\(\int x \sinh ^7(a+b x^2) \, dx\) [22]

Optimal result

Integrand size = 12, antiderivative size = 67 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=-\frac {\cosh \left (a+b x^2\right )}{2 b}+\frac {\cosh ^3\left (a+b x^2\right )}{2 b}-\frac {3 \cosh ^5\left (a+b x^2\right )}{10 b}+\frac {\cosh ^7\left (a+b x^2\right )}{14 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5428, 2713} \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=\frac {\cosh ^7\left (a+b x^2\right )}{14 b}-\frac {3 \cosh ^5\left (a+b x^2\right )}{10 b}+\frac {\cosh ^3\left (a+b x^2\right )}{2 b}-\frac {\cosh \left (a+b x^2\right )}{2 b} \]

[In]

Int[x*Sinh[a + b*x^2]^7,x]

[Out]

-1/2*Cosh[a + b*x^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 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 5428

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]))

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \sinh ^7(a+b x) \, dx,x,x^2\right ) \\ & = -\frac {\text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=-\frac {35 \cosh \left (a+b x^2\right )}{128 b}+\frac {7 \cosh \left (3 \left (a+b x^2\right )\right )}{128 b}-\frac {7 \cosh \left (5 \left (a+b x^2\right )\right )}{640 b}+\frac {\cosh \left (7 \left (a+b x^2\right )\right )}{896 b} \]

[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] (verified)

Time = 1.92 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {\left (-\frac {16}{35}+\frac {\sinh \left (x^{2} b +a \right )^{6}}{7}-\frac {6 \sinh \left (x^{2} b +a \right )^{4}}{35}+\frac {8 \sinh \left (x^{2} b +a \right )^{2}}{35}\right ) \cosh \left (x^{2} b +a \right )}{2 b}\)	\(52\)
default	\(\frac {\left (-\frac {16}{35}+\frac {\sinh \left (x^{2} b +a \right )^{6}}{7}-\frac {6 \sinh \left (x^{2} b +a \right )^{4}}{35}+\frac {8 \sinh \left (x^{2} b +a \right )^{2}}{35}\right ) \cosh \left (x^{2} b +a \right )}{2 b}\)	\(52\)
parallelrisch	\(\frac {-1024+245 \cosh \left (3 x^{2} b +3 a \right )-1225 \cosh \left (x^{2} b +a \right )-49 \cosh \left (5 x^{2} b +5 a \right )+5 \cosh \left (7 x^{2} b +7 a \right )}{4480 b}\)	\(57\)
risch	\(\frac {{\mathrm e}^{7 x^{2} b +7 a}}{1792 b}-\frac {7 \,{\mathrm e}^{5 x^{2} b +5 a}}{1280 b}+\frac {7 \,{\mathrm e}^{3 x^{2} b +3 a}}{256 b}-\frac {35 \,{\mathrm e}^{x^{2} b +a}}{256 b}-\frac {35 \,{\mathrm e}^{-x^{2} b -a}}{256 b}+\frac {7 \,{\mathrm e}^{-3 x^{2} b -3 a}}{256 b}-\frac {7 \,{\mathrm e}^{-5 x^{2} b -5 a}}{1280 b}+\frac {{\mathrm e}^{-7 x^{2} b -7 a}}{1792 b}\)	\(127\)

[In]

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

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 154 vs. \(2 (59) = 118\).

Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 2.30 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=\frac {5 \, \cosh \left (b x^{2} + a\right )^{7} + 35 \, \cosh \left (b x^{2} + a\right ) \sinh \left (b x^{2} + a\right )^{6} - 49 \, \cosh \left (b x^{2} + a\right )^{5} + 35 \, {\left (5 \, \cosh \left (b x^{2} + a\right )^{3} - 7 \, \cosh \left (b x^{2} + a\right )\right )} \sinh \left (b x^{2} + a\right )^{4} + 245 \, \cosh \left (b x^{2} + a\right )^{3} + 35 \, {\left (3 \, \cosh \left (b x^{2} + a\right )^{5} - 14 \, \cosh \left (b x^{2} + a\right )^{3} + 21 \, \cosh \left (b x^{2} + a\right )\right )} \sinh \left (b x^{2} + a\right )^{2} - 1225 \, \cosh \left (b x^{2} + a\right )}{4480 \, b} \]

[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] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=\begin {cases} \frac {\sinh ^{6}{\left (a + b x^{2} \right )} \cosh {\left (a + b x^{2} \right )}}{2 b} - \frac {\sinh ^{4}{\left (a + b x^{2} \right )} \cosh ^{3}{\left (a + b x^{2} \right )}}{b} + \frac {4 \sinh ^{2}{\left (a + b x^{2} \right )} \cosh ^{5}{\left (a + b x^{2} \right )}}{5 b} - \frac {8 \cosh ^{7}{\left (a + b x^{2} \right )}}{35 b} & \text {for}\: b \neq 0 \\\frac {x^{2} \sinh ^{7}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]

[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))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (59) = 118\).

Time = 0.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.88 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=\frac {e^{\left (7 \, b x^{2} + 7 \, a\right )}}{1792 \, b} - \frac {7 \, e^{\left (5 \, b x^{2} + 5 \, a\right )}}{1280 \, b} + \frac {7 \, e^{\left (3 \, b x^{2} + 3 \, a\right )}}{256 \, b} - \frac {35 \, e^{\left (b x^{2} + a\right )}}{256 \, b} - \frac {35 \, e^{\left (-b x^{2} - a\right )}}{256 \, b} + \frac {7 \, e^{\left (-3 \, b x^{2} - 3 \, a\right )}}{256 \, b} - \frac {7 \, e^{\left (-5 \, b x^{2} - 5 \, a\right )}}{1280 \, b} + \frac {e^{\left (-7 \, b x^{2} - 7 \, a\right )}}{1792 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.61 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=-\frac {{\left (1225 \, e^{\left (6 \, b x^{2} + 6 \, a\right )} - 245 \, e^{\left (4 \, b x^{2} + 4 \, a\right )} + 49 \, e^{\left (2 \, b x^{2} + 2 \, a\right )} - 5\right )} e^{\left (-7 \, b x^{2} - 7 \, a\right )} - 5 \, e^{\left (7 \, b x^{2} + 7 \, a\right )} + 49 \, e^{\left (5 \, b x^{2} + 5 \, a\right )} - 245 \, e^{\left (3 \, b x^{2} + 3 \, a\right )} + 1225 \, e^{\left (b x^{2} + a\right )}}{8960 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int x \sinh ^7\left (a+b x^2\right ) \, dx=-\frac {-5\,{\mathrm {cosh}\left (b\,x^2+a\right )}^7+21\,{\mathrm {cosh}\left (b\,x^2+a\right )}^5-35\,{\mathrm {cosh}\left (b\,x^2+a\right )}^3+35\,\mathrm {cosh}\left (b\,x^2+a\right )}{70\,b} \]

[In]

int(x*sinh(a + b*x^2)^7,x)

[Out]

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