3.22 \(\int x \sinh ^7(a+b x^2) \, dx\)

Optimal. Leaf size=67 \[ \frac{\cosh ^7\left (a+b x^2\right )}{14 b}-\frac{3 \cosh ^5\left (a+b x^2\right )}{10 b}+\frac{\cosh ^3\left (a+b x^2\right )}{2 b}-\frac{\cosh \left (a+b x^2\right )}{2 b} \]

[Out]

-Cosh[a + b*x^2]/(2*b) + Cosh[a + b*x^2]^3/(2*b) - (3*Cosh[a + b*x^2]^5)/(10*b) + Cosh[a + b*x^2]^7/(14*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0473215, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5320, 2633} \[ \frac{\cosh ^7\left (a+b x^2\right )}{14 b}-\frac{3 \cosh ^5\left (a+b x^2\right )}{10 b}+\frac{\cosh ^3\left (a+b x^2\right )}{2 b}-\frac{\cosh \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Sinh[a + b*x^2]^7,x]

[Out]

-Cosh[a + b*x^2]/(2*b) + Cosh[a + b*x^2]^3/(2*b) - (3*Cosh[a + b*x^2]^5)/(10*b) + Cosh[a + b*x^2]^7/(14*b)

Rule 5320

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

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 x \sinh ^7\left (a+b x^2\right ) \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \sinh ^7(a+b x) \, dx,x,x^2\right )\\ &=-\frac{\operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,\cosh \left (a+b x^2\right )\right )}{2 b}\\ &=-\frac{\cosh \left (a+b x^2\right )}{2 b}+\frac{\cosh ^3\left (a+b x^2\right )}{2 b}-\frac{3 \cosh ^5\left (a+b x^2\right )}{10 b}+\frac{\cosh ^7\left (a+b x^2\right )}{14 b}\\ \end{align*}

Mathematica [A]  time = 0.0248622, size = 67, normalized size = 1. \[ -\frac{35 \cosh \left (a+b x^2\right )}{128 b}+\frac{7 \cosh \left (3 \left (a+b x^2\right )\right )}{128 b}-\frac{7 \cosh \left (5 \left (a+b x^2\right )\right )}{640 b}+\frac{\cosh \left (7 \left (a+b x^2\right )\right )}{896 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sinh[a + b*x^2]^7,x]

[Out]

(-35*Cosh[a + b*x^2])/(128*b) + (7*Cosh[3*(a + b*x^2)])/(128*b) - (7*Cosh[5*(a + b*x^2)])/(640*b) + Cosh[7*(a
+ b*x^2)]/(896*b)

________________________________________________________________________________________

Maple [A]  time = 0.045, size = 52, normalized size = 0.8 \begin{align*}{\frac{\cosh \left ( b{x}^{2}+a \right ) }{2\,b} \left ( -{\frac{16}{35}}+{\frac{ \left ( \sinh \left ( b{x}^{2}+a \right ) \right ) ^{6}}{7}}-{\frac{6\, \left ( \sinh \left ( b{x}^{2}+a \right ) \right ) ^{4}}{35}}+{\frac{8\, \left ( \sinh \left ( b{x}^{2}+a \right ) \right ) ^{2}}{35}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sinh(b*x^2+a)^7,x)

[Out]

1/2/b*(-16/35+1/7*sinh(b*x^2+a)^6-6/35*sinh(b*x^2+a)^4+8/35*sinh(b*x^2+a)^2)*cosh(b*x^2+a)

________________________________________________________________________________________

Maxima [B]  time = 1.02966, size = 170, normalized size = 2.54 \begin{align*} \frac{e^{\left (7 \, b x^{2} + 7 \, a\right )}}{1792 \, b} - \frac{7 \, e^{\left (5 \, b x^{2} + 5 \, a\right )}}{1280 \, b} + \frac{7 \, e^{\left (3 \, b x^{2} + 3 \, a\right )}}{256 \, b} - \frac{35 \, e^{\left (b x^{2} + a\right )}}{256 \, b} - \frac{35 \, e^{\left (-b x^{2} - a\right )}}{256 \, b} + \frac{7 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{256 \, b} - \frac{7 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )}}{1280 \, b} + \frac{e^{\left (-7 \, b x^{2} - 7 \, a\right )}}{1792 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^7,x, algorithm="maxima")

[Out]

1/1792*e^(7*b*x^2 + 7*a)/b - 7/1280*e^(5*b*x^2 + 5*a)/b + 7/256*e^(3*b*x^2 + 3*a)/b - 35/256*e^(b*x^2 + a)/b -
35/256*e^(-b*x^2 - a)/b + 7/256*e^(-3*b*x^2 - 3*a)/b - 7/1280*e^(-5*b*x^2 - 5*a)/b + 1/1792*e^(-7*b*x^2 - 7*a
)/b

________________________________________________________________________________________

Fricas [B]  time = 1.68794, size = 398, normalized size = 5.94 \begin{align*} \frac{5 \, \cosh \left (b x^{2} + a\right )^{7} + 35 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{6} - 49 \, \cosh \left (b x^{2} + a\right )^{5} + 35 \,{\left (5 \, \cosh \left (b x^{2} + a\right )^{3} - 7 \, \cosh \left (b x^{2} + a\right )\right )} \sinh \left (b x^{2} + a\right )^{4} + 245 \, \cosh \left (b x^{2} + a\right )^{3} + 35 \,{\left (3 \, \cosh \left (b x^{2} + a\right )^{5} - 14 \, \cosh \left (b x^{2} + a\right )^{3} + 21 \, \cosh \left (b x^{2} + a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} - 1225 \, \cosh \left (b x^{2} + a\right )}{4480 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^7,x, algorithm="fricas")

[Out]

1/4480*(5*cosh(b*x^2 + a)^7 + 35*cosh(b*x^2 + a)*sinh(b*x^2 + a)^6 - 49*cosh(b*x^2 + a)^5 + 35*(5*cosh(b*x^2 +
a)^3 - 7*cosh(b*x^2 + a))*sinh(b*x^2 + a)^4 + 245*cosh(b*x^2 + a)^3 + 35*(3*cosh(b*x^2 + a)^5 - 14*cosh(b*x^2
+ a)^3 + 21*cosh(b*x^2 + a))*sinh(b*x^2 + a)^2 - 1225*cosh(b*x^2 + a))/b

________________________________________________________________________________________

Sympy [A]  time = 13.3207, size = 94, normalized size = 1.4 \begin{align*} \begin{cases} \frac{\sinh ^{6}{\left (a + b x^{2} \right )} \cosh{\left (a + b x^{2} \right )}}{2 b} - \frac{\sinh ^{4}{\left (a + b x^{2} \right )} \cosh ^{3}{\left (a + b x^{2} \right )}}{b} + \frac{4 \sinh ^{2}{\left (a + b x^{2} \right )} \cosh ^{5}{\left (a + b x^{2} \right )}}{5 b} - \frac{8 \cosh ^{7}{\left (a + b x^{2} \right )}}{35 b} & \text{for}\: b \neq 0 \\\frac{x^{2} \sinh ^{7}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x**2+a)**7,x)

[Out]

Piecewise((sinh(a + b*x**2)**6*cosh(a + b*x**2)/(2*b) - sinh(a + b*x**2)**4*cosh(a + b*x**2)**3/b + 4*sinh(a +
b*x**2)**2*cosh(a + b*x**2)**5/(5*b) - 8*cosh(a + b*x**2)**7/(35*b), Ne(b, 0)), (x**2*sinh(a)**7/2, True))

________________________________________________________________________________________

Giac [A]  time = 1.3114, size = 146, normalized size = 2.18 \begin{align*} -\frac{{\left (1225 \, e^{\left (6 \, b x^{2} + 6 \, a\right )} - 245 \, e^{\left (4 \, b x^{2} + 4 \, a\right )} + 49 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} - 5\right )} e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 5 \, e^{\left (7 \, b x^{2} + 7 \, a\right )} + 49 \, e^{\left (5 \, b x^{2} + 5 \, a\right )} - 245 \, e^{\left (3 \, b x^{2} + 3 \, a\right )} + 1225 \, e^{\left (b x^{2} + a\right )}}{8960 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sinh(b*x^2+a)^7,x, algorithm="giac")

[Out]

-1/8960*((1225*e^(6*b*x^2 + 6*a) - 245*e^(4*b*x^2 + 4*a) + 49*e^(2*b*x^2 + 2*a) - 5)*e^(-7*b*x^2 - 7*a) - 5*e^
(7*b*x^2 + 7*a) + 49*e^(5*b*x^2 + 5*a) - 245*e^(3*b*x^2 + 3*a) + 1225*e^(b*x^2 + a))/b