3.889 \(\int e^{a+b x} \text {sech}^2(c+d x) \, dx\)

Optimal. Leaf size=56 \[ \frac {4 e^{a+b x+2 (c+d x)} \, _2F_1\left (2,\frac {b}{2 d}+1;\frac {b}{2 d}+2;-e^{2 (c+d x)}\right )}{b+2 d} \]

[Out]

4*exp(b*x+2*d*x+a+2*c)*hypergeom([2, 1+1/2*b/d],[2+1/2*b/d],-exp(2*d*x+2*c))/(b+2*d)

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Rubi [A]  time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {5492} \[ \frac {4 e^{a+b x+2 (c+d x)} \, _2F_1\left (2,\frac {b}{2 d}+1;\frac {b}{2 d}+2;-e^{2 (c+d x)}\right )}{b+2 d} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sech[c + d*x]^2,x]

[Out]

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

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}^2(c+d x) \, dx &=\frac {4 e^{a+b x+2 (c+d x)} \, _2F_1\left (2,1+\frac {b}{2 d};2+\frac {b}{2 d};-e^{2 (c+d x)}\right )}{b+2 d}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 1.00 \[ \frac {4 e^{a+b x+2 (c+d x)} \, _2F_1\left (2,\frac {b}{2 d}+1;\frac {b}{2 d}+2;-e^{2 (c+d x)}\right )}{b+2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (e^{\left (b x + a\right )} \operatorname {sech}\left (d x + c\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(e^(b*x + a)*sech(d*x + c)^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (b x + a\right )} \operatorname {sech}\left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(e^(b*x + a)*sech(d*x + c)^2, x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{b x +a} \mathrm {sech}\left (d x +c \right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {e}}^{a+b\,x}}{{\mathrm {cosh}\left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(a + b*x)/cosh(c + d*x)^2,x)

[Out]

int(exp(a + b*x)/cosh(c + d*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(a)*Integral(exp(b*x)*sech(c + d*x)**2, x)

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