3.42 \(\int \sec ^{-1}(c e^{a+b x}) \, dx\)

Optimal. Leaf size=85 \[ \frac{i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{i \sec ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\sec ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

[Out]

((I/2)*ArcSec[c*E^(a + b*x)]^2)/b - (ArcSec[c*E^(a + b*x)]*Log[1 + E^((2*I)*ArcSec[c*E^(a + b*x)])])/b + ((I/2
)*PolyLog[2, -E^((2*I)*ArcSec[c*E^(a + b*x)])])/b

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Rubi [A]  time = 0.081309, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {2282, 5218, 4626, 3719, 2190, 2279, 2391} \[ \frac{i \text{PolyLog}\left (2,-e^{2 i \sec ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{i \sec ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\sec ^{-1}\left (c e^{a+b x}\right ) \log \left (1+e^{2 i \sec ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSec[c*E^(a + b*x)],x]

[Out]

((I/2)*ArcSec[c*E^(a + b*x)]^2)/b - (ArcSec[c*E^(a + b*x)]*Log[1 + E^((2*I)*ArcSec[c*E^(a + b*x)])])/b + ((I/2
)*PolyLog[2, -E^((2*I)*ArcSec[c*E^(a + b*x)])])/b

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 5218

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCos[x/c])/x, x], x, 1/x] /; Fre
eQ[{a, b, c}, x]

Rule 4626

Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> -Subst[Int[(a + b*x)^n/Cot[x], x], x, ArcCos[c
*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*(m + 1)), x
] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \sec ^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sec ^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\cos ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,e^{-a-b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x \tan (x) \, dx,x,\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1+e^{2 i x}} \, dx,x,\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{\operatorname{Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ &=\frac{i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1+e^{2 i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{i \text{Li}_2\left (-e^{2 i \cos ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.849665, size = 280, normalized size = 3.29 \[ x \sec ^{-1}\left (c e^{a+b x}\right )-\frac{e^{-a-b x} \left (-4 \sqrt{1-c^2 e^{2 (a+b x)}} \text{PolyLog}\left (2,\frac{1}{2} \left (1-\sqrt{1-c^2 e^{2 (a+b x)}}\right )\right )+\sqrt{1-c^2 e^{2 (a+b x)}} \left (\log ^2\left (c^2 e^{2 (a+b x)}\right )+2 \log ^2\left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right )-4 \log \left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right ) \log \left (c^2 e^{2 (a+b x)}\right )\right )+4 \sqrt{c^2 e^{2 (a+b x)}-1} \left (2 b x-\log \left (c^2 e^{2 (a+b x)}\right )\right ) \tan ^{-1}\left (\sqrt{c^2 e^{2 (a+b x)}-1}\right )\right )}{8 b c \sqrt{1-\frac{e^{-2 (a+b x)}}{c^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSec[c*E^(a + b*x)],x]

[Out]

x*ArcSec[c*E^(a + b*x)] - (E^(-a - b*x)*(4*Sqrt[-1 + c^2*E^(2*(a + b*x))]*ArcTan[Sqrt[-1 + c^2*E^(2*(a + b*x))
]]*(2*b*x - Log[c^2*E^(2*(a + b*x))]) + Sqrt[1 - c^2*E^(2*(a + b*x))]*(Log[c^2*E^(2*(a + b*x))]^2 - 4*Log[c^2*
E^(2*(a + b*x))]*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2] + 2*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2]^2)
- 4*Sqrt[1 - c^2*E^(2*(a + b*x))]*PolyLog[2, (1 - Sqrt[1 - c^2*E^(2*(a + b*x))])/2]))/(8*b*c*Sqrt[1 - 1/(c^2*E
^(2*(a + b*x)))])

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Maple [A]  time = 0.392, size = 116, normalized size = 1.4 \begin{align*}{\frac{{\frac{i}{2}} \left ({\rm arcsec} \left (c{{\rm e}^{bx+a}}\right ) \right ) ^{2}}{b}}-{\frac{{\rm arcsec} \left (c{{\rm e}^{bx+a}}\right )}{b}\ln \left ( 1+ \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+i\sqrt{1-{\frac{1}{{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}}}} \right ) ^{2} \right ) }+{\frac{{\frac{i}{2}}}{b}{\it polylog} \left ( 2,- \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+i\sqrt{1-{\frac{1}{{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}}}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsec(c*exp(b*x+a)),x)

[Out]

1/2*I*arcsec(c*exp(b*x+a))^2/b-arcsec(c*exp(b*x+a))*ln(1+(1/c/exp(b*x+a)+I*(1-1/c^2/exp(b*x+a)^2)^(1/2))^2)/b+
1/2*I*polylog(2,-(1/c/exp(b*x+a)+I*(1-1/c^2/exp(b*x+a)^2)^(1/2))^2)/b

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(c*exp(b*x+a)),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(c*exp(b*x+a)),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asec(c*exp(b*x+a)),x)

[Out]

Integral(asec(c*exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsec(c*exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(arcsec(c*e^(b*x + a)), x)