3.20 \(\int \sinh (c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=70 \[ -\frac {b^2 (a+b) \text {sech}^3(c+d x)}{d}+\frac {(a+b)^3 \cosh (c+d x)}{d}+\frac {3 b (a+b)^2 \text {sech}(c+d x)}{d}+\frac {b^3 \text {sech}^5(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 70, 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 = {3664, 270} \[ -\frac {b^2 (a+b) \text {sech}^3(c+d x)}{d}+\frac {(a+b)^3 \cosh (c+d x)}{d}+\frac {3 b (a+b)^2 \text {sech}(c+d x)}{d}+\frac {b^3 \text {sech}^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^3*Cosh[c + d*x])/d + (3*b*(a + b)^2*Sech[c + d*x])/d - (b^2*(a + b)*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 3664

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

Rubi steps

\begin {align*} \int \sinh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\left (a+b-b x^2\right )^3}{x^2} \, dx,x,\text {sech}(c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-3 b (a+b)^2+\frac {(a+b)^3}{x^2}+3 b^2 (a+b) x^2-b^3 x^4\right ) \, dx,x,\text {sech}(c+d x)\right )}{d}\\ &=\frac {(a+b)^3 \cosh (c+d x)}{d}+\frac {3 b (a+b)^2 \text {sech}(c+d x)}{d}-\frac {b^2 (a+b) \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 = 0.88, size = 63, normalized size = 0.90 \[ \frac {b \text {sech}(c+d x) \left (-5 b (a+b) \text {sech}^2(c+d x)+15 (a+b)^2+b^2 \text {sech}^4(c+d x)\right )+5 (a+b)^3 \cosh (c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.64, size = 383, normalized size = 5.47 \[ \frac {5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{6} + 5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{6} + 30 \, {\left (a^{3} + 5 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 15 \, {\left (2 \, a^{3} + 10 \, a^{2} b + 14 \, a b^{2} + 6 \, b^{3} + 5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{4} + 50 \, a^{3} + 330 \, a^{2} b + 430 \, a b^{2} + 182 \, b^{3} + 5 \, {\left (15 \, a^{3} + 93 \, a^{2} b + 125 \, a b^{2} + 47 \, b^{3}\right )} \cosh \left (d x + c\right )^{2} + 5 \, {\left (15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 15 \, a^{3} + 93 \, a^{2} b + 125 \, a b^{2} + 47 \, b^{3} + 36 \, {\left (a^{3} + 5 \, a^{2} b + 7 \, a b^{2} + 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{2}}{10 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^6 + 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^6 +
30*(a^3 + 5*a^2*b + 7*a*b^2 + 3*b^3)*cosh(d*x + c)^4 + 15*(2*a^3 + 10*a^2*b + 14*a*b^2 + 6*b^3 + 5*(a^3 + 3*a^
2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 50*a^3 + 330*a^2*b + 430*a*b^2 + 182*b^3 + 5*(15*a^3 +
93*a^2*b + 125*a*b^2 + 47*b^3)*cosh(d*x + c)^2 + 5*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 15*a
^3 + 93*a^2*b + 125*a*b^2 + 47*b^3 + 36*(a^3 + 5*a^2*b + 7*a*b^2 + 3*b^3)*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 = 0.39, size = 322, normalized size = 4.60 \[ \frac {5 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-d x - c\right )} + 5 \, {\left (a^{3} e^{\left (d x + 14 \, c\right )} + 3 \, a^{2} b e^{\left (d x + 14 \, c\right )} + 3 \, a b^{2} e^{\left (d x + 14 \, c\right )} + b^{3} e^{\left (d x + 14 \, c\right )}\right )} e^{\left (-13 \, c\right )} + \frac {4 \, {\left (15 \, a^{2} b e^{\left (9 \, d x + 9 \, c\right )} + 30 \, a b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 15 \, b^{3} e^{\left (9 \, d x + 9 \, c\right )} + 60 \, a^{2} b e^{\left (7 \, d x + 7 \, c\right )} + 100 \, a b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 40 \, b^{3} e^{\left (7 \, d x + 7 \, c\right )} + 90 \, a^{2} b e^{\left (5 \, d x + 5 \, c\right )} + 140 \, a b^{2} e^{\left (5 \, d x + 5 \, c\right )} + 66 \, b^{3} e^{\left (5 \, d x + 5 \, c\right )} + 60 \, a^{2} b e^{\left (3 \, d x + 3 \, c\right )} + 100 \, a b^{2} e^{\left (3 \, d x + 3 \, c\right )} + 40 \, b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 15 \, a^{2} b e^{\left (d x + c\right )} + 30 \, a b^{2} e^{\left (d x + c\right )} + 15 \, b^{3} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{10 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-d*x - c) + 5*(a^3*e^(d*x + 14*c) + 3*a^2*b*e^(d*x + 14*c) + 3*a*b^
2*e^(d*x + 14*c) + b^3*e^(d*x + 14*c))*e^(-13*c) + 4*(15*a^2*b*e^(9*d*x + 9*c) + 30*a*b^2*e^(9*d*x + 9*c) + 15
*b^3*e^(9*d*x + 9*c) + 60*a^2*b*e^(7*d*x + 7*c) + 100*a*b^2*e^(7*d*x + 7*c) + 40*b^3*e^(7*d*x + 7*c) + 90*a^2*
b*e^(5*d*x + 5*c) + 140*a*b^2*e^(5*d*x + 5*c) + 66*b^3*e^(5*d*x + 5*c) + 60*a^2*b*e^(3*d*x + 3*c) + 100*a*b^2*
e^(3*d*x + 3*c) + 40*b^3*e^(3*d*x + 3*c) + 15*a^2*b*e^(d*x + c) + 30*a*b^2*e^(d*x + c) + 15*b^3*e^(d*x + c))/(
e^(2*d*x + 2*c) + 1)^5)/d

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maple [B]  time = 0.23, size = 170, normalized size = 2.43 \[ \frac {a^{3} \cosh \left (d x +c \right )+3 a^{2} b \left (\frac {\sinh ^{2}\left (d x +c \right )}{\cosh \left (d x +c \right )}+\frac {2}{\cosh \left (d x +c \right )}\right )+3 a \,b^{2} \left (\frac {\sinh ^{4}\left (d x +c \right )}{\cosh \left (d x +c \right )^{3}}+\frac {4 \left (\sinh ^{2}\left (d x +c \right )\right )}{\cosh \left (d x +c \right )^{3}}+\frac {8}{3 \cosh \left (d x +c \right )^{3}}\right )+b^{3} \left (\frac {\sinh ^{6}\left (d x +c \right )}{\cosh \left (d x +c \right )^{5}}+\frac {6 \left (\sinh ^{4}\left (d x +c \right )\right )}{\cosh \left (d x +c \right )^{5}}+\frac {8 \left (\sinh ^{2}\left (d x +c \right )\right )}{\cosh \left (d x +c \right )^{5}}+\frac {16}{5 \cosh \left (d x +c \right )^{5}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*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)^4/cosh(d*x+c)^3+4*
sinh(d*x+c)^2/cosh(d*x+c)^3+8/3/cosh(d*x+c)^3)+b^3*(sinh(d*x+c)^6/cosh(d*x+c)^5+6*sinh(d*x+c)^4/cosh(d*x+c)^5+
8*sinh(d*x+c)^2/cosh(d*x+c)^5+16/5/cosh(d*x+c)^5))

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maxima [B]  time = 0.33, size = 321, normalized size = 4.59 \[ \frac {1}{10} \, b^{3} {\left (\frac {5 \, e^{\left (-d x - c\right )}}{d} + \frac {85 \, e^{\left (-2 \, d x - 2 \, c\right )} + 210 \, e^{\left (-4 \, d x - 4 \, c\right )} + 314 \, e^{\left (-6 \, d x - 6 \, c\right )} + 185 \, e^{\left (-8 \, d x - 8 \, c\right )} + 65 \, e^{\left (-10 \, d x - 10 \, c\right )} + 5}{d {\left (e^{\left (-d x - c\right )} + 5 \, e^{\left (-3 \, d x - 3 \, c\right )} + 10 \, e^{\left (-5 \, d x - 5 \, c\right )} + 10 \, e^{\left (-7 \, d x - 7 \, c\right )} + 5 \, e^{\left (-9 \, d x - 9 \, c\right )} + e^{\left (-11 \, d x - 11 \, c\right )}\right )}}\right )} + \frac {1}{2} \, a b^{2} {\left (\frac {3 \, e^{\left (-d x - c\right )}}{d} + \frac {33 \, e^{\left (-2 \, d x - 2 \, c\right )} + 41 \, e^{\left (-4 \, d x - 4 \, c\right )} + 27 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-d x - c\right )} + 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} + e^{\left (-7 \, d x - 7 \, c\right )}\right )}}\right )} + \frac {3}{2} \, a^{2} b {\left (\frac {e^{\left (-d x - c\right )}}{d} + \frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-d x - c\right )} + e^{\left (-3 \, d x - 3 \, c\right )}\right )}}\right )} + \frac {a^{3} \cosh \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*b^3*(5*e^(-d*x - c)/d + (85*e^(-2*d*x - 2*c) + 210*e^(-4*d*x - 4*c) + 314*e^(-6*d*x - 6*c) + 185*e^(-8*d*
x - 8*c) + 65*e^(-10*d*x - 10*c) + 5)/(d*(e^(-d*x - c) + 5*e^(-3*d*x - 3*c) + 10*e^(-5*d*x - 5*c) + 10*e^(-7*d
*x - 7*c) + 5*e^(-9*d*x - 9*c) + e^(-11*d*x - 11*c)))) + 1/2*a*b^2*(3*e^(-d*x - c)/d + (33*e^(-2*d*x - 2*c) +
41*e^(-4*d*x - 4*c) + 27*e^(-6*d*x - 6*c) + 3)/(d*(e^(-d*x - c) + 3*e^(-3*d*x - 3*c) + 3*e^(-5*d*x - 5*c) + e^
(-7*d*x - 7*c)))) + 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)))
) + a^3*cosh(d*x + c)/d

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mupad [B]  time = 1.25, size = 308, normalized size = 4.40 \[ \frac {{\mathrm {e}}^{c+d\,x}\,{\left (a+b\right )}^3}{2\,d}+\frac {{\mathrm {e}}^{-c-d\,x}\,{\left (a+b\right )}^3}{2\,d}+\frac {6\,{\mathrm {e}}^{c+d\,x}\,\left (a^2\,b+2\,a\,b^2+b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {64\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}+\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (9\,b^3+5\,a\,b^2\right )}{5\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {32\,b^3\,{\mathrm {e}}^{c+d\,x}}{5\,d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {8\,{\mathrm {e}}^{c+d\,x}\,\left (b^3+a\,b^2\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(c + d*x)*(a + b)^3)/(2*d) + (exp(- c - d*x)*(a + b)^3)/(2*d) + (6*exp(c + d*x)*(2*a*b^2 + a^2*b + b^3))/(
d*(exp(2*c + 2*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)*(5*a*b^2 + 9*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) + 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)) - (8*exp(c + d*x)*(a*b^2 + b^3))/(d*(2*exp(2*c +
2*d*x) + exp(4*c + 4*d*x) + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \sinh {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*sinh(c + d*x), x)

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