3.914 \(\int e^{a+b x} \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=91 \[ -\frac {e^{-4 a-4 b x}}{128 b}-\frac {e^{-2 a-2 b x}}{64 b}-\frac {e^{2 a+2 b x}}{32 b}+\frac {e^{4 a+4 b x}}{128 b}+\frac {e^{6 a+6 b x}}{192 b}-\frac {x}{16} \]

[Out]

-1/128*exp(-4*b*x-4*a)/b-1/64*exp(-2*b*x-2*a)/b-1/32*exp(2*b*x+2*a)/b+1/128*exp(4*b*x+4*a)/b+1/192*exp(6*b*x+6
*a)/b-1/16*x

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Rubi [A]  time = 0.08, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2282, 12, 446, 88} \[ -\frac {e^{-4 a-4 b x}}{128 b}-\frac {e^{-2 a-2 b x}}{64 b}-\frac {e^{2 a+2 b x}}{32 b}+\frac {e^{4 a+4 b x}}{128 b}+\frac {e^{6 a+6 b x}}{192 b}-\frac {x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(-4*a - 4*b*x)/(128*b) - E^(-2*a - 2*b*x)/(64*b) - E^(2*a + 2*b*x)/(32*b) + E^(4*a + 4*b*x)/(128*b) + E^(6*
a + 6*b*x)/(192*b) - x/16

Rule 12

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && 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) \sinh ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2 \left (1+x^2\right )^3}{32 x^5} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2 \left (1+x^2\right )^3}{x^5} \, dx,x,e^{a+b x}\right )}{32 b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(1-x)^2 (1+x)^3}{x^3} \, dx,x,e^{2 a+2 b x}\right )}{64 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2+\frac {1}{x^3}+\frac {1}{x^2}-\frac {2}{x}+x+x^2\right ) \, dx,x,e^{2 a+2 b x}\right )}{64 b}\\ &=-\frac {e^{-4 a-4 b x}}{128 b}-\frac {e^{-2 a-2 b x}}{64 b}-\frac {e^{2 a+2 b x}}{32 b}+\frac {e^{4 a+4 b x}}{128 b}+\frac {e^{6 a+6 b x}}{192 b}-\frac {x}{16}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 67, normalized size = 0.74 \[ -\frac {3 e^{-4 (a+b x)}+6 e^{-2 (a+b x)}+12 e^{2 (a+b x)}-3 e^{4 (a+b x)}-2 e^{6 (a+b x)}+24 b x}{384 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/384*(3/E^(4*(a + b*x)) + 6/E^(2*(a + b*x)) + 12*E^(2*(a + b*x)) - 3*E^(4*(a + b*x)) - 2*E^(6*(a + b*x)) + 2
4*b*x)/b

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fricas [B]  time = 0.47, size = 165, normalized size = 1.81 \[ -\frac {\cosh \left (b x + a\right )^{5} + 5 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 5 \, \sinh \left (b x + a\right )^{5} - {\left (50 \, \cosh \left (b x + a\right )^{2} + 9\right )} \sinh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right )^{3} + {\left (10 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} + 12 \, {\left (2 \, b x + 1\right )} \cosh \left (b x + a\right ) - {\left (25 \, \cosh \left (b x + a\right )^{4} + 24 \, b x + 27 \, \cosh \left (b x + a\right )^{2} - 12\right )} \sinh \left (b x + a\right )}{384 \, {\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(cosh(b*x + a)^5 + 5*cosh(b*x + a)*sinh(b*x + a)^4 - 5*sinh(b*x + a)^5 - (50*cosh(b*x + a)^2 + 9)*sinh(
b*x + a)^3 + 3*cosh(b*x + a)^3 + (10*cosh(b*x + a)^3 + 9*cosh(b*x + a))*sinh(b*x + a)^2 + 12*(2*b*x + 1)*cosh(
b*x + a) - (25*cosh(b*x + a)^4 + 24*b*x + 27*cosh(b*x + a)^2 - 12)*sinh(b*x + a))/(b*cosh(b*x + a) - b*sinh(b*
x + a))

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giac [A]  time = 0.14, size = 81, normalized size = 0.89 \[ -\frac {24 \, b x - 3 \, {\left (6 \, e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-4 \, b x - 4 \, a\right )} + 24 \, a - 2 \, e^{\left (6 \, b x + 6 \, a\right )} - 3 \, e^{\left (4 \, b x + 4 \, a\right )} + 12 \, e^{\left (2 \, b x + 2 \, a\right )}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384*(24*b*x - 3*(6*e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) - 1)*e^(-4*b*x - 4*a) + 24*a - 2*e^(6*b*x + 6*a) - 3
*e^(4*b*x + 4*a) + 12*e^(2*b*x + 2*a))/b

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maple [A]  time = 0.38, size = 84, normalized size = 0.92 \[ \frac {\frac {\left (\cosh ^{4}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )}{6}-\frac {\left (\cosh ^{4}\left (b x +a \right )\right )}{12}+\frac {\sinh \left (b x +a \right ) \left (\cosh ^{5}\left (b x +a \right )\right )}{6}-\frac {\left (\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{4}+\frac {3 \cosh \left (b x +a \right )}{8}\right ) \sinh \left (b x +a \right )}{6}-\frac {b x}{16}-\frac {a}{16}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(1/6*cosh(b*x+a)^4*sinh(b*x+a)^2-1/12*cosh(b*x+a)^4+1/6*sinh(b*x+a)*cosh(b*x+a)^5-1/6*(1/4*cosh(b*x+a)^3+3
/8*cosh(b*x+a))*sinh(b*x+a)-1/16*b*x-1/16*a)

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maxima [A]  time = 0.33, size = 77, normalized size = 0.85 \[ -\frac {{\left (2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{128 \, b} - \frac {b x + a}{16 \, b} + \frac {2 \, e^{\left (6 \, b x + 6 \, a\right )} + 3 \, e^{\left (4 \, b x + 4 \, a\right )} - 12 \, e^{\left (2 \, b x + 2 \, a\right )}}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/128*(2*e^(2*b*x + 2*a) + 1)*e^(-4*b*x - 4*a)/b - 1/16*(b*x + a)/b + 1/384*(2*e^(6*b*x + 6*a) + 3*e^(4*b*x +
4*a) - 12*e^(2*b*x + 2*a))/b

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mupad [B]  time = 0.60, size = 65, normalized size = 0.71 \[ -\frac {6\,{\mathrm {e}}^{-2\,a-2\,b\,x}+12\,{\mathrm {e}}^{2\,a+2\,b\,x}+3\,{\mathrm {e}}^{-4\,a-4\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{6\,a+6\,b\,x}+24\,b\,x}{384\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*exp(- 2*a - 2*b*x) + 12*exp(2*a + 2*b*x) + 3*exp(- 4*a - 4*b*x) - 3*exp(4*a + 4*b*x) - 2*exp(6*a + 6*b*x)
+ 24*b*x)/(384*b)

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sympy [A]  time = 170.93, size = 325, normalized size = 3.57 \[ \begin {cases} \frac {x e^{a} e^{b x} \sinh ^{5}{\left (a + b x \right )}}{16} - \frac {x e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{16} - \frac {x e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{8} + \frac {x e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8} + \frac {x e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{16} - \frac {x e^{a} e^{b x} \cosh ^{5}{\left (a + b x \right )}}{16} - \frac {13 e^{a} e^{b x} \sinh ^{5}{\left (a + b x \right )}}{96 b} + \frac {7 e^{a} e^{b x} \sinh ^{4}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{96 b} + \frac {e^{a} e^{b x} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} - \frac {e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{6 b} + \frac {e^{a} e^{b x} \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{96 b} + \frac {5 e^{a} e^{b x} \cosh ^{5}{\left (a + b x \right )}}{96 b} & \text {for}\: b \neq 0 \\x e^{a} \sinh ^{2}{\relax (a )} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(a)*exp(b*x)*sinh(a + b*x)**5/16 - x*exp(a)*exp(b*x)*sinh(a + b*x)**4*cosh(a + b*x)/16 - x*exp
(a)*exp(b*x)*sinh(a + b*x)**3*cosh(a + b*x)**2/8 + x*exp(a)*exp(b*x)*sinh(a + b*x)**2*cosh(a + b*x)**3/8 + x*e
xp(a)*exp(b*x)*sinh(a + b*x)*cosh(a + b*x)**4/16 - x*exp(a)*exp(b*x)*cosh(a + b*x)**5/16 - 13*exp(a)*exp(b*x)*
sinh(a + b*x)**5/(96*b) + 7*exp(a)*exp(b*x)*sinh(a + b*x)**4*cosh(a + b*x)/(96*b) + exp(a)*exp(b*x)*sinh(a + b
*x)**3*cosh(a + b*x)**2/(3*b) - exp(a)*exp(b*x)*sinh(a + b*x)**2*cosh(a + b*x)**3/(6*b) + exp(a)*exp(b*x)*sinh
(a + b*x)*cosh(a + b*x)**4/(96*b) + 5*exp(a)*exp(b*x)*cosh(a + b*x)**5/(96*b), Ne(b, 0)), (x*exp(a)*sinh(a)**2
*cosh(a)**3, True))

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