Integrand size = 24, antiderivative size = 194 \[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\frac {2 e^{3 d+3 e x} F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )}+\frac {2 e^{3 d+3 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right )}{3 e+b c \log (F)}-\frac {2 b c e^{3 d+3 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right ) \log (F)}{e (3 e+b c \log (F))} \] Output:
2*exp(3*e*x+3*d)*F^(c*(b*x+a))/e/(1-exp(2*e*x+2*d))+2*exp(3*e*x+3*d)*F^(c* (b*x+a))*hypergeom([1, 3/2+1/2*b*c*ln(F)/e],[5/2+1/2*b*c*ln(F)/e],-exp(2*e *x+2*d))/(b*c*ln(F)+3*e)-2*b*c*exp(3*e*x+3*d)*F^(c*(b*x+a))*hypergeom([1, 3/2+1/2*b*c*ln(F)/e],[5/2+1/2*b*c*ln(F)/e],exp(2*e*x+2*d))*ln(F)/e/(b*c*ln (F)+3*e)
Time = 1.21 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.93 \[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (2 e e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right )+2 b c e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) \log (F)+F^{\frac {b c (d+e x)}{e}} \text {csch}(d+e x) (e+b c \log (F))\right )}{e (e+b c \log (F))} \] Input:
Integrate[F^(c*(a + b*x))*Csch[d + e*x]^2*Sech[d + e*x],x]
Output:
-((F^(c*(a - (b*d)/e))*(2*e*E^(((d + e*x)*(e + b*c*Log[F]))/e)*Hypergeomet ric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x)) ] + 2*b*c*E^(((d + e*x)*(e + b*c*Log[F]))/e)*Hypergeometric2F1[1, (e + b*c *Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, E^(2*(d + e*x))]*Log[F] + F^((b*c* (d + e*x))/e)*Csch[d + e*x]*(e + b*c*Log[F])))/(e*(e + b*c*Log[F])))
Time = 0.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.75, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6037, 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.
\(\displaystyle \int \text {csch}^2(d+e x) \text {sech}(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6037 |
\(\displaystyle \int \left (\frac {4 e^{3 d+3 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^2}-\frac {4 e^{3 d+3 e x} F^{a c+b c x}}{e^{4 (d+e x)}-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 e^{3 d+3 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} \left (\frac {b c \log (F)}{e}+3\right ),\frac {1}{4} \left (\frac {b c \log (F)}{e}+7\right ),e^{4 (d+e x)}\right )}{b c \log (F)+3 e}+\frac {4 e^{3 d+3 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+5\right ),e^{2 (d+e x)}\right )}{b c \log (F)+3 e}\) |
Input:
Int[F^(c*(a + b*x))*Csch[d + e*x]^2*Sech[d + e*x],x]
Output:
(4*E^(3*d + 3*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (3 + (b*c*Log[F])/ e)/4, (7 + (b*c*Log[F])/e)/4, E^(4*(d + e*x))])/(3*e + b*c*Log[F]) + (4*E^ (3*d + 3*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (3 + (b*c*Log[F])/e)/2, (5 + (b*c*Log[F])/e)/2, E^(2*(d + e*x))])/(3*e + b*c*Log[F])
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*(G_)[(d_.) + (e_.)*(x_)]^(m_.)*(H_)[( d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^(c*(a + b*x)), G[d + e*x]^m*H[d + e*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IGtQ[ m, 0] && IGtQ[n, 0] && HyperbolicQ[G] && HyperbolicQ[H]
\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right )^{2} \operatorname {sech}\left (e x +d \right )d x\]
Input:
int(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x)
Output:
int(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x)
\[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{2} \operatorname {sech}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*csch(e*x + d)^2*sech(e*x + d), x)
\[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}^{2}{\left (d + e x \right )} \operatorname {sech}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*csch(e*x+d)**2*sech(e*x+d),x)
Output:
Integral(F**(c*(a + b*x))*csch(d + e*x)**2*sech(d + e*x), x)
\[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{2} \operatorname {sech}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x, algorithm="maxima")
Output:
8*(2*F^(a*c)*e*e^(e*x + d) + (F^(a*c)*b*c*e^(3*d)*log(F) - 5*F^(a*c)*e*e^( 3*d))*e^(3*e*x))*F^(b*c*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*log(F) + 15*e^2 + ( b^2*c^2*e^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F) + 15*e^2*e^(6*d))*e^(6*e *x) - (b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e^(4*d)*log(F) + 15*e^2*e^(4*d)) *e^(4*e*x) - (b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d)*log(F) + 15*e^2*e ^(2*d))*e^(2*e*x)) + 8*integrate(-2*((3*F^(a*c)*b*c*e*e^(3*d)*log(F) - 13* F^(a*c)*e^2*e^(3*d))*e^(3*e*x) + (F^(a*c)*b*c*e*e^d*log(F) + F^(a*c)*e^2*e ^d)*e^(e*x))*F^(b*c*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*log(F) + 15*e^2 - (b^2* c^2*e^(10*d)*log(F)^2 - 8*b*c*e*e^(10*d)*log(F) + 15*e^2*e^(10*d))*e^(10*e *x) + (b^2*c^2*e^(8*d)*log(F)^2 - 8*b*c*e*e^(8*d)*log(F) + 15*e^2*e^(8*d)) *e^(8*e*x) + 2*(b^2*c^2*e^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F) + 15*e^2 *e^(6*d))*e^(6*e*x) - 2*(b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e^(4*d)*log(F) + 15*e^2*e^(4*d))*e^(4*e*x) - (b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d) *log(F) + 15*e^2*e^(2*d))*e^(2*e*x)), x)
\[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{2} \operatorname {sech}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*csch(e*x + d)^2*sech(e*x + d), x)
Timed out. \[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{\mathrm {cosh}\left (d+e\,x\right )\,{\mathrm {sinh}\left (d+e\,x\right )}^2} \,d x \] Input:
int(F^(c*(a + b*x))/(cosh(d + e*x)*sinh(d + e*x)^2),x)
Output:
int(F^(c*(a + b*x))/(cosh(d + e*x)*sinh(d + e*x)^2), x)
\[ \int F^{c (a+b x)} \text {csch}^2(d+e x) \text {sech}(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {csch}\left (e x +d \right )^{2} \mathrm {sech}\left (e x +d \right )d x \right ) \] Input:
int(F^(c*(b*x+a))*csch(e*x+d)^2*sech(e*x+d),x)
Output:
f**(a*c)*int(f**(b*c*x)*csch(d + e*x)**2*sech(d + e*x),x)