Integrand size = 22, antiderivative size = 61 \[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\frac {2 e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{e},2+\frac {b c \log (F)}{e},-e^{d+e x}\right )}{f (e+b c \log (F))} \] Output:
2*exp(e*x+d)*F^(c*(b*x+a))*hypergeom([2, 1+b*c*ln(F)/e],[2+b*c*ln(F)/e],-e xp(e*x+d))/f/(e+b*c*ln(F))
Time = 0.04 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\frac {2 e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{e},2+\frac {b c \log (F)}{e},-e^{d+e x}\right )}{e f+b c f \log (F)} \] Input:
Integrate[F^(c*(a + b*x))/(f + f*Cosh[d + e*x]),x]
Output:
(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^(d + e*x)])/(e*f + b*c*f*Log[F])
Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6020, 6015}
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.
\(\displaystyle \int \frac {F^{c (a+b x)}}{f \cosh (d+e x)+f} \, dx\) |
\(\Big \downarrow \) 6020 |
\(\displaystyle \frac {\int F^{c (a+b x)} \text {sech}^2\left (\frac {d}{2}+\frac {e x}{2}\right )dx}{2 f}\) |
\(\Big \downarrow \) 6015 |
\(\displaystyle \frac {2 e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{e}+1,\frac {b c \log (F)}{e}+2,-e^{d+e x}\right )}{f (b c \log (F)+e)}\) |
Input:
Int[F^(c*(a + b*x))/(f + f*Cosh[d + e*x]),x]
Output:
(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^(d + e*x)])/(f*(e + b*c*Log[F]))
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Sym bol] :> Simp[2^n*E^(n*(d + e*x))*(F^(c*(a + b*x))/(e*n + b*c*Log[F]))*Hyper geometric2F1[n, n/2 + b*c*(Log[F]/(2*e)), 1 + n/2 + b*c*(Log[F]/(2*e)), -E^ (2*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
Int[(Cosh[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_. )*(x_))), x_Symbol] :> Simp[2^n*g^n Int[F^(c*(a + b*x))*Cosh[d/2 + e*(x/2 )]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f - g, 0] && ILtQ[n, 0]
\[\int \frac {F^{c \left (b x +a \right )}}{f +f \cosh \left (e x +d \right )}d x\]
Input:
int(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x)
Output:
int(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x)
\[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cosh \left (e x + d\right ) + f} \,d x } \] Input:
integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)/(f*cosh(e*x + d) + f), x)
\[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\frac {\int \frac {F^{a c + b c x}}{\cosh {\left (d + e x \right )} + 1}\, dx}{f} \] Input:
integrate(F**(c*(b*x+a))/(f+f*cosh(e*x+d)),x)
Output:
Integral(F**(a*c + b*c*x)/(cosh(d + e*x) + 1), x)/f
\[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cosh \left (e x + d\right ) + f} \,d x } \] Input:
integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x, algorithm="maxima")
Output:
4*F^(a*c)*b*c*e*integrate(F^(b*c*x)/(b^2*c^2*f*log(F)^2 - 3*b*c*e*f*log(F) + 2*e^2*f + (b^2*c^2*f*e^(3*d)*log(F)^2 - 3*b*c*e*f*e^(3*d)*log(F) + 2*e^ 2*f*e^(3*d))*e^(3*e*x) + 3*(b^2*c^2*f*e^(2*d)*log(F)^2 - 3*b*c*e*f*e^(2*d) *log(F) + 2*e^2*f*e^(2*d))*e^(2*e*x) + 3*(b^2*c^2*f*e^d*log(F)^2 - 3*b*c*e *f*e^d*log(F) + 2*e^2*f*e^d)*e^(e*x)), x)*log(F) - 2*(2*F^(a*c)*e - (F^(a* c)*b*c*e^d*log(F) - 2*F^(a*c)*e*e^d)*e^(e*x))*F^(b*c*x)/(b^2*c^2*f*log(F)^ 2 - 3*b*c*e*f*log(F) + 2*e^2*f + (b^2*c^2*f*e^(2*d)*log(F)^2 - 3*b*c*e*f*e ^(2*d)*log(F) + 2*e^2*f*e^(2*d))*e^(2*e*x) + 2*(b^2*c^2*f*e^d*log(F)^2 - 3 *b*c*e*f*e^d*log(F) + 2*e^2*f*e^d)*e^(e*x))
\[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cosh \left (e x + d\right ) + f} \,d x } \] Input:
integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)/(f*cosh(e*x + d) + f), x)
Timed out. \[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{f+f\,\mathrm {cosh}\left (d+e\,x\right )} \,d x \] Input:
int(F^(c*(a + b*x))/(f + f*cosh(d + e*x)),x)
Output:
int(F^(c*(a + b*x))/(f + f*cosh(d + e*x)), x)
\[ \int \frac {F^{c (a+b x)}}{f+f \cosh (d+e x)} \, dx=\frac {f^{a c} \left (\int \frac {f^{b c x}}{\cosh \left (e x +d \right )+1}d x \right )}{f} \] Input:
int(F^(c*(b*x+a))/(f+f*cosh(e*x+d)),x)
Output:
(f**(a*c)*int(f**(b*c*x)/(cosh(d + e*x) + 1),x))/f