3.142 \(\int \sinh ^3(c+d x) (a+b \sinh ^3(c+d x)) \, dx\)

Optimal. Leaf size=99 \[ \frac{a \cosh ^3(c+d x)}{3 d}-\frac{a \cosh (c+d x)}{d}+\frac{b \sinh ^5(c+d x) \cosh (c+d x)}{6 d}-\frac{5 b \sinh ^3(c+d x) \cosh (c+d x)}{24 d}+\frac{5 b \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{5 b x}{16} \]

[Out]

(-5*b*x)/16 - (a*Cosh[c + d*x])/d + (a*Cosh[c + d*x]^3)/(3*d) + (5*b*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (5*
b*Cosh[c + d*x]*Sinh[c + d*x]^3)/(24*d) + (b*Cosh[c + d*x]*Sinh[c + d*x]^5)/(6*d)

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Rubi [A]  time = 0.108472, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3220, 2633, 2635, 8} \[ \frac{a \cosh ^3(c+d x)}{3 d}-\frac{a \cosh (c+d x)}{d}+\frac{b \sinh ^5(c+d x) \cosh (c+d x)}{6 d}-\frac{5 b \sinh ^3(c+d x) \cosh (c+d x)}{24 d}+\frac{5 b \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{5 b x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*x)/16 - (a*Cosh[c + d*x])/d + (a*Cosh[c + d*x]^3)/(3*d) + (5*b*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (5*
b*Cosh[c + d*x]*Sinh[c + d*x]^3)/(24*d) + (b*Cosh[c + d*x]*Sinh[c + d*x]^5)/(6*d)

Rule 3220

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.), x_Symbol] :> Int[ExpandTr
ig[sin[e + f*x]^m*(a + b*sin[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, e, f}, x] && IntegersQ[m, p] && (EqQ[n, 4]
|| GtQ[p, 0] || (EqQ[p, -1] && IntegerQ[n]))

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \sinh ^3(c+d x) \left (a+b \sinh ^3(c+d x)\right ) \, dx &=i \int \left (-i a \sinh ^3(c+d x)-i b \sinh ^6(c+d x)\right ) \, dx\\ &=a \int \sinh ^3(c+d x) \, dx+b \int \sinh ^6(c+d x) \, dx\\ &=\frac{b \cosh (c+d x) \sinh ^5(c+d x)}{6 d}-\frac{1}{6} (5 b) \int \sinh ^4(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{a \cosh (c+d x)}{d}+\frac{a \cosh ^3(c+d x)}{3 d}-\frac{5 b \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac{b \cosh (c+d x) \sinh ^5(c+d x)}{6 d}+\frac{1}{8} (5 b) \int \sinh ^2(c+d x) \, dx\\ &=-\frac{a \cosh (c+d x)}{d}+\frac{a \cosh ^3(c+d x)}{3 d}+\frac{5 b \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{5 b \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac{b \cosh (c+d x) \sinh ^5(c+d x)}{6 d}-\frac{1}{16} (5 b) \int 1 \, dx\\ &=-\frac{5 b x}{16}-\frac{a \cosh (c+d x)}{d}+\frac{a \cosh ^3(c+d x)}{3 d}+\frac{5 b \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{5 b \cosh (c+d x) \sinh ^3(c+d x)}{24 d}+\frac{b \cosh (c+d x) \sinh ^5(c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.129329, size = 66, normalized size = 0.67 \[ \frac{-144 a \cosh (c+d x)+16 a \cosh (3 (c+d x))+b (45 \sinh (2 (c+d x))-9 \sinh (4 (c+d x))+\sinh (6 (c+d x))-60 c-60 d x)}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-144*a*Cosh[c + d*x] + 16*a*Cosh[3*(c + d*x)] + b*(-60*c - 60*d*x + 45*Sinh[2*(c + d*x)] - 9*Sinh[4*(c + d*x)
] + Sinh[6*(c + d*x)]))/(192*d)

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Maple [A]  time = 0.015, size = 72, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( b \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \right ) +a \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b*((1/6*sinh(d*x+c)^5-5/24*sinh(d*x+c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5/16*d*x-5/16*c)+a*(-2/3+1/3*sinh(
d*x+c)^2)*cosh(d*x+c))

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Maxima [A]  time = 1.11576, size = 193, normalized size = 1.95 \begin{align*} -\frac{1}{384} \, b{\left (\frac{{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac{120 \,{\left (d x + c\right )}}{d} + \frac{45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac{1}{24} \, a{\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*b*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*c) - 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x -
2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d) + 1/24*a*(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)

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Fricas [A]  time = 1.88429, size = 375, normalized size = 3.79 \begin{align*} \frac{3 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + 8 \, a \cosh \left (d x + c\right )^{3} + 24 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \,{\left (5 \, b \cosh \left (d x + c\right )^{3} - 9 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} - 30 \, b d x - 72 \, a \cosh \left (d x + c\right ) + 3 \,{\left (b \cosh \left (d x + c\right )^{5} - 6 \, b \cosh \left (d x + c\right )^{3} + 15 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*b*cosh(d*x + c)*sinh(d*x + c)^5 + 8*a*cosh(d*x + c)^3 + 24*a*cosh(d*x + c)*sinh(d*x + c)^2 + 2*(5*b*co
sh(d*x + c)^3 - 9*b*cosh(d*x + c))*sinh(d*x + c)^3 - 30*b*d*x - 72*a*cosh(d*x + c) + 3*(b*cosh(d*x + c)^5 - 6*
b*cosh(d*x + c)^3 + 15*b*cosh(d*x + c))*sinh(d*x + c))/d

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Sympy [A]  time = 4.72493, size = 194, normalized size = 1.96 \begin{align*} \begin{cases} \frac{a \sinh ^{2}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{d} - \frac{2 a \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac{5 b x \sinh ^{6}{\left (c + d x \right )}}{16} - \frac{15 b x \sinh ^{4}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{16} + \frac{15 b x \sinh ^{2}{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{16} - \frac{5 b x \cosh ^{6}{\left (c + d x \right )}}{16} + \frac{11 b \sinh ^{5}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{16 d} - \frac{5 b \sinh ^{3}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{6 d} + \frac{5 b \sinh{\left (c + d x \right )} \cosh ^{5}{\left (c + d x \right )}}{16 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\left (c \right )}\right ) \sinh ^{3}{\left (c \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*sinh(c + d*x)**2*cosh(c + d*x)/d - 2*a*cosh(c + d*x)**3/(3*d) + 5*b*x*sinh(c + d*x)**6/16 - 15*b*
x*sinh(c + d*x)**4*cosh(c + d*x)**2/16 + 15*b*x*sinh(c + d*x)**2*cosh(c + d*x)**4/16 - 5*b*x*cosh(c + d*x)**6/
16 + 11*b*sinh(c + d*x)**5*cosh(c + d*x)/(16*d) - 5*b*sinh(c + d*x)**3*cosh(c + d*x)**3/(6*d) + 5*b*sinh(c + d
*x)*cosh(c + d*x)**5/(16*d), Ne(d, 0)), (x*(a + b*sinh(c)**3)*sinh(c)**3, True))

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Giac [A]  time = 1.15322, size = 177, normalized size = 1.79 \begin{align*} -\frac{120 \,{\left (d x + c\right )} b - b e^{\left (6 \, d x + 6 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 16 \, a e^{\left (3 \, d x + 3 \, c\right )} - 45 \, b e^{\left (2 \, d x + 2 \, c\right )} + 144 \, a e^{\left (d x + c\right )} +{\left (144 \, a e^{\left (5 \, d x + 5 \, c\right )} + 45 \, b e^{\left (4 \, d x + 4 \, c\right )} - 16 \, a e^{\left (3 \, d x + 3 \, c\right )} - 9 \, b e^{\left (2 \, d x + 2 \, c\right )} + b\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(120*(d*x + c)*b - b*e^(6*d*x + 6*c) + 9*b*e^(4*d*x + 4*c) - 16*a*e^(3*d*x + 3*c) - 45*b*e^(2*d*x + 2*c
) + 144*a*e^(d*x + c) + (144*a*e^(5*d*x + 5*c) + 45*b*e^(4*d*x + 4*c) - 16*a*e^(3*d*x + 3*c) - 9*b*e^(2*d*x +
2*c) + b)*e^(-6*d*x - 6*c))/d