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

Optimal. Leaf size=83 \[ \frac {b e^{c+d x} \sinh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )}-\frac {d e^{c+d x} \cosh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}-\frac {e^{c+d x}}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5509, 2194, 5475} \[ \frac {b e^{c+d x} \sinh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )}-\frac {d e^{c+d x} \cosh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}-\frac {e^{c+d x}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(c + d*x)/(8*d) - (d*E^(c + d*x)*Cosh[4*a + 4*b*x])/(8*(16*b^2 - d^2)) + (b*E^(c + d*x)*Sinh[4*a + 4*b*x])/
(2*(16*b^2 - 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 5475

Int[Cosh[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> -Simp[(b*c*Log[F]*F^(c*(a + b*x))
*Cosh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sinh[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 ^2(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac {1}{8} e^{c+d x}+\frac {1}{8} e^{c+d x} \cosh (4 a+4 b x)\right ) \, dx\\ &=-\left (\frac {1}{8} \int e^{c+d x} \, dx\right )+\frac {1}{8} \int e^{c+d x} \cosh (4 a+4 b x) \, dx\\ &=-\frac {e^{c+d x}}{8 d}-\frac {d e^{c+d x} \cosh (4 a+4 b x)}{8 \left (16 b^2-d^2\right )}+\frac {b e^{c+d x} \sinh (4 a+4 b x)}{2 \left (16 b^2-d^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 58, normalized size = 0.70 \[ \frac {e^{c+d x} \left (d^2 \cosh (4 (a+b x))-4 b d \sinh (4 (a+b x))+16 b^2-d^2\right )}{8 \left (d^3-16 b^2 d\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c + d*x)*(16*b^2 - d^2 + d^2*Cosh[4*(a + b*x)] - 4*b*d*Sinh[4*(a + b*x)]))/(8*(-16*b^2*d + d^3))

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fricas [B]  time = 0.55, size = 303, normalized size = 3.65 \[ \frac {16 \, b d \cosh \left (b x + a\right )^{3} \cosh \left (d x + c\right ) \sinh \left (b x + a\right ) - 6 \, d^{2} \cosh \left (b x + a\right )^{2} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{2} + 16 \, b d \cosh \left (b x + a\right ) \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{3} - d^{2} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{4} - {\left (d^{2} \cosh \left (b x + a\right )^{4} + 16 \, b^{2} - d^{2}\right )} \cosh \left (d x + c\right ) - {\left (d^{2} \cosh \left (b x + a\right )^{4} - 16 \, b d \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right ) + 6 \, d^{2} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} - 16 \, b d \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + d^{2} \sinh \left (b x + a\right )^{4} + 16 \, b^{2} - d^{2}\right )} \sinh \left (d x + c\right )}{8 \, {\left ({\left (16 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (16 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (16 \, b^{2} d - d^{3}\right )} \sinh \left (b x + a\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(16*b*d*cosh(b*x + a)^3*cosh(d*x + c)*sinh(b*x + a) - 6*d^2*cosh(b*x + a)^2*cosh(d*x + c)*sinh(b*x + a)^2
+ 16*b*d*cosh(b*x + a)*cosh(d*x + c)*sinh(b*x + a)^3 - d^2*cosh(d*x + c)*sinh(b*x + a)^4 - (d^2*cosh(b*x + a)^
4 + 16*b^2 - d^2)*cosh(d*x + c) - (d^2*cosh(b*x + a)^4 - 16*b*d*cosh(b*x + a)^3*sinh(b*x + a) + 6*d^2*cosh(b*x
+ a)^2*sinh(b*x + a)^2 - 16*b*d*cosh(b*x + a)*sinh(b*x + a)^3 + d^2*sinh(b*x + a)^4 + 16*b^2 - d^2)*sinh(d*x
+ c))/((16*b^2*d - d^3)*cosh(b*x + a)^4 - 2*(16*b^2*d - d^3)*cosh(b*x + a)^2*sinh(b*x + a)^2 + (16*b^2*d - d^3
)*sinh(b*x + a)^4)

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giac [A]  time = 0.13, size = 58, normalized size = 0.70 \[ \frac {e^{\left (4 \, b x + d x + 4 \, a + c\right )}}{16 \, {\left (4 \, b + d\right )}} - \frac {e^{\left (-4 \, b x + d x - 4 \, a + c\right )}}{16 \, {\left (4 \, b - d\right )}} - \frac {e^{\left (d x + c\right )}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 124, normalized size = 1.49 \[ -\frac {\sinh \left (d x +c \right )}{8 d}+\frac {\sinh \left (\left (4 b -d \right ) x +4 a -c \right )}{64 b -16 d}+\frac {\sinh \left (\left (4 b +d \right ) x +4 a +c \right )}{64 b +16 d}-\frac {\cosh \left (d x +c \right )}{8 d}-\frac {\cosh \left (\left (4 b -d \right ) x +4 a -c \right )}{16 \left (4 b -d \right )}+\frac {\cosh \left (\left (4 b +d \right ) x +4 a +c \right )}{64 b +16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sinh(d*x+c)/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(d*x+c)/d-1/1
6*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)^2*sinh(b*x+a)^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(3-d/b>0)', see `assume?` for m
ore details)Is 3-d/b equal to -1?

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mupad [B]  time = 2.73, size = 96, normalized size = 1.16 \[ -\frac {\frac {d^2\,{\mathrm {e}}^{c+d\,x}\,\left (\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}}{2}+\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{2}\right )}{8}+\frac {b\,d\,{\mathrm {e}}^{c+d\,x}\,\left (\frac {{\mathrm {e}}^{-4\,a-4\,b\,x}}{2}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}}{2}\right )}{2}}{16\,b^2\,d-d^3}-\frac {{\mathrm {e}}^{c+d\,x}}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((d^2*exp(c + d*x)*(exp(- 4*a - 4*b*x)/2 + exp(4*a + 4*b*x)/2))/8 + (b*d*exp(c + d*x)*(exp(- 4*a - 4*b*x)/2
- exp(4*a + 4*b*x)/2))/2)/(16*b^2*d - d^3) - exp(c + d*x)/(8*d)

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sympy [A]  time = 88.85, size = 819, normalized size = 9.87 \[ \begin {cases} x e^{c} \sinh ^{2}{\relax (a )} \cosh ^{2}{\relax (a )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {x e^{c} e^{d x} \sinh ^{4}{\left (a - \frac {d x}{4} \right )}}{16} + \frac {x e^{c} e^{d x} \sinh ^{3}{\left (a - \frac {d x}{4} \right )} \cosh {\left (a - \frac {d x}{4} \right )}}{4} + \frac {3 x e^{c} e^{d x} \sinh ^{2}{\left (a - \frac {d x}{4} \right )} \cosh ^{2}{\left (a - \frac {d x}{4} \right )}}{8} + \frac {x e^{c} e^{d x} \sinh {\left (a - \frac {d x}{4} \right )} \cosh ^{3}{\left (a - \frac {d x}{4} \right )}}{4} + \frac {x e^{c} e^{d x} \cosh ^{4}{\left (a - \frac {d x}{4} \right )}}{16} - \frac {e^{c} e^{d x} \sinh ^{4}{\left (a - \frac {d x}{4} \right )}}{6 d} - \frac {5 e^{c} e^{d x} \sinh ^{3}{\left (a - \frac {d x}{4} \right )} \cosh {\left (a - \frac {d x}{4} \right )}}{12 d} - \frac {5 e^{c} e^{d x} \sinh {\left (a - \frac {d x}{4} \right )} \cosh ^{3}{\left (a - \frac {d x}{4} \right )}}{12 d} - \frac {e^{c} e^{d x} \cosh ^{4}{\left (a - \frac {d x}{4} \right )}}{6 d} & \text {for}\: b = - \frac {d}{4} \\\frac {x e^{c} e^{d x} \sinh ^{4}{\left (a + \frac {d x}{4} \right )}}{16} - \frac {x e^{c} e^{d x} \sinh ^{3}{\left (a + \frac {d x}{4} \right )} \cosh {\left (a + \frac {d x}{4} \right )}}{4} + \frac {3 x e^{c} e^{d x} \sinh ^{2}{\left (a + \frac {d x}{4} \right )} \cosh ^{2}{\left (a + \frac {d x}{4} \right )}}{8} - \frac {x e^{c} e^{d x} \sinh {\left (a + \frac {d x}{4} \right )} \cosh ^{3}{\left (a + \frac {d x}{4} \right )}}{4} + \frac {x e^{c} e^{d x} \cosh ^{4}{\left (a + \frac {d x}{4} \right )}}{16} - \frac {e^{c} e^{d x} \sinh ^{4}{\left (a + \frac {d x}{4} \right )}}{6 d} + \frac {5 e^{c} e^{d x} \sinh ^{3}{\left (a + \frac {d x}{4} \right )} \cosh {\left (a + \frac {d x}{4} \right )}}{12 d} + \frac {5 e^{c} e^{d x} \sinh {\left (a + \frac {d x}{4} \right )} \cosh ^{3}{\left (a + \frac {d x}{4} \right )}}{12 d} - \frac {e^{c} e^{d x} \cosh ^{4}{\left (a + \frac {d x}{4} \right )}}{6 d} & \text {for}\: b = \frac {d}{4} \\\left (- \frac {x \sinh ^{4}{\left (a + b x \right )}}{8} + \frac {x \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{4} - \frac {x \cosh ^{4}{\left (a + b x \right )}}{8} + \frac {\sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{8 b} + \frac {\sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{8 b}\right ) e^{c} & \text {for}\: d = 0 \\- \frac {2 b^{2} e^{c} e^{d x} \sinh ^{4}{\left (a + b x \right )}}{16 b^{2} d - d^{3}} + \frac {4 b^{2} e^{c} e^{d x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b^{2} d - d^{3}} - \frac {2 b^{2} e^{c} e^{d x} \cosh ^{4}{\left (a + b x \right )}}{16 b^{2} d - d^{3}} + \frac {2 b d e^{c} e^{d x} \sinh ^{3}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{16 b^{2} d - d^{3}} + \frac {2 b d e^{c} e^{d x} \sinh {\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{16 b^{2} d - d^{3}} - \frac {d^{2} e^{c} e^{d x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b^{2} d - d^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*exp(c)*sinh(a)**2*cosh(a)**2, Eq(b, 0) & Eq(d, 0)), (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) - 5*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/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) + 5*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*sinh(a + b*x)**4/8 + x*sinh(a + b*x)**2*cosh(a + b*x)**2/4 - x*
cosh(a + b*x)**4/8 + sinh(a + b*x)**3*cosh(a + b*x)/(8*b) + sinh(a + b*x)*cosh(a + b*x)**3/(8*b))*exp(c), Eq(d
, 0)), (-2*b**2*exp(c)*exp(d*x)*sinh(a + b*x)**4/(16*b**2*d - d**3) + 4*b**2*exp(c)*exp(d*x)*sinh(a + b*x)**2*
cosh(a + b*x)**2/(16*b**2*d - d**3) - 2*b**2*exp(c)*exp(d*x)*cosh(a + b*x)**4/(16*b**2*d - d**3) + 2*b*d*exp(c
)*exp(d*x)*sinh(a + b*x)**3*cosh(a + b*x)/(16*b**2*d - d**3) + 2*b*d*exp(c)*exp(d*x)*sinh(a + b*x)*cosh(a + b*
x)**3/(16*b**2*d - d**3) - d**2*exp(c)*exp(d*x)*sinh(a + b*x)**2*cosh(a + b*x)**2/(16*b**2*d - d**3), True))

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