Integrand size = 18, antiderivative size = 74 \[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\frac {4 e^{2 d+2 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (4+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right )}{2 e+b c \log (F)} \] Output:
4*exp(2*e*x+2*d)*F^(c*(b*x+a))*hypergeom([2, 1+1/2*b*c*ln(F)/e],[2+1/2*b*c *ln(F)/e],-exp(2*e*x+2*d))/(2*e+b*c*ln(F))
Time = 0.01 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95 \[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\frac {4 e^{2 (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{2 e},2+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right )}{2 e+b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*Sech[d + e*x]^2,x]
Output:
(4*E^(2*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/( 2*e), 2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])/(2*e + b*c*Log[F])
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.95, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {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.
\(\displaystyle \int \text {sech}^2(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6015 |
\(\displaystyle \frac {4 e^{2 (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{2 e}+1,\frac {b c \log (F)}{2 e}+2,-e^{2 (d+e x)}\right )}{b c \log (F)+2 e}\) |
Input:
Int[F^(c*(a + b*x))*Sech[d + e*x]^2,x]
Output:
(4*E^(2*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/( 2*e), 2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])/(2*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]
\[\int F^{c \left (b x +a \right )} \operatorname {sech}\left (e x +d \right )^{2}d x\]
Input:
int(F^(c*(b*x+a))*sech(e*x+d)^2,x)
Output:
int(F^(c*(b*x+a))*sech(e*x+d)^2,x)
\[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sech(e*x+d)^2,x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*sech(e*x + d)^2, x)
\[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {sech}^{2}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*sech(e*x+d)**2,x)
Output:
Integral(F**(c*(a + b*x))*sech(d + e*x)**2, x)
\[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sech(e*x+d)^2,x, algorithm="maxima")
Output:
16*F^(a*c)*b*c*e*integrate(F^(b*c*x)/(b^2*c^2*log(F)^2 - 6*b*c*e*log(F) + 8*e^2 + (b^2*c^2*e^(6*d)*log(F)^2 - 6*b*c*e*e^(6*d)*log(F) + 8*e^2*e^(6*d) )*e^(6*e*x) + 3*(b^2*c^2*e^(4*d)*log(F)^2 - 6*b*c*e*e^(4*d)*log(F) + 8*e^2 *e^(4*d))*e^(4*e*x) + 3*(b^2*c^2*e^(2*d)*log(F)^2 - 6*b*c*e*e^(2*d)*log(F) + 8*e^2*e^(2*d))*e^(2*e*x)), x)*log(F) - 4*(4*F^(a*c)*e - (F^(a*c)*b*c*e^ (2*d)*log(F) - 4*F^(a*c)*e*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^2*c^2*log(F)^2 - 6*b*c*e*log(F) + 8*e^2 + (b^2*c^2*e^(4*d)*log(F)^2 - 6*b*c*e*e^(4*d)*lo g(F) + 8*e^2*e^(4*d))*e^(4*e*x) + 2*(b^2*c^2*e^(2*d)*log(F)^2 - 6*b*c*e*e^ (2*d)*log(F) + 8*e^2*e^(2*d))*e^(2*e*x))
\[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sech(e*x+d)^2,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*sech(e*x + d)^2, x)
Timed out. \[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^2} \,d x \] Input:
int(F^(c*(a + b*x))/cosh(d + e*x)^2,x)
Output:
int(F^(c*(a + b*x))/cosh(d + e*x)^2, x)
\[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {sech}\left (e x +d \right )^{2}d x \right ) \] Input:
int(F^(c*(b*x+a))*sech(e*x+d)^2,x)
Output:
f**(a*c)*int(f**(b*c*x)*sech(d + e*x)**2,x)