Integrand size = 25, antiderivative size = 162 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=-\frac {e^{-2 c (a+b x)} \text {sech}(a c+b c x)}{16 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {3 e^{2 c (a+b x)} \text {sech}(a c+b c x)}{16 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {e^{4 c (a+b x)} \text {sech}(a c+b c x)}{32 b c \sqrt {\text {sech}^2(a c+b c x)}}+\frac {3 x \text {sech}(a c+b c x)}{8 \sqrt {\text {sech}^2(a c+b c x)}} \] Output:
-1/16*sech(b*c*x+a*c)/b/c/exp(2*c*(b*x+a))/(sech(b*c*x+a*c)^2)^(1/2)+3/16* exp(2*c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+1/32*exp(4* c*(b*x+a))*sech(b*c*x+a*c)/b/c/(sech(b*c*x+a*c)^2)^(1/2)+3/8*x*sech(b*c*x+ a*c)/(sech(b*c*x+a*c)^2)^(1/2)
Time = 0.07 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.50 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=\frac {\left (-\frac {1}{16} e^{-2 c (a+b x)}+\frac {3}{16} e^{2 c (a+b x)}+\frac {1}{32} e^{4 c (a+b x)}+\frac {3 b c x}{8}\right ) \text {sech}^3(c (a+b x))}{b c \text {sech}^2(c (a+b x))^{3/2}} \] Input:
Integrate[E^(c*(a + b*x))/(Sech[a*c + b*c*x]^2)^(3/2),x]
Output:
((-1/16*1/E^(2*c*(a + b*x)) + (3*E^(2*c*(a + b*x)))/16 + E^(4*c*(a + b*x)) /32 + (3*b*c*x)/8)*Sech[c*(a + b*x)]^3)/(b*c*(Sech[c*(a + b*x)]^2)^(3/2))
Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.46, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {7271, 2720, 27, 243, 49, 2009}
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.
\(\displaystyle \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 7271 |
\(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{c (a+b x)} \cosh ^3(a c+b x c)dx}{\sqrt {\text {sech}^2(a c+b c x)}}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\text {sech}(a c+b c x) \int \frac {1}{8} e^{-3 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^3de^{c (a+b x)}}{b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{-3 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^3de^{c (a+b x)}}{8 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\text {sech}(a c+b c x) \int e^{-2 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^3de^{2 c (a+b x)}}{16 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\text {sech}(a c+b c x) \int \left (3+e^{-2 c (a+b x)}+3 e^{-c (a+b x)}+e^{2 c (a+b x)}\right )de^{2 c (a+b x)}}{16 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-e^{-c (a+b x)}+\frac {7}{2} e^{2 c (a+b x)}+3 \log \left (e^{2 c (a+b x)}\right )\right ) \text {sech}(a c+b c x)}{16 b c \sqrt {\text {sech}^2(a c+b c x)}}\) |
Input:
Int[E^(c*(a + b*x))/(Sech[a*c + b*c*x]^2)^(3/2),x]
Output:
((-E^(-(c*(a + b*x))) + (7*E^(2*c*(a + b*x)))/2 + 3*Log[E^(2*c*(a + b*x))] )*Sech[a*c + b*c*x])/(16*b*c*Sqrt[Sech[a*c + b*c*x]^2])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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])
Time = 0.57 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.33
method | result | size |
risch | \(\frac {3 x \,{\mathrm e}^{c \left (b x +a \right )}}{8 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {{\mathrm e}^{5 c \left (b x +a \right )}}{32 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}+\frac {3 \,{\mathrm e}^{3 c \left (b x +a \right )}}{16 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}-\frac {{\mathrm e}^{-c \left (b x +a \right )}}{16 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {{\mathrm e}^{2 c \left (b x +a \right )}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}}\) | \(216\) |
Input:
int(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
3/8*x/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/ 2)*exp(c*(b*x+a))+1/32/b/c/(1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2* exp(2*c*(b*x+a)))^(1/2)*exp(5*c*(b*x+a))+3/16/b/c/(1+exp(2*c*(b*x+a)))/(1/ (1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp(3*c*(b*x+a))-1/16/b/c/( 1+exp(2*c*(b*x+a)))/(1/(1+exp(2*c*(b*x+a)))^2*exp(2*c*(b*x+a)))^(1/2)*exp( -c*(b*x+a))
Time = 0.19 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.78 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=-\frac {\cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{2} - 3 \, \sinh \left (b c x + a c\right )^{3} - 6 \, {\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) + 3 \, {\left (4 \, b c x - 3 \, \cosh \left (b c x + a c\right )^{2} - 2\right )} \sinh \left (b c x + a c\right )}{32 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \] Input:
integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")
Output:
-1/32*(cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^2 - 3*s inh(b*c*x + a*c)^3 - 6*(2*b*c*x + 1)*cosh(b*c*x + a*c) + 3*(4*b*c*x - 3*co sh(b*c*x + a*c)^2 - 2)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sin h(b*c*x + a*c))
\[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=e^{a c} \int \frac {e^{b c x}}{\left (\operatorname {sech}^{2}{\left (a c + b c x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)**2)**(3/2),x)
Output:
exp(a*c)*Integral(exp(b*c*x)/(sech(a*c + b*c*x)**2)**(3/2), x)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.46 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=\frac {3 \, {\left (b c x + a c\right )}}{8 \, b c} + \frac {e^{\left (4 \, b c x + 4 \, a c\right )}}{32 \, b c} + \frac {3 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{16 \, b c} - \frac {e^{\left (-2 \, b c x - 2 \, a c\right )}}{16 \, b c} \] Input:
integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")
Output:
3/8*(b*c*x + a*c)/(b*c) + 1/32*e^(4*b*c*x + 4*a*c)/(b*c) + 3/16*e^(2*b*c*x + 2*a*c)/(b*c) - 1/16*e^(-2*b*c*x - 2*a*c)/(b*c)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.44 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=\frac {{\left (12 \, b c x e^{\left (2 \, a c\right )} - 2 \, {\left (3 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-2 \, b c x\right )} + e^{\left (4 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (2 \, b c x + 4 \, a c\right )}\right )} e^{\left (-2 \, a c\right )}}{32 \, b c} \] Input:
integrate(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")
Output:
1/32*(12*b*c*x*e^(2*a*c) - 2*(3*e^(2*b*c*x + 2*a*c) + 1)*e^(-2*b*c*x) + e^ (4*b*c*x + 6*a*c) + 6*e^(2*b*c*x + 4*a*c))*e^(-2*a*c)/(b*c)
Timed out. \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=\int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left (\frac {1}{{\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2}\right )}^{3/2}} \,d x \] Input:
int(exp(c*(a + b*x))/(1/cosh(a*c + b*c*x)^2)^(3/2),x)
Output:
int(exp(c*(a + b*x))/(1/cosh(a*c + b*c*x)^2)^(3/2), x)
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.41 \[ \int \frac {e^{c (a+b x)}}{\text {sech}^2(a c+b c x)^{3/2}} \, dx=\frac {e^{6 b c x +6 a c}+6 e^{4 b c x +4 a c}+12 e^{2 b c x +2 a c} b c x -2}{32 e^{2 b c x +2 a c} b c} \] Input:
int(exp(c*(b*x+a))/(sech(b*c*x+a*c)^2)^(3/2),x)
Output:
(e**(6*a*c + 6*b*c*x) + 6*e**(4*a*c + 4*b*c*x) + 12*e**(2*a*c + 2*b*c*x)*b *c*x - 2)/(32*e**(2*a*c + 2*b*c*x)*b*c)