3.879 \(\int e^{a+b x} \sinh ^2(c+d x) \, dx\)

Optimal. Leaf size=88 \[ \frac {b e^{a+b x} \sinh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \sinh (c+d x) \cosh (c+d x)}{b^2-4 d^2}+\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )} \]

[Out]

2*d^2*exp(b*x+a)/b/(b^2-4*d^2)-2*d*exp(b*x+a)*cosh(d*x+c)*sinh(d*x+c)/(b^2-4*d^2)+b*exp(b*x+a)*sinh(d*x+c)^2/(
b^2-4*d^2)

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Rubi [A]  time = 0.04, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5476, 2194} \[ \frac {b e^{a+b x} \sinh ^2(c+d x)}{b^2-4 d^2}-\frac {2 d e^{a+b x} \sinh (c+d x) \cosh (c+d x)}{b^2-4 d^2}+\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sinh[c + d*x]^2,x]

[Out]

(2*d^2*E^(a + b*x))/(b*(b^2 - 4*d^2)) - (2*d*E^(a + b*x)*Cosh[c + d*x]*Sinh[c + d*x])/(b^2 - 4*d^2) + (b*E^(a
+ b*x)*Sinh[c + d*x]^2)/(b^2 - 4*d^2)

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 5476

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a +
b*x))*Sinh[d + e*x]^n)/(e^2*n^2 - b^2*c^2*Log[F]^2), x] + (-Dist[(n*(n - 1)*e^2)/(e^2*n^2 - b^2*c^2*Log[F]^2),
Int[F^(c*(a + b*x))*Sinh[d + e*x]^(n - 2), x], x] + Simp[(e*n*F^(c*(a + b*x))*Cosh[d + e*x]*Sinh[d + e*x]^(n
- 1))/(e^2*n^2 - b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 - b^2*c^2*Log[F]^2, 0]
&& GtQ[n, 1]

Rubi steps

\begin {align*} \int e^{a+b x} \sinh ^2(c+d x) \, dx &=-\frac {2 d e^{a+b x} \cosh (c+d x) \sinh (c+d x)}{b^2-4 d^2}+\frac {b e^{a+b x} \sinh ^2(c+d x)}{b^2-4 d^2}+\frac {\left (2 d^2\right ) \int e^{a+b x} \, dx}{b^2-4 d^2}\\ &=\frac {2 d^2 e^{a+b x}}{b \left (b^2-4 d^2\right )}-\frac {2 d e^{a+b x} \cosh (c+d x) \sinh (c+d x)}{b^2-4 d^2}+\frac {b e^{a+b x} \sinh ^2(c+d x)}{b^2-4 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 58, normalized size = 0.66 \[ \frac {e^{a+b x} \left (b^2 \cosh (2 (c+d x))-b^2-2 b d \sinh (2 (c+d x))+4 d^2\right )}{2 \left (b^3-4 b d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a + b*x)*(-b^2 + 4*d^2 + b^2*Cosh[2*(c + d*x)] - 2*b*d*Sinh[2*(c + d*x)]))/(2*(b^3 - 4*b*d^2))

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fricas [A]  time = 0.45, size = 149, normalized size = 1.69 \[ \frac {b^{2} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} + {\left (b^{2} \cosh \left (b x + a\right ) + b^{2} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{2} - {\left (b^{2} - 4 \, d^{2}\right )} \cosh \left (b x + a\right ) + {\left (b^{2} \cosh \left (d x + c\right )^{2} - b^{2} + 4 \, d^{2}\right )} \sinh \left (b x + a\right ) - 4 \, {\left (b d \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + b d \cosh \left (d x + c\right ) \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left (b^{3} - 4 \, b d^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 56, normalized size = 0.64 \[ \frac {e^{\left (b x + 2 \, d x + a + 2 \, c\right )}}{4 \, {\left (b + 2 \, d\right )}} + \frac {e^{\left (b x - 2 \, d x + a - 2 \, c\right )}}{4 \, {\left (b - 2 \, d\right )}} - \frac {e^{\left (b x + a\right )}}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.23, size = 112, normalized size = 1.27 \[ -\frac {\sinh \left (b x +a \right )}{2 b}+\frac {\sinh \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\sinh \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d}-\frac {\cosh \left (b x +a \right )}{2 b}+\frac {\cosh \left (a -2 c +\left (b -2 d \right ) x \right )}{4 b -8 d}+\frac {\cosh \left (a +2 c +\left (b +2 d \right ) x \right )}{4 b +8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sinh(b*x+a)/b+1/4*sinh(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*sinh(a+2*c+(b+2*d)*x)/(b+2*d)-1/2*cosh(b*x+a)/b+1/4*c
osh(a-2*c+(b-2*d)*x)/(b-2*d)+1/4*cosh(a+2*c+(b+2*d)*x)/(b+2*d)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-(2*d)/b>0)', see `assume?` fo
r more details)Is -(2*d)/b equal to -1?

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mupad [B]  time = 0.24, size = 105, normalized size = 1.19 \[ -\frac {2\,d^2\,{\mathrm {e}}^{a+b\,x}-b^2\,\left (\frac {{\mathrm {e}}^{a+b\,x}}{2}-{\mathrm {e}}^{a+b\,x}\,\left (\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{4}+\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{4}\right )\right )+b\,d\,{\mathrm {e}}^{a+b\,x}\,\left (\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{2}-\frac {{\mathrm {e}}^{2\,c+2\,d\,x}}{2}\right )}{4\,b\,d^2-b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)*sinh(c + d*x)^2,x)

[Out]

-(2*d^2*exp(a + b*x) - b^2*(exp(a + b*x)/2 - exp(a + b*x)*(exp(- 2*c - 2*d*x)/4 + exp(2*c + 2*d*x)/4)) + b*d*e
xp(a + b*x)*(exp(- 2*c - 2*d*x)/2 - exp(2*c + 2*d*x)/2))/(4*b*d^2 - b^3)

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sympy [A]  time = 9.12, size = 428, normalized size = 4.86 \[ \begin {cases} x e^{a} \sinh ^{2}{\relax (c )} & \text {for}\: b = 0 \wedge d = 0 \\\left (\frac {x \sinh ^{2}{\left (c + d x \right )}}{2} - \frac {x \cosh ^{2}{\left (c + d x \right )}}{2} + \frac {\sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2 d}\right ) e^{a} & \text {for}\: b = 0 \\\frac {x e^{a} e^{- 2 d x} \sinh ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{- 2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{- 2 d x} \cosh ^{2}{\left (c + d x \right )}}{4} - \frac {e^{a} e^{- 2 d x} \sinh ^{2}{\left (c + d x \right )}}{2 d} - \frac {e^{a} e^{- 2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} & \text {for}\: b = - 2 d \\\frac {x e^{a} e^{2 d x} \sinh ^{2}{\left (c + d x \right )}}{4} - \frac {x e^{a} e^{2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{2} + \frac {x e^{a} e^{2 d x} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {e^{a} e^{2 d x} \sinh ^{2}{\left (c + d x \right )}}{2 d} - \frac {e^{a} e^{2 d x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{4 d} & \text {for}\: b = 2 d \\\frac {b^{2} e^{a} e^{b x} \sinh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 b d e^{a} e^{b x} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{b^{3} - 4 b d^{2}} - \frac {2 d^{2} e^{a} e^{b x} \sinh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} + \frac {2 d^{2} e^{a} e^{b x} \cosh ^{2}{\left (c + d x \right )}}{b^{3} - 4 b d^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(a)*sinh(c)**2, Eq(b, 0) & Eq(d, 0)), ((x*sinh(c + d*x)**2/2 - x*cosh(c + d*x)**2/2 + sinh(c +
d*x)*cosh(c + d*x)/(2*d))*exp(a), Eq(b, 0)), (x*exp(a)*exp(-2*d*x)*sinh(c + d*x)**2/4 + x*exp(a)*exp(-2*d*x)*
sinh(c + d*x)*cosh(c + d*x)/2 + x*exp(a)*exp(-2*d*x)*cosh(c + d*x)**2/4 - exp(a)*exp(-2*d*x)*sinh(c + d*x)**2/
(2*d) - exp(a)*exp(-2*d*x)*sinh(c + d*x)*cosh(c + d*x)/(4*d), Eq(b, -2*d)), (x*exp(a)*exp(2*d*x)*sinh(c + d*x)
**2/4 - x*exp(a)*exp(2*d*x)*sinh(c + d*x)*cosh(c + d*x)/2 + x*exp(a)*exp(2*d*x)*cosh(c + d*x)**2/4 + exp(a)*ex
p(2*d*x)*sinh(c + d*x)**2/(2*d) - exp(a)*exp(2*d*x)*sinh(c + d*x)*cosh(c + d*x)/(4*d), Eq(b, 2*d)), (b**2*exp(
a)*exp(b*x)*sinh(c + d*x)**2/(b**3 - 4*b*d**2) - 2*b*d*exp(a)*exp(b*x)*sinh(c + d*x)*cosh(c + d*x)/(b**3 - 4*b
*d**2) - 2*d**2*exp(a)*exp(b*x)*sinh(c + d*x)**2/(b**3 - 4*b*d**2) + 2*d**2*exp(a)*exp(b*x)*cosh(c + d*x)**2/(
b**3 - 4*b*d**2), True))

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