\(\int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx\) [153]

Optimal result

Integrand size = 25, antiderivative size = 56 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=\frac {2 e^{4 c (a+b x)} \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 270} \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=\frac {2 e^{4 c (a+b x)} \cosh (a c+b c x) \sqrt {\text {sech}^2(a c+b c x)}}{b c \left (e^{2 c (a+b x)}+1\right )^2} \]

[In]

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

[Out]

(2*E^(4*c*(a + b*x))*Cosh[a*c + b*c*x]*Sqrt[Sech[a*c + b*c*x]^2])/(b*c*(1 + E^(2*c*(a + b*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 270

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 2320

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 6852

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.79 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=\frac {e^{3 c (a+b x)} \sqrt {\text {sech}^2(c (a+b x))}}{b c+b c e^{2 c (a+b x)}} \]

[In]

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

[Out]

(E^(3*c*(a + b*x))*Sqrt[Sech[c*(a + b*x)]^2])/(b*c + b*c*E^(2*c*(a + b*x)))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.36 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.68

method	result	size
default	\(\frac {\operatorname {csgn}\left (\operatorname {sech}\left (c \left (b x +a \right )\right )\right ) \left (\frac {\tanh \left (c \left (b x +a \right )\right )^{2}}{2}+\tanh \left (c \left (b x +a \right )\right )\right )}{c b}\)	\(38\)
risch	\(-\frac {2 \left (2 \,{\mathrm e}^{2 c \left (b x +a \right )}+1\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{-c \left (b x +a \right )}}{b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}\)	\(69\)

[In]

int(exp(c*(b*x+a))*(sech(b*c*x+a*c)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

csgn(sech(c*(b*x+a)))/c/b*(1/2*tanh(c*(b*x+a))^2+tanh(c*(b*x+a)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (52) = 104\).

Time = 0.25 (sec) , antiderivative size = 120, normalized size of antiderivative = 2.14 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=-\frac {2 \, {\left (3 \, \cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )}}{b c \cosh \left (b c x + a c\right )^{3} + 3 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} + b c \sinh \left (b c x + a c\right )^{3} + 3 \, b c \cosh \left (b c x + a c\right ) + {\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )} \]

[In]

integrate(exp(c*(b*x+a))*(sech(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
))

Sympy [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.50 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=-\frac {4 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} - \frac {2}{b c {\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

[In]

integrate(exp(c*(b*x+a))*(sech(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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.68 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=-\frac {2 \, {\left (2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}}{b c {\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{2}} \]

[In]

integrate(exp(c*(b*x+a))*(sech(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)

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.39 \[ \int e^{c (a+b x)} \text {sech}^2(a c+b c x)^{3/2} \, dx=-\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}\,\left (2\,{\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )\,\sqrt {\frac {1}{{\left (\frac {{\mathrm {e}}^{a\,c+b\,c\,x}}{2}+\frac {{\mathrm {e}}^{-a\,c-b\,c\,x}}{2}\right )}^2}}}{b\,c\,\left ({\mathrm {e}}^{2\,a\,c+2\,b\,c\,x}+1\right )} \]

[In]

int(exp(c*(a + b*x))*(1/cosh(a*c + b*c*x)^2)^(3/2),x)

[Out]

-(exp(- a*c - b*c*x)*(2*exp(2*a*c + 2*b*c*x) + 1)*(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))