3.4.32 \(\int \frac {e^{c (a+b x)}}{\sqrt {\sinh ^2(a c+b c x)}} \, dx\) [332]

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]

ln(1-exp(2*c*(b*x+a)))*sinh(b*c*x+a*c)/b/c/(sinh(b*c*x+a*c)^2)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 12, 266} \begin {gather*} \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 {gather*}

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 12

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

Rule 266

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]

Rule 2320

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 6852

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])

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) \text {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)) \text {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*}

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Mathematica [A]
time = 0.04, size = 44, normalized size = 0.96 \begin {gather*} \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))}} \end {gather*}

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 [A]
time = 9.83, size = 68, normalized size = 1.48

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{2 b c x}-{\mathrm e}^{-2 a c}\right ) \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{-c \left (b x +a \right )}}{c b \sqrt {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}}\)	\(68\)

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,method=_RETURNVERBOSE)

[Out]

ln(exp(2*b*c*x)-exp(-2*a*c))/c/b*(exp(2*c*(b*x+a))-1)/((exp(2*c*(b*x+a))-1)^2*exp(-2*c*(b*x+a)))^(1/2)*exp(-c*
(b*x+a))

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Maxima [A]
time = 0.49, size = 39, normalized size = 0.85 \begin {gather*} \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 {gather*}

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 = 0.38, size = 42, normalized size = 0.91 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a c} \int \frac {e^{b c x}}{\sqrt {\sinh ^{2}{\left (a c + b c x \right )}}}\, dx \end {gather*}

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 = 0.43, size = 85, normalized size = 1.85 \begin {gather*} \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 {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{\sqrt {{\mathrm {sinh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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