\(\int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx\) [327]

Optimal result

Integrand size = 18, antiderivative size = 122 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=-\frac {F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x)}{2 e}-\frac {b c F^{c (a+b x)} \text {csch}(d+e x) \log (F)}{2 e^2}+\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) (e-b c \log (F))}{e^2} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(281\) vs. \(2(122)=244\).

Time = 15.84 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.30 \[ \int F^{c (a+b x)} \text {csch}^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-e \text {csch}^2\left (\frac {1}{2} (d+e x)\right )-4 b c \text {csch}(d) \log (F)+\text {csch}(d) \left (-\frac {4 e^2}{b c \log (F)}+4 b c \log (F)\right )+\frac {4 \left (1-\left (1+e^d\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},-e^{d+e x}\right )\right ) \left (e^2-b^2 c^2 \log ^2(F)\right )}{b c \left (1+e^d\right ) \log (F)}+\frac {4 \left (1+\left (-1+e^d\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},e^{d+e x}\right )\right ) \left (e^2-b^2 c^2 \log ^2(F)\right )}{b c \left (-1+e^d\right ) \log (F)}-e \text {sech}^2\left (\frac {1}{2} (d+e x)\right )+2 b c \text {csch}\left (\frac {d}{2}\right ) \text {csch}\left (\frac {1}{2} (d+e x)\right ) \log (F) \sinh \left (\frac {e x}{2}\right )+2 b c \log (F) \text {sech}\left (\frac {d}{2}\right ) \text {sech}\left (\frac {1}{2} (d+e x)\right ) \sinh \left (\frac {e x}{2}\right )\right )}{8 e^2} \] Input:

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

Output:

(F^(c*(a + b*x))*(-(e*Csch[(d + e*x)/2]^2) - 4*b*c*Csch[d]*Log[F] + Csch[d 
]*((-4*e^2)/(b*c*Log[F]) + 4*b*c*Log[F]) + (4*(1 - (1 + E^d)*Hypergeometri 
c2F1[1, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, -E^(d + e*x)])*(e^2 - b^2*c^2* 
Log[F]^2))/(b*c*(1 + E^d)*Log[F]) + (4*(1 + (-1 + E^d)*Hypergeometric2F1[1 
, (b*c*Log[F])/e, 1 + (b*c*Log[F])/e, E^(d + e*x)])*(e^2 - b^2*c^2*Log[F]^ 
2))/(b*c*(-1 + E^d)*Log[F]) - e*Sech[(d + e*x)/2]^2 + 2*b*c*Csch[d/2]*Csch 
[(d + e*x)/2]*Log[F]*Sinh[(e*x)/2] + 2*b*c*Log[F]*Sech[d/2]*Sech[(d + e*x) 
/2]*Sinh[(e*x)/2]))/(8*e^2)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.12, 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.

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

\(\Big \downarrow \) 6014

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

\(\Big \downarrow \) 6016

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

Input:

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

Output:

-1/2*(F^(c*(a + b*x))*Coth[d + e*x]*Csch[d + e*x])/e - (b*c*F^(c*(a + b*x) 
)*Csch[d + e*x]*Log[F])/(2*e^2) + (E^(d + e*x)*F^(c*(a + b*x))*Hypergeomet 
ric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, E^(2*(d + e*x))] 
*(1 - (b^2*c^2*Log[F]^2)/e^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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

48*(F^(a*c)*b*c*e*e^d*log(F) + F^(a*c)*e^2*e^d)*integrate(e^(b*c*x*log(F) 
+ e*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*log(F) + 15*e^2 + (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) - 4*(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) + 6*(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 
) - 4*(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) - 8*(6*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*l 
og(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) + 3*(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) - 3*(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))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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