\(\int F^{c (a+b x)} \coth ^2(d+e x) \, dx\) [71]

Optimal result

Integrand size = 18, antiderivative size = 101 \[ \int F^{c (a+b x)} \coth ^2(d+e x) \, dx=\frac {2 F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )}-\frac {2 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 d+2 e x}\right )}{e}+\frac {F^{c (a+b x)}}{b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.91 \[ \int F^{c (a+b x)} \coth ^2(d+e x) \, dx=F^{c (a+b x)} \left (\frac {2}{e-e e^{2 d}}-\frac {2 \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right )}{e}+\frac {1}{b c \log (F)}+\frac {\text {csch}(d) \text {csch}(d+e x) \sinh (e x)}{e}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.32, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6008, 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.

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

\(\Big \downarrow \) 6008

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 6008

Int[Coth[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Sym 
bol] :> Int[ExpandIntegrand[F^(c*(a + b*x))*((1 + E^(2*(d + e*x)))^n/(-1 + 
E^(2*(d + e*x)))^n), x], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

16*F^(a*c)*b*c*e*integrate(-F^(b*c*x)/(b^2*c^2*log(F)^2 - 6*b*c*e*log(F) + 
 8*e^2 - (b^2*c^2*e^(6*d)*log(F)^2 - 6*b*c*e*e^(6*d)*log(F) + 8*e^2*e^(6*d 
))*e^(6*e*x) + 3*(b^2*c^2*e^(4*d)*log(F)^2 - 6*b*c*e*e^(4*d)*log(F) + 8*e^ 
2*e^(4*d))*e^(4*e*x) - 3*(b^2*c^2*e^(2*d)*log(F)^2 - 6*b*c*e*e^(2*d)*log(F 
) + 8*e^2*e^(2*d))*e^(2*e*x)), x)*log(F) + (F^(a*c)*b^2*c^2*log(F)^2 + 10* 
F^(a*c)*b*c*e*log(F) + 8*F^(a*c)*e^2 + (F^(a*c)*b^2*c^2*e^(4*d)*log(F)^2 - 
 6*F^(a*c)*b*c*e*e^(4*d)*log(F) + 8*F^(a*c)*e^2*e^(4*d))*e^(4*e*x) + 2*(F^ 
(a*c)*b^2*c^2*e^(2*d)*log(F)^2 - 2*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 - 6*b^2*c^2*e*log(F) 
^2 + 8*b*c*e^2*log(F) + (b^3*c^3*e^(4*d)*log(F)^3 - 6*b^2*c^2*e*e^(4*d)*lo 
g(F)^2 + 8*b*c*e^2*e^(4*d)*log(F))*e^(4*e*x) - 2*(b^3*c^3*e^(2*d)*log(F)^3 
 - 6*b^2*c^2*e*e^(2*d)*log(F)^2 + 8*b*c*e^2*e^(2*d)*log(F))*e^(2*e*x))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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