3.157 \(\int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx\)

Optimal. Leaf size=250 \[ -\frac {e^{-4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {sech}(a c+b c x)}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {sech}(a c+b c x)}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {sech}(a c+b c x)}{192 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 x \text {sech}(a c+b c x)}{16 \sqrt {\text {sech}^2(a c+b c x)}} \]

[Out]

-1/128*sech(b*c*x+a*c)/b/c/exp(4*c*(b*x+a))/(sech(b*c*x+a*c)^2)^(1/2)-5/64*sech(b*c*x+a*c)/b/c/exp(2*c*(b*x+a)
)/(sech(b*c*x+a*c)^2)^(1/2)+5/32*exp(2*c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+5/128*exp(4*c*
(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+1/192*exp(6*c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+
a*c)^2)^(1/2)+5/16*x*sech(b*c*x+a*c)/(sech(b*c*x+a*c)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6720, 2282, 12, 266, 43} \[ -\frac {e^{-4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {sech}(a c+b c x)}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {sech}(a c+b c x)}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {sech}(a c+b c x)}{192 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 x \text {sech}(a c+b c x)}{16 \sqrt {\text {sech}^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))/(Sech[a*c + b*c*x]^2)^(5/2),x]

[Out]

-Sech[a*c + b*c*x]/(128*b*c*E^(4*c*(a + b*x))*Sqrt[Sech[a*c + b*c*x]^2]) - (5*Sech[a*c + b*c*x])/(64*b*c*E^(2*
c*(a + b*x))*Sqrt[Sech[a*c + b*c*x]^2]) + (5*E^(2*c*(a + b*x))*Sech[a*c + b*c*x])/(32*b*c*Sqrt[Sech[a*c + b*c*
x]^2]) + (5*E^(4*c*(a + b*x))*Sech[a*c + b*c*x])/(128*b*c*Sqrt[Sech[a*c + b*c*x]^2]) + (E^(6*c*(a + b*x))*Sech
[a*c + b*c*x])/(192*b*c*Sqrt[Sech[a*c + b*c*x]^2]) + (5*x*Sech[a*c + b*c*x])/(16*Sqrt[Sech[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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)}}{\text {sech}^2(a c+b c x)^{5/2}} \, dx &=\frac {\text {sech}(a c+b c x) \int e^{c (a+b x)} \cosh ^5(a c+b c x) \, dx}{\sqrt {\text {sech}^2(a c+b c x)}}\\ &=\frac {\text {sech}(a c+b c x) \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^5}{32 x^5} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt {\text {sech}^2(a c+b c x)}}\\ &=\frac {\text {sech}(a c+b c x) \operatorname {Subst}\left (\int \frac {\left (1+x^2\right )^5}{x^5} \, dx,x,e^{c (a+b x)}\right )}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}\\ &=\frac {\text {sech}(a c+b c x) \operatorname {Subst}\left (\int \frac {(1+x)^5}{x^3} \, dx,x,e^{2 c (a+b x)}\right )}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}\\ &=\frac {\text {sech}(a c+b c x) \operatorname {Subst}\left (\int \left (10+\frac {1}{x^3}+\frac {5}{x^2}+\frac {10}{x}+5 x+x^2\right ) \, dx,x,e^{2 c (a+b x)}\right )}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}\\ &=-\frac {e^{-4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}-\frac {5 e^{-2 c (a+b x)} \text {sech}(a c+b c x)}{64 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{2 c (a+b x)} \text {sech}(a c+b c x)}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 e^{4 c (a+b x)} \text {sech}(a c+b c x)}{128 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {e^{6 c (a+b x)} \text {sech}(a c+b c x)}{192 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {5 x \text {sech}(a c+b c x)}{16 \sqrt {\text {sech}^2(a c+b c x)}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 106, normalized size = 0.42 \[ \frac {\left (-\frac {1}{2} e^{-4 c (a+b x)}-5 e^{-2 c (a+b x)}+10 e^{2 c (a+b x)}+\frac {5}{2} e^{4 c (a+b x)}+\frac {1}{3} e^{6 c (a+b x)}+20 b c x\right ) \text {sech}^5(c (a+b x))}{64 b c \text {sech}^2(c (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))/(Sech[a*c + b*c*x]^2)^(5/2),x]

[Out]

((-1/2*1/E^(4*c*(a + b*x)) - 5/E^(2*c*(a + b*x)) + 10*E^(2*c*(a + b*x)) + (5*E^(4*c*(a + b*x)))/2 + E^(6*c*(a
+ b*x))/3 + 20*b*c*x)*Sech[c*(a + b*x)]^5)/(64*b*c*(Sech[c*(a + b*x)]^2)^(5/2))

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fricas [A]  time = 0.43, size = 218, normalized size = 0.87 \[ -\frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - 5 \, \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{2} + 9\right )} \sinh \left (b c x + a c\right )^{3} + 15 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (5 \, \cosh \left (b c x + a c\right )^{4} - 24 \, b c x + 27 \, \cosh \left (b c x + a c\right )^{2} + 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/384*(cosh(b*c*x + a*c)^5 + 5*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 - 5*sinh(b*c*x + a*c)^5 - 5*(10*cosh(b*c
*x + a*c)^2 + 9)*sinh(b*c*x + a*c)^3 + 15*cosh(b*c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 9*cosh(b*c*x + a*c)
)*sinh(b*c*x + a*c)^2 - 60*(2*b*c*x + 1)*cosh(b*c*x + a*c) - 5*(5*cosh(b*c*x + a*c)^4 - 24*b*c*x + 27*cosh(b*c
*x + a*c)^2 + 12)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b*c*x + a*c))

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giac [A]  time = 0.12, size = 110, normalized size = 0.44 \[ \frac {{\left (120 \, b c x e^{\left (-a c\right )} - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-4 \, b c x - 5 \, a c\right )} + {\left (2 \, e^{\left (6 \, b c x + 20 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 18 \, a c\right )} + 60 \, e^{\left (2 \, b c x + 16 \, a c\right )}\right )} e^{\left (-15 \, a c\right )}\right )} e^{\left (a c\right )}}{384 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(5/2),x, algorithm="giac")

[Out]

1/384*(120*b*c*x*e^(-a*c) - 3*(30*e^(4*b*c*x + 4*a*c) + 10*e^(2*b*c*x + 2*a*c) + 1)*e^(-4*b*c*x - 5*a*c) + (2*
e^(6*b*c*x + 20*a*c) + 15*e^(4*b*c*x + 18*a*c) + 60*e^(2*b*c*x + 16*a*c))*e^(-15*a*c))*e^(a*c)/(b*c)

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maple [A]  time = 0.75, size = 326, normalized size = 1.30 \[ \frac {5 x \,{\mathrm e}^{c \left (b x +a \right )}}{16 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {{\mathrm e}^{7 c \left (b x +a \right )}}{192 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{5 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {5 \,{\mathrm e}^{3 c \left (b x +a \right )}}{32 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}-\frac {5 \,{\mathrm e}^{-c \left (b x +a \right )}}{64 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}-\frac {{\mathrm e}^{-3 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(5/2),x)

[Out]

5/16*x/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp(c*(b*x+a))+1/192/b/c/(1+exp(
2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp(7*c*(b*x+a))+5/128/b/c/(1+exp(2*c*(b*x+a))
)/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp(5*c*(b*x+a))+5/32/b/c/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2
*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp(3*c*(b*x+a))-5/64/b/c/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^
2*exp(2*c*(b*x+a)))^(1/2)*exp(-c*(b*x+a))-1/128/b/c/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*
x+a)))^(1/2)*exp(-3*c*(b*x+a))

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maxima [A]  time = 0.32, size = 112, normalized size = 0.45 \[ \frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} + \frac {e^{\left (6 \, b c x + 6 \, a c\right )}}{192 \, b c} + \frac {5 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{128 \, b c} + \frac {5 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{32 \, b c} - \frac {5 \, e^{\left (-2 \, b c x - 2 \, a c\right )}}{64 \, b c} - \frac {e^{\left (-4 \, b c x - 4 \, a c\right )}}{128 \, b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(5/2),x, algorithm="maxima")

[Out]

5/16*(b*c*x + a*c)/(b*c) + 1/192*e^(6*b*c*x + 6*a*c)/(b*c) + 5/128*e^(4*b*c*x + 4*a*c)/(b*c) + 5/32*e^(2*b*c*x
+ 2*a*c)/(b*c) - 5/64*e^(-2*b*c*x - 2*a*c)/(b*c) - 1/128*e^(-4*b*c*x - 4*a*c)/(b*c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left (\frac {1}{{\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(a + b*x))/(1/cosh(a*c + b*c*x)^2)^(5/2),x)

[Out]

int(exp(c*(a + b*x))/(1/cosh(a*c + b*c*x)^2)^(5/2), x)

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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(c*(b*x+a))/(sech(b*c*x+a*c)**2)**(5/2),x)

[Out]

Timed out

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