### 3.873 $$\int e^{2 (a+b x)} \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=40 $-\frac{e^{2 a+2 b x}}{4 b}+\frac{e^{4 a+4 b x}}{16 b}+\frac{x}{4}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.035378, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {2282, 12, 266, 43} $-\frac{e^{2 a+2 b x}}{4 b}+\frac{e^{4 a+4 b x}}{16 b}+\frac{x}{4}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0191898, size = 32, normalized size = 0.8 $\frac{-4 e^{2 (a+b x)}+e^{4 (a+b x)}+4 b x}{16 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 61, normalized size = 1.5 \begin{align*}{\frac{x}{4}}-{\frac{\sinh \left ( 2\,bx+2\,a \right ) }{4\,b}}+{\frac{\sinh \left ( 4\,bx+4\,a \right ) }{16\,b}}-{\frac{\cosh \left ( 2\,bx+2\,a \right ) }{4\,b}}+{\frac{\cosh \left ( 4\,bx+4\,a \right ) }{16\,b}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x-1/4*sinh(2*b*x+2*a)/b+1/16/b*sinh(4*b*x+4*a)-1/4*cosh(2*b*x+2*a)/b+1/16*cosh(4*b*x+4*a)/b

________________________________________________________________________________________

Maxima [A]  time = 1.25528, size = 50, normalized size = 1.25 \begin{align*} \frac{1}{4} \, x - \frac{{\left (4 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{16 \, b} + \frac{a}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 1.80609, size = 254, normalized size = 6.35 \begin{align*} \frac{{\left (4 \, b x + 1\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (4 \, b x - 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (4 \, b x + 1\right )} \sinh \left (b x + a\right )^{2} - 4}{16 \,{\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{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((4*b*x + 1)*cosh(b*x + a)^2 - 2*(4*b*x - 1)*cosh(b*x + a)*sinh(b*x + a) + (4*b*x + 1)*sinh(b*x + a)^2 -
4)/(b*cosh(b*x + a)^2 - 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

________________________________________________________________________________________

Sympy [A]  time = 7.28974, size = 139, normalized size = 3.48 \begin{align*} \begin{cases} \frac{x e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )}}{4} - \frac{x e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{2} + \frac{x e^{2 a} e^{2 b x} \cosh ^{2}{\left (a + b x \right )}}{4} + \frac{e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )}}{2 b} - \frac{e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.14819, size = 41, normalized size = 1.02 \begin{align*} \frac{4 \, b x + e^{\left (4 \, b x + 4 \, a\right )} - 4 \, e^{\left (2 \, b x + 2 \, a\right )}}{16 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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