\(\int e^{a+b x} \sinh ^3(c+d x) \, dx\) [878]

Optimal result

Integrand size = 16, antiderivative size = 139 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=-\frac {6 d^3 e^{a+b x} \cosh (c+d x)}{b^4-10 b^2 d^2+9 d^4}+\frac {6 b d^2 e^{a+b x} \sinh (c+d x)}{b^4-10 b^2 d^2+9 d^4}-\frac {3 d e^{a+b x} \cosh (c+d x) \sinh ^2(c+d x)}{b^2-9 d^2}+\frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2} \]

[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)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5584, 5582} \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=\frac {b e^{a+b x} \sinh ^3(c+d x)}{b^2-9 d^2}-\frac {3 d e^{a+b x} \sinh ^2(c+d x) \cosh (c+d x)}{b^2-9 d^2}+\frac {6 b d^2 e^{a+b x} \sinh (c+d x)}{b^4-10 b^2 d^2+9 d^4}-\frac {6 d^3 e^{a+b x} \cosh (c+d x)}{b^4-10 b^2 d^2+9 d^4} \]

[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 5582

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 5584

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*} \text {integral}& = -\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] (verified)

Time = 0.38 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=\frac {e^{a+b x} \left (3 d \left (b^2-9 d^2\right ) \cosh (c+d x)+\left (-3 b^2 d+3 d^3\right ) \cosh (3 (c+d x))+2 b \left (-b^2+13 d^2+\left (b^2-d^2\right ) \cosh (2 (c+d x))\right ) \sinh (c+d x)\right )}{4 \left (b^4-10 b^2 d^2+9 d^4\right )} \]

[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))

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.61

method	result	size
risch	\(\frac {{\mathrm e}^{b x +3 d x +a +3 c}}{8 b +24 d}-\frac {3 \,{\mathrm e}^{b x +d x +a +c}}{8 \left (b +d \right )}+\frac {3 \,{\mathrm e}^{b x -d x +a -c}}{8 \left (b -d \right )}-\frac {{\mathrm e}^{b x -3 d x +a -3 c}}{8 \left (b -3 d \right )}\)	\(85\)
parallelrisch	\(\frac {{\mathrm e}^{b x +a} \left (\left (-3 b^{2} d +3 d^{3}\right ) \cosh \left (3 d x +3 c \right )+\left (b^{3}-b \,d^{2}\right ) \sinh \left (3 d x +3 c \right )-3 \left (b -3 d \right ) \left (b +3 d \right ) \left (b \sinh \left (d x +c \right )-d \cosh \left (d x +c \right )\right )\right )}{4 b^{4}-40 b^{2} d^{2}+36 d^{4}}\)	\(102\)
default	\(-\frac {\sinh \left (a -3 c +\left (b -3 d \right ) x \right )}{8 \left (b -3 d \right )}+\frac {3 \sinh \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \sinh \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}+\frac {\sinh \left (a +3 c +\left (b +3 d \right ) x \right )}{8 b +24 d}-\frac {\cosh \left (a -3 c +\left (b -3 d \right ) x \right )}{8 \left (b -3 d \right )}+\frac {3 \cosh \left (a -c +\left (b -d \right ) x \right )}{8 \left (b -d \right )}-\frac {3 \cosh \left (a +c +\left (b +d \right ) x \right )}{8 \left (b +d \right )}+\frac {\cosh \left (a +3 c +\left (b +3 d \right ) x \right )}{8 b +24 d}\)	\(166\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (135) = 270\).

Time = 0.25 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.27 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=-\frac {3 \, {\left (b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{3} - {\left ({\left (b^{3} - b d^{2}\right )} \cosh \left (b x + a\right ) + {\left (b^{3} - b d^{2}\right )} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{3} - 3 \, {\left (b^{2} d - 9 \, d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + 9 \, {\left ({\left (b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) + {\left (b^{2} d - d^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )^{2} + 3 \, {\left ({\left (b^{2} d - d^{3}\right )} \cosh \left (d x + c\right )^{3} - {\left (b^{2} d - 9 \, d^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (b x + a\right ) - 3 \, {\left ({\left (b^{3} - b d^{2}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right )^{2} - {\left (b^{3} - 9 \, b d^{2}\right )} \cosh \left (b x + a\right ) - {\left (b^{3} - 9 \, b d^{2} - {\left (b^{3} - b d^{2}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{4 \, {\left (b^{4} - 10 \, b^{2} d^{2} + 9 \, d^{4}\right )}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 976 vs. \(2 (131) = 262\).

Time = 2.35 (sec) , antiderivative size = 976, normalized size of antiderivative = 7.02 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=\text {Too large to display} \]

[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))

Maxima [F(-2)]

Exception generated. \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=\text {Exception raised: ValueError} \]

[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 detai

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.60 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=\frac {e^{\left (b x + 3 \, d x + a + 3 \, c\right )}}{8 \, {\left (b + 3 \, d\right )}} - \frac {3 \, e^{\left (b x + d x + a + c\right )}}{8 \, {\left (b + d\right )}} + \frac {3 \, e^{\left (b x - d x + a - c\right )}}{8 \, {\left (b - d\right )}} - \frac {e^{\left (b x - 3 \, d x + a - 3 \, c\right )}}{8 \, {\left (b - 3 \, d\right )}} \]

[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)

Mupad [B] (verification not implemented)

Time = 2.92 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int e^{a+b x} \sinh ^3(c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (-b^3\,{\mathrm {sinh}\left (c+d\,x\right )}^3+3\,b^2\,d\,\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {sinh}\left (c+d\,x\right )}^2-6\,b\,d^2\,{\mathrm {cosh}\left (c+d\,x\right )}^2\,\mathrm {sinh}\left (c+d\,x\right )+7\,b\,d^2\,{\mathrm {sinh}\left (c+d\,x\right )}^3+6\,d^3\,{\mathrm {cosh}\left (c+d\,x\right )}^3-9\,d^3\,\mathrm {cosh}\left (c+d\,x\right )\,{\mathrm {sinh}\left (c+d\,x\right )}^2\right )}{b^4-10\,b^2\,d^2+9\,d^4} \]

[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)