3.878 \(\int e^{a+b x} \sinh ^3(c+d x) \, dx\)

Optimal. Leaf size=139 \[ \frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2}-\frac {3 d e^{a+b x} \sinh ^2(c+d x) \cosh (c+d x)}{b^2-9 d^2}+\frac {6 b d^2 e^{a+b x} \sinh (c+d x)}{b^4-10 b^2 d^2+9 d^4}-\frac {6 d^3 e^{a+b x} \cosh (c+d x)}{b^4-10 b^2 d^2+9 d^4} \]

[Out]

-6*d^3*exp(b*x+a)*cosh(d*x+c)/(b^4-10*b^2*d^2+9*d^4)+6*b*d^2*exp(b*x+a)*sinh(d*x+c)/(b^4-10*b^2*d^2+9*d^4)-3*d
*exp(b*x+a)*cosh(d*x+c)*sinh(d*x+c)^2/(b^2-9*d^2)+b*exp(b*x+a)*sinh(d*x+c)^3/(b^2-9*d^2)

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Rubi [A]  time = 0.06, antiderivative size = 139, 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, 5474} \[ \frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2}+\frac {6 b d^2 e^{a+b x} \sinh (c+d x)}{-10 b^2 d^2+b^4+9 d^4}-\frac {6 d^3 e^{a+b x} \cosh (c+d x)}{-10 b^2 d^2+b^4+9 d^4}-\frac {3 d e^{a+b x} \sinh ^2(c+d x) \cosh (c+d x)}{b^2-9 d^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sinh[c + d*x]^3,x]

[Out]

(-6*d^3*E^(a + b*x)*Cosh[c + d*x])/(b^4 - 10*b^2*d^2 + 9*d^4) + (6*b*d^2*E^(a + b*x)*Sinh[c + d*x])/(b^4 - 10*
b^2*d^2 + 9*d^4) - (3*d*E^(a + b*x)*Cosh[c + d*x]*Sinh[c + d*x]^2)/(b^2 - 9*d^2) + (b*E^(a + b*x)*Sinh[c + d*x
]^3)/(b^2 - 9*d^2)

Rule 5474

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

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 ^3(c+d x) \, dx &=-\frac {3 d e^{a+b x} \cosh (c+d x) \sinh ^2(c+d x)}{b^2-9 d^2}+\frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2}+\frac {\left (6 d^2\right ) \int e^{a+b x} \sinh (c+d x) \, dx}{b^2-9 d^2}\\ &=-\frac {6 d^3 e^{a+b x} \cosh (c+d x)}{b^4-10 b^2 d^2+9 d^4}+\frac {6 b d^2 e^{a+b x} \sinh (c+d x)}{b^4-10 b^2 d^2+9 d^4}-\frac {3 d e^{a+b x} \cosh (c+d x) \sinh ^2(c+d x)}{b^2-9 d^2}+\frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 108, normalized size = 0.78 \[ \frac {e^{a+b x} \left (\left (3 d^3-3 b^2 d\right ) \cosh (3 (c+d x))+3 d \left (b^2-9 d^2\right ) \cosh (c+d x)+2 b \sinh (c+d x) \left (\left (b^2-d^2\right ) \cosh (2 (c+d x))-b^2+13 d^2\right )\right )}{4 \left (b^4-10 b^2 d^2+9 d^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(a + b*x)*Sinh[c + d*x]^3,x]

[Out]

(E^(a + b*x)*(3*d*(b^2 - 9*d^2)*Cosh[c + d*x] + (-3*b^2*d + 3*d^3)*Cosh[3*(c + d*x)] + 2*b*(-b^2 + 13*d^2 + (b
^2 - d^2)*Cosh[2*(c + d*x)])*Sinh[c + d*x]))/(4*(b^4 - 10*b^2*d^2 + 9*d^4))

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fricas [B]  time = 0.45, size = 316, normalized size = 2.27 \[ -\frac {3 \, {\left (b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{3} - {\left ({\left (b^{3} - b d^{2}\right )} \cosh \left (b x + a\right ) + {\left (b^{3} - b d^{2}\right )} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{3} - 3 \, {\left (b^{2} d - 9 \, d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + 9 \, {\left ({\left (b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + {\left (b^{2} d - d^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{2} + 3 \, {\left ({\left (b^{2} d - d^{3}\right )} \cosh \left (d x + c\right )^{3} - {\left (b^{2} d - 9 \, d^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right ) - 3 \, {\left ({\left (b^{3} - b d^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} - {\left (b^{3} - 9 \, b d^{2}\right )} \cosh \left (b x + a\right ) - {\left (b^{3} - 9 \, b d^{2} - {\left (b^{3} - b d^{2}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{4 \, {\left (b^{4} - 10 \, b^{2} d^{2} + 9 \, d^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(3*(b^2*d - d^3)*cosh(b*x + a)*cosh(d*x + c)^3 - ((b^3 - b*d^2)*cosh(b*x + a) + (b^3 - b*d^2)*sinh(b*x +
a))*sinh(d*x + c)^3 - 3*(b^2*d - 9*d^3)*cosh(b*x + a)*cosh(d*x + c) + 9*((b^2*d - d^3)*cosh(b*x + a)*cosh(d*x
+ c) + (b^2*d - d^3)*cosh(d*x + c)*sinh(b*x + a))*sinh(d*x + c)^2 + 3*((b^2*d - d^3)*cosh(d*x + c)^3 - (b^2*d
- 9*d^3)*cosh(d*x + c))*sinh(b*x + a) - 3*((b^3 - b*d^2)*cosh(b*x + a)*cosh(d*x + c)^2 - (b^3 - 9*b*d^2)*cosh(
b*x + a) - (b^3 - 9*b*d^2 - (b^3 - b*d^2)*cosh(d*x + c)^2)*sinh(b*x + a))*sinh(d*x + c))/(b^4 - 10*b^2*d^2 + 9
*d^4)

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giac [A]  time = 0.12, size = 84, normalized size = 0.60 \[ \frac {e^{\left (b x + 3 \, d x + a + 3 \, c\right )}}{8 \, {\left (b + 3 \, d\right )}} - \frac {3 \, e^{\left (b x + d x + a + c\right )}}{8 \, {\left (b + d\right )}} + \frac {3 \, e^{\left (b x - d x + a - c\right )}}{8 \, {\left (b - d\right )}} - \frac {e^{\left (b x - 3 \, d x + a - 3 \, c\right )}}{8 \, {\left (b - 3 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*e^(b*x + 3*d*x + a + 3*c)/(b + 3*d) - 3/8*e^(b*x + d*x + a + c)/(b + d) + 3/8*e^(b*x - d*x + a - c)/(b - d
) - 1/8*e^(b*x - 3*d*x + a - 3*c)/(b - 3*d)

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maple [A]  time = 0.33, size = 166, normalized size = 1.19 \[ -\frac {\sinh \left (a -3 c +\left (b -3 d \right ) x \right )}{8 \left (b -3 d \right )}+\frac {3 \sinh \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \sinh \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}+\frac {\sinh \left (a +3 c +\left (b +3 d \right ) x \right )}{8 b +24 d}-\frac {\cosh \left (a -3 c +\left (b -3 d \right ) x \right )}{8 \left (b -3 d \right )}+\frac {3 \cosh \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \cosh \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}+\frac {\cosh \left (a +3 c +\left (b +3 d \right ) x \right )}{8 b +24 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sinh(d*x+c)^3,x)

[Out]

-1/8*sinh(a-3*c+(b-3*d)*x)/(b-3*d)+3/8*sinh(a-c+(b-d)*x)/(b-d)-3/8*sinh(a+c+(b+d)*x)/(b+d)+1/8*sinh(a+3*c+(b+3
*d)*x)/(b+3*d)-1/8*cosh(a-3*c+(b-3*d)*x)/(b-3*d)+3/8*cosh(a-c+(b-d)*x)/(b-d)-3/8*cosh(a+c+(b+d)*x)/(b+d)+1/8*c
osh(a+3*c+(b+3*d)*x)/(b+3*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)^3,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(-(3*d)/b>0)', see `assume?` fo
r more details)Is -(3*d)/b equal to -1?

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mupad [B]  time = 2.32, size = 127, normalized size = 0.91 \[ -\frac {{\mathrm {e}}^{a+b\,x}\,\left (-b^3\,{\mathrm {sinh}\left (c+d\,x\right )}^3+3\,b^2\,d\,\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {sinh}\left (c+d\,x\right )}^2-6\,b\,d^2\,{\mathrm {cosh}\left (c+d\,x\right )}^2\,\mathrm {sinh}\left (c+d\,x\right )+7\,b\,d^2\,{\mathrm {sinh}\left (c+d\,x\right )}^3+6\,d^3\,{\mathrm {cosh}\left (c+d\,x\right )}^3-9\,d^3\,\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {sinh}\left (c+d\,x\right )}^2\right )}{b^4-10\,b^2\,d^2+9\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)*sinh(c + d*x)^3,x)

[Out]

-(exp(a + b*x)*(6*d^3*cosh(c + d*x)^3 - b^3*sinh(c + d*x)^3 - 9*d^3*cosh(c + d*x)*sinh(c + d*x)^2 + 7*b*d^2*si
nh(c + d*x)^3 - 6*b*d^2*cosh(c + d*x)^2*sinh(c + d*x) + 3*b^2*d*cosh(c + d*x)*sinh(c + d*x)^2))/(b^4 + 9*d^4 -
10*b^2*d^2)

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sympy [A]  time = 44.61, size = 976, normalized size = 7.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sinh(d*x+c)**3,x)

[Out]

Piecewise((x*exp(a)*sinh(c)**3, Eq(b, 0) & Eq(d, 0)), (x*exp(a)*exp(-3*d*x)*sinh(c + d*x)**3/8 + 3*x*exp(a)*ex
p(-3*d*x)*sinh(c + d*x)**2*cosh(c + d*x)/8 + 3*x*exp(a)*exp(-3*d*x)*sinh(c + d*x)*cosh(c + d*x)**2/8 + x*exp(a
)*exp(-3*d*x)*cosh(c + d*x)**3/8 - 3*exp(a)*exp(-3*d*x)*sinh(c + d*x)**3/(8*d) - exp(a)*exp(-3*d*x)*sinh(c + d
*x)**2*cosh(c + d*x)/(4*d) + exp(a)*exp(-3*d*x)*cosh(c + d*x)**3/(24*d), Eq(b, -3*d)), (3*x*exp(a)*exp(-d*x)*s
inh(c + d*x)**3/8 + 3*x*exp(a)*exp(-d*x)*sinh(c + d*x)**2*cosh(c + d*x)/8 - 3*x*exp(a)*exp(-d*x)*sinh(c + d*x)
*cosh(c + d*x)**2/8 - 3*x*exp(a)*exp(-d*x)*cosh(c + d*x)**3/8 + exp(a)*exp(-d*x)*sinh(c + d*x)**3/(8*d) + 3*ex
p(a)*exp(-d*x)*sinh(c + d*x)**2*cosh(c + d*x)/(4*d) - 3*exp(a)*exp(-d*x)*cosh(c + d*x)**3/(8*d), Eq(b, -d)), (
3*x*exp(a)*exp(d*x)*sinh(c + d*x)**3/8 - 3*x*exp(a)*exp(d*x)*sinh(c + d*x)**2*cosh(c + d*x)/8 - 3*x*exp(a)*exp
(d*x)*sinh(c + d*x)*cosh(c + d*x)**2/8 + 3*x*exp(a)*exp(d*x)*cosh(c + d*x)**3/8 - exp(a)*exp(d*x)*sinh(c + d*x
)**3/(8*d) + 3*exp(a)*exp(d*x)*sinh(c + d*x)**2*cosh(c + d*x)/(4*d) - 3*exp(a)*exp(d*x)*cosh(c + d*x)**3/(8*d)
, Eq(b, d)), (x*exp(a)*exp(3*d*x)*sinh(c + d*x)**3/8 - 3*x*exp(a)*exp(3*d*x)*sinh(c + d*x)**2*cosh(c + d*x)/8
+ 3*x*exp(a)*exp(3*d*x)*sinh(c + d*x)*cosh(c + d*x)**2/8 - x*exp(a)*exp(3*d*x)*cosh(c + d*x)**3/8 + 3*exp(a)*e
xp(3*d*x)*sinh(c + d*x)**3/(8*d) - exp(a)*exp(3*d*x)*sinh(c + d*x)**2*cosh(c + d*x)/(4*d) + exp(a)*exp(3*d*x)*
cosh(c + d*x)**3/(24*d), Eq(b, 3*d)), (b**3*exp(a)*exp(b*x)*sinh(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4) -
3*b**2*d*exp(a)*exp(b*x)*sinh(c + d*x)**2*cosh(c + d*x)/(b**4 - 10*b**2*d**2 + 9*d**4) - 7*b*d**2*exp(a)*exp(b
*x)*sinh(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4) + 6*b*d**2*exp(a)*exp(b*x)*sinh(c + d*x)*cosh(c + d*x)**2/
(b**4 - 10*b**2*d**2 + 9*d**4) + 9*d**3*exp(a)*exp(b*x)*sinh(c + d*x)**2*cosh(c + d*x)/(b**4 - 10*b**2*d**2 +
9*d**4) - 6*d**3*exp(a)*exp(b*x)*cosh(c + d*x)**3/(b**4 - 10*b**2*d**2 + 9*d**4), True))

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