3.20 \(\int (a+b \text{sech}^2(c+d x))^3 \sinh (c+d x) \, dx\)

Optimal. Leaf size=64 \[ -\frac{3 a^2 b \text{sech}(c+d x)}{d}+\frac{a^3 \cosh (c+d x)}{d}-\frac{a b^2 \text{sech}^3(c+d x)}{d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d} \]

[Out]

(a^3*Cosh[c + d*x])/d - (3*a^2*b*Sech[c + d*x])/d - (a*b^2*Sech[c + d*x]^3)/d - (b^3*Sech[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0592691, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {4133, 270} \[ -\frac{3 a^2 b \text{sech}(c+d x)}{d}+\frac{a^3 \cosh (c+d x)}{d}-\frac{a b^2 \text{sech}^3(c+d x)}{d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^3*Sinh[c + d*x],x]

[Out]

(a^3*Cosh[c + d*x])/d - (3*a^2*b*Sech[c + d*x])/d - (a*b^2*Sech[c + d*x]^3)/d - (b^3*Sech[c + d*x]^5)/(5*d)

Rule 4133

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

Rule 270

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

Rubi steps

\begin{align*} \int \left (a+b \text{sech}^2(c+d x)\right )^3 \sinh (c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b+a x^2\right )^3}{x^6} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (a^3+\frac{b^3}{x^6}+\frac{3 a b^2}{x^4}+\frac{3 a^2 b}{x^2}\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{a^3 \cosh (c+d x)}{d}-\frac{3 a^2 b \text{sech}(c+d x)}{d}-\frac{a b^2 \text{sech}^3(c+d x)}{d}-\frac{b^3 \text{sech}^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.247026, size = 93, normalized size = 1.45 \[ \frac{8 \text{sech}^5(c+d x) \left (a \cosh ^2(c+d x)+b\right )^3 \left (-15 a^2 b \cosh ^4(c+d x)+5 a^3 \cosh ^6(c+d x)-5 a b^2 \cosh ^2(c+d x)-b^3\right )}{5 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sech[c + d*x]^2)^3*Sinh[c + d*x],x]

[Out]

(8*(b + a*Cosh[c + d*x]^2)^3*(-b^3 - 5*a*b^2*Cosh[c + d*x]^2 - 15*a^2*b*Cosh[c + d*x]^4 + 5*a^3*Cosh[c + d*x]^
6)*Sech[c + d*x]^5)/(5*d*(a + 2*b + a*Cosh[2*(c + d*x)])^3)

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Maple [A]  time = 0.019, size = 58, normalized size = 0.9 \begin{align*} -{\frac{1}{d} \left ({\frac{{b}^{3} \left ({\rm sech} \left (dx+c\right ) \right ) ^{5}}{5}}+a{b}^{2} \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}+3\,{a}^{2}b{\rm sech} \left (dx+c\right )-{\frac{{a}^{3}}{{\rm sech} \left (dx+c\right )}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c)^2)^3*sinh(d*x+c),x)

[Out]

-1/d*(1/5*b^3*sech(d*x+c)^5+a*b^2*sech(d*x+c)^3+3*a^2*b*sech(d*x+c)-a^3/sech(d*x+c))

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Maxima [A]  time = 1.05862, size = 127, normalized size = 1.98 \begin{align*} \frac{a^{3} \cosh \left (d x + c\right )}{d} - \frac{6 \, a^{2} b}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} - \frac{8 \, a b^{2}}{d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3}} - \frac{32 \, b^{3}}{5 \, d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c)^2)^3*sinh(d*x+c),x, algorithm="maxima")

[Out]

a^3*cosh(d*x + c)/d - 6*a^2*b/(d*(e^(d*x + c) + e^(-d*x - c))) - 8*a*b^2/(d*(e^(d*x + c) + e^(-d*x - c))^3) -
32/5*b^3/(d*(e^(d*x + c) + e^(-d*x - c))^5)

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Fricas [B]  time = 2.52159, size = 702, normalized size = 10.97 \begin{align*} \frac{5 \, a^{3} \cosh \left (d x + c\right )^{6} + 5 \, a^{3} \sinh \left (d x + c\right )^{6} + 30 \,{\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{4} + 15 \,{\left (5 \, a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} - 4 \, a^{2} b\right )} \sinh \left (d x + c\right )^{4} + 50 \, a^{3} - 180 \, a^{2} b - 80 \, a b^{2} - 32 \, b^{3} + 5 \,{\left (15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2} + 5 \,{\left (15 \, a^{3} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2} + 36 \,{\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{10 \,{\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \,{\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, 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)^3*sinh(d*x+c),x, algorithm="fricas")

[Out]

1/10*(5*a^3*cosh(d*x + c)^6 + 5*a^3*sinh(d*x + c)^6 + 30*(a^3 - 2*a^2*b)*cosh(d*x + c)^4 + 15*(5*a^3*cosh(d*x
+ c)^2 + 2*a^3 - 4*a^2*b)*sinh(d*x + c)^4 + 50*a^3 - 180*a^2*b - 80*a*b^2 - 32*b^3 + 5*(15*a^3 - 48*a^2*b - 16
*a*b^2)*cosh(d*x + c)^2 + 5*(15*a^3*cosh(d*x + c)^4 + 15*a^3 - 48*a^2*b - 16*a*b^2 + 36*(a^3 - 2*a^2*b)*cosh(d
*x + c)^2)*sinh(d*x + c)^2)/(d*cosh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(
2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*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)**3*sinh(d*x+c),x)

[Out]

Timed out

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Giac [A]  time = 1.19049, size = 138, normalized size = 2.16 \begin{align*} \frac{a^{3}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}}{2 \, d} - \frac{2 \,{\left (15 \, a^{2} b{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4} + 20 \, a b^{2}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 16 \, b^{3}\right )}}{5 \, d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c)^2)^3*sinh(d*x+c),x, algorithm="giac")

[Out]

1/2*a^3*(e^(d*x + c) + e^(-d*x - c))/d - 2/5*(15*a^2*b*(e^(d*x + c) + e^(-d*x - c))^4 + 20*a*b^2*(e^(d*x + c)
+ e^(-d*x - c))^2 + 16*b^3)/(d*(e^(d*x + c) + e^(-d*x - c))^5)