\(\int F^{c (a+b x)} \coth ^4(d+e x) \, dx\) [73]

Optimal result

Integrand size = 18, antiderivative size = 219 \[ \int F^{c (a+b x)} \coth ^4(d+e x) \, dx=\frac {8 F^{c (a+b x)}}{3 e \left (1-e^{2 d+2 e x}\right )^3}+\frac {F^{c (a+b x)}}{b c \log (F)}-\frac {2 F^{c (a+b x)} (6 e+b c \log (F))}{3 e^2 \left (1-e^{2 d+2 e x}\right )^2}-\frac {F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 d+2 e x}\right ) \left (8 e^2+b^2 c^2 \log ^2(F)\right )}{3 e^3}+\frac {F^{c (a+b x)} \left (12 e^2+2 b c e \log (F)+b^2 c^2 \log ^2(F)\right )}{3 e^3 \left (1-e^{2 d+2 e x}\right )} \] Output:

8/3*F^(c*(b*x+a))/e/(1-exp(2*e*x+2*d))^3+F^(c*(b*x+a))/b/c/ln(F)-2/3*F^(c* 
(b*x+a))*(6*e+b*c*ln(F))/e^2/(1-exp(2*e*x+2*d))^2-1/3*F^(c*(b*x+a))*hyperg 
eom([1, 1/2*b*c*ln(F)/e],[1+1/2*b*c*ln(F)/e],exp(2*e*x+2*d))*(8*e^2+b^2*c^ 
2*ln(F)^2)/e^3+1/3*F^(c*(b*x+a))*(12*e^2+2*b*c*e*ln(F)+b^2*c^2*ln(F)^2)/e^ 
3/(1-exp(2*e*x+2*d))
 

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.81 \[ \int F^{c (a+b x)} \coth ^4(d+e x) \, dx=\frac {1}{6} F^{c (a+b x)} \left (\frac {6}{b c \log (F)}-\frac {\text {csch}^2(d+e x) (2 e \coth (d)+b c \log (F))}{e^2}-\frac {2 \left (1+\left (-1+e^{2 d}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right )\right ) \left (8 e^2+b^2 c^2 \log ^2(F)\right )}{e^3 \left (-1+e^{2 d}\right )}+\frac {2 \text {csch}(d) \text {csch}^3(d+e x) \sinh (e x)}{e}+\frac {\text {csch}(d) \text {csch}(d+e x) \left (8 e^2+b^2 c^2 \log ^2(F)\right ) \sinh (e x)}{e^3}\right ) \] Input:

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

Output:

(F^(c*(a + b*x))*(6/(b*c*Log[F]) - (Csch[d + e*x]^2*(2*e*Coth[d] + b*c*Log 
[F]))/e^2 - (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))])*(8*e^2 + b^2*c^2*Log[F]^2))/(e 
^3*(-1 + E^(2*d))) + (2*Csch[d]*Csch[d + e*x]^3*Sinh[e*x])/e + (Csch[d]*Cs 
ch[d + e*x]*(8*e^2 + b^2*c^2*Log[F]^2)*Sinh[e*x])/e^3))/6
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 245, 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 = {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.

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

\(\Big \downarrow \) 6008

\(\displaystyle \int \left (\frac {8 F^{c (a+b x)}}{e^{2 (d+e x)}-1}+\frac {24 F^{c (a+b x)}}{\left (e^{2 (d+e x)}-1\right )^2}+\frac {32 F^{c (a+b x)}}{\left (e^{2 (d+e x)}-1\right )^3}+\frac {16 F^{c (a+b x)}}{\left (e^{2 (d+e x)}-1\right )^4}+F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {8 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}+\frac {24 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}-\frac {32 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}+\frac {16 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (4,\frac {b c \log (F)}{2 e},\frac {b c \log (F)}{2 e}+1,e^{2 (d+e x)}\right )}{b c \log (F)}+\frac {F^{c (a+b x)}}{b c \log (F)}\)

Input:

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

Output:

F^(c*(a + b*x))/(b*c*Log[F]) - (8*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]) + 
(24*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]) - (32*F^(c*(a + b*x))*Hypergeome 
tric2F1[3, (b*c*Log[F])/(2*e), 1 + (b*c*Log[F])/(2*e), E^(2*(d + e*x))])/( 
b*c*Log[F]) + (16*F^(c*(a + b*x))*Hypergeometric2F1[4, (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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

128*(F^(a*c)*b^3*c^3*e*log(F)^3 + 8*F^(a*c)*b*c*e^3*log(F))*integrate(-F^( 
b*c*x)/(b^4*c^4*log(F)^4 - 20*b^3*c^3*e*log(F)^3 + 140*b^2*c^2*e^2*log(F)^ 
2 - 400*b*c*e^3*log(F) + 384*e^4 - (b^4*c^4*e^(10*d)*log(F)^4 - 20*b^3*c^3 
*e*e^(10*d)*log(F)^3 + 140*b^2*c^2*e^2*e^(10*d)*log(F)^2 - 400*b*c*e^3*e^( 
10*d)*log(F) + 384*e^4*e^(10*d))*e^(10*e*x) + 5*(b^4*c^4*e^(8*d)*log(F)^4 
- 20*b^3*c^3*e*e^(8*d)*log(F)^3 + 140*b^2*c^2*e^2*e^(8*d)*log(F)^2 - 400*b 
*c*e^3*e^(8*d)*log(F) + 384*e^4*e^(8*d))*e^(8*e*x) - 10*(b^4*c^4*e^(6*d)*l 
og(F)^4 - 20*b^3*c^3*e*e^(6*d)*log(F)^3 + 140*b^2*c^2*e^2*e^(6*d)*log(F)^2 
 - 400*b*c*e^3*e^(6*d)*log(F) + 384*e^4*e^(6*d))*e^(6*e*x) + 10*(b^4*c^4*e 
^(4*d)*log(F)^4 - 20*b^3*c^3*e*e^(4*d)*log(F)^3 + 140*b^2*c^2*e^2*e^(4*d)* 
log(F)^2 - 400*b*c*e^3*e^(4*d)*log(F) + 384*e^4*e^(4*d))*e^(4*e*x) - 5*(b^ 
4*c^4*e^(2*d)*log(F)^4 - 20*b^3*c^3*e*e^(2*d)*log(F)^3 + 140*b^2*c^2*e^2*e 
^(2*d)*log(F)^2 - 400*b*c*e^3*e^(2*d)*log(F) + 384*e^4*e^(2*d))*e^(2*e*x)) 
, x) + (F^(a*c)*b^4*c^4*log(F)^4 + 108*F^(a*c)*b^3*c^3*e*log(F)^3 + 140*F^ 
(a*c)*b^2*c^2*e^2*log(F)^2 + 624*F^(a*c)*b*c*e^3*log(F) + 384*F^(a*c)*e^4 
+ (F^(a*c)*b^4*c^4*e^(8*d)*log(F)^4 - 20*F^(a*c)*b^3*c^3*e*e^(8*d)*log(F)^ 
3 + 140*F^(a*c)*b^2*c^2*e^2*e^(8*d)*log(F)^2 - 400*F^(a*c)*b*c*e^3*e^(8*d) 
*log(F) + 384*F^(a*c)*e^4*e^(8*d))*e^(8*e*x) + 4*(F^(a*c)*b^4*c^4*e^(6*d)* 
log(F)^4 - 16*F^(a*c)*b^3*c^3*e*e^(6*d)*log(F)^3 + 68*F^(a*c)*b^2*c^2*e^2* 
e^(6*d)*log(F)^2 + 16*F^(a*c)*b*c*e^3*e^(6*d)*log(F) - 384*F^(a*c)*e^4*...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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