - 24*a*b + 8*b^2)*(d*x + c)/d - 1/64*(18*a^2*e^(4*d*x + 4*c) - 144*a*b*e^(4*d*x + 4*c) + 48*b^2*e^(
4*d*x + 4*c) - 8*a^2*e^(2*d*x + 2*c) + 16*a*b*e^(2*d*x + 2*c) + a^2)*e^(-4*d*x - 4*c)/d + 1/64*(a^2*d*e^(4*d*x
+ 4*c) - 8*a^2*d*e^(2*d*x + 2*c) + 16*a*b*d*e^(2*d*x + 2*c))/d^2 - 4/3*(3*a*b*e^(4*d*x + 4*c) - 3*b^2*e^(4*d*
x + 4*c) + 6*a*b*e^(2*d*x + 2*c) - 3*b^2*e^(2*d*x + 2*c) + 3*a*b - 2*b^2)/(d*(e^(2*d*x + 2*c) + 1)^3)