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

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

[Out]

exp(c*(b*x+a))*coth(b*c*x+a*c)*(tanh(b*c*x+a*c)^2)^(1/2)/b/c-2*arctan(exp(c*(b*x+a)))*coth(b*c*x+a*c)*(tanh(b*
c*x+a*c)^2)^(1/2)/b/c

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Rubi [A]  time = 0.15, 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, 203} \[ \frac {e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}-\frac {2 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c} \]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[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 e^{c (a+b x)} \sqrt {\tanh ^2(a c+b c x)} \, dx &=\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \tanh (a c+b c x) \, dx\\ &=\frac {\left (\coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {-1+x^2}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {\left (2 \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac {e^{c (a+b x)} \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}-\frac {2 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt {\tanh ^2(a c+b c x)}}{b c}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 51, normalized size = 0.61 \[ \frac {\left (e^{c (a+b x)}-2 \tan ^{-1}\left (e^{c (a+b x)}\right )\right ) \sqrt {\tanh ^2(c (a+b x))} \coth (c (a+b x))}{b c} \]

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

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fricas [A]  time = 0.62, size = 53, normalized size = 0.64 \[ -\frac {2 \, \arctan \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) - \cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}{b c} \]

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]

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

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giac [A]  time = 0.15, size = 60, normalized size = 0.72 \[ -\frac {2 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} \]

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]

-(2*arctan(e^(b*c*x + a*c))*sgn(e^(2*b*c*x + 2*a*c) - 1) - e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(b*c)

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maple [C]  time = 0.74, size = 218, normalized size = 2.63 \[ \frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}+\frac {i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-i\right )}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b}-\frac {i \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+i\right )}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) c b} \]

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

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maxima [A]  time = 0.42, size = 35, normalized size = 0.42 \[ -\frac {2 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{b c} + \frac {e^{\left (b c x + a c\right )}}{b c} \]

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="maxima")

[Out]

-2*arctan(e^(b*c*x + a*c))/(b*c) + e^(b*c*x + a*c)/(b*c)

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

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 \[ e^{a c} \int \sqrt {\tanh ^{2}{\left (a c + b c x \right )}} e^{b c x}\, dx \]

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

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