3.296 \(\int \frac {e^{c (a+b x)}}{\cosh ^2(a c+b c x)^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ \frac {2 e^{4 c (a+b x)} \cosh (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2 \sqrt {\cosh ^2(a c+b c x)}} \]

[Out]

2*exp(4*c*(b*x+a))*cosh(b*c*x+a*c)/b/c/(1+exp(2*c*(b*x+a)))^2/(cosh(b*c*x+a*c)^2)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6720, 2282, 12, 264} \[ \frac {2 e^{4 c (a+b x)} \cosh (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2 \sqrt {\cosh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(4*c*(a + b*x))*Cosh[a*c + b*c*x])/(b*c*(1 + E^(2*c*(a + b*x)))^2*Sqrt[Cosh[a*c + b*c*x]^2])

Rule 12

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

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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]]]

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {e^{c (a+b x)}}{\cosh ^2(a c+b c x)^{3/2}} \, dx &=\frac {\cosh (a c+b c x) \int e^{c (a+b x)} \text {sech}^3(a c+b c x) \, dx}{\sqrt {\cosh ^2(a c+b c x)}}\\ &=\frac {\cosh (a c+b c x) \operatorname {Subst}\left (\int \frac {8 x^3}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}}\\ &=\frac {(8 \cosh (a c+b c x)) \operatorname {Subst}\left (\int \frac {x^3}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\cosh ^2(a c+b c x)}}\\ &=\frac {2 e^{4 c (a+b x)} \cosh (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt {\cosh ^2(a c+b c x)}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 46, normalized size = 0.82 \[ \frac {4 e^{5 c (a+b x)} \sqrt {\cosh ^2(c (a+b x))}}{b c \left (e^{2 c (a+b x)}+1\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*E^(5*c*(a + b*x))*Sqrt[Cosh[c*(a + b*x)]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^3)

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fricas [B]  time = 0.49, size = 120, normalized size = 2.14 \[ -\frac {2 \, {\left (3 \, \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )}}{b c \cosh \left (b c x + a c\right )^{3} + 3 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} + b c \sinh \left (b c x + a c\right )^{3} + 3 \, b c \cosh \left (b c x + a c\right ) + {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(3*cosh(b*c*x + a*c) + sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^3 + 3*b*c*cosh(b*c*x + a*c)*sinh(b*c*x + a
*c)^2 + b*c*sinh(b*c*x + a*c)^3 + 3*b*c*cosh(b*c*x + a*c) + (3*b*c*cosh(b*c*x + a*c)^2 + b*c)*sinh(b*c*x + a*c
))

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giac [A]  time = 0.12, size = 38, normalized size = 0.68 \[ -\frac {2 \, {\left (2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}}{b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*e^(2*b*c*x + 2*a*c) + 1)/(b*c*(e^(2*b*c*x + 2*a*c) + 1)^2)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {8 \,{\mathrm e}^{c \left (b x +a \right )}}{\left (2 \cosh \left (2 b c x +2 a c \right )+2\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 84, normalized size = 1.50 \[ -\frac {4 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac {2}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.94, size = 76, normalized size = 1.36 \[ -\frac {4\,{\mathrm {e}}^{a\,c+b\,c\,x}\,\left (2\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )\,\sqrt {{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}+\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}{b\,c\,{\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*exp(a*c + b*c*x)*(2*exp(2*a*c + 2*b*c*x) + 1)*((exp(a*c + b*c*x)/2 + exp(- a*c - b*c*x)/2)^2)^(1/2))/(b*c*
(exp(2*a*c + 2*b*c*x) + 1)^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{a c} \int \frac {e^{b c x}}{\left (\cosh ^{2}{\left (a c + b c x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(a*c)*Integral(exp(b*c*x)/(cosh(a*c + b*c*x)**2)**(3/2), x)

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