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

Optimal. Leaf size=137 \[ -\frac {d e^{c+d x} \sinh (2 a+2 b x)}{4 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}+\frac {b e^{c+d x} \cosh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )}+\frac {b e^{c+d x} \cosh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )} \]

[Out]

1/2*b*exp(d*x+c)*cosh(2*b*x+2*a)/(4*b^2-d^2)+1/2*b*exp(d*x+c)*cosh(4*b*x+4*a)/(16*b^2-d^2)-1/4*d*exp(d*x+c)*si
nh(2*b*x+2*a)/(4*b^2-d^2)-1/8*d*exp(d*x+c)*sinh(4*b*x+4*a)/(16*b^2-d^2)

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Rubi [A]  time = 0.09, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {5509, 5474} \[ -\frac {d e^{c+d x} \sinh (2 a+2 b x)}{4 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}+\frac {b e^{c+d x} \cosh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )}+\frac {b e^{c+d x} \cosh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*E^(c + d*x)*Cosh[2*a + 2*b*x])/(2*(4*b^2 - d^2)) + (b*E^(c + d*x)*Cosh[4*a + 4*b*x])/(2*(16*b^2 - d^2)) - (
d*E^(c + d*x)*Sinh[2*a + 2*b*x])/(4*(4*b^2 - d^2)) - (d*E^(c + d*x)*Sinh[4*a + 4*b*x])/(8*(16*b^2 - 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 5509

Int[Cosh[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol]
:> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sinh[d + e*x]^m*Cosh[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e,
f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int e^{c+d x} \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\int \left (\frac {1}{4} e^{c+d x} \sinh (2 a+2 b x)+\frac {1}{8} e^{c+d x} \sinh (4 a+4 b x)\right ) \, dx\\ &=\frac {1}{8} \int e^{c+d x} \sinh (4 a+4 b x) \, dx+\frac {1}{4} \int e^{c+d x} \sinh (2 a+2 b x) \, dx\\ &=\frac {b e^{c+d x} \cosh (2 a+2 b x)}{2 \left (4 b^2-d^2\right )}+\frac {b e^{c+d x} \cosh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (2 a+2 b x)}{4 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.02, size = 86, normalized size = 0.63 \[ \frac {1}{8} e^{c+d x} \left (\frac {4 b \cosh (2 (a+b x))-2 d \sinh (2 (a+b x))}{4 b^2-d^2}+\frac {4 b \cosh (4 (a+b x))-d \sinh (4 (a+b x))}{16 b^2-d^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c + d*x)*((4*b*Cosh[2*(a + b*x)] - 2*d*Sinh[2*(a + b*x)])/(4*b^2 - d^2) + (4*b*Cosh[4*(a + b*x)] - d*Sinh[
4*(a + b*x)])/(16*b^2 - d^2)))/8

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fricas [B]  time = 0.44, size = 501, normalized size = 3.66 \[ -\frac {{\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{3} - {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{4} - {\left (16 \, b^{3} - b d^{2} + 6 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{2} + {\left ({\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{3} + {\left (16 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) - {\left ({\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{4} + {\left (16 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right ) - {\left ({\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{4} - {\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + {\left (4 \, b^{3} - b d^{2}\right )} \sinh \left (b x + a\right )^{4} + {\left (16 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2} + {\left (16 \, b^{3} - b d^{2} + 6 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2}\right )} \sinh \left (b x + a\right )^{2} - {\left ({\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{3} + {\left (16 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )\right )} \sinh \left (d x + c\right )}{2 \, {\left ({\left (64 \, b^{4} - 20 \, b^{2} d^{2} + d^{4}\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (64 \, b^{4} - 20 \, b^{2} d^{2} + d^{4}\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (64 \, b^{4} - 20 \, b^{2} d^{2} + d^{4}\right )} \sinh \left (b x + a\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((4*b^2*d - d^3)*cosh(b*x + a)*cosh(d*x + c)*sinh(b*x + a)^3 - (4*b^3 - b*d^2)*cosh(d*x + c)*sinh(b*x + a
)^4 - (16*b^3 - b*d^2 + 6*(4*b^3 - b*d^2)*cosh(b*x + a)^2)*cosh(d*x + c)*sinh(b*x + a)^2 + ((4*b^2*d - d^3)*co
sh(b*x + a)^3 + (16*b^2*d - d^3)*cosh(b*x + a))*cosh(d*x + c)*sinh(b*x + a) - ((4*b^3 - b*d^2)*cosh(b*x + a)^4
+ (16*b^3 - b*d^2)*cosh(b*x + a)^2)*cosh(d*x + c) - ((4*b^3 - b*d^2)*cosh(b*x + a)^4 - (4*b^2*d - d^3)*cosh(b
*x + a)*sinh(b*x + a)^3 + (4*b^3 - b*d^2)*sinh(b*x + a)^4 + (16*b^3 - b*d^2)*cosh(b*x + a)^2 + (16*b^3 - b*d^2
+ 6*(4*b^3 - b*d^2)*cosh(b*x + a)^2)*sinh(b*x + a)^2 - ((4*b^2*d - d^3)*cosh(b*x + a)^3 + (16*b^2*d - d^3)*co
sh(b*x + a))*sinh(b*x + a))*sinh(d*x + c))/((64*b^4 - 20*b^2*d^2 + d^4)*cosh(b*x + a)^4 - 2*(64*b^4 - 20*b^2*d
^2 + d^4)*cosh(b*x + a)^2*sinh(b*x + a)^2 + (64*b^4 - 20*b^2*d^2 + d^4)*sinh(b*x + a)^4)

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giac [A]  time = 0.13, size = 93, normalized size = 0.68 \[ \frac {e^{\left (4 \, b x + d x + 4 \, a + c\right )}}{16 \, {\left (4 \, b + d\right )}} + \frac {e^{\left (2 \, b x + d x + 2 \, a + c\right )}}{8 \, {\left (2 \, b + d\right )}} + \frac {e^{\left (-2 \, b x + d x - 2 \, a + c\right )}}{8 \, {\left (2 \, b - d\right )}} + \frac {e^{\left (-4 \, b x + d x - 4 \, a + c\right )}}{16 \, {\left (4 \, b - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*e^(4*b*x + d*x + 4*a + c)/(4*b + d) + 1/8*e^(2*b*x + d*x + 2*a + c)/(2*b + d) + 1/8*e^(-2*b*x + d*x - 2*a
+ c)/(2*b - d) + 1/16*e^(-4*b*x + d*x - 4*a + c)/(4*b - d)

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maple [A]  time = 0.43, size = 202, normalized size = 1.47 \[ -\frac {\sinh \left (2 a -c +\left (2 b -d \right ) x \right )}{8 \left (2 b -d \right )}+\frac {\sinh \left (2 a +c +\left (2 b +d \right ) x \right )}{16 b +8 d}-\frac {\sinh \left (\left (4 b -d \right ) x +4 a -c \right )}{16 \left (4 b -d \right )}+\frac {\sinh \left (\left (4 b +d \right ) x +4 a +c \right )}{64 b +16 d}+\frac {\cosh \left (2 a -c +\left (2 b -d \right ) x \right )}{16 b -8 d}+\frac {\cosh \left (2 a +c +\left (2 b +d \right ) x \right )}{16 b +8 d}+\frac {\cosh \left (\left (4 b -d \right ) x +4 a -c \right )}{64 b -16 d}+\frac {\cosh \left (\left (4 b +d \right ) x +4 a +c \right )}{64 b +16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sinh(2*a-c+(2*b-d)*x)/(2*b-d)+1/8*sinh(2*a+c+(2*b+d)*x)/(2*b+d)-1/16/(4*b-d)*sinh((4*b-d)*x+4*a-c)+1/16/(
4*b+d)*sinh((4*b+d)*x+4*a+c)+1/8*cosh(2*a-c+(2*b-d)*x)/(2*b-d)+1/8*cosh(2*a+c+(2*b+d)*x)/(2*b+d)+1/16*cosh((4*
b-d)*x+4*a-c)/(4*b-d)+1/16*cosh((4*b+d)*x+4*a+c)/(4*b+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(d*x+c)*cosh(b*x+a)^3*sinh(b*x+a),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(1-d/b>0)', see `assume?` for m
ore details)Is 1-d/b equal to -1?

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mupad [B]  time = 2.68, size = 163, normalized size = 1.19 \[ -\frac {b^3\,\left (6\,{\mathrm {e}}^{c+d\,x}-16\,{\mathrm {cosh}\left (a+b\,x\right )}^4\,{\mathrm {e}}^{c+d\,x}\right )+b^2\,d\,\left (4\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (a+b\,x\right )\,{\mathrm {cosh}\left (a+b\,x\right )}^3+6\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (a+b\,x\right )\,\mathrm {cosh}\left (a+b\,x\right )\right )-b\,d^2\,\left (3\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\mathrm {e}}^{c+d\,x}-4\,{\mathrm {cosh}\left (a+b\,x\right )}^4\,{\mathrm {e}}^{c+d\,x}\right )-d^3\,{\mathrm {cosh}\left (a+b\,x\right )}^3\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (a+b\,x\right )}{64\,b^4-20\,b^2\,d^2+d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^3*(6*exp(c + d*x) - 16*cosh(a + b*x)^4*exp(c + d*x)) + b^2*d*(6*cosh(a + b*x)*exp(c + d*x)*sinh(a + b*x) +
4*cosh(a + b*x)^3*exp(c + d*x)*sinh(a + b*x)) - b*d^2*(3*cosh(a + b*x)^2*exp(c + d*x) - 4*cosh(a + b*x)^4*exp
(c + d*x)) - d^3*cosh(a + b*x)^3*exp(c + d*x)*sinh(a + b*x))/(64*b^4 + d^4 - 20*b^2*d^2)

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sympy [A]  time = 143.71, size = 1295, normalized size = 9.45 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(c)*sinh(a)*cosh(a)**3, Eq(b, 0) & Eq(d, 0)), (-x*exp(c)*exp(d*x)*sinh(a - d*x/2)**4/8 - x*exp
(c)*exp(d*x)*sinh(a - d*x/2)**3*cosh(a - d*x/2)/4 + x*exp(c)*exp(d*x)*sinh(a - d*x/2)*cosh(a - d*x/2)**3/4 + x
*exp(c)*exp(d*x)*cosh(a - d*x/2)**4/8 + exp(c)*exp(d*x)*sinh(a - d*x/2)**4/(8*d) - exp(c)*exp(d*x)*sinh(a - d*
x/2)**2*cosh(a - d*x/2)**2/(2*d) - exp(c)*exp(d*x)*sinh(a - d*x/2)*cosh(a - d*x/2)**3/(3*d) - 7*exp(c)*exp(d*x
)*cosh(a - d*x/2)**4/(24*d), Eq(b, -d/2)), (x*exp(c)*exp(d*x)*sinh(a - d*x/4)**4/16 + x*exp(c)*exp(d*x)*sinh(a
- d*x/4)**3*cosh(a - d*x/4)/4 + 3*x*exp(c)*exp(d*x)*sinh(a - d*x/4)**2*cosh(a - d*x/4)**2/8 + x*exp(c)*exp(d*
x)*sinh(a - d*x/4)*cosh(a - d*x/4)**3/4 + x*exp(c)*exp(d*x)*cosh(a - d*x/4)**4/16 - exp(c)*exp(d*x)*sinh(a - d
*x/4)**4/(6*d) - 5*exp(c)*exp(d*x)*sinh(a - d*x/4)**3*cosh(a - d*x/4)/(12*d) + 11*exp(c)*exp(d*x)*sinh(a - d*x
/4)*cosh(a - d*x/4)**3/(12*d) + exp(c)*exp(d*x)*cosh(a - d*x/4)**4/(6*d), Eq(b, -d/4)), (-x*exp(c)*exp(d*x)*si
nh(a + d*x/4)**4/16 + x*exp(c)*exp(d*x)*sinh(a + d*x/4)**3*cosh(a + d*x/4)/4 - 3*x*exp(c)*exp(d*x)*sinh(a + d*
x/4)**2*cosh(a + d*x/4)**2/8 + x*exp(c)*exp(d*x)*sinh(a + d*x/4)*cosh(a + d*x/4)**3/4 - x*exp(c)*exp(d*x)*cosh
(a + d*x/4)**4/16 + exp(c)*exp(d*x)*sinh(a + d*x/4)**4/(6*d) - 5*exp(c)*exp(d*x)*sinh(a + d*x/4)**3*cosh(a + d
*x/4)/(12*d) + 11*exp(c)*exp(d*x)*sinh(a + d*x/4)*cosh(a + d*x/4)**3/(12*d) - exp(c)*exp(d*x)*cosh(a + d*x/4)*
*4/(6*d), Eq(b, d/4)), (x*exp(c)*exp(d*x)*sinh(a + d*x/2)**4/8 - x*exp(c)*exp(d*x)*sinh(a + d*x/2)**3*cosh(a +
d*x/2)/4 + x*exp(c)*exp(d*x)*sinh(a + d*x/2)*cosh(a + d*x/2)**3/4 - x*exp(c)*exp(d*x)*cosh(a + d*x/2)**4/8 -
exp(c)*exp(d*x)*sinh(a + d*x/2)**4/(8*d) + exp(c)*exp(d*x)*sinh(a + d*x/2)**2*cosh(a + d*x/2)**2/(2*d) - exp(c
)*exp(d*x)*sinh(a + d*x/2)*cosh(a + d*x/2)**3/(3*d) + 7*exp(c)*exp(d*x)*cosh(a + d*x/2)**4/(24*d), Eq(b, d/2))
, (-6*b**3*exp(c)*exp(d*x)*sinh(a + b*x)**4/(64*b**4 - 20*b**2*d**2 + d**4) + 12*b**3*exp(c)*exp(d*x)*sinh(a +
b*x)**2*cosh(a + b*x)**2/(64*b**4 - 20*b**2*d**2 + d**4) + 10*b**3*exp(c)*exp(d*x)*cosh(a + b*x)**4/(64*b**4
- 20*b**2*d**2 + d**4) + 6*b**2*d*exp(c)*exp(d*x)*sinh(a + b*x)**3*cosh(a + b*x)/(64*b**4 - 20*b**2*d**2 + d**
4) - 10*b**2*d*exp(c)*exp(d*x)*sinh(a + b*x)*cosh(a + b*x)**3/(64*b**4 - 20*b**2*d**2 + d**4) - 3*b*d**2*exp(c
)*exp(d*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(64*b**4 - 20*b**2*d**2 + d**4) - b*d**2*exp(c)*exp(d*x)*cosh(a +
b*x)**4/(64*b**4 - 20*b**2*d**2 + d**4) + d**3*exp(c)*exp(d*x)*sinh(a + b*x)*cosh(a + b*x)**3/(64*b**4 - 20*b
**2*d**2 + d**4), True))

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