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)*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)

________________________________________________________________________________________

Rubi [A]  time = 0.0500383, antiderivative size = 103, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.146712, 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)

________________________________________________________________________________________

Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{bx+a}} \left ({\rm sech} \left (dx+c\right ) \right ) ^{3}\, dx \end{align*}

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)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -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}} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (e^{\left (b x + a\right )} \operatorname{sech}\left (d x + c\right )^{3}, x\right ) \end{align*}

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)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a} \int e^{b x} \operatorname{sech}^{3}{\left (c + d x \right )}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (b x + a\right )} \operatorname{sech}\left (d x + c\right )^{3}\,{d x} \end{align*}

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)