3.2.43 \(\int \sinh ^2(c+d x) (a+b \sinh ^3(c+d x)) \, dx\) [143]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3299, 2715, 8, 2713} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*x) + (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]*
Sinh[c + d*x])/(2*d)

Rule 8

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

Rule 2713

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 2715

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] && Integ
erQ[2*n]

Rule 3299

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]))

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 \text {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*}

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Mathematica [A]
time = 0.07, size = 79, normalized size = 1.13 \begin {gather*} \frac {a (-c-d x)}{2 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}+\frac {a \sinh (2 (c+d x))}{4 d} \end {gather*}

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 = 0.98, size = 63, normalized size = 0.90

method	result	size
default	\(-\frac {a x}{2}+\frac {5 b \cosh \left (d x +c \right )}{8 d}-\frac {5 b \cosh \left (3 d x +3 c \right )}{48 d}+\frac {b \cosh \left (5 d x +5 c \right )}{80 d}+\frac {a \sinh \left (2 d x +2 c \right )}{4 d}\)	\(63\)
risch	\(-\frac {a x}{2}+\frac {b \,{\mathrm e}^{5 d x +5 c}}{160 d}-\frac {5 b \,{\mathrm e}^{3 d x +3 c}}{96 d}+\frac {a \,{\mathrm e}^{2 d x +2 c}}{8 d}+\frac {5 b \,{\mathrm e}^{d x +c}}{16 d}+\frac {5 b \,{\mathrm e}^{-d x -c}}{16 d}-\frac {a \,{\mathrm e}^{-2 d x -2 c}}{8 d}-\frac {5 b \,{\mathrm e}^{-3 d x -3 c}}{96 d}+\frac {b \,{\mathrm e}^{-5 d x -5 c}}{160 d}\)	\(123\)

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,method=_RETURNVERBOSE)

[Out]

-1/2*a*x+5/8*b*cosh(d*x+c)/d-5/48*b/d*cosh(3*d*x+3*c)+1/80*b/d*cosh(5*d*x+5*c)+1/4*a*sinh(2*d*x+2*c)/d

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Maxima [A]
time = 0.28, size = 120, normalized size = 1.71 \begin {gather*} -\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 {gather*}

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 = 0.44, size = 105, normalized size = 1.50 \begin {gather*} \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 {gather*}

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 = 0.27, size = 117, normalized size = 1.67 \begin {gather*} \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 {gather*}

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 = 0.42, size = 122, normalized size = 1.74 \begin {gather*} -\frac {1}{2} \, a x + \frac {b e^{\left (5 \, d x + 5 \, c\right )}}{160 \, d} - \frac {5 \, b e^{\left (3 \, d x + 3 \, c\right )}}{96 \, d} + \frac {a e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac {5 \, b e^{\left (d x + c\right )}}{16 \, d} + \frac {5 \, b e^{\left (-d x - c\right )}}{16 \, d} - \frac {a e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {5 \, b e^{\left (-3 \, d x - 3 \, c\right )}}{96 \, d} + \frac {b e^{\left (-5 \, d x - 5 \, c\right )}}{160 \, d} \end {gather*}

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]

-1/2*a*x + 1/160*b*e^(5*d*x + 5*c)/d - 5/96*b*e^(3*d*x + 3*c)/d + 1/8*a*e^(2*d*x + 2*c)/d + 5/16*b*e^(d*x + c)
/d + 5/16*b*e^(-d*x - c)/d - 1/8*a*e^(-2*d*x - 2*c)/d - 5/96*b*e^(-3*d*x - 3*c)/d + 1/160*b*e^(-5*d*x - 5*c)/d

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Mupad [B]
time = 0.12, size = 55, normalized size = 0.79 \begin {gather*} \frac {b\,\mathrm {cosh}\left (c+d\,x\right )-\frac {2\,b\,{\mathrm {cosh}\left (c+d\,x\right )}^3}{3}+\frac {b\,{\mathrm {cosh}\left (c+d\,x\right )}^5}{5}+\frac {a\,\mathrm {cosh}\left (c+d\,x\right )\,\mathrm {sinh}\left (c+d\,x\right )}{2}}{d}-\frac {a\,x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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