\(\int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx\) [213]

Optimal result

Integrand size = 25, antiderivative size = 83 \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c}-\frac {2 \text {arctanh}\left (e^{c (a+b x)}\right ) \sqrt {\coth ^2(a c+b c x)} \tanh (a c+b c x)}{b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6852, 2320, 396, 212} \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {e^{c (a+b x)} \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c}-\frac {2 \text {arctanh}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt {\coth ^2(a c+b c x)}}{b c} \]

[In]

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

[Out]

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

Rule 212

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

Rule 396

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {\left (e^{c (a+b x)}-2 \text {arctanh}\left (e^{c (a+b x)}\right )\right ) \sqrt {\coth ^2(c (a+b x))} \tanh (c (a+b x))}{b c} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(212\) vs. \(2(77)=154\).

Time = 0.77 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.57

method	result	size
risch	\(\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, {\mathrm e}^{c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}+\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}{\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}\)	\(213\)

[In]

int(exp(c*(b*x+a))*(coth(b*c*x+a*c)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {\cosh \left (b c x + a c\right ) - \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) + 1\right ) + \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) - 1\right ) + \sinh \left (b c x + a c\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*(coth(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)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {e^{\left (b c x + a c\right )}}{b c} - \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \]

[In]

integrate(exp(c*(b*x+a))*(coth(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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13 \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\frac {\frac {e^{\left (b c x + a c\right )}}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} + \frac {\log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right )}{\mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}}{b c} \]

[In]

integrate(exp(c*(b*x+a))*(coth(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)

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \sqrt {\coth ^2(a c+b c x)} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,\sqrt {{\mathrm {coth}\left (a\,c+b\,c\,x\right )}^2} \,d x \]

[In]

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

[Out]

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