Integrand size = 24, antiderivative size = 166 \[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=-\frac {2 e^{d+e x} F^{c (a+b x)}}{e \left (1-e^{2 d+2 e x}\right )^2}+\frac {e^{d+e x} F^{c (a+b x)} (e+b c \log (F))}{e^2 \left (1-e^{2 d+2 e x}\right )}-\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right ) \left (e^2+b^2 c^2 \log ^2(F)\right )}{e^2 (e+b c \log (F))} \] Output:
-2*exp(e*x+d)*F^(c*(b*x+a))/e/(1-exp(2*e*x+2*d))^2+exp(e*x+d)*F^(c*(b*x+a) )*(e+b*c*ln(F))/e^2/(1-exp(2*e*x+2*d))-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))*(e^2+b^2*c^ 2*ln(F)^2)/e^2/(e+b*c*ln(F))
Time = 2.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.87 \[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=-\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (F^{\frac {b c (d+e x)}{e}} \text {csch}(d+e x) (e+b c \log (F)) (e \coth (d+e x)+b c \log (F))+2 e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),e^{2 (d+e x)}\right ) \left (e^2+b^2 c^2 \log ^2(F)\right )\right )}{2 e^2 (e+b c \log (F))} \] Input:
Integrate[F^(c*(a + b*x))*Coth[d + e*x]^2*Csch[d + e*x],x]
Output:
-1/2*(F^(c*(a - (b*d)/e))*(F^((b*c*(d + e*x))/e)*Csch[d + e*x]*(e + b*c*Lo g[F])*(e*Coth[d + e*x] + b*c*Log[F]) + 2*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))]*(e^2 + b^2*c^2*Log[F]^2)))/(e^2*(e + b*c*Log[F]))
Time = 0.54 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.22, 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.
\(\displaystyle \int \coth ^2(d+e x) \text {csch}(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6037 |
\(\displaystyle \int \left (\frac {2 e^{d+e x} F^{a c+b c x}}{e^{2 (d+e x)}-1}+\frac {8 e^{d+e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^2}+\frac {8 e^{d+e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}-1\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 e^{d+e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),e^{2 (d+e x)}\right )}{b c \log (F)+e}+\frac {8 e^{d+e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),e^{2 (d+e x)}\right )}{b c \log (F)+e}-\frac {8 e^{d+e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right ),e^{2 (d+e x)}\right )}{b c \log (F)+e}\) |
Input:
Int[F^(c*(a + b*x))*Coth[d + e*x]^2*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]) + (8*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]) - (8*E^(d + e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[3, (e + b*c*Log[F])/(2*e), (3 + (b*c*Log[F])/e )/2, E^(2*(d + e*x))])/(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]
\[\int F^{c \left (b x +a \right )} \coth \left (e x +d \right )^{2} \operatorname {csch}\left (e x +d \right )d x\]
Input:
int(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x)
Output:
int(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x)
\[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \coth \left (e x + d\right )^{2} \operatorname {csch}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*coth(e*x + d)^2*csch(e*x + d), x)
\[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=\int F^{c \left (a + b x\right )} \coth ^{2}{\left (d + e x \right )} \operatorname {csch}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*coth(e*x+d)**2*csch(e*x+d),x)
Output:
Integral(F**(c*(a + b*x))*coth(d + e*x)**2*csch(d + e*x), x)
\[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \coth \left (e x + d\right )^{2} \operatorname {csch}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x, algorithm="maxima")
Output:
48*(F^(a*c)*b^2*c^2*e*e^d*log(F)^2 + F^(a*c)*e^3*e^d)*integrate(e^(b*c*x*l og(F) + e*x)/(b^3*c^3*log(F)^3 - 9*b^2*c^2*e*log(F)^2 + 23*b*c*e^2*log(F) - 15*e^3 + (b^3*c^3*e^(8*d)*log(F)^3 - 9*b^2*c^2*e*e^(8*d)*log(F)^2 + 23*b *c*e^2*e^(8*d)*log(F) - 15*e^3*e^(8*d))*e^(8*e*x) - 4*(b^3*c^3*e^(6*d)*log (F)^3 - 9*b^2*c^2*e*e^(6*d)*log(F)^2 + 23*b*c*e^2*e^(6*d)*log(F) - 15*e^3* e^(6*d))*e^(6*e*x) + 6*(b^3*c^3*e^(4*d)*log(F)^3 - 9*b^2*c^2*e*e^(4*d)*log (F)^2 + 23*b*c*e^2*e^(4*d)*log(F) - 15*e^3*e^(4*d))*e^(4*e*x) - 4*(b^3*c^3 *e^(2*d)*log(F)^3 - 9*b^2*c^2*e*e^(2*d)*log(F)^2 + 23*b*c*e^2*e^(2*d)*log( F) - 15*e^3*e^(2*d))*e^(2*e*x)), x) - 2*((F^(a*c)*b^2*c^2*e^(5*d)*log(F)^2 - 8*F^(a*c)*b*c*e*e^(5*d)*log(F) + 15*F^(a*c)*e^2*e^(5*d))*e^(5*e*x) + 2* (F^(a*c)*b^2*c^2*e^(3*d)*log(F)^2 - 3*F^(a*c)*b*c*e*e^(3*d)*log(F) - 10*F^ (a*c)*e^2*e^(3*d))*e^(3*e*x) + (F^(a*c)*b^2*c^2*e^d*log(F)^2 + 14*F^(a*c)* b*c*e*e^d*log(F) + 9*F^(a*c)*e^2*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 9*b^2*c^2*e*log(F)^2 + 23*b*c*e^2*log(F) - 15*e^3 - (b^3*c^3*e^(6*d)*lo g(F)^3 - 9*b^2*c^2*e*e^(6*d)*log(F)^2 + 23*b*c*e^2*e^(6*d)*log(F) - 15*e^3 *e^(6*d))*e^(6*e*x) + 3*(b^3*c^3*e^(4*d)*log(F)^3 - 9*b^2*c^2*e*e^(4*d)*lo g(F)^2 + 23*b*c*e^2*e^(4*d)*log(F) - 15*e^3*e^(4*d))*e^(4*e*x) - 3*(b^3*c^ 3*e^(2*d)*log(F)^3 - 9*b^2*c^2*e*e^(2*d)*log(F)^2 + 23*b*c*e^2*e^(2*d)*log (F) - 15*e^3*e^(2*d))*e^(2*e*x))
\[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \coth \left (e x + d\right )^{2} \operatorname {csch}\left (e x + d\right ) \,d x } \] Input:
integrate(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*coth(e*x + d)^2*csch(e*x + d), x)
Timed out. \[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}\,{\mathrm {coth}\left (d+e\,x\right )}^2}{\mathrm {sinh}\left (d+e\,x\right )} \,d x \] Input:
int((F^(c*(a + b*x))*coth(d + e*x)^2)/sinh(d + e*x),x)
Output:
int((F^(c*(a + b*x))*coth(d + e*x)^2)/sinh(d + e*x), x)
\[ \int F^{c (a+b x)} \coth ^2(d+e x) \text {csch}(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \coth \left (e x +d \right )^{2} \mathrm {csch}\left (e x +d \right )d x \right ) \] Input:
int(F^(c*(b*x+a))*coth(e*x+d)^2*csch(e*x+d),x)
Output:
f**(a*c)*int(f**(b*c*x)*coth(d + e*x)**2*csch(d + e*x),x)