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

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

[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)

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Rubi [A]  time = 0.0630272, antiderivative size = 139, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.488643, size = 108, normalized size = 0.78 $\frac{e^{a+b x} \left (3 d \left (b^2-9 d^2\right ) \cosh (c+d x)+\left (3 d^3-3 b^2 d\right ) \cosh (3 (c+d x))+2 b \sinh (c+d x) \left (\left (b^2-d^2\right ) \cosh (2 (c+d x))-b^2+13 d^2\right )\right )}{4 \left (-10 b^2 d^2+b^4+9 d^4\right )}$

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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Maple [A]  time = 0.028, size = 166, normalized size = 1.2 \begin{align*} -{\frac{\sinh \left ( a-3\,c+ \left ( b-3\,d \right ) x \right ) }{8\,b-24\,d}}+{\frac{3\,\sinh \left ( a-c+ \left ( b-d \right ) x \right ) }{8\,b-8\,d}}-{\frac{3\,\sinh \left ( a+c+ \left ( b+d \right ) x \right ) }{8\,b+8\,d}}+{\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\,b-24\,d}}+{\frac{3\,\cosh \left ( a-c+ \left ( b-d \right ) x \right ) }{8\,b-8\,d}}-{\frac{3\,\cosh \left ( a+c+ \left ( b+d \right ) x \right ) }{8\,b+8\,d}}+{\frac{\cosh \left ( a+3\,c+ \left ( b+3\,d \right ) x \right ) }{8\,b+24\,d}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 1.79787, size = 752, normalized size = 5.41 \begin{align*} -\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 )}} \end{align*}

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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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)*sinh(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.13631, size = 113, normalized size = 0.81 \begin{align*} \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 )}} \end{align*}

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)