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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 64, normalized size of antiderivative = 1.00, 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 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]

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
]

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*}

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Mathematica [A]  time = 0.26, size = 93, normalized size = 1.45 \[ \frac {8 \text {sech}^5(c+d x) \left (a \cosh ^2(c+d x)+b\right )^3 \left (5 a^3 \cosh ^6(c+d x)-15 a^2 b \cosh ^4(c+d x)-5 a b^2 \cosh ^2(c+d x)-b^3\right )}{5 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

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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fricas [B]  time = 0.39, size = 276, normalized size = 4.31 \[ \frac {5 \, a^{3} \cosh \left (d x + c\right )^{6} + 5 \, a^{3} \sinh \left (d x + c\right )^{6} + 30 \, {\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{4} + 15 \, {\left (5 \, a^{3} \cosh \left (d x + c\right )^{2} + 2 \, a^{3} - 4 \, a^{2} b\right )} \sinh \left (d x + c\right )^{4} + 50 \, a^{3} - 180 \, a^{2} b - 80 \, a b^{2} - 32 \, b^{3} + 5 \, {\left (15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2} + 5 \, {\left (15 \, a^{3} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} - 48 \, a^{2} b - 16 \, a b^{2} + 36 \, {\left (a^{3} - 2 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{10 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.19, size = 101, normalized size = 1.58 \[ \frac {5 \, a^{3} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} - \frac {4 \, {\left (15 \, a^{2} b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4} + 20 \, a b^{2} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 16 \, b^{3}\right )}}{{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}}}{10 \, d} \]

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]

1/10*(5*a^3*(e^(d*x + c) + e^(-d*x - c)) - 4*(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)/(e^(d*x + c) + e^(-d*x - c))^5)/d

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maple [A]  time = 0.13, size = 58, normalized size = 0.91 \[ -\frac {\frac {b^{3} \mathrm {sech}\left (d x +c \right )^{5}}{5}+a \,b^{2} \mathrm {sech}\left (d x +c \right )^{3}+3 a^{2} b \,\mathrm {sech}\left (d x +c \right )-\frac {a^{3}}{\mathrm {sech}\left (d x +c \right )}}{d} \]

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 = 0.32, size = 94, normalized size = 1.47 \[ \frac {a^{3} \cosh \left (d x + c\right )}{d} - \frac {6 \, a^{2} b}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}} - \frac {8 \, a b^{2}}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3}} - \frac {32 \, b^{3}}{5 \, d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}} \]

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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mupad [B]  time = 1.50, size = 288, normalized size = 4.50 \[ \frac {a^3\,{\mathrm {e}}^{c+d\,x}}{2\,d}+\frac {a^3\,{\mathrm {e}}^{-c-d\,x}}{2\,d}+\frac {64\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (5\,a\,b^2-4\,b^3\right )}{5\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {32\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {6\,a^2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {8\,a\,b^2\,{\mathrm {e}}^{c+d\,x}}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*exp(c + d*x))/(2*d) + (a^3*exp(- c - d*x))/(2*d) + (64*b^3*exp(c + d*x))/(5*d*(4*exp(2*c + 2*d*x) + 6*exp
(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) + (8*exp(c + d*x)*(5*a*b^2 - 4*b^3))/(5*d*(3*exp(2
*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1)) - (32*b^3*exp(c + d*x))/(5*d*(5*exp(2*c + 2*d*x) + 1
0*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1)) - (6*a^2*b*exp(c + d*
x))/(d*(exp(2*c + 2*d*x) + 1)) - (8*a*b^2*exp(c + d*x))/(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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