3.932 \(\int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=100 \[ -\frac {e^{-3 a-3 b x}}{96 b}-\frac {e^{-a-b x}}{32 b}-\frac {e^{a+b x}}{16 b}-\frac {e^{3 a+3 b x}}{48 b}+\frac {e^{5 a+5 b x}}{160 b}+\frac {e^{7 a+7 b x}}{224 b} \]

[Out]

-1/96*exp(-3*b*x-3*a)/b-1/32*exp(-b*x-a)/b-1/16*exp(b*x+a)/b-1/48*exp(3*b*x+3*a)/b+1/160*exp(5*b*x+5*a)/b+1/22
4*exp(7*b*x+7*a)/b

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {2282, 12, 448} \[ -\frac {e^{-3 a-3 b x}}{96 b}-\frac {e^{-a-b x}}{32 b}-\frac {e^{a+b x}}{16 b}-\frac {e^{3 a+3 b x}}{48 b}+\frac {e^{5 a+5 b x}}{160 b}+\frac {e^{7 a+7 b x}}{224 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

-E^(-3*a - 3*b*x)/(96*b) - E^(-a - b*x)/(32*b) - E^(a + b*x)/(16*b) - E^(3*a + 3*b*x)/(48*b) + E^(5*a + 5*b*x)
/(160*b) + E^(7*a + 7*b*x)/(224*b)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {align*} \int e^{2 (a+b x)} \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2 \left (1+x^2\right )^3}{32 x^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2 \left (1+x^2\right )^3}{x^4} \, dx,x,e^{a+b x}\right )}{32 b}\\ &=\frac {\operatorname {Subst}\left (\int \left (-2+\frac {1}{x^4}+\frac {1}{x^2}-2 x^2+x^4+x^6\right ) \, dx,x,e^{a+b x}\right )}{32 b}\\ &=-\frac {e^{-3 a-3 b x}}{96 b}-\frac {e^{-a-b x}}{32 b}-\frac {e^{a+b x}}{16 b}-\frac {e^{3 a+3 b x}}{48 b}+\frac {e^{5 a+5 b x}}{160 b}+\frac {e^{7 a+7 b x}}{224 b}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 73, normalized size = 0.73 \[ \frac {e^{-3 (a+b x)} \left (-105 e^{2 (a+b x)}-210 e^{4 (a+b x)}-70 e^{6 (a+b x)}+21 e^{8 (a+b x)}+15 e^{10 (a+b x)}-35\right )}{3360 b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*(a + b*x))*Cosh[a + b*x]^3*Sinh[a + b*x]^2,x]

[Out]

(-35 - 105*E^(2*(a + b*x)) - 210*E^(4*(a + b*x)) - 70*E^(6*(a + b*x)) + 21*E^(8*(a + b*x)) + 15*E^(10*(a + b*x
)))/(3360*b*E^(3*(a + b*x)))

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fricas [B]  time = 0.46, size = 176, normalized size = 1.76 \[ -\frac {10 \, \cosh \left (b x + a\right )^{5} + 50 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 25 \, \sinh \left (b x + a\right )^{5} - {\left (250 \, \cosh \left (b x + a\right )^{2} + 63\right )} \sinh \left (b x + a\right )^{3} + 42 \, \cosh \left (b x + a\right )^{3} + 2 \, {\left (50 \, \cosh \left (b x + a\right )^{3} + 63 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - {\left (125 \, \cosh \left (b x + a\right )^{4} + 189 \, \cosh \left (b x + a\right )^{2} + 70\right )} \sinh \left (b x + a\right ) + 140 \, \cosh \left (b x + a\right )}{1680 \, {\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/1680*(10*cosh(b*x + a)^5 + 50*cosh(b*x + a)*sinh(b*x + a)^4 - 25*sinh(b*x + a)^5 - (250*cosh(b*x + a)^2 + 6
3)*sinh(b*x + a)^3 + 42*cosh(b*x + a)^3 + 2*(50*cosh(b*x + a)^3 + 63*cosh(b*x + a))*sinh(b*x + a)^2 - (125*cos
h(b*x + a)^4 + 189*cosh(b*x + a)^2 + 70)*sinh(b*x + a) + 140*cosh(b*x + a))/(b*cosh(b*x + a)^2 - 2*b*cosh(b*x
+ a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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giac [A]  time = 0.14, size = 80, normalized size = 0.80 \[ -\frac {35 \, {\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} - {\left (15 \, e^{\left (7 \, b x + 28 \, a\right )} + 21 \, e^{\left (5 \, b x + 26 \, a\right )} - 70 \, e^{\left (3 \, b x + 24 \, a\right )} - 210 \, e^{\left (b x + 22 \, a\right )}\right )} e^{\left (-21 \, a\right )}}{3360 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="giac")

[Out]

-1/3360*(35*(3*e^(2*b*x + 2*a) + 1)*e^(-3*b*x - 3*a) - (15*e^(7*b*x + 28*a) + 21*e^(5*b*x + 26*a) - 70*e^(3*b*
x + 24*a) - 210*e^(b*x + 22*a))*e^(-21*a))/b

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maple [A]  time = 0.23, size = 108, normalized size = 1.08 \[ -\frac {\sinh \left (b x +a \right )}{32 b}-\frac {\sinh \left (3 b x +3 a \right )}{96 b}+\frac {\sinh \left (5 b x +5 a \right )}{160 b}+\frac {\sinh \left (7 b x +7 a \right )}{224 b}-\frac {3 \cosh \left (b x +a \right )}{32 b}-\frac {\cosh \left (3 b x +3 a \right )}{32 b}+\frac {\cosh \left (5 b x +5 a \right )}{160 b}+\frac {\cosh \left (7 b x +7 a \right )}{224 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^2,x)

[Out]

-1/32*sinh(b*x+a)/b-1/96/b*sinh(3*b*x+3*a)+1/160/b*sinh(5*b*x+5*a)+1/224/b*sinh(7*b*x+7*a)-3/32*cosh(b*x+a)/b-
1/32*cosh(3*b*x+3*a)/b+1/160*cosh(5*b*x+5*a)/b+1/224*cosh(7*b*x+7*a)/b

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maxima [A]  time = 0.32, size = 76, normalized size = 0.76 \[ \frac {{\left (21 \, e^{\left (-2 \, b x - 2 \, a\right )} - 70 \, e^{\left (-4 \, b x - 4 \, a\right )} - 210 \, e^{\left (-6 \, b x - 6 \, a\right )} + 15\right )} e^{\left (7 \, b x + 7 \, a\right )}}{3360 \, b} - \frac {3 \, e^{\left (-b x - a\right )} + e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)^3*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

1/3360*(21*e^(-2*b*x - 2*a) - 70*e^(-4*b*x - 4*a) - 210*e^(-6*b*x - 6*a) + 15)*e^(7*b*x + 7*a)/b - 1/96*(3*e^(
-b*x - a) + e^(-3*b*x - 3*a))/b

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mupad [B]  time = 0.60, size = 69, normalized size = 0.69 \[ -\frac {210\,{\mathrm {e}}^{a+b\,x}+105\,{\mathrm {e}}^{-a-b\,x}+35\,{\mathrm {e}}^{-3\,a-3\,b\,x}+70\,{\mathrm {e}}^{3\,a+3\,b\,x}-21\,{\mathrm {e}}^{5\,a+5\,b\,x}-15\,{\mathrm {e}}^{7\,a+7\,b\,x}}{3360\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^3*exp(2*a + 2*b*x)*sinh(a + b*x)^2,x)

[Out]

-(210*exp(a + b*x) + 105*exp(- a - b*x) + 35*exp(- 3*a - 3*b*x) + 70*exp(3*a + 3*b*x) - 21*exp(5*a + 5*b*x) -
15*exp(7*a + 7*b*x))/(3360*b)

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sympy [A]  time = 159.71, size = 197, normalized size = 1.97 \[ \begin {cases} \frac {2 e^{2 a} e^{2 b x} \sinh ^{5}{\left (a + b x \right )}}{105 b} - \frac {4 e^{2 a} e^{2 b x} \sinh ^{4}{\left (a + b x \right )} \cosh {\left (a + b x \right )}}{105 b} - \frac {e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{105 b} + \frac {2 e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{35 b} + \frac {8 e^{2 a} e^{2 b x} \sinh {\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{35 b} - \frac {4 e^{2 a} e^{2 b x} \cosh ^{5}{\left (a + b x \right )}}{35 b} & \text {for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\relax (a )} \cosh ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)**3*sinh(b*x+a)**2,x)

[Out]

Piecewise((2*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**5/(105*b) - 4*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**4*cosh(a + b*
x)/(105*b) - exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3*cosh(a + b*x)**2/(105*b) + 2*exp(2*a)*exp(2*b*x)*sinh(a + b*
x)**2*cosh(a + b*x)**3/(35*b) + 8*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**4/(35*b) - 4*exp(2*a)*exp(2
*b*x)*cosh(a + b*x)**5/(35*b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**2*cosh(a)**3, True))

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