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

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

[Out]

1/2*sinh(b*x^2+a)/b+1/2*sinh(b*x^2+a)^3/b+3/10*sinh(b*x^2+a)^5/b+1/14*sinh(b*x^2+a)^7/b

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Rubi [A]  time = 0.04, antiderivative size = 67, normalized size of antiderivative = 1.00, 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 = {5321, 2633} \[ \frac {\sinh ^7\left (a+b x^2\right )}{14 b}+\frac {3 \sinh ^5\left (a+b x^2\right )}{10 b}+\frac {\sinh ^3\left (a+b x^2\right )}{2 b}+\frac {\sinh \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[a + b*x^2]^7,x]

[Out]

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

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 5321

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[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]))

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 67, normalized size = 1.00 \[ \frac {\sinh ^7\left (a+b x^2\right )}{14 b}+\frac {3 \sinh ^5\left (a+b x^2\right )}{10 b}+\frac {\sinh ^3\left (a+b x^2\right )}{2 b}+\frac {\sinh \left (a+b x^2\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[a + b*x^2]^7,x]

[Out]

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

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fricas [B]  time = 1.01, size = 128, normalized size = 1.91 \[ \frac {5 \, \sinh \left (b x^{2} + a\right )^{7} + 7 \, {\left (15 \, \cosh \left (b x^{2} + a\right )^{2} + 7\right )} \sinh \left (b x^{2} + a\right )^{5} + 35 \, {\left (5 \, \cosh \left (b x^{2} + a\right )^{4} + 14 \, \cosh \left (b x^{2} + a\right )^{2} + 7\right )} \sinh \left (b x^{2} + a\right )^{3} + 35 \, {\left (\cosh \left (b x^{2} + a\right )^{6} + 7 \, \cosh \left (b x^{2} + a\right )^{4} + 21 \, \cosh \left (b x^{2} + a\right )^{2} + 35\right )} \sinh \left (b x^{2} + a\right )}{4480 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4480*(5*sinh(b*x^2 + a)^7 + 7*(15*cosh(b*x^2 + a)^2 + 7)*sinh(b*x^2 + a)^5 + 35*(5*cosh(b*x^2 + a)^4 + 14*co
sh(b*x^2 + a)^2 + 7)*sinh(b*x^2 + a)^3 + 35*(cosh(b*x^2 + a)^6 + 7*cosh(b*x^2 + a)^4 + 21*cosh(b*x^2 + a)^2 +
35)*sinh(b*x^2 + a))/b

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giac [A]  time = 0.14, size = 108, normalized size = 1.61 \[ -\frac {{\left (1225 \, e^{\left (6 \, b x^{2} + 6 \, a\right )} + 245 \, e^{\left (4 \, b x^{2} + 4 \, a\right )} + 49 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} + 5\right )} e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 5 \, e^{\left (7 \, b x^{2} + 7 \, a\right )} - 49 \, e^{\left (5 \, b x^{2} + 5 \, a\right )} - 245 \, e^{\left (3 \, b x^{2} + 3 \, a\right )} - 1225 \, e^{\left (b x^{2} + a\right )}}{8960 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(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

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maple [A]  time = 0.17, size = 52, normalized size = 0.78 \[ \frac {\left (\frac {16}{35}+\frac {\left (\cosh ^{6}\left (b \,x^{2}+a \right )\right )}{7}+\frac {6 \left (\cosh ^{4}\left (b \,x^{2}+a \right )\right )}{35}+\frac {8 \left (\cosh ^{2}\left (b \,x^{2}+a \right )\right )}{35}\right ) \sinh \left (b \,x^{2}+a \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.30, size = 126, normalized size = 1.88 \[ \frac {e^{\left (7 \, b x^{2} + 7 \, a\right )}}{1792 \, b} + \frac {7 \, e^{\left (5 \, b x^{2} + 5 \, a\right )}}{1280 \, b} + \frac {7 \, e^{\left (3 \, b x^{2} + 3 \, a\right )}}{256 \, b} + \frac {35 \, e^{\left (b x^{2} + a\right )}}{256 \, b} - \frac {35 \, e^{\left (-b x^{2} - a\right )}}{256 \, b} - \frac {7 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{256 \, b} - \frac {7 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )}}{1280 \, b} - \frac {e^{\left (-7 \, b x^{2} - 7 \, a\right )}}{1792 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(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

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mupad [B]  time = 0.99, size = 52, normalized size = 0.78 \[ \frac {5\,{\mathrm {sinh}\left (b\,x^2+a\right )}^7+21\,{\mathrm {sinh}\left (b\,x^2+a\right )}^5+35\,{\mathrm {sinh}\left (b\,x^2+a\right )}^3+35\,\mathrm {sinh}\left (b\,x^2+a\right )}{70\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(a + b*x^2)^7,x)

[Out]

(35*sinh(a + b*x^2) + 35*sinh(a + b*x^2)^3 + 21*sinh(a + b*x^2)^5 + 5*sinh(a + b*x^2)^7)/(70*b)

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sympy [A]  time = 8.16, size = 94, normalized size = 1.40 \[ \begin {cases} - \frac {8 \sinh ^{7}{\left (a + b x^{2} \right )}}{35 b} + \frac {4 \sinh ^{5}{\left (a + b x^{2} \right )} \cosh ^{2}{\left (a + b x^{2} \right )}}{5 b} - \frac {\sinh ^{3}{\left (a + b x^{2} \right )} \cosh ^{4}{\left (a + b x^{2} \right )}}{b} + \frac {\sinh {\left (a + b x^{2} \right )} \cosh ^{6}{\left (a + b x^{2} \right )}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \cosh ^{7}{\relax (a )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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