Integrand size = 24, antiderivative size = 181 \[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\frac {2 e^{-d-e x} F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )}-\frac {5 e^{-d-e x} F^{c (a+b x)}}{2 (e-b c \log (F))}+\frac {2 b c e^{-d-e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (-1+\frac {b c \log (F)}{e}\right ),\frac {e+b c \log (F)}{2 e},e^{2 d+2 e x}\right ) \log (F)}{e (e-b c \log (F))}+\frac {e^{d+e x} F^{c (a+b x)}}{2 (e+b c \log (F))} \] Output:
2*exp(-e*x-d)*F^(c*(b*x+a))/e/(1-exp(2*e*x+2*d))-5*exp(-e*x-d)*F^(c*(b*x+a ))/(2*e-2*b*c*ln(F))+2*b*c*exp(-e*x-d)*F^(c*(b*x+a))*hypergeom([1, -1/2+1/ 2*b*c*ln(F)/e],[1/2*(e+b*c*ln(F))/e],exp(2*e*x+2*d))*ln(F)/e/(e-b*c*ln(F)) +exp(e*x+d)*F^(c*(b*x+a))/(2*e+2*b*c*ln(F))
Time = 1.83 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.01 \[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \text {csch}(d+e x) \left (-4 b c e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) \log (F) (e-b c \log (F)) \sinh (d+e x)+F^{\frac {b c (d+e x)}{e}} \left (-3 e^2+e^2 \cosh (2 (d+e x))+2 b^2 c^2 \log ^2(F)-b c e \log (F) \sinh (2 (d+e x))\right )\right )}{2 e (e-b c \log (F)) (e+b c \log (F))} \] Input:
Integrate[F^(c*(a + b*x))*Cosh[d + e*x]*Coth[d + e*x]^2,x]
Output:
(F^(c*(a - (b*d)/e))*Csch[d + e*x]*(-4*b*c*E^(((d + e*x)*(e + b*c*Log[F])) /e)*Hypergeometric2F1[1, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e)/2, E ^(2*(d + e*x))]*Log[F]*(e - b*c*Log[F])*Sinh[d + e*x] + F^((b*c*(d + e*x)) /e)*(-3*e^2 + e^2*Cosh[2*(d + e*x)] + 2*b^2*c^2*Log[F]^2 - b*c*e*Log[F]*Si nh[2*(d + e*x)])))/(2*e*(e - b*c*Log[F])*(e + b*c*Log[F]))
Time = 0.89 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.16, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.
\(\displaystyle \int \cosh (d+e x) \coth ^2(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6037 |
\(\displaystyle \int \left (\frac {5}{2} e^{-d-e x} F^{a c+b c x}+\frac {1}{2} e^{2 (d+e x)-d-e x} F^{a c+b c x}+\frac {6 e^{-d-e x} F^{a c+b c x}}{e^{2 (d+e x)}-1}+\frac {4 e^{-d-e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 e^{-d-e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (\frac {b c \log (F)}{e}-1\right ),\frac {e+b c \log (F)}{2 e},e^{2 (d+e x)}\right )}{e-b c \log (F)}-\frac {4 e^{-d-e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (\frac {b c \log (F)}{e}-1\right ),\frac {e+b c \log (F)}{2 e},e^{2 (d+e x)}\right )}{e-b c \log (F)}-\frac {5 F^{a c} e^{-x (e-b c \log (F))-d}}{2 (e-b c \log (F))}+\frac {F^{a c} e^{x (b c \log (F)+e)+d}}{2 (b c \log (F)+e)}\) |
Input:
Int[F^(c*(a + b*x))*Cosh[d + e*x]*Coth[d + e*x]^2,x]
Output:
(-5*E^(-d - x*(e - b*c*Log[F]))*F^(a*c))/(2*(e - b*c*Log[F])) + (6*E^(-d - e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[1, (-1 + (b*c*Log[F])/e)/2, (e + b *c*Log[F])/(2*e), E^(2*(d + e*x))])/(e - b*c*Log[F]) - (4*E^(-d - e*x)*F^( a*c + b*c*x)*Hypergeometric2F1[2, (-1 + (b*c*Log[F])/e)/2, (e + b*c*Log[F] )/(2*e), E^(2*(d + e*x))])/(e - b*c*Log[F]) + (E^(d + x*(e + b*c*Log[F]))* F^(a*c))/(2*(e + b*c*Log[F]))
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]
\[\int F^{c \left (b x +a \right )} \cosh \left (e x +d \right ) \coth \left (e x +d \right )^{2}d x\]
Input:
int(F^(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)^2,x)
Output:
int(F^(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)^2,x)
\[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cosh \left (e x + d\right ) \coth \left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)^2,x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*cosh(e*x + d)*coth(e*x + d)^2, x)
\[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \cosh {\left (d + e x \right )} \coth ^{2}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)**2,x)
Output:
Integral(F**(c*(a + b*x))*cosh(d + e*x)*coth(d + e*x)**2, x)
\[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cosh \left (e x + d\right ) \coth \left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cosh(e*x+d)*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*e^(7*d)*log(F)^2 - 8*b*c*e* e^(7*d)*log(F) + 15*e^2*e^(7*d))*e^(7*e*x) - 3*(b^2*c^2*e^(5*d)*log(F)^2 - 8*b*c*e*e^(5*d)*log(F) + 15*e^2*e^(5*d))*e^(5*e*x) + 3*(b^2*c^2*e^(3*d)*l og(F)^2 - 8*b*c*e*e^(3*d)*log(F) + 15*e^2*e^(3*d))*e^(3*e*x) - (b^2*c^2*e^ d*log(F)^2 - 8*b*c*e*e^d*log(F) + 15*e^2*e^d)*e^(e*x)), x)*log(F) + 1/2*(F ^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b^2*c^2*e*log(F)^2 + 39*F^(a*c)*b*c*e ^2*log(F) + 15*F^(a*c)*e^3 + (F^(a*c)*b^3*c^3*e^(6*d)*log(F)^3 - 9*F^(a*c) *b^2*c^2*e*e^(6*d)*log(F)^2 + 23*F^(a*c)*b*c*e^2*e^(6*d)*log(F) - 15*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 - 17*F^(a* c)*b^2*c^2*e*e^(4*d)*log(F)^2 - 11*F^(a*c)*b*c*e^2*e^(4*d)*log(F) + 105*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 + F^(a* c)*b^2*c^2*e*e^(2*d)*log(F)^2 - 59*F^(a*c)*b*c*e^2*e^(2*d)*log(F) - 105*F^ (a*c)*e^3*e^(2*d))*e^(2*e*x))*F^(b*c*x)/((b^4*c^4*e^(5*d)*log(F)^4 - 8*b^3 *c^3*e*e^(5*d)*log(F)^3 + 14*b^2*c^2*e^2*e^(5*d)*log(F)^2 + 8*b*c*e^3*e^(5 *d)*log(F) - 15*e^4*e^(5*d))*e^(5*e*x) - 2*(b^4*c^4*e^(3*d)*log(F)^4 - 8*b ^3*c^3*e*e^(3*d)*log(F)^3 + 14*b^2*c^2*e^2*e^(3*d)*log(F)^2 + 8*b*c*e^3*e^ (3*d)*log(F) - 15*e^4*e^(3*d))*e^(3*e*x) + (b^4*c^4*e^d*log(F)^4 - 8*b^3*c ^3*e*e^d*log(F)^3 + 14*b^2*c^2*e^2*e^d*log(F)^2 + 8*b*c*e^3*e^d*log(F) - 1 5*e^4*e^d)*e^(e*x))
\[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cosh \left (e x + d\right ) \coth \left (e x + d\right )^{2} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)^2,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*cosh(e*x + d)*coth(e*x + d)^2, x)
Timed out. \[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,\mathrm {cosh}\left (d+e\,x\right )\,{\mathrm {coth}\left (d+e\,x\right )}^2 \,d x \] Input:
int(F^(c*(a + b*x))*cosh(d + e*x)*coth(d + e*x)^2,x)
Output:
int(F^(c*(a + b*x))*cosh(d + e*x)*coth(d + e*x)^2, x)
\[ \int F^{c (a+b x)} \cosh (d+e x) \coth ^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \cosh \left (e x +d \right ) \coth \left (e x +d \right )^{2}d x \right ) \] Input:
int(F^(c*(b*x+a))*cosh(e*x+d)*coth(e*x+d)^2,x)
Output:
f**(a*c)*int(f**(b*c*x)*cosh(d + e*x)*coth(d + e*x)**2,x)