\(\int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx\) [129]

Optimal result

Integrand size = 26, antiderivative size = 301 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=-\frac {4 e^{5 d+5 e x} F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )^2 \left (1+e^{2 d+2 e x}\right )}+\frac {2 b c e^{5 d+5 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (7+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) \log (F)}{e (5 e+b c \log (F))}+\frac {e^{5 d+5 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (5+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (7+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right ) \left (3 e^2-b^2 c^2 \log ^2(F)\right )}{e^2 (5 e+b c \log (F))}-\frac {e^{5 d+5 e x} F^{c (a+b x)} \left (3-\frac {b c \log (F)}{e}+e^{2 d+2 e x} \left (1-\frac {b c \log (F)}{e}\right )\right )}{e \left (1-e^{4 d+4 e x}\right )} \] Output:

-4*exp(5*e*x+5*d)*F^(c*(b*x+a))/e/(1-exp(2*e*x+2*d))^2/(1+exp(2*e*x+2*d))+ 
2*b*c*exp(5*e*x+5*d)*F^(c*(b*x+a))*hypergeom([1, 5/2+1/2*b*c*ln(F)/e],[7/2 
+1/2*b*c*ln(F)/e],-exp(2*e*x+2*d))*ln(F)/e/(b*c*ln(F)+5*e)+exp(5*e*x+5*d)* 
F^(c*(b*x+a))*hypergeom([1, 5/2+1/2*b*c*ln(F)/e],[7/2+1/2*b*c*ln(F)/e],exp 
(2*e*x+2*d))*(3*e^2-b^2*c^2*ln(F)^2)/e^2/(b*c*ln(F)+5*e)-exp(5*e*x+5*d)*F^ 
(c*(b*x+a))*(3-b*c*ln(F)/e+exp(2*e*x+2*d)*(1-b*c*ln(F)/e))/e/(1-exp(4*e*x+ 
4*d))
 

Mathematica [A] (verified)

Time = 3.16 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.79 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (8 b c e e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) \log (F)+4 e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) \left (3 e^2-b^2 c^2 \log ^2(F)\right )-F^{\frac {b c (d+e x)}{e}} \text {csch}^2(d+e x) (e+b c \log (F)) \text {sech}(d+e x) (-e+3 e \cosh (2 (d+e x))+b c \log (F) \sinh (2 (d+e x)))\right )}{4 e^2 (e+b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Csch[d + e*x]^3*Sech[d + e*x]^2,x]
 

Output:

(F^(c*(a - (b*d)/e))*(8*b*c*e*E^(((d + e*x)*(e + b*c*Log[F]))/e)*Hypergeom 
etric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x 
))]*Log[F] + 4*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))]*(3*e^2 - b^2 
*c^2*Log[F]^2) - F^((b*c*(d + e*x))/e)*Csch[d + e*x]^2*(e + b*c*Log[F])*Se 
ch[d + e*x]*(-e + 3*e*Cosh[2*(d + e*x)] + b*c*Log[F]*Sinh[2*(d + e*x)])))/ 
(4*e^2*(e + b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, 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.

\(\displaystyle \int \text {csch}^3(d+e x) \text {sech}^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 6037

\(\displaystyle \int \left (\frac {12 e^{5 d+5 e x} F^{a c+b c x}}{e^{4 (d+e x)}-1}-\frac {8 e^{5 d+5 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^2}-\frac {4 e^{5 d+5 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}+1\right )^2}+\frac {8 e^{5 d+5 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {12 e^{5 d+5 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} \left (\frac {b c \log (F)}{e}+5\right ),\frac {1}{4} \left (\frac {b c \log (F)}{e}+9\right ),e^{4 (d+e x)}\right )}{b c \log (F)+5 e}-\frac {4 e^{5 d+5 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (\frac {b c \log (F)}{e}+5\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+7\right ),-e^{2 (d+e x)}\right )}{b c \log (F)+5 e}-\frac {8 e^{5 d+5 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (\frac {b c \log (F)}{e}+5\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+7\right ),e^{2 (d+e x)}\right )}{b c \log (F)+5 e}-\frac {8 e^{5 d+5 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (\frac {b c \log (F)}{e}+5\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+7\right ),e^{2 (d+e x)}\right )}{b c \log (F)+5 e}\)

Input:

Int[F^(c*(a + b*x))*Csch[d + e*x]^3*Sech[d + e*x]^2,x]
 

Output:

(-12*E^(5*d + 5*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (5 + (b*c*Log[F] 
)/e)/4, (9 + (b*c*Log[F])/e)/4, E^(4*(d + e*x))])/(5*e + b*c*Log[F]) - (4* 
E^(5*d + 5*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (5 + (b*c*Log[F])/e)/ 
2, (7 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x))])/(5*e + b*c*Log[F]) - (8*E^(5 
*d + 5*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[2, (5 + (b*c*Log[F])/e)/2, ( 
7 + (b*c*Log[F])/e)/2, E^(2*(d + e*x))])/(5*e + b*c*Log[F]) - (8*E^(5*d + 
5*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[3, (5 + (b*c*Log[F])/e)/2, (7 + ( 
b*c*Log[F])/e)/2, E^(2*(d + e*x))])/(5*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]

\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right )^{3} \operatorname {sech}\left (e x +d \right )^{2}d x\]

Input:

int(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x)
 

Output:

int(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csch(e*x + d)^3*sech(e*x + d)^2, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}^{3}{\left (d + e x \right )} \operatorname {sech}^{2}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*csch(e*x+d)**3*sech(e*x+d)**2,x)
 

Output:

Integral(F**(c*(a + b*x))*csch(d + e*x)**3*sech(d + e*x)**2, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x, algorithm="maxima")
 

Output:

-32*((F^(a*c)*b^2*c^2*e^(5*d)*log(F)^2 - 16*F^(a*c)*b*c*e*e^(5*d)*log(F) + 
 63*F^(a*c)*e^2*e^(5*d))*e^(5*e*x) + 2*(F^(a*c)*b*c*e*e^(3*d)*log(F) - 9*F 
^(a*c)*e^2*e^(3*d))*e^(3*e*x) + 2*(5*F^(a*c)*b*c*e*e^d*log(F) - 33*F^(a*c) 
*e^2*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 21*b^2*c^2*e*log(F)^2 + 1 
43*b*c*e^2*log(F) - 315*e^3 - (b^3*c^3*e^(10*d)*log(F)^3 - 21*b^2*c^2*e*e^ 
(10*d)*log(F)^2 + 143*b*c*e^2*e^(10*d)*log(F) - 315*e^3*e^(10*d))*e^(10*e* 
x) + (b^3*c^3*e^(8*d)*log(F)^3 - 21*b^2*c^2*e*e^(8*d)*log(F)^2 + 143*b*c*e 
^2*e^(8*d)*log(F) - 315*e^3*e^(8*d))*e^(8*e*x) + 2*(b^3*c^3*e^(6*d)*log(F) 
^3 - 21*b^2*c^2*e*e^(6*d)*log(F)^2 + 143*b*c*e^2*e^(6*d)*log(F) - 315*e^3* 
e^(6*d))*e^(6*e*x) - 2*(b^3*c^3*e^(4*d)*log(F)^3 - 21*b^2*c^2*e*e^(4*d)*lo 
g(F)^2 + 143*b*c*e^2*e^(4*d)*log(F) - 315*e^3*e^(4*d))*e^(4*e*x) - (b^3*c^ 
3*e^(2*d)*log(F)^3 - 21*b^2*c^2*e*e^(2*d)*log(F)^2 + 143*b*c*e^2*e^(2*d)*l 
og(F) - 315*e^3*e^(2*d))*e^(2*e*x)) + 32*integrate(2*((F^(a*c)*b^2*c^2*e*e 
^(3*d)*log(F)^2 + 4*F^(a*c)*b*c*e^2*e^(3*d)*log(F) - 93*F^(a*c)*e^3*e^(3*d 
))*e^(3*e*x) + (5*F^(a*c)*b^2*c^2*e*e^d*log(F)^2 - 28*F^(a*c)*b*c*e^2*e^d* 
log(F) - 33*F^(a*c)*e^3*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 21*b^2 
*c^2*e*log(F)^2 + 143*b*c*e^2*log(F) - 315*e^3 + (b^3*c^3*e^(14*d)*log(F)^ 
3 - 21*b^2*c^2*e*e^(14*d)*log(F)^2 + 143*b*c*e^2*e^(14*d)*log(F) - 315*e^3 
*e^(14*d))*e^(14*e*x) - (b^3*c^3*e^(12*d)*log(F)^3 - 21*b^2*c^2*e*e^(12*d) 
*log(F)^2 + 143*b*c*e^2*e^(12*d)*log(F) - 315*e^3*e^(12*d))*e^(12*e*x) ...
 

Giac [F]

\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{3} \operatorname {sech}\left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csch(e*x + d)^3*sech(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^2\,{\mathrm {sinh}\left (d+e\,x\right )}^3} \,d x \] Input:

int(F^(c*(a + b*x))/(cosh(d + e*x)^2*sinh(d + e*x)^3),x)
 

Output:

int(F^(c*(a + b*x))/(cosh(d + e*x)^2*sinh(d + e*x)^3), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \text {sech}^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {csch}\left (e x +d \right )^{3} \mathrm {sech}\left (e x +d \right )^{2}d x \right ) \] Input:

int(F^(c*(b*x+a))*csch(e*x+d)^3*sech(e*x+d)^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csch(d + e*x)**3*sech(d + e*x)**2,x)