Integrand size = 24, antiderivative size = 249 \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=-\frac {2 e^{4 d+4 e x} F^{c (a+b x)}}{e \left (1+e^{2 d+2 e x}\right )^2}-\frac {e^{4 d+4 e x} F^{c (a+b x)} (2 e-b c \log (F))}{e^2 \left (1+e^{2 d+2 e x}\right )}-\frac {2 e^{4 d+4 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (6+\frac {b c \log (F)}{e}\right ),e^{2 d+2 e x}\right )}{4 e+b c \log (F)}+\frac {e^{4 d+4 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (4+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (6+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) \left (2-\frac {b^2 c^2 \log ^2(F)}{e^2}\right )}{4 e+b c \log (F)} \] Output:
-2*exp(4*e*x+4*d)*F^(c*(b*x+a))/e/(1+exp(2*e*x+2*d))^2-exp(4*e*x+4*d)*F^(c *(b*x+a))*(2*e-b*c*ln(F))/e^2/(1+exp(2*e*x+2*d))-2*exp(4*e*x+4*d)*F^(c*(b* x+a))*hypergeom([1, 2+1/2*b*c*ln(F)/e],[3+1/2*b*c*ln(F)/e],exp(2*e*x+2*d)) /(b*c*ln(F)+4*e)+exp(4*e*x+4*d)*F^(c*(b*x+a))*hypergeom([1, 2+1/2*b*c*ln(F )/e],[3+1/2*b*c*ln(F)/e],-exp(2*e*x+2*d))*(2-b^2*c^2*ln(F)^2/e^2)/(b*c*ln( F)+4*e)
Time = 0.89 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.61 \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (-4 e^2 \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},e^{2 (d+e x)}\right )+\operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \left (4 e^2-2 b^2 c^2 \log ^2(F)\right )+b c \log (F) \left (e \text {sech}^2(d+e x)-b c \log (F) (-1+\tanh (d+e x))\right )\right )}{2 b c e^2 \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*Csch[d + e*x]*Sech[d + e*x]^3,x]
Output:
(F^(c*(a + b*x))*(-4*e^2*Hypergeometric2F1[1, (b*c*Log[F])/(2*e), 1 + (b*c *Log[F])/(2*e), E^(2*(d + e*x))] + Hypergeometric2F1[1, (b*c*Log[F])/(2*e) , 1 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))]*(4*e^2 - 2*b^2*c^2*Log[F]^2) + b*c*Log[F]*(e*Sech[d + e*x]^2 - b*c*Log[F]*(-1 + Tanh[d + e*x]))))/(2*b*c *e^2*Log[F])
Time = 1.03 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.77, 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.
\(\displaystyle \int \text {csch}(d+e x) \text {sech}^3(d+e x) F^{c (a+b x)} \, dx\) |
\(\Big \downarrow \) 6037 |
\(\displaystyle \int \left (\frac {4 e^{4 d+4 e x} F^{a c+b c x}}{e^{4 (d+e x)}-1}-\frac {4 e^{4 d+4 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}+1\right )^2}-\frac {8 e^{4 d+4 e x} F^{a c+b c x}}{\left (e^{2 (d+e x)}+1\right )^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (1,-\frac {b c \log (F)}{4 e},1-\frac {b c \log (F)}{4 e},e^{-4 (d+e x)}\right )}{b c \log (F)}-\frac {4 F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (2,-\frac {b c \log (F)}{2 e},1-\frac {b c \log (F)}{2 e},-e^{-2 (d+e x)}\right )}{b c \log (F)}-\frac {8 e^{4 d+4 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (\frac {b c \log (F)}{e}+4\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+6\right ),-e^{2 (d+e x)}\right )}{b c \log (F)+4 e}\) |
Input:
Int[F^(c*(a + b*x))*Csch[d + e*x]*Sech[d + e*x]^3,x]
Output:
(4*F^(a*c + b*c*x)*Hypergeometric2F1[1, -1/4*(b*c*Log[F])/e, 1 - (b*c*Log[ F])/(4*e), E^(-4*(d + e*x))])/(b*c*Log[F]) - (4*F^(a*c + b*c*x)*Hypergeome tric2F1[2, -1/2*(b*c*Log[F])/e, 1 - (b*c*Log[F])/(2*e), -E^(-2*(d + e*x))] )/(b*c*Log[F]) - (8*E^(4*d + 4*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[3, ( 4 + (b*c*Log[F])/e)/2, (6 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x))])/(4*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]
\[\int F^{c \left (b x +a \right )} \operatorname {csch}\left (e x +d \right ) \operatorname {sech}\left (e x +d \right )^{3}d x\]
Input:
int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x)
Output:
int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x)
\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*csch(e*x + d)*sech(e*x + d)^3, x)
\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\int F^{c \left (a + b x\right )} \operatorname {csch}{\left (d + e x \right )} \operatorname {sech}^{3}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)**3,x)
Output:
Integral(F**(c*(a + b*x))*csch(d + e*x)*sech(d + e*x)**3, x)
\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x, algorithm="maxima")
Output:
-16*(8*F^(a*c)*b*c*e*log(F) - 32*F^(a*c)*e^2 + (F^(a*c)*b^2*c^2*e^(4*d)*lo g(F)^2 - 14*F^(a*c)*b*c*e*e^(4*d)*log(F) + 48*F^(a*c)*e^2*e^(4*d))*e^(4*e* x) - 4*(F^(a*c)*b*c*e*e^(2*d)*log(F) - 8*F^(a*c)*e^2*e^(2*d))*e^(2*e*x))*F ^(b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2*e*log(F)^2 + 104*b*c*e^2*log(F) - 192*e^3 - (b^3*c^3*e^(8*d)*log(F)^3 - 18*b^2*c^2*e*e^(8*d)*log(F)^2 + 104* b*c*e^2*e^(8*d)*log(F) - 192*e^3*e^(8*d))*e^(8*e*x) - 2*(b^3*c^3*e^(6*d)*l og(F)^3 - 18*b^2*c^2*e*e^(6*d)*log(F)^2 + 104*b*c*e^2*e^(6*d)*log(F) - 192 *e^3*e^(6*d))*e^(6*e*x) + 2*(b^3*c^3*e^(2*d)*log(F)^3 - 18*b^2*c^2*e*e^(2* d)*log(F)^2 + 104*b*c*e^2*e^(2*d)*log(F) - 192*e^3*e^(2*d))*e^(2*e*x)) - 1 6*integrate(-4*(2*F^(a*c)*b^2*c^2*e*log(F)^2 - 8*F^(a*c)*b*c*e^2*log(F) - (F^(a*c)*b^2*c^2*e*e^(2*d)*log(F)^2 + 2*F^(a*c)*b*c*e^2*e^(2*d)*log(F) - 4 8*F^(a*c)*e^3*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 18*b^2*c^2 *e*log(F)^2 + 104*b*c*e^2*log(F) - 192*e^3 + (b^3*c^3*e^(12*d)*log(F)^3 - 18*b^2*c^2*e*e^(12*d)*log(F)^2 + 104*b*c*e^2*e^(12*d)*log(F) - 192*e^3*e^( 12*d))*e^(12*e*x) + 2*(b^3*c^3*e^(10*d)*log(F)^3 - 18*b^2*c^2*e*e^(10*d)*l og(F)^2 + 104*b*c*e^2*e^(10*d)*log(F) - 192*e^3*e^(10*d))*e^(10*e*x) - (b^ 3*c^3*e^(8*d)*log(F)^3 - 18*b^2*c^2*e*e^(8*d)*log(F)^2 + 104*b*c*e^2*e^(8* d)*log(F) - 192*e^3*e^(8*d))*e^(8*e*x) - 4*(b^3*c^3*e^(6*d)*log(F)^3 - 18* b^2*c^2*e*e^(6*d)*log(F)^2 + 104*b*c*e^2*e^(6*d)*log(F) - 192*e^3*e^(6*d)) *e^(6*e*x) - (b^3*c^3*e^(4*d)*log(F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 ...
\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {csch}\left (e x + d\right ) \operatorname {sech}\left (e x + d\right )^{3} \,d x } \] Input:
integrate(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*csch(e*x + d)*sech(e*x + d)^3, x)
Timed out. \[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^3\,\mathrm {sinh}\left (d+e\,x\right )} \,d x \] Input:
int(F^(c*(a + b*x))/(cosh(d + e*x)^3*sinh(d + e*x)),x)
Output:
int(F^(c*(a + b*x))/(cosh(d + e*x)^3*sinh(d + e*x)), x)
\[ \int F^{c (a+b x)} \text {csch}(d+e x) \text {sech}^3(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {csch}\left (e x +d \right ) \mathrm {sech}\left (e x +d \right )^{3}d x \right ) \] Input:
int(F^(c*(b*x+a))*csch(e*x+d)*sech(e*x+d)^3,x)
Output:
f**(a*c)*int(f**(b*c*x)*csch(d + e*x)*sech(d + e*x)**3,x)