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

Optimal. Leaf size=137 \[ \frac {3 d e^{c+d x} \sinh (2 a+2 b x)}{32 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (6 a+6 b x)}{32 \left (36 b^2-d^2\right )}-\frac {3 b e^{c+d x} \cosh (2 a+2 b x)}{16 \left (4 b^2-d^2\right )}+\frac {3 b e^{c+d x} \cosh (6 a+6 b x)}{16 \left (36 b^2-d^2\right )} \]

[Out]

-3/16*b*exp(d*x+c)*cosh(2*b*x+2*a)/(4*b^2-d^2)+3/16*b*exp(d*x+c)*cosh(6*b*x+6*a)/(36*b^2-d^2)+3/32*d*exp(d*x+c
)*sinh(2*b*x+2*a)/(4*b^2-d^2)-1/32*d*exp(d*x+c)*sinh(6*b*x+6*a)/(36*b^2-d^2)

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Rubi [A]  time = 0.11, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {5509, 5474} \[ \frac {3 d e^{c+d x} \sinh (2 a+2 b x)}{32 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (6 a+6 b x)}{32 \left (36 b^2-d^2\right )}-\frac {3 b e^{c+d x} \cosh (2 a+2 b x)}{16 \left (4 b^2-d^2\right )}+\frac {3 b e^{c+d x} \cosh (6 a+6 b x)}{16 \left (36 b^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*E^(c + d*x)*Cosh[2*a + 2*b*x])/(16*(4*b^2 - d^2)) + (3*b*E^(c + d*x)*Cosh[6*a + 6*b*x])/(16*(36*b^2 - d^
2)) + (3*d*E^(c + d*x)*Sinh[2*a + 2*b*x])/(32*(4*b^2 - d^2)) - (d*E^(c + d*x)*Sinh[6*a + 6*b*x])/(32*(36*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 ^3(a+b x) \, dx &=\int \left (-\frac {3}{32} e^{c+d x} \sinh (2 a+2 b x)+\frac {1}{32} e^{c+d x} \sinh (6 a+6 b x)\right ) \, dx\\ &=\frac {1}{32} \int e^{c+d x} \sinh (6 a+6 b x) \, dx-\frac {3}{32} \int e^{c+d x} \sinh (2 a+2 b x) \, dx\\ &=-\frac {3 b e^{c+d x} \cosh (2 a+2 b x)}{16 \left (4 b^2-d^2\right )}+\frac {3 b e^{c+d x} \cosh (6 a+6 b x)}{16 \left (36 b^2-d^2\right )}+\frac {3 d e^{c+d x} \sinh (2 a+2 b x)}{32 \left (4 b^2-d^2\right )}-\frac {d e^{c+d x} \sinh (6 a+6 b x)}{32 \left (36 b^2-d^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 113, normalized size = 0.82 \[ \frac {e^{c+d x} \left (6 \left (4 b^3-b d^2\right ) \cosh (6 (a+b x))+6 b \left (d^2-36 b^2\right ) \cosh (2 (a+b x))+2 d \sinh (2 (a+b x)) \left (\left (d^2-4 b^2\right ) \cosh (4 (a+b x))+52 b^2-d^2\right )\right )}{32 \left (144 b^4-40 b^2 d^2+d^4\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c + d*x)*(6*b*(-36*b^2 + d^2)*Cosh[2*(a + b*x)] + 6*(4*b^3 - b*d^2)*Cosh[6*(a + b*x)] + 2*d*(52*b^2 - d^2
+ (-4*b^2 + d^2)*Cosh[4*(a + b*x)])*Sinh[2*(a + b*x)]))/(32*(144*b^4 - 40*b^2*d^2 + d^4))

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fricas [B]  time = 0.48, size = 676, normalized size = 4.93 \[ -\frac {10 \, {\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{3} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{3} - 45 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{4} + 3 \, {\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 )^{5} - 3 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{6} - 3 \, {\left (15 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{4} - 36 \, b^{3} + b d^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (b x + a\right )^{2} + 3 \, {\left ({\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{5} - {\left (36 \, 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 ) - 3 \, {\left ({\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{6} - {\left (36 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2}\right )} \cosh \left (d x + c\right ) - {\left (3 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{6} - 10 \, {\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{3} \sinh \left (b x + a\right )^{3} + 45 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} - 3 \, {\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{5} + 3 \, {\left (4 \, b^{3} - b d^{2}\right )} \sinh \left (b x + a\right )^{6} - 3 \, {\left (36 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{2} + 3 \, {\left (15 \, {\left (4 \, b^{3} - b d^{2}\right )} \cosh \left (b x + a\right )^{4} - 36 \, b^{3} + b d^{2}\right )} \sinh \left (b x + a\right )^{2} - 3 \, {\left ({\left (4 \, b^{2} d - d^{3}\right )} \cosh \left (b x + a\right )^{5} - {\left (36 \, 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 )}{16 \, {\left ({\left (144 \, b^{4} - 40 \, b^{2} d^{2} + d^{4}\right )} \cosh \left (b x + a\right )^{6} - 3 \, {\left (144 \, b^{4} - 40 \, b^{2} d^{2} + d^{4}\right )} \cosh \left (b x + a\right )^{4} \sinh \left (b x + a\right )^{2} + 3 \, {\left (144 \, b^{4} - 40 \, b^{2} d^{2} + d^{4}\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{4} - {\left (144 \, b^{4} - 40 \, b^{2} d^{2} + d^{4}\right )} \sinh \left (b x + a\right )^{6}\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)^3,x, algorithm="fricas")

[Out]

-1/16*(10*(4*b^2*d - d^3)*cosh(b*x + a)^3*cosh(d*x + c)*sinh(b*x + a)^3 - 45*(4*b^3 - b*d^2)*cosh(b*x + a)^2*c
osh(d*x + c)*sinh(b*x + a)^4 + 3*(4*b^2*d - d^3)*cosh(b*x + a)*cosh(d*x + c)*sinh(b*x + a)^5 - 3*(4*b^3 - b*d^
2)*cosh(d*x + c)*sinh(b*x + a)^6 - 3*(15*(4*b^3 - b*d^2)*cosh(b*x + a)^4 - 36*b^3 + b*d^2)*cosh(d*x + c)*sinh(
b*x + a)^2 + 3*((4*b^2*d - d^3)*cosh(b*x + a)^5 - (36*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(b*x + a)^6 - (36*b^3 - b*d^2)*cosh(b*x + a)^2)*cosh(d*x + c) - (3*(4*b^3 - b*d^2)*co
sh(b*x + a)^6 - 10*(4*b^2*d - d^3)*cosh(b*x + a)^3*sinh(b*x + a)^3 + 45*(4*b^3 - b*d^2)*cosh(b*x + a)^2*sinh(b
*x + a)^4 - 3*(4*b^2*d - d^3)*cosh(b*x + a)*sinh(b*x + a)^5 + 3*(4*b^3 - b*d^2)*sinh(b*x + a)^6 - 3*(36*b^3 -
b*d^2)*cosh(b*x + a)^2 + 3*(15*(4*b^3 - b*d^2)*cosh(b*x + a)^4 - 36*b^3 + b*d^2)*sinh(b*x + a)^2 - 3*((4*b^2*d
- d^3)*cosh(b*x + a)^5 - (36*b^2*d - d^3)*cosh(b*x + a))*sinh(b*x + a))*sinh(d*x + c))/((144*b^4 - 40*b^2*d^2
+ d^4)*cosh(b*x + a)^6 - 3*(144*b^4 - 40*b^2*d^2 + d^4)*cosh(b*x + a)^4*sinh(b*x + a)^2 + 3*(144*b^4 - 40*b^2
*d^2 + d^4)*cosh(b*x + a)^2*sinh(b*x + a)^4 - (144*b^4 - 40*b^2*d^2 + d^4)*sinh(b*x + a)^6)

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giac [A]  time = 0.13, size = 93, normalized size = 0.68 \[ \frac {e^{\left (6 \, b x + d x + 6 \, a + c\right )}}{64 \, {\left (6 \, b + d\right )}} - \frac {3 \, e^{\left (2 \, b x + d x + 2 \, a + c\right )}}{64 \, {\left (2 \, b + d\right )}} - \frac {3 \, e^{\left (-2 \, b x + d x - 2 \, a + c\right )}}{64 \, {\left (2 \, b - d\right )}} + \frac {e^{\left (-6 \, b x + d x - 6 \, a + c\right )}}{64 \, {\left (6 \, b - d\right )}} \]

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)^3,x, algorithm="giac")

[Out]

1/64*e^(6*b*x + d*x + 6*a + c)/(6*b + d) - 3/64*e^(2*b*x + d*x + 2*a + c)/(2*b + d) - 3/64*e^(-2*b*x + d*x - 2
*a + c)/(2*b - d) + 1/64*e^(-6*b*x + d*x - 6*a + c)/(6*b - d)

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maple [A]  time = 0.39, size = 202, normalized size = 1.47 \[ \frac {3 \sinh \left (2 a -c +\left (2 b -d \right ) x \right )}{64 \left (2 b -d \right )}-\frac {3 \sinh \left (2 a +c +\left (2 b +d \right ) x \right )}{64 \left (2 b +d \right )}-\frac {\sinh \left (\left (6 b -d \right ) x +6 a -c \right )}{64 \left (6 b -d \right )}+\frac {\sinh \left (\left (6 b +d \right ) x +6 a +c \right )}{384 b +64 d}-\frac {3 \cosh \left (2 a -c +\left (2 b -d \right ) x \right )}{64 \left (2 b -d \right )}-\frac {3 \cosh \left (2 a +c +\left (2 b +d \right ) x \right )}{64 \left (2 b +d \right )}+\frac {\cosh \left (\left (6 b -d \right ) x +6 a -c \right )}{384 b -64 d}+\frac {\cosh \left (\left (6 b +d \right ) x +6 a +c \right )}{384 b +64 d} \]

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)^3,x)

[Out]

3/64*sinh(2*a-c+(2*b-d)*x)/(2*b-d)-3/64*sinh(2*a+c+(2*b+d)*x)/(2*b+d)-1/64/(6*b-d)*sinh((6*b-d)*x+6*a-c)+1/64/
(6*b+d)*sinh((6*b+d)*x+6*a+c)-3/64*cosh(2*a-c+(2*b-d)*x)/(2*b-d)-3/64*cosh(2*a+c+(2*b+d)*x)/(2*b+d)+1/64*cosh(
(6*b-d)*x+6*a-c)/(6*b-d)+1/64*cosh((6*b+d)*x+6*a+c)/(6*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)^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(1-d/b>0)', see `assume?` for m
ore details)Is 1-d/b equal to -1?

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mupad [B]  time = 0.96, size = 182, normalized size = 1.33 \[ -\frac {b^3\,\left (\frac {27\,{\mathrm {e}}^{c+d\,x}\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{4}-\frac {3\,{\mathrm {e}}^{c+d\,x}\,\mathrm {cosh}\left (6\,a+6\,b\,x\right )}{4}\right )+d^3\,\left (\frac {3\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{32}-\frac {{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (6\,a+6\,b\,x\right )}{32}\right )-b^2\,d\,\left (\frac {27\,{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (2\,a+2\,b\,x\right )}{8}-\frac {{\mathrm {e}}^{c+d\,x}\,\mathrm {sinh}\left (6\,a+6\,b\,x\right )}{8}\right )-b\,d^2\,\left (\frac {3\,{\mathrm {e}}^{c+d\,x}\,\mathrm {cosh}\left (2\,a+2\,b\,x\right )}{16}-\frac {3\,{\mathrm {e}}^{c+d\,x}\,\mathrm {cosh}\left (6\,a+6\,b\,x\right )}{16}\right )}{144\,b^4-40\,b^2\,d^2+d^4} \]

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)^3,x)

[Out]

-(b^3*((27*exp(c + d*x)*cosh(2*a + 2*b*x))/4 - (3*exp(c + d*x)*cosh(6*a + 6*b*x))/4) + d^3*((3*exp(c + d*x)*si
nh(2*a + 2*b*x))/32 - (exp(c + d*x)*sinh(6*a + 6*b*x))/32) - b^2*d*((27*exp(c + d*x)*sinh(2*a + 2*b*x))/8 - (e
xp(c + d*x)*sinh(6*a + 6*b*x))/8) - b*d^2*((3*exp(c + d*x)*cosh(2*a + 2*b*x))/16 - (3*exp(c + d*x)*cosh(6*a +
6*b*x))/16))/(144*b^4 + d^4 - 40*b^2*d^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**3,x)

[Out]

Timed out

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