3.265 \(\int e^{a+b x} \cosh ^3(a+b x) \, dx\)

Optimal. Leaf size=57 \[ -\frac {e^{-2 a-2 b x}}{16 b}+\frac {3 e^{2 a+2 b x}}{16 b}+\frac {e^{4 a+4 b x}}{32 b}+\frac {3 x}{8} \]

[Out]

-1/16*exp(-2*b*x-2*a)/b+3/16*exp(2*b*x+2*a)/b+1/32*exp(4*b*x+4*a)/b+3/8*x

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Rubi [A]  time = 0.04, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2282, 12, 266, 43} \[ -\frac {e^{-2 a-2 b x}}{16 b}+\frac {3 e^{2 a+2 b x}}{16 b}+\frac {e^{4 a+4 b x}}{32 b}+\frac {3 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Cosh[a + b*x]^3,x]

[Out]

-E^(-2*a - 2*b*x)/(16*b) + (3*E^(2*a + 2*b*x))/(16*b) + E^(4*a + 4*b*x)/(32*b) + (3*x)/8

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int e^{a+b x} \cosh ^3(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{8 x^3} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^3} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(1+x)^3}{x^2} \, dx,x,e^{2 a+2 b x}\right )}{16 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (3+\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,e^{2 a+2 b x}\right )}{16 b}\\ &=-\frac {e^{-2 a-2 b x}}{16 b}+\frac {3 e^{2 a+2 b x}}{16 b}+\frac {e^{4 a+4 b x}}{32 b}+\frac {3 x}{8}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 47, normalized size = 0.82 \[ \frac {-e^{-2 (a+b x)}+3 e^{2 (a+b x)}+\frac {1}{2} e^{4 (a+b x)}+6 b x}{16 b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Cosh[a + b*x]^3,x]

[Out]

(-E^(-2*(a + b*x)) + 3*E^(2*(a + b*x)) + E^(4*(a + b*x))/2 + 6*b*x)/(16*b)

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fricas [B]  time = 0.48, size = 95, normalized size = 1.67 \[ -\frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 3 \, \sinh \left (b x + a\right )^{3} - 6 \, {\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) + 3 \, {\left (4 \, b x - 3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right )}{32 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/32*(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 - 3*sinh(b*x + a)^3 - 6*(2*b*x + 1)*cosh(b*x + a) + 3
*(4*b*x - 3*cosh(b*x + a)^2 - 2)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*x + a))

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giac [A]  time = 0.14, size = 57, normalized size = 1.00 \[ \frac {12 \, b x - 2 \, {\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )} + 12 \, a + e^{\left (4 \, b x + 4 \, a\right )} + 6 \, e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3,x, algorithm="giac")

[Out]

1/32*(12*b*x - 2*(3*e^(2*b*x + 2*a) + 1)*e^(-2*b*x - 2*a) + 12*a + e^(4*b*x + 4*a) + 6*e^(2*b*x + 2*a))/b

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maple [A]  time = 0.23, size = 49, normalized size = 0.86 \[ \frac {\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{4}+\left (\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )+\frac {3 b x}{8}+\frac {3 a}{8}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 53, normalized size = 0.93 \[ \frac {3 \, {\left (b x + a\right )}}{8 \, b} + \frac {e^{\left (4 \, b x + 4 \, a\right )}}{32 \, b} + \frac {3 \, e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b} - \frac {e^{\left (-2 \, b x - 2 \, a\right )}}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3,x, algorithm="maxima")

[Out]

3/8*(b*x + a)/b + 1/32*e^(4*b*x + 4*a)/b + 3/16*e^(2*b*x + 2*a)/b - 1/16*e^(-2*b*x - 2*a)/b

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mupad [B]  time = 0.26, size = 42, normalized size = 0.74 \[ \frac {3\,x}{8}+\frac {\frac {3\,{\mathrm {e}}^{2\,a+2\,b\,x}}{16}-\frac {{\mathrm {e}}^{-2\,a-2\,b\,x}}{16}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{32}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*exp(a + b*x),x)

[Out]

(3*x)/8 + ((3*exp(2*a + 2*b*x))/16 - exp(- 2*a - 2*b*x)/16 + exp(4*a + 4*b*x)/32)/b

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sympy [A]  time = 15.95, size = 207, normalized size = 3.63 \[ \begin {cases} \frac {3 x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8} - \frac {3 x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8} - \frac {3 x e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} + \frac {3 x e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8} - \frac {5 e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )}}{8 b} + \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{4 b} + \frac {e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b} - \frac {3 e^{a} e^{b x} \cosh ^{3}{\left (a + b x \right )}}{8 b} & \text {for}\: b \neq 0 \\x e^{a} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)**3,x)

[Out]

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

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