3.890 \(\int e^{a+b x} \text {sech}^3(c+d x) \, dx\)

Optimal. Leaf size=103 \[ -\frac {(b-d) e^{a+b x+c+d x} \, _2F_1\left (1,\frac {b+d}{2 d};\frac {1}{2} \left (\frac {b}{d}+3\right );-e^{2 (c+d x)}\right )}{d^2}+\frac {b e^{a+b x} \text {sech}(c+d x)}{2 d^2}+\frac {e^{a+b x} \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]

[Out]

-(b-d)*exp(b*x+d*x+a+c)*hypergeom([1, 1/2*(b+d)/d],[3/2+1/2*b/d],-exp(2*d*x+2*c))/d^2+1/2*b*exp(b*x+a)*sech(d*
x+c)/d^2+1/2*exp(b*x+a)*sech(d*x+c)*tanh(d*x+c)/d

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Rubi [A]  time = 0.05, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {5490, 5492} \[ -\frac {(b-d) e^{a+b x+c+d x} \, _2F_1\left (1,\frac {b+d}{2 d};\frac {1}{2} \left (\frac {b}{d}+3\right );-e^{2 (c+d x)}\right )}{d^2}+\frac {b e^{a+b x} \text {sech}(c+d x)}{2 d^2}+\frac {e^{a+b x} \tanh (c+d x) \text {sech}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[c + d*x]^3,x]

[Out]

-(((b - d)*E^(a + c + b*x + d*x)*Hypergeometric2F1[1, (b + d)/(2*d), (3 + b/d)/2, -E^(2*(c + d*x))])/d^2) + (b
*E^(a + b*x)*Sech[c + d*x])/(2*d^2) + (E^(a + b*x)*Sech[c + d*x]*Tanh[c + d*x])/(2*d)

Rule 5490

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sech[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1)*
(n - 2)), Int[F^(c*(a + b*x))*Sech[d + e*x]^(n - 2), x], x] + Simp[(F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*Sinh
[d + e*x])/(e*(n - 1)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ
[n, 1] && NeQ[n, 2]

Rule 5492

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(n*(d + e*x))*F
^(c*(a + b*x))*Hypergeometric2F1[n, n/2 + (b*c*Log[F])/(2*e), 1 + n/2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])
/(e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin {align*} \int e^{a+b x} \text {sech}^3(c+d x) \, dx &=\frac {b e^{a+b x} \text {sech}(c+d x)}{2 d^2}+\frac {e^{a+b x} \text {sech}(c+d x) \tanh (c+d x)}{2 d}+\frac {1}{2} \left (1-\frac {b^2}{d^2}\right ) \int e^{a+b x} \text {sech}(c+d x) \, dx\\ &=-\frac {(b-d) e^{a+c+b x+d x} \, _2F_1\left (1,\frac {b+d}{2 d};\frac {1}{2} \left (3+\frac {b}{d}\right );-e^{2 (c+d x)}\right )}{d^2}+\frac {b e^{a+b x} \text {sech}(c+d x)}{2 d^2}+\frac {e^{a+b x} \text {sech}(c+d x) \tanh (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 80, normalized size = 0.78 \[ \frac {e^{a+b x} \left (\text {sech}(c+d x) (b+d \tanh (c+d x))-2 (b-d) e^{c+d x} \, _2F_1\left (1,\frac {b+d}{2 d};\frac {1}{2} \left (\frac {b}{d}+3\right );-e^{2 (c+d x)}\right )\right )}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(a + b*x)*(-2*(b - d)*E^(c + d*x)*Hypergeometric2F1[1, (b + d)/(2*d), (3 + b/d)/2, -E^(2*(c + d*x))] + Sech
[c + d*x]*(b + d*Tanh[c + d*x])))/(2*d^2)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (b x + a\right )} \operatorname {sech}\left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(d*x+c)^3,x, algorithm="fricas")

[Out]

integral(e^(b*x + a)*sech(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (b x + a\right )} \operatorname {sech}\left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(d*x+c)^3,x, algorithm="giac")

[Out]

integrate(e^(b*x + a)*sech(d*x + c)^3, x)

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{b x +a} \mathrm {sech}\left (d x +c \right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*sech(d*x+c)^3,x)

[Out]

int(exp(b*x+a)*sech(d*x+c)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -8 \, {\left (b^{2} e^{c} - d^{2} e^{c}\right )} \int \frac {e^{\left (b x + d x + a\right )}}{8 \, {\left (d^{2} e^{\left (2 \, d x + 2 \, c\right )} + d^{2}\right )}}\,{d x} + \frac {{\left (b e^{\left (3 \, c\right )} + d e^{\left (3 \, c\right )}\right )} e^{\left (b x + 3 \, d x + a\right )} + {\left (b e^{c} - d e^{c}\right )} e^{\left (b x + d x + a\right )}}{d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 2 \, d^{2} e^{\left (2 \, d x + 2 \, c\right )} + d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*sech(d*x+c)^3,x, algorithm="maxima")

[Out]

-8*(b^2*e^c - d^2*e^c)*integrate(1/8*e^(b*x + d*x + a)/(d^2*e^(2*d*x + 2*c) + d^2), x) + ((b*e^(3*c) + d*e^(3*
c))*e^(b*x + 3*d*x + a) + (b*e^c - d*e^c)*e^(b*x + d*x + a))/(d^2*e^(4*d*x + 4*c) + 2*d^2*e^(2*d*x + 2*c) + d^
2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {cosh}\left (c+d\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(c + d*x)^3,x)

[Out]

int(exp(a + b*x)/cosh(c + d*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(a)*Integral(exp(b*x)*sech(c + d*x)**3, x)

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