\(\int F^{c (a+b x)} \coth ^3(d+e x) \, dx\) [72]

Optimal result

Integrand size = 18, antiderivative size = 157 \[ \int F^{c (a+b x)} \coth ^3(d+e x) \, dx=-\frac {2 F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )^2}+\frac {F^{c (a+b x)}}{b c \log (F)}+\frac {F^{c (a+b x)} (2 e+b c \log (F))}{e^2 \left (1-e^{2 d+2 e x}\right )}-F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 d+2 e x}\right ) \left (\frac {2}{b c \log (F)}+\frac {b c \log (F)}{e^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94 \[ \int F^{c (a+b x)} \coth ^3(d+e x) \, dx=\frac {1}{2} F^{c (a+b x)} \left (-\frac {\text {csch}^2(d+e x)}{e}+\frac {2 \coth (d)}{b c \log (F)}-\frac {2 \left (1+\left (-1+e^{2 d}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right )\right ) \left (2 e^2+b^2 c^2 \log ^2(F)\right )}{b c e^2 \left (-1+e^{2 d}\right ) \log (F)}+\frac {b c \text {csch}(d) \text {csch}(d+e x) \log (F) \sinh (e x)}{e^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.20, 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.

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

\(\Big \downarrow \) 6008

\(\displaystyle \int \left (\frac {6 F^{c (a+b x)}}{e^{2 (d+e x)}-1}+\frac {12 F^{c (a+b x)}}{\left (e^{2 (d+e x)}-1\right )^2}+\frac {8 F^{c (a+b x)}}{\left (e^{2 (d+e x)}-1\right )^3}+F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}+\frac {12 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}-\frac {8 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}+\frac {F^{c (a+b x)}}{b c \log (F)}\)

Input:

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

Output:

F^(c*(a + b*x))/(b*c*Log[F]) - (6*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]) + 
(12*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]) - (8*F^(c*(a + b*x))*Hypergeomet 
ric2F1[3, (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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

48*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2*F^(a*c)*e^3)*integrate(F^(b*c*x)/(b^3*c 
^3*log(F)^3 - 12*b^2*c^2*e*log(F)^2 + 44*b*c*e^2*log(F) - 48*e^3 + (b^3*c^ 
3*e^(8*d)*log(F)^3 - 12*b^2*c^2*e*e^(8*d)*log(F)^2 + 44*b*c*e^2*e^(8*d)*lo 
g(F) - 48*e^3*e^(8*d))*e^(8*e*x) - 4*(b^3*c^3*e^(6*d)*log(F)^3 - 12*b^2*c^ 
2*e*e^(6*d)*log(F)^2 + 44*b*c*e^2*e^(6*d)*log(F) - 48*e^3*e^(6*d))*e^(6*e* 
x) + 6*(b^3*c^3*e^(4*d)*log(F)^3 - 12*b^2*c^2*e*e^(4*d)*log(F)^2 + 44*b*c* 
e^2*e^(4*d)*log(F) - 48*e^3*e^(4*d))*e^(4*e*x) - 4*(b^3*c^3*e^(2*d)*log(F) 
^3 - 12*b^2*c^2*e*e^(2*d)*log(F)^2 + 44*b*c*e^2*e^(2*d)*log(F) - 48*e^3*e^ 
(2*d))*e^(2*e*x)), x) - (F^(a*c)*b^3*c^3*log(F)^3 + 36*F^(a*c)*b^2*c^2*e*l 
og(F)^2 + 44*F^(a*c)*b*c*e^2*log(F) + 48*F^(a*c)*e^3 + (F^(a*c)*b^3*c^3*e^ 
(6*d)*log(F)^3 - 12*F^(a*c)*b^2*c^2*e*e^(6*d)*log(F)^2 + 44*F^(a*c)*b*c*e^ 
2*e^(6*d)*log(F) - 48*F^(a*c)*e^3*e^(6*d))*e^(6*e*x) + 3*(F^(a*c)*b^3*c^3* 
e^(4*d)*log(F)^3 - 8*F^(a*c)*b^2*c^2*e*e^(4*d)*log(F)^2 + 4*F^(a*c)*b*c*e^ 
2*e^(4*d)*log(F) + 48*F^(a*c)*e^3*e^(4*d))*e^(4*e*x) + 3*(F^(a*c)*b^3*c^3* 
e^(2*d)*log(F)^3 - 28*F^(a*c)*b*c*e^2*e^(2*d)*log(F) - 48*F^(a*c)*e^3*e^(2 
*d))*e^(2*e*x))*F^(b*c*x)/(b^4*c^4*log(F)^4 - 12*b^3*c^3*e*log(F)^3 + 44*b 
^2*c^2*e^2*log(F)^2 - 48*b*c*e^3*log(F) - (b^4*c^4*e^(6*d)*log(F)^4 - 12*b 
^3*c^3*e*e^(6*d)*log(F)^3 + 44*b^2*c^2*e^2*e^(6*d)*log(F)^2 - 48*b*c*e^3*e 
^(6*d)*log(F))*e^(6*e*x) + 3*(b^4*c^4*e^(4*d)*log(F)^4 - 12*b^3*c^3*e*e^(4 
*d)*log(F)^3 + 44*b^2*c^2*e^2*e^(4*d)*log(F)^2 - 48*b*c*e^3*e^(4*d)*log...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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