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

Optimal. Leaf size=99 \[ \frac {a^3 \cosh ^3(c+d x)}{3 d}-\frac {a^2 (a-3 b) \cosh (c+d x)}{d}+\frac {b^2 (3 a-b) \text {sech}^3(c+d x)}{3 d}+\frac {3 a b (a-b) \text {sech}(c+d x)}{d}+\frac {b^3 \text {sech}^5(c+d x)}{5 d} \]

[Out]

-a^2*(a-3*b)*cosh(d*x+c)/d+1/3*a^3*cosh(d*x+c)^3/d+3*a*(a-b)*b*sech(d*x+c)/d+1/3*(3*a-b)*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 = 0.11, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4133, 448} \[ -\frac {a^2 (a-3 b) \cosh (c+d x)}{d}+\frac {a^3 \cosh ^3(c+d x)}{3 d}+\frac {b^2 (3 a-b) \text {sech}^3(c+d x)}{3 d}+\frac {3 a b (a-b) \text {sech}(c+d x)}{d}+\frac {b^3 \text {sech}^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 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 ^3(c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right ) \left (b+a x^2\right )^3}{x^6} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (a^2 (a-3 b)+\frac {b^3}{x^6}+\frac {(3 a-b) b^2}{x^4}+\frac {3 a (a-b) b}{x^2}-a^3 x^2\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {a^2 (a-3 b) \cosh (c+d x)}{d}+\frac {a^3 \cosh ^3(c+d x)}{3 d}+\frac {3 a (a-b) b \text {sech}(c+d x)}{d}+\frac {(3 a-b) b^2 \text {sech}^3(c+d x)}{3 d}+\frac {b^3 \text {sech}^5(c+d x)}{5 d}\\ \end {align*}

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Mathematica [A]  time = 1.25, size = 119, normalized size = 1.20 \[ \frac {4 \text {sech}^5(c+d x) \left (a \cosh ^2(c+d x)+b\right )^3 \left (5 a^2 \cosh ^6(c+d x) (a \cosh (2 (c+d x))-5 a+18 b)+10 b^2 (3 a-b) \cosh ^2(c+d x)+90 a b (a-b) \cosh ^4(c+d x)+6 b^3\right )}{15 d (a \cosh (2 (c+d x))+a+2 b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(b + a*Cosh[c + d*x]^2)^3*(6*b^3 + 10*(3*a - b)*b^2*Cosh[c + d*x]^2 + 90*a*(a - b)*b*Cosh[c + d*x]^4 + 5*a^
2*Cosh[c + d*x]^6*(-5*a + 18*b + a*Cosh[2*(c + d*x)]))*Sech[c + d*x]^5)/(15*d*(a + 2*b + a*Cosh[2*(c + d*x)])^
3)

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fricas [B]  time = 0.41, size = 403, normalized size = 4.07 \[ \frac {5 \, a^{3} \cosh \left (d x + c\right )^{8} + 5 \, a^{3} \sinh \left (d x + c\right )^{8} - 20 \, {\left (a^{3} - 9 \, a^{2} b\right )} \cosh \left (d x + c\right )^{6} + 20 \, {\left (7 \, a^{3} \cosh \left (d x + c\right )^{2} - a^{3} + 9 \, a^{2} b\right )} \sinh \left (d x + c\right )^{6} - 20 \, {\left (11 \, a^{3} - 90 \, a^{2} b + 36 \, a b^{2}\right )} \cosh \left (d x + c\right )^{4} + 10 \, {\left (35 \, a^{3} \cosh \left (d x + c\right )^{4} - 22 \, a^{3} + 180 \, a^{2} b - 72 \, a b^{2} - 30 \, {\left (a^{3} - 9 \, a^{2} b\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{4} - 425 \, a^{3} + 3960 \, a^{2} b - 1200 \, a b^{2} + 64 \, b^{3} - 20 \, {\left (31 \, a^{3} - 279 \, a^{2} b + 96 \, a b^{2} + 16 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 20 \, {\left (7 \, a^{3} \cosh \left (d x + c\right )^{6} - 15 \, {\left (a^{3} - 9 \, a^{2} b\right )} \cosh \left (d x + c\right )^{4} - 31 \, a^{3} + 279 \, a^{2} b - 96 \, a b^{2} - 16 \, b^{3} - 6 \, {\left (11 \, a^{3} - 90 \, a^{2} b + 36 \, a b^{2}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{120 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(5*a^3*cosh(d*x + c)^8 + 5*a^3*sinh(d*x + c)^8 - 20*(a^3 - 9*a^2*b)*cosh(d*x + c)^6 + 20*(7*a^3*cosh(d*x
+ c)^2 - a^3 + 9*a^2*b)*sinh(d*x + c)^6 - 20*(11*a^3 - 90*a^2*b + 36*a*b^2)*cosh(d*x + c)^4 + 10*(35*a^3*cosh
(d*x + c)^4 - 22*a^3 + 180*a^2*b - 72*a*b^2 - 30*(a^3 - 9*a^2*b)*cosh(d*x + c)^2)*sinh(d*x + c)^4 - 425*a^3 +
3960*a^2*b - 1200*a*b^2 + 64*b^3 - 20*(31*a^3 - 279*a^2*b + 96*a*b^2 + 16*b^3)*cosh(d*x + c)^2 + 20*(7*a^3*cos
h(d*x + c)^6 - 15*(a^3 - 9*a^2*b)*cosh(d*x + c)^4 - 31*a^3 + 279*a^2*b - 96*a*b^2 - 16*b^3 - 6*(11*a^3 - 90*a^
2*b + 36*a*b^2)*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 [B]  time = 0.21, size = 193, normalized size = 1.95 \[ \frac {5 \, a^{3} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3} - 60 \, a^{3} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} + 180 \, a^{2} b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )} + \frac {16 \, {\left (45 \, a^{2} b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4} - 45 \, a b^{2} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4} + 60 \, a b^{2} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 20 \, b^{3} {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 48 \, b^{3}\right )}}{{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(5*a^3*(e^(d*x + c) + e^(-d*x - c))^3 - 60*a^3*(e^(d*x + c) + e^(-d*x - c)) + 180*a^2*b*(e^(d*x + c) + e
^(-d*x - c)) + 16*(45*a^2*b*(e^(d*x + c) + e^(-d*x - c))^4 - 45*a*b^2*(e^(d*x + c) + e^(-d*x - c))^4 + 60*a*b^
2*(e^(d*x + c) + e^(-d*x - c))^2 - 20*b^3*(e^(d*x + c) + e^(-d*x - c))^2 + 48*b^3)/(e^(d*x + c) + e^(-d*x - c)
)^5)/d

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maple [A]  time = 0.37, size = 130, normalized size = 1.31 \[ \frac {a^{3} \left (-\frac {2}{3}+\frac {\left (\sinh ^{2}\left (d x +c \right )\right )}{3}\right ) \cosh \left (d x +c \right )+3 a^{2} b \left (\frac {\sinh ^{2}\left (d x +c \right )}{\cosh \left (d x +c \right )}+\frac {2}{\cosh \left (d x +c \right )}\right )+3 a \,b^{2} \left (-\frac {\sinh ^{2}\left (d x +c \right )}{\cosh \left (d x +c \right )^{3}}-\frac {2}{3 \cosh \left (d x +c \right )^{3}}\right )+b^{3} \left (-\frac {\sinh ^{2}\left (d x +c \right )}{3 \cosh \left (d x +c \right )^{5}}-\frac {2}{15 \cosh \left (d x +c \right )^{5}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(-2/3+1/3*sinh(d*x+c)^2)*cosh(d*x+c)+3*a^2*b*(sinh(d*x+c)^2/cosh(d*x+c)+2/cosh(d*x+c))+3*a*b^2*(-sinh
(d*x+c)^2/cosh(d*x+c)^3-2/3/cosh(d*x+c)^3)+b^3*(-1/3*sinh(d*x+c)^2/cosh(d*x+c)^5-2/15/cosh(d*x+c)^5))

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maxima [B]  time = 0.34, size = 489, normalized size = 4.94 \[ \frac {1}{24} \, a^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} - \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} + \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac {3}{2} \, a^{2} b {\left (\frac {e^{\left (-d x - c\right )}}{d} + \frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-d x - c\right )} + e^{\left (-3 \, d x - 3 \, c\right )}\right )}}\right )} - 2 \, a b^{2} {\left (\frac {3 \, e^{\left (-d x - 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 {2 \, e^{\left (-3 \, d x - 3 \, 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 {3 \, e^{\left (-5 \, d x - 5 \, 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 )}}\right )} - \frac {8}{15} \, b^{3} {\left (\frac {5 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac {2 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\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)^3,x, algorithm="maxima")

[Out]

1/24*a^3*(e^(3*d*x + 3*c)/d - 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d + e^(-3*d*x - 3*c)/d) + 3/2*a^2*b*(e^(-d*x -
c)/d + (5*e^(-2*d*x - 2*c) + 1)/(d*(e^(-d*x - c) + e^(-3*d*x - 3*c)))) - 2*a*b^2*(3*e^(-d*x - c)/(d*(3*e^(-2*d
*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 2*e^(-3*d*x - 3*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*
d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 3*e^(-5*d*x - 5*c)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d
*x - 6*c) + 1))) - 8/15*b^3*(5*e^(-3*d*x - 3*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x -
6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) - 2*e^(-5*d*x - 5*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*
x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 5*e^(-7*d*x - 7*c)/(d*(5*e^(-
2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)))

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mupad [B]  time = 0.33, size = 348, normalized size = 3.52 \[ \frac {a^3\,{\mathrm {e}}^{-3\,c-3\,d\,x}}{24\,d}+\frac {a^3\,{\mathrm {e}}^{3\,c+3\,d\,x}}{24\,d}-\frac {3\,a^2\,{\mathrm {e}}^{-c-d\,x}\,\left (a-4\,b\right )}{8\,d}+\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (3\,a\,b^2-b^3\right )}{3\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {64\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (15\,a\,b^2-17\,b^3\right )}{15\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {32\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {6\,{\mathrm {e}}^{c+d\,x}\,\left (a\,b^2-a^2\,b\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {3\,a^2\,{\mathrm {e}}^{c+d\,x}\,\left (a-4\,b\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^3*exp(- 3*c - 3*d*x))/(24*d) + (a^3*exp(3*c + 3*d*x))/(24*d) - (3*a^2*exp(- c - d*x)*(a - 4*b))/(8*d) + (8*
exp(c + d*x)*(3*a*b^2 - b^3))/(3*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) - (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)*(15*a*
b^2 - 17*b^3))/(15*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) + 10*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*exp(c + d*x)*(a*b^2 - a^2*b))/(d*(exp(2*c + 2*d*x) + 1)) - (3*a^2*exp(c + d*x)*(a - 4*b))/(8*d)

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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)**3,x)

[Out]

Timed out

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