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

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 8, antiderivative size = 46 \[ \int \cosh ^4(a+b x) \, dx=\frac {3 x}{8}+\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh (a+b x)}{4 b} \]

[Out]

3/8*x+3/8*cosh(b*x+a)*sinh(b*x+a)/b+1/4*cosh(b*x+a)^3*sinh(b*x+a)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 8} \[ \int \cosh ^4(a+b x) \, dx=\frac {\sinh (a+b x) \cosh ^3(a+b x)}{4 b}+\frac {3 \sinh (a+b x) \cosh (a+b x)}{8 b}+\frac {3 x}{8} \]

[In]

Int[Cosh[a + b*x]^4,x]

[Out]

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

Rule 8

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

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3}{4} \int \cosh ^2(a+b x) \, dx \\ & = \frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh (a+b x)}{4 b}+\frac {3 \int 1 \, dx}{8} \\ & = \frac {3 x}{8}+\frac {3 \cosh (a+b x) \sinh (a+b x)}{8 b}+\frac {\cosh ^3(a+b x) \sinh (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.72 \[ \int \cosh ^4(a+b x) \, dx=\frac {12 (a+b x)+8 \sinh (2 (a+b x))+\sinh (4 (a+b x))}{32 b} \]

[In]

Integrate[Cosh[a + b*x]^4,x]

[Out]

(12*(a + b*x) + 8*Sinh[2*(a + b*x)] + Sinh[4*(a + b*x)])/(32*b)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67

method result size
parallelrisch \(\frac {12 b x +8 \sinh \left (2 b x +2 a \right )+\sinh \left (4 b x +4 a \right )}{32 b}\) \(31\)
derivativedivides \(\frac {\left (\frac {\cosh \left (b x +a \right )^{3}}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}}{b}\) \(39\)
default \(\frac {\left (\frac {\cosh \left (b x +a \right )^{3}}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}}{b}\) \(39\)
risch \(\frac {3 x}{8}+\frac {{\mathrm e}^{4 b x +4 a}}{64 b}+\frac {{\mathrm e}^{2 b x +2 a}}{8 b}-\frac {{\mathrm e}^{-2 b x -2 a}}{8 b}-\frac {{\mathrm e}^{-4 b x -4 a}}{64 b}\) \(61\)

[In]

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

[Out]

1/32*(12*b*x+8*sinh(2*b*x+2*a)+sinh(4*b*x+4*a))/b

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int \cosh ^4(a+b x) \, dx=\frac {\cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 3 \, b x + {\left (\cosh \left (b x + a\right )^{3} + 4 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{8 \, b} \]

[In]

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

[Out]

1/8*(cosh(b*x + a)*sinh(b*x + a)^3 + 3*b*x + (cosh(b*x + a)^3 + 4*cosh(b*x + a))*sinh(b*x + a))/b

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 95 vs. \(2 (41) = 82\).

Time = 0.16 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.07 \[ \int \cosh ^4(a+b x) \, dx=\begin {cases} \frac {3 x \sinh ^{4}{\left (a + b x \right )}}{8} - \frac {3 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac {3 x \cosh ^{4}{\left (a + b x \right )}}{8} - \frac {3 \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b} + \frac {5 \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x \cosh ^{4}{\left (a \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \cosh ^4(a+b x) \, dx=\frac {3}{8} \, x + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{8 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} - \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} \]

[In]

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

[Out]

3/8*x + 1/64*e^(4*b*x + 4*a)/b + 1/8*e^(2*b*x + 2*a)/b - 1/8*e^(-2*b*x - 2*a)/b - 1/64*e^(-4*b*x - 4*a)/b

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.30 \[ \int \cosh ^4(a+b x) \, dx=\frac {3}{8} \, x + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{64 \, b} + \frac {e^{\left (2 \, b x + 2 \, a\right )}}{8 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, b} - \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{64 \, b} \]

[In]

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

[Out]

3/8*x + 1/64*e^(4*b*x + 4*a)/b + 1/8*e^(2*b*x + 2*a)/b - 1/8*e^(-2*b*x - 2*a)/b - 1/64*e^(-4*b*x - 4*a)/b

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.67 \[ \int \cosh ^4(a+b x) \, dx=\frac {3\,x}{8}+\frac {\frac {\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{4}+\frac {\mathrm {sinh}\left (4\,a+4\,b\,x\right )}{32}}{b} \]

[In]

int(cosh(a + b*x)^4,x)

[Out]

(3*x)/8 + (sinh(2*a + 2*b*x)/4 + sinh(4*a + 4*b*x)/32)/b