3.10.31 \(\int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx\) [931]

Optimal. Leaf size=57 \[ \frac {e^{-4 a-4 b x}}{256 b}-\frac {3 e^{4 a+4 b x}}{256 b}+\frac {e^{8 a+8 b x}}{512 b}+\frac {3 x}{64} \]

[Out]

1/256*exp(-4*b*x-4*a)/b-3/256*exp(4*b*x+4*a)/b+1/512*exp(8*b*x+8*a)/b+3/64*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 = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2320, 12, 272, 45} \begin {gather*} \frac {e^{-4 a-4 b x}}{256 b}-\frac {3 e^{4 a+4 b x}}{256 b}+\frac {e^{8 a+8 b x}}{512 b}+\frac {3 x}{64} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

E^(-4*a - 4*b*x)/(256*b) - (3*E^(4*a + 4*b*x))/(256*b) + E^(8*a + 8*b*x)/(512*b) + (3*x)/64

Rule 12

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

Rule 45

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 272

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 2320

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^{2 (a+b x)} \cosh ^3(a+b x) \sinh ^3(a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^4\right )^3}{64 x^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\left (-1+x^4\right )^3}{x^5} \, dx,x,e^{a+b x}\right )}{64 b}\\ &=\frac {\text {Subst}\left (\int \frac {(-1+x)^3}{x^2} \, dx,x,e^{4 a+4 b x}\right )}{256 b}\\ &=\frac {\text {Subst}\left (\int \left (-3-\frac {1}{x^2}+\frac {3}{x}+x\right ) \, dx,x,e^{4 a+4 b x}\right )}{256 b}\\ &=\frac {e^{-4 a-4 b x}}{256 b}-\frac {3 e^{4 a+4 b x}}{256 b}+\frac {e^{8 a+8 b x}}{512 b}+\frac {3 x}{64}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 45, normalized size = 0.79 \begin {gather*} \frac {e^{-4 (a+b x)}-3 e^{4 (a+b x)}+\frac {1}{2} e^{8 (a+b x)}+12 b x}{256 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^3,x]

[Out]

(E^(-4*(a + b*x)) - 3*E^(4*(a + b*x)) + E^(8*(a + b*x))/2 + 12*b*x)/(256*b)

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Maple [A]
time = 1.10, size = 61, normalized size = 1.07

method	result	size
risch	\(\frac {{\mathrm e}^{-4 b x -4 a}}{256 b}-\frac {3 \,{\mathrm e}^{4 b x +4 a}}{256 b}+\frac {{\mathrm e}^{8 b x +8 a}}{512 b}+\frac {3 x}{64}\)	\(47\)
default	\(\frac {3 x}{64}-\frac {\sinh \left (4 b x +4 a \right )}{64 b}+\frac {\sinh \left (8 b x +8 a \right )}{512 b}-\frac {\cosh \left (4 b x +4 a \right )}{128 b}+\frac {\cosh \left (8 b x +8 a \right )}{512 b}\)	\(61\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

3/64*x-1/64/b*sinh(4*b*x+4*a)+1/512/b*sinh(8*b*x+8*a)-1/128*cosh(4*b*x+4*a)/b+1/512*cosh(8*b*x+8*a)/b

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Maxima [A]
time = 0.28, size = 52, normalized size = 0.91 \begin {gather*} -\frac {{\left (6 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (8 \, b x + 8 \, a\right )}}{512 \, b} + \frac {3 \, {\left (b x + a\right )}}{64 \, b} + \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/512*(6*e^(-4*b*x - 4*a) - 1)*e^(8*b*x + 8*a)/b + 3/64*(b*x + a)/b + 1/256*e^(-4*b*x - 4*a)/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (46) = 92\).
time = 0.35, size = 186, normalized size = 3.26 \begin {gather*} \frac {3 \, \cosh \left (b x + a\right )^{6} - 20 \, \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 45 \, \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} - 6 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \, \sinh \left (b x + a\right )^{6} + 6 \, {\left (4 \, b x - 1\right )} \cosh \left (b x + a\right )^{2} + 3 \, {\left (15 \, \cosh \left (b x + a\right )^{4} + 8 \, b x - 2\right )} \sinh \left (b x + a\right )^{2} - 6 \, {\left (\cosh \left (b x + a\right )^{5} + 2 \, {\left (4 \, b x + 1\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{512 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/512*(3*cosh(b*x + a)^6 - 20*cosh(b*x + a)^3*sinh(b*x + a)^3 + 45*cosh(b*x + a)^2*sinh(b*x + a)^4 - 6*cosh(b*
x + a)*sinh(b*x + a)^5 + 3*sinh(b*x + a)^6 + 6*(4*b*x - 1)*cosh(b*x + a)^2 + 3*(15*cosh(b*x + a)^4 + 8*b*x - 2
)*sinh(b*x + a)^2 - 6*(cosh(b*x + a)^5 + 2*(4*b*x + 1)*cosh(b*x + a))*sinh(b*x + a))/(b*cosh(b*x + a)^2 - 2*b*
cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (48) = 96\).
time = 17.51, size = 382, normalized size = 6.70 \begin {gather*} \begin {cases} \frac {3 x e^{2 a} e^{2 b x} \sinh ^{6}{\left (a + b x \right )}}{64} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{5}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{32} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{4}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{64} + \frac {3 x e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{16} - \frac {3 x e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{64} - \frac {3 x e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{32} + \frac {3 x e^{2 a} e^{2 b x} \cosh ^{6}{\left (a + b x \right )}}{64} + \frac {3 e^{2 a} e^{2 b x} \sinh ^{6}{\left (a + b x \right )}}{32 b} - \frac {15 e^{2 a} e^{2 b x} \sinh ^{5}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{64 b} + \frac {13 e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{32 b} - \frac {15 e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh ^{5}{\left (a + b x \right )}}{64 b} + \frac {3 e^{2 a} e^{2 b x} \cosh ^{6}{\left (a + b x \right )}}{32 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{3}{\left (a \right )} \cosh ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)**3*sinh(b*x+a)**3,x)

[Out]

Piecewise((3*x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**6/64 - 3*x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**5*cosh(a + b*x
)/32 - 3*x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**4*cosh(a + b*x)**2/64 + 3*x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3
*cosh(a + b*x)**3/16 - 3*x*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**2*cosh(a + b*x)**4/64 - 3*x*exp(2*a)*exp(2*b*x)*
sinh(a + b*x)*cosh(a + b*x)**5/32 + 3*x*exp(2*a)*exp(2*b*x)*cosh(a + b*x)**6/64 + 3*exp(2*a)*exp(2*b*x)*sinh(a
+ b*x)**6/(32*b) - 15*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**5*cosh(a + b*x)/(64*b) + 13*exp(2*a)*exp(2*b*x)*sinh
(a + b*x)**3*cosh(a + b*x)**3/(32*b) - 15*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**5/(64*b) + 3*exp(2*
a)*exp(2*b*x)*cosh(a + b*x)**6/(32*b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**3*cosh(a)**3, True))

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Giac [A]
time = 0.43, size = 46, normalized size = 0.81 \begin {gather*} \frac {3}{64} \, x + \frac {e^{\left (8 \, b x + 8 \, a\right )}}{512 \, b} - \frac {3 \, e^{\left (4 \, b x + 4 \, a\right )}}{256 \, b} + \frac {e^{\left (-4 \, b x - 4 \, a\right )}}{256 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

3/64*x + 1/512*e^(8*b*x + 8*a)/b - 3/256*e^(4*b*x + 4*a)/b + 1/256*e^(-4*b*x - 4*a)/b

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Mupad [B]
time = 1.93, size = 46, normalized size = 0.81 \begin {gather*} \frac {3\,x}{64}+\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}}{256\,b}-\frac {3\,{\mathrm {e}}^{4\,a+4\,b\,x}}{256\,b}+\frac {{\mathrm {e}}^{8\,a+8\,b\,x}}{512\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*exp(2*a + 2*b*x)*sinh(a + b*x)^3,x)

[Out]

(3*x)/64 + exp(- 4*a - 4*b*x)/(256*b) - (3*exp(4*a + 4*b*x))/(256*b) + exp(8*a + 8*b*x)/(512*b)

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