\(\int F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x) \, dx\) [100]

Optimal result

Integrand size = 22, antiderivative size = 112 \[ \int F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x) \, dx=\frac {2 e^{d+e x} F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )}-\frac {2 b c e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right ) \log (F)}{e (e+b c \log (F))} \] Output:

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

Mathematica [A] (verified)

Time = 10.44 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.21 \[ \int F^{c (a+b x)} \coth (d+e x) \text {csch}(d+e x) \, dx=\frac {F^{c (a+b x)} \left (\operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},-e^{d+e x}\right )-\operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{e},1+\frac {b c \log (F)}{e},e^{d+e x}\right )-\frac {2 e^d}{\left (\left (1+e^d\right ) \cosh \left (\frac {e x}{2}\right )+\left (-1+e^d\right ) \sinh \left (\frac {e x}{2}\right )\right ) \left (\left (-1+e^d\right ) \cosh \left (\frac {e x}{2}\right )+\left (1+e^d\right ) \sinh \left (\frac {e x}{2}\right )\right )}\right )}{e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.21, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, 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.

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

\(\Big \downarrow \) 6037

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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