\(\int x \cosh ^4(a+b x) \, dx\) [25]

Optimal result

Integrand size = 10, antiderivative size = 80 \[ \int x \cosh ^4(a+b x) \, dx=\frac {3 x^2}{16}-\frac {3 \cosh ^2(a+b x)}{16 b^2}-\frac {\cosh ^4(a+b x)}{16 b^2}+\frac {3 x \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {x \cosh ^3(a+b x) \sinh (a+b x)}{4 b} \]

[Out]

3/16*x^2-3/16*cosh(b*x+a)^2/b^2-1/16*cosh(b*x+a)^4/b^2+3/8*x*cosh(b*x+a)*sinh(b*x+a)/b+1/4*x*cosh(b*x+a)^3*sin
h(b*x+a)/b

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3391, 30} \[ \int x \cosh ^4(a+b x) \, dx=-\frac {\cosh ^4(a+b x)}{16 b^2}-\frac {3 \cosh ^2(a+b x)}{16 b^2}+\frac {x \sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3 x \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac {3 x^2}{16} \]

[In]

Int[x*Cosh[a + b*x]^4,x]

[Out]

(3*x^2)/16 - (3*Cosh[a + b*x]^2)/(16*b^2) - Cosh[a + b*x]^4/(16*b^2) + (3*x*Cosh[a + b*x]*Sinh[a + b*x])/(8*b)
+ (x*Cosh[a + b*x]^3*Sinh[a + b*x])/(4*b)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\cosh ^4(a+b x)}{16 b^2}+\frac {x \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3}{4} \int x \cosh ^2(a+b x) \, dx \\ & = -\frac {3 \cosh ^2(a+b x)}{16 b^2}-\frac {\cosh ^4(a+b x)}{16 b^2}+\frac {3 x \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {x \cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3 \int x \, dx}{8} \\ & = \frac {3 x^2}{16}-\frac {3 \cosh ^2(a+b x)}{16 b^2}-\frac {\cosh ^4(a+b x)}{16 b^2}+\frac {3 x \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {x \cosh ^3(a+b x) \sinh (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int x \cosh ^4(a+b x) \, dx=-\frac {16 \cosh (2 (a+b x))+\cosh (4 (a+b x))-4 b x (6 b x+8 \sinh (2 (a+b x))+\sinh (4 (a+b x)))}{128 b^2} \]

[In]

Integrate[x*Cosh[a + b*x]^4,x]

[Out]

-1/128*(16*Cosh[2*(a + b*x)] + Cosh[4*(a + b*x)] - 4*b*x*(6*b*x + 8*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)]))/b^
2

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.80

method	result	size
parallelrisch	\(\frac {24 x^{2} b^{2}+4 b x \sinh \left (4 b x +4 a \right )+32 b x \sinh \left (2 b x +2 a \right )-\cosh \left (4 b x +4 a \right )-16 \cosh \left (2 b x +2 a \right )+17}{128 b^{2}}\)	\(64\)
risch	\(\frac {3 x^{2}}{16}+\frac {\left (4 b x -1\right ) {\mathrm e}^{4 b x +4 a}}{256 b^{2}}+\frac {\left (2 b x -1\right ) {\mathrm e}^{2 b x +2 a}}{16 b^{2}}-\frac {\left (2 b x +1\right ) {\mathrm e}^{-2 b x -2 a}}{16 b^{2}}-\frac {\left (4 b x +1\right ) {\mathrm e}^{-4 b x -4 a}}{256 b^{2}}\)	\(87\)
derivativedivides	\(\frac {\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{4}+\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}+\frac {3 \left (b x +a \right )^{2}}{16}-\frac {\cosh \left (b x +a \right )^{4}}{16}-\frac {3 \cosh \left (b x +a \right )^{2}}{16}-a \left (\left (\frac {\cosh \left (b x +a \right )^{3}}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{2}}\)	\(112\)
default	\(\frac {\frac {\left (b x +a \right ) \sinh \left (b x +a \right ) \cosh \left (b x +a \right )^{3}}{4}+\frac {3 \left (b x +a \right ) \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{8}+\frac {3 \left (b x +a \right )^{2}}{16}-\frac {\cosh \left (b x +a \right )^{4}}{16}-\frac {3 \cosh \left (b x +a \right )^{2}}{16}-a \left (\left (\frac {\cosh \left (b x +a \right )^{3}}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}\right )}{b^{2}}\)	\(112\)

[In]

int(x*cosh(b*x+a)^4,x,method=_RETURNVERBOSE)

[Out]

1/128*(24*x^2*b^2+4*b*x*sinh(4*b*x+4*a)+32*b*x*sinh(2*b*x+2*a)-cosh(4*b*x+4*a)-16*cosh(2*b*x+2*a)+17)/b^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.42 \[ \int x \cosh ^4(a+b x) \, dx=\frac {16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 24 \, b^{2} x^{2} - \cosh \left (b x + a\right )^{4} - \sinh \left (b x + a\right )^{4} - 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} + 8\right )} \sinh \left (b x + a\right )^{2} - 16 \, \cosh \left (b x + a\right )^{2} + 16 \, {\left (b x \cosh \left (b x + a\right )^{3} + 4 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{128 \, b^{2}} \]

[In]

integrate(x*cosh(b*x+a)^4,x, algorithm="fricas")

[Out]

1/128*(16*b*x*cosh(b*x + a)*sinh(b*x + a)^3 + 24*b^2*x^2 - cosh(b*x + a)^4 - sinh(b*x + a)^4 - 2*(3*cosh(b*x +
a)^2 + 8)*sinh(b*x + a)^2 - 16*cosh(b*x + a)^2 + 16*(b*x*cosh(b*x + a)^3 + 4*b*x*cosh(b*x + a))*sinh(b*x + a)
)/b^2

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.72 \[ \int x \cosh ^4(a+b x) \, dx=\begin {cases} \frac {3 x^{2} \sinh ^{4}{\left (a + b x \right )}}{16} - \frac {3 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} + \frac {3 x^{2} \cosh ^{4}{\left (a + b x \right )}}{16} - \frac {3 x \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b} + \frac {5 x \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} + \frac {3 \sinh ^{4}{\left (a + b x \right )}}{32 b^{2}} - \frac {5 \cosh ^{4}{\left (a + b x \right )}}{32 b^{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \cosh ^{4}{\left (a \right )}}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(x*cosh(b*x+a)**4,x)

[Out]

Piecewise((3*x**2*sinh(a + b*x)**4/16 - 3*x**2*sinh(a + b*x)**2*cosh(a + b*x)**2/8 + 3*x**2*cosh(a + b*x)**4/1
6 - 3*x*sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + 5*x*sinh(a + b*x)*cosh(a + b*x)**3/(8*b) + 3*sinh(a + b*x)**4/(
32*b**2) - 5*cosh(a + b*x)**4/(32*b**2), Ne(b, 0)), (x**2*cosh(a)**4/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.20 \[ \int x \cosh ^4(a+b x) \, dx=\frac {3}{16} \, x^{2} + \frac {{\left (4 \, b x e^{\left (4 \, a\right )} - e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x e^{\left (2 \, a\right )} - e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{16 \, b^{2}} - \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} - \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \]

[In]

integrate(x*cosh(b*x+a)^4,x, algorithm="maxima")

[Out]

3/16*x^2 + 1/256*(4*b*x*e^(4*a) - e^(4*a))*e^(4*b*x)/b^2 + 1/16*(2*b*x*e^(2*a) - e^(2*a))*e^(2*b*x)/b^2 - 1/16
*(2*b*x + 1)*e^(-2*b*x - 2*a)/b^2 - 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int x \cosh ^4(a+b x) \, dx=\frac {3}{16} \, x^{2} + \frac {{\left (4 \, b x - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b^{2}} + \frac {{\left (2 \, b x - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b^{2}} - \frac {{\left (2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b^{2}} - \frac {{\left (4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b^{2}} \]

[In]

integrate(x*cosh(b*x+a)^4,x, algorithm="giac")

[Out]

3/16*x^2 + 1/256*(4*b*x - 1)*e^(4*b*x + 4*a)/b^2 + 1/16*(2*b*x - 1)*e^(2*b*x + 2*a)/b^2 - 1/16*(2*b*x + 1)*e^(
-2*b*x - 2*a)/b^2 - 1/256*(4*b*x + 1)*e^(-4*b*x - 4*a)/b^2

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int x \cosh ^4(a+b x) \, dx=\frac {3\,x^2}{16}-\frac {\frac {3\,{\mathrm {cosh}\left (a+b\,x\right )}^2}{16}+\frac {{\mathrm {cosh}\left (a+b\,x\right )}^4}{16}-b\,\left (\frac {x\,\mathrm {sinh}\left (a+b\,x\right )\,{\mathrm {cosh}\left (a+b\,x\right )}^3}{4}+\frac {3\,x\,\mathrm {sinh}\left (a+b\,x\right )\,\mathrm {cosh}\left (a+b\,x\right )}{8}\right )}{b^2} \]

[In]

int(x*cosh(a + b*x)^4,x)

[Out]

(3*x^2)/16 - ((3*cosh(a + b*x)^2)/16 + cosh(a + b*x)^4/16 - b*((x*cosh(a + b*x)^3*sinh(a + b*x))/4 + (3*x*cosh
(a + b*x)*sinh(a + b*x))/8))/b^2