### 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]

-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

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Rubi [A]  time = 0.0793523, antiderivative size = 91, normalized size of antiderivative = 1., 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 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]]]

Rule 12

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

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 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]))

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*}

Mathematica [A]  time = 0.102667, 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]

-(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)

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Maple [A]  time = 0.01, size = 102, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{6}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{12}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{12}}+{\frac{\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{5}}{6}}-{\frac{\sinh \left ( bx+a \right ) }{6} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{4}}+{\frac{3\,\cosh \left ( bx+a \right ) }{8}} \right ) }-{\frac{bx}{16}}-{\frac{a}{16}} \right ) } \end{align*}

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)^2*sinh(b*x+a)^2-1/12*cosh(b*x+a)^2+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 = 1.03309, size = 104, normalized size = 1.14 \begin{align*} -\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} \end{align*}

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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Fricas [B]  time = 1.47983, size = 458, normalized size = 5.03 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.17899, size = 109, normalized size = 1.2 \begin{align*} -\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} \end{align*}

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