3.332 \(\int \frac{e^{c (a+b x)}}{\sqrt{\sinh ^2(a c+b c x)}} \, dx\)

Optimal. Leaf size=46 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt{\sinh ^2(a c+b c x)}} \]

[Out]

(Log[1 - E^(2*c*(a + b*x))]*Sinh[a*c + b*c*x])/(b*c*Sqrt[Sinh[a*c + b*c*x]^2])

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Rubi [A]  time = 0.129109, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6720, 2282, 12, 260} \[ \frac{\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt{\sinh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[1 - E^(2*c*(a + b*x))]*Sinh[a*c + b*c*x])/(b*c*Sqrt[Sinh[a*c + b*c*x]^2])

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])

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{e^{c (a+b x)}}{\sqrt{\sinh ^2(a c+b c x)}} \, dx &=\frac{\sinh (a c+b c x) \int e^{c (a+b x)} \text{csch}(a c+b c x) \, dx}{\sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{\sinh (a c+b c x) \operatorname{Subst}\left (\int \frac{2 x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{(2 \sinh (a c+b c x)) \operatorname{Subst}\left (\int \frac{x}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ &=\frac{\log \left (1-e^{2 c (a+b x)}\right ) \sinh (a c+b c x)}{b c \sqrt{\sinh ^2(a c+b c x)}}\\ \end{align*}

Mathematica [A]  time = 0.050542, size = 44, normalized size = 0.96 \[ \frac{\log \left (1-e^{2 c (a+b x)}\right ) \sinh (c (a+b x))}{b c \sqrt{\sinh ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Log[1 - E^(2*c*(a + b*x))]*Sinh[c*(a + b*x)])/(b*c*Sqrt[Sinh[c*(a + b*x)]^2])

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{c \left ( bx+a \right ) }}{\frac{1}{\sqrt{ \left ( \sinh \left ( bcx+ac \right ) \right ) ^{2}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.60244, size = 53, normalized size = 1.15 \begin{align*} \frac{\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac{\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="maxima")

[Out]

log(e^(b*c*x + a*c) + 1)/(b*c) + log(e^(b*c*x + a*c) - 1)/(b*c)

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Fricas [A]  time = 1.82143, size = 97, normalized size = 2.11 \begin{align*} \frac{\log \left (\frac{2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="fricas")

[Out]

log(2*sinh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c)))/(b*c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a c} \int \frac{e^{b c x}}{\sqrt{\sinh ^{2}{\left (a c + b c x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sinh(b*c*x+a*c)**2)**(1/2),x)

[Out]

exp(a*c)*Integral(exp(b*c*x)/sqrt(sinh(a*c + b*c*x)**2), x)

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Giac [A]  time = 1.21016, size = 115, normalized size = 2.5 \begin{align*} \frac{\log \left (e^{\left (b c x\right )} + e^{\left (-a c\right )}\right ) \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right ) + \log \left ({\left | e^{\left (b c x\right )} - e^{\left (-a c\right )} \right |}\right ) \mathrm{sgn}\left (e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}\right )}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(sinh(b*c*x+a*c)^2)^(1/2),x, algorithm="giac")

[Out]

(log(e^(b*c*x) + e^(-a*c))*sgn(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) + log(abs(e^(b*c*x) - e^(-a*c)))*sgn(e^(b*c
*x + a*c) - e^(-b*c*x - a*c)))/(b*c)