\(\int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx\) [328]

Optimal result

Integrand size = 18, antiderivative size = 131 \[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=-\frac {F^{c (a+b x)} \coth (d+e x) \text {csch}^2(d+e x)}{3 e}-\frac {b c F^{c (a+b x)} \text {csch}^2(d+e x) \log (F)}{6 e^2}-\frac {2 e^{2 (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1+\frac {b c \log (F)}{2 e},2+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right ) (2 e-b c \log (F))}{3 e^2} \] Output:

-1/3*F^(c*(b*x+a))*coth(e*x+d)*csch(e*x+d)^2/e-1/6*b*c*F^(c*(b*x+a))*csch( 
e*x+d)^2*ln(F)/e^2-2/3*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))/e^2
 

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.24 \[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-e \text {csch}^2(d+e x) (2 e \coth (d)+b c \log (F))-\frac {2 \left (1+\left (-1+e^{2 d}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right )\right ) \left (-4 e^2+b^2 c^2 \log ^2(F)\right )}{-1+e^{2 d}}+2 e^2 \text {csch}(d) \text {csch}^3(d+e x) \sinh (e x)-\text {csch}(d) \text {csch}(d+e x) \left (4 e^2-b^2 c^2 \log ^2(F)\right ) \sinh (e x)\right )}{6 e^3} \] Input:

Integrate[F^(c*(a + b*x))*Csch[d + e*x]^4,x]
 

Output:

(F^(c*(a + b*x))*(-(e*Csch[d + e*x]^2*(2*e*Coth[d] + b*c*Log[F])) - (2*(1 
+ (-1 + E^(2*d))*Hypergeometric2F1[1, (b*c*Log[F])/(2*e), 1 + (b*c*Log[F]) 
/(2*e), E^(2*(d + e*x))])*(-4*e^2 + b^2*c^2*Log[F]^2))/(-1 + E^(2*d)) + 2* 
e^2*Csch[d]*Csch[d + e*x]^3*Sinh[e*x] - Csch[d]*Csch[d + e*x]*(4*e^2 - b^2 
*c^2*Log[F]^2)*Sinh[e*x]))/(6*e^3)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6014, 6016}

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.

\(\displaystyle \int \text {csch}^4(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 6014

\(\displaystyle -\frac {1}{6} \left (4-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {csch}^2(d+e x)dx-\frac {b c \log (F) \text {csch}^2(d+e x) F^{c (a+b x)}}{6 e^2}-\frac {\coth (d+e x) \text {csch}^2(d+e x) F^{c (a+b x)}}{3 e}\)

\(\Big \downarrow \) 6016

\(\displaystyle -\frac {2 e^{2 (d+e x)} F^{c (a+b x)} \left (4-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{2 e}+1,\frac {b c \log (F)}{2 e}+2,e^{2 (d+e x)}\right )}{3 (b c \log (F)+2 e)}-\frac {b c \log (F) \text {csch}^2(d+e x) F^{c (a+b x)}}{6 e^2}-\frac {\coth (d+e x) \text {csch}^2(d+e x) F^{c (a+b x)}}{3 e}\)

Input:

Int[F^(c*(a + b*x))*Csch[d + e*x]^4,x]
 

Output:

-1/3*(F^(c*(a + b*x))*Coth[d + e*x]*Csch[d + e*x]^2)/e - (b*c*F^(c*(a + b* 
x))*Csch[d + e*x]^2*Log[F])/(6*e^2) - (2*E^(2*(d + e*x))*F^(c*(a + b*x))*H 
ypergeometric2F1[2, 1 + (b*c*Log[F])/(2*e), 2 + (b*c*Log[F])/(2*e), E^(2*( 
d + e*x))]*(4 - (b^2*c^2*Log[F]^2)/e^2))/(3*(2*e + b*c*Log[F]))
 

Defintions of rubi rules used

rule 6014

Int[Csch[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symb 
ol] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Csch[d + e*x]^(n - 2)/(e^2*(n - 
1)*(n - 2))), x] + (-Simp[F^(c*(a + b*x))*Csch[d + e*x]^(n - 1)*(Cosh[d + e 
*x]/(e*(n - 1))), x] - Simp[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1) 
*(n - 2))   Int[F^(c*(a + b*x))*Csch[d + e*x]^(n - 2), x], x]) /; FreeQ[{F, 
 a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ[n, 1 
] && NeQ[n, 2]
 

rule 6016

Int[Csch[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Sym 
bol] :> Simp[(-2)^n*E^(n*(d + e*x))*(F^(c*(a + b*x))/(e*n + b*c*Log[F]))*Hy 
pergeometric2F1[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]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right )^{4}d x\]

Input:

int(F^(c*(b*x+a))*csch(e*x+d)^4,x)
 

Output:

int(F^(c*(b*x+a))*csch(e*x+d)^4,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^4,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*csch(e*x + d)^4, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}^{4}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*csch(e*x+d)**4,x)
 

Output:

Integral(F**(c*(a + b*x))*csch(d + e*x)**4, x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^4,x, algorithm="maxima")
 

Output:

128*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2*F^(a*c)*b*c*e^2*log(F))*integrate(-F^( 
b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2*e*log(F)^2 + 104*b*c*e^2*log(F) - 19 
2*e^3 - (b^3*c^3*e^(10*d)*log(F)^3 - 18*b^2*c^2*e*e^(10*d)*log(F)^2 + 104* 
b*c*e^2*e^(10*d)*log(F) - 192*e^3*e^(10*d))*e^(10*e*x) + 5*(b^3*c^3*e^(8*d 
)*log(F)^3 - 18*b^2*c^2*e*e^(8*d)*log(F)^2 + 104*b*c*e^2*e^(8*d)*log(F) - 
192*e^3*e^(8*d))*e^(8*e*x) - 10*(b^3*c^3*e^(6*d)*log(F)^3 - 18*b^2*c^2*e*e 
^(6*d)*log(F)^2 + 104*b*c*e^2*e^(6*d)*log(F) - 192*e^3*e^(6*d))*e^(6*e*x) 
+ 10*(b^3*c^3*e^(4*d)*log(F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e 
^2*e^(4*d)*log(F) - 192*e^3*e^(4*d))*e^(4*e*x) - 5*(b^3*c^3*e^(2*d)*log(F) 
^3 - 18*b^2*c^2*e*e^(2*d)*log(F)^2 + 104*b*c*e^2*e^(2*d)*log(F) - 192*e^3* 
e^(2*d))*e^(2*e*x)), x) + 16*(8*F^(a*c)*b*c*e*log(F) + 16*F^(a*c)*e^2 + (F 
^(a*c)*b^2*c^2*e^(4*d)*log(F)^2 - 14*F^(a*c)*b*c*e*e^(4*d)*log(F) + 48*F^( 
a*c)*e^2*e^(4*d))*e^(4*e*x) + 8*(F^(a*c)*b*c*e*e^(2*d)*log(F) - 8*F^(a*c)* 
e^2*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2*e*log(F)^ 
2 + 104*b*c*e^2*log(F) - 192*e^3 + (b^3*c^3*e^(8*d)*log(F)^3 - 18*b^2*c^2* 
e*e^(8*d)*log(F)^2 + 104*b*c*e^2*e^(8*d)*log(F) - 192*e^3*e^(8*d))*e^(8*e* 
x) - 4*(b^3*c^3*e^(6*d)*log(F)^3 - 18*b^2*c^2*e*e^(6*d)*log(F)^2 + 104*b*c 
*e^2*e^(6*d)*log(F) - 192*e^3*e^(6*d))*e^(6*e*x) + 6*(b^3*c^3*e^(4*d)*log( 
F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e^2*e^(4*d)*log(F) - 192*e^ 
3*e^(4*d))*e^(4*e*x) - 4*(b^3*c^3*e^(2*d)*log(F)^3 - 18*b^2*c^2*e*e^(2*...
 

Giac [F]

\[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right )^{4} \,d x } \] Input:

integrate(F^(c*(b*x+a))*csch(e*x+d)^4,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*csch(e*x + d)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {sinh}\left (d+e\,x\right )}^4} \,d x \] Input:

int(F^(c*(a + b*x))/sinh(d + e*x)^4,x)
 

Output:

int(F^(c*(a + b*x))/sinh(d + e*x)^4, x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \text {csch}^4(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {csch}\left (e x +d \right )^{4}d x \right ) \] Input:

int(F^(c*(b*x+a))*csch(e*x+d)^4,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*csch(d + e*x)**4,x)