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

Optimal. Leaf size=83 \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}} \]

[Out]

exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)-2*arctanh(exp(c*(b*x+a)))*tanh(b*c*x+a*c)/b/c/(ta
nh(b*c*x+a*c)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 83, 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 = {6720, 2282, 388, 206} \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(c*(a + b*x))*Tanh[a*c + b*c*x])/(b*c*Sqrt[Tanh[a*c + b*c*x]^2]) - (2*ArcTanh[E^(c*(a + b*x))]*Tanh[a*c + b
*c*x])/(b*c*Sqrt[Tanh[a*c + b*c*x]^2])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 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])

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 51, normalized size = 0.61 \[ \frac {\left (e^{c (a+b x)}-2 \tanh ^{-1}\left (e^{c (a+b x)}\right )\right ) \tanh (c (a+b x))}{b c \sqrt {\tanh ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((E^(c*(a + b*x)) - 2*ArcTanh[E^(c*(a + b*x))])*Tanh[c*(a + b*x)])/(b*c*Sqrt[Tanh[c*(a + b*x)]^2])

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fricas [A]  time = 0.49, size = 70, normalized size = 0.84 \[ \frac {\cosh \left (b c x + a c\right ) - \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) + 1\right ) + \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) - 1\right ) + \sinh \left (b c x + a c\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(b*c*x + a*c) - log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) + 1) + log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c
) - 1) + sinh(b*c*x + a*c))/(b*c)

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giac [A]  time = 0.21, size = 88, normalized size = 1.06 \[ \frac {e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - log(e^(b*c*x + a*c) + 1)*sgn(e^(2*b*c*x + 2*a*c) - 1) + log(ab
s(e^(b*c*x + a*c) - 1))*sgn(e^(2*b*c*x + 2*a*c) - 1))/(b*c)

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maple [B]  time = 1.02, size = 213, normalized size = 2.57 \[ \frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}+\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left (1+{\mathrm e}^{c \left (b x +a \right )}\right )}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.75, size = 56, normalized size = 0.67 \[ \frac {e^{\left (b c x + a c\right )}}{b c} - \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{\sqrt {{\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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