\(\int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx\) [127]

Optimal result

Integrand size = 24, antiderivative size = 192 \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=-\frac {2 e^{3 d+3 e x} F^{c (a+b x)}}{e \left (1+e^{2 d+2 e x}\right )}-\frac {2 e^{3 d+3 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right )}{3 e+b c \log (F)}+\frac {2 b c e^{3 d+3 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) \log (F)}{e (3 e+b c \log (F))} \] 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*l 
n(F)+3*e)
 

Mathematica [A] (verified)

Time = 0.93 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.93 \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (-2 e e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right )-2 b c e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) \log (F)+F^{\frac {b c (d+e x)}{e}} (e+b c \log (F)) \text {sech}(d+e x)\right )}{e (e+b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Csch[d + e*x]*Sech[d + e*x]^2,x]
 

Output:

(F^(c*(a - (b*d)/e))*(-2*e*E^(((d + e*x)*(e + b*c*Log[F]))/e)*Hypergeometr 
ic2F1[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*L 
og[F])/(2*e), (3 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x))]*Log[F] + F^((b*c*( 
d + e*x))/e)*(e + b*c*Log[F])*Sech[d + e*x]))/(e*(e + b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.77, 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.

\(\displaystyle \int \text {csch}(d+e x) \text {sech}^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 6037

\(\displaystyle \int \left (\frac {4 e^{3 d+3 e x} F^{a c+b c x}}{e^{4 (d+e x)}-1}-\frac {4 e^{3 d+3 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}+1\right )^2}\right )dx\)

\(\Big \downarrow \) 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]*Sech[d + e*x]^2,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])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6037

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]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right ) \operatorname {sech}\left (e x +d \right )^{2}d x\]

Input:

int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,x)
 

Output:

int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csch(e*x + d)*sech(e*x + d)^2, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}{\left (d + e x \right )} \operatorname {sech}^{2}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)**2,x)
 

Output:

Integral(F**(c*(a + b*x))*csch(d + e*x)*sech(d + e*x)**2, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,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)
 

Giac [F]

\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csch(e*x + d)*sech(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^2\,\mathrm {sinh}\left (d+e\,x\right )} \,d x \] Input:

int(F^(c*(a + b*x))/(cosh(d + e*x)^2*sinh(d + e*x)),x)
 

Output:

int(F^(c*(a + b*x))/(cosh(d + e*x)^2*sinh(d + e*x)), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {csch}\left (e x +d \right ) \mathrm {sech}\left (e x +d \right )^{2}d x \right ) \] Input:

int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csch(d + e*x)*sech(d + e*x)**2,x)