Integrand size = 25, antiderivative size = 83 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {2 \text {arctanh}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}} \] Output:
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/(tanh(b*c*x+a*c)^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.61 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {\left (e^{c (a+b x)}-2 \text {arctanh}\left (e^{c (a+b x)}\right )\right ) \tanh (c (a+b x))}{b c \sqrt {\tanh ^2(c (a+b x))}} \] Input:
Integrate[E^(c*(a + b*x))/Sqrt[Tanh[a*c + b*c*x]^2],x]
Output:
((E^(c*(a + b*x)) - 2*ArcTanh[E^(c*(a + b*x))])*Tanh[c*(a + b*x)])/(b*c*Sq rt[Tanh[c*(a + b*x)]^2])
Time = 0.39 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.64, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7271, 2720, 25, 299, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth (a c+b x c)dx}{\sqrt {\tanh ^2(a c+b c x)}}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\tanh (a c+b c x) \int -\frac {1+e^{2 c (a+b x)}}{1-e^{2 c (a+b x)}}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\tanh (a c+b c x) \int \frac {1+e^{2 c (a+b x)}}{1-e^{2 c (a+b x)}}de^{c (a+b x)}}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\tanh (a c+b c x) \left (e^{c (a+b x)}-2 \int \frac {1}{1-e^{2 c (a+b x)}}de^{c (a+b x)}\right )}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (e^{c (a+b x)}-2 \text {arctanh}\left (e^{c (a+b x)}\right )\right ) \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}\) |
Input:
Int[E^(c*(a + b*x))/Sqrt[Tanh[a*c + b*c*x]^2],x]
Output:
((E^(c*(a + b*x)) - 2*ArcTanh[E^(c*(a + b*x))])*Tanh[a*c + b*c*x])/(b*c*Sq rt[Tanh[a*c + b*c*x]^2])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) && !(Eq Q[v, x] && EqQ[m, 1])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58
method | result | size |
default | \(\frac {\operatorname {csgn}\left (\tanh \left (c \left (b x +a \right )\right )\right ) \left (\sinh \left (b c x +a c \right )+\cosh \left (b c x +a c \right )-2 \,\operatorname {arctanh}\left ({\mathrm e}^{b c x +a c}\right )\right )}{c b}\) | \(48\) |
risch | \(\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) b c}+\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}+1\right )}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}\) | \(213\) |
Input:
int(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
csgn(tanh(c*(b*x+a)))/c/b*(sinh(b*c*x+a*c)+cosh(b*c*x+a*c)-2*arctanh(exp(b *c*x+a*c)))
Time = 0.09 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {\cosh \left (b c x + a c\right ) - \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) + 1\right ) + \log \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right ) - 1\right ) + \sinh \left (b c x + a c\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(1/2),x, algorithm="fricas")
Output:
(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)
\[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=e^{a c} \int \frac {e^{b c x}}{\sqrt {\tanh ^{2}{\left (a c + b c x \right )}}}\, dx \] Input:
integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)**2)**(1/2),x)
Output:
exp(a*c)*Integral(exp(b*c*x)/sqrt(tanh(a*c + b*c*x)**2), x)
Time = 0.14 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {e^{\left (b c x + a c\right )}}{b c} - \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(1/2),x, algorithm="maxima")
Output:
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)
Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} - \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} + \frac {\log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{b c} \] Input:
integrate(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(1/2),x, algorithm="giac")
Output:
e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) - log(e^(b*c*x + a*c) + 1)*sgn(e^(2*b*c*x + 2*a*c) - 1)/(b*c) + log(abs(e^(b*c*x + a*c) - 1))*sgn (e^(2*b*c*x + 2*a*c) - 1)/(b*c)
Timed out. \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{\sqrt {{\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2}} \,d x \] Input:
int(exp(c*(a + b*x))/(tanh(a*c + b*c*x)^2)^(1/2),x)
Output:
int(exp(c*(a + b*x))/(tanh(a*c + b*c*x)^2)^(1/2), x)
Time = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.70 \[ \int \frac {e^{c (a+b x)}}{\sqrt {\tanh ^2(a c+b c x)}} \, dx=\frac {e^{b c x +a c}-\mathrm {log}\left (e^{b c x +2 a c}+e^{a c}\right )+\mathrm {log}\left (e^{b c x +2 a c}-e^{a c}\right )}{b c} \] Input:
int(exp(c*(b*x+a))/(tanh(b*c*x+a*c)^2)^(1/2),x)
Output:
(e**(a*c + b*c*x) - log(e**(2*a*c + b*c*x) + e**(a*c)) + log(e**(2*a*c + b *c*x) - e**(a*c)))/(b*c)