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

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

[Out]

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

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Rubi [A]  time = 0.0708939, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3220, 2635, 8, 2633} \[ \frac{a \sinh (c+d x) \cosh (c+d x)}{2 d}-\frac{a x}{2}+\frac{b \cosh ^5(c+d x)}{5 d}-\frac{2 b \cosh ^3(c+d x)}{3 d}+\frac{b \cosh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0954682, size = 79, normalized size = 1.13 \[ \frac{a (-c-d x)}{2 d}+\frac{a \sinh (2 (c+d x))}{4 d}+\frac{5 b \cosh (c+d x)}{8 d}-\frac{5 b \cosh (3 (c+d x))}{48 d}+\frac{b \cosh (5 (c+d x))}{80 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-c - d*x))/(2*d) + (5*b*Cosh[c + d*x])/(8*d) - (5*b*Cosh[3*(c + d*x)])/(48*d) + (b*Cosh[5*(c + d*x)])/(80*
d) + (a*Sinh[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.013, size = 60, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( b \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \cosh \left ( dx+c \right ) +a \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b*(8/15+1/5*sinh(d*x+c)^4-4/15*sinh(d*x+c)^2)*cosh(d*x+c)+a*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c))

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Maxima [A]  time = 1.11831, size = 162, normalized size = 2.31 \begin{align*} -\frac{1}{8} \, a{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + \frac{1}{480} \, b{\left (\frac{3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} - \frac{25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{150 \, e^{\left (d x + c\right )}}{d} + \frac{150 \, e^{\left (-d x - c\right )}}{d} - \frac{25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} + \frac{3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 1/480*b*(3*e^(5*d*x + 5*c)/d - 25*e^(3*d*x + 3*c)/d +
150*e^(d*x + c)/d + 150*e^(-d*x - c)/d - 25*e^(-3*d*x - 3*c)/d + 3*e^(-5*d*x - 5*c)/d)

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Fricas [A]  time = 2.0538, size = 302, normalized size = 4.31 \begin{align*} \frac{3 \, b \cosh \left (d x + c\right )^{5} + 15 \, b \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - 25 \, b \cosh \left (d x + c\right )^{3} - 120 \, a d x + 120 \, a \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + 15 \,{\left (2 \, b \cosh \left (d x + c\right )^{3} - 5 \, b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 150 \, b \cosh \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*b*cosh(d*x + c)^5 + 15*b*cosh(d*x + c)*sinh(d*x + c)^4 - 25*b*cosh(d*x + c)^3 - 120*a*d*x + 120*a*cos
h(d*x + c)*sinh(d*x + c) + 15*(2*b*cosh(d*x + c)^3 - 5*b*cosh(d*x + c))*sinh(d*x + c)^2 + 150*b*cosh(d*x + c))
/d

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Sympy [A]  time = 2.30679, size = 117, normalized size = 1.67 \begin{align*} \begin{cases} \frac{a x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac{a x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac{a \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{2 d} + \frac{b \sinh ^{4}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{d} - \frac{4 b \sinh ^{2}{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{3 d} + \frac{8 b \cosh ^{5}{\left (c + d x \right )}}{15 d} & \text{for}\: d \neq 0 \\x \left (a + b \sinh ^{3}{\left (c \right )}\right ) \sinh ^{2}{\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)**2*(a+b*sinh(d*x+c)**3),x)

[Out]

Piecewise((a*x*sinh(c + d*x)**2/2 - a*x*cosh(c + d*x)**2/2 + a*sinh(c + d*x)*cosh(c + d*x)/(2*d) + b*sinh(c +
d*x)**4*cosh(c + d*x)/d - 4*b*sinh(c + d*x)**2*cosh(c + d*x)**3/(3*d) + 8*b*cosh(c + d*x)**5/(15*d), Ne(d, 0))
, (x*(a + b*sinh(c)**3)*sinh(c)**2, True))

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Giac [A]  time = 1.16479, size = 149, normalized size = 2.13 \begin{align*} -\frac{240 \,{\left (d x + c\right )} a - 3 \, b e^{\left (5 \, d x + 5 \, c\right )} + 25 \, b e^{\left (3 \, d x + 3 \, c\right )} - 60 \, a e^{\left (2 \, d x + 2 \, c\right )} - 150 \, b e^{\left (d x + c\right )} -{\left (150 \, b e^{\left (4 \, d x + 4 \, c\right )} - 60 \, a e^{\left (3 \, d x + 3 \, c\right )} - 25 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/480*(240*(d*x + c)*a - 3*b*e^(5*d*x + 5*c) + 25*b*e^(3*d*x + 3*c) - 60*a*e^(2*d*x + 2*c) - 150*b*e^(d*x + c
) - (150*b*e^(4*d*x + 4*c) - 60*a*e^(3*d*x + 3*c) - 25*b*e^(2*d*x + 2*c) + 3*b)*e^(-5*d*x - 5*c))/d