\(\int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx\) [131]

Optimal result

Integrand size = 24, antiderivative size = 164 \[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\frac {2 e^{d+e x} F^{c (a+b x)}}{e \left (1+e^{2 d+2 e x}\right )^2}-\frac {e^{d+e x} F^{c (a+b x)} (e+b c \log (F))}{e^2 \left (1+e^{2 d+2 e x}\right )}+\frac {e^{d+e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) \left (e^2+b^2 c^2 \log ^2(F)\right )}{e^2 (e+b c \log (F))} \] 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))
 

Mathematica [A] (verified)

Time = 0.88 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.90 \[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left (2 e^{\frac {(d+e x) (e+b c \log (F))}{e}} \operatorname {Hypergeometric2F1}\left (1,\frac {e+b c \log (F)}{2 e},\frac {1}{2} \left (3+\frac {b c \log (F)}{e}\right ),-e^{2 (d+e x)}\right ) \left (e^2+b^2 c^2 \log ^2(F)\right )-F^{\frac {b c (d+e x)}{e}} (e+b c \log (F)) \text {sech}(d+e x) (b c \log (F)+e \tanh (d+e x))\right )}{2 e^2 (e+b c \log (F))} \] Input:

Integrate[F^(c*(a + b*x))*Sech[d + e*x]*Tanh[d + e*x]^2,x]
 

Output:

(F^(c*(a - (b*d)/e))*(2*E^(((d + e*x)*(e + b*c*Log[F]))/e)*Hypergeometric2 
F1[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) - F^((b*c*(d + e*x))/e)*(e + b*c*Log[F])*Sech[d + e 
*x]*(b*c*Log[F] + e*Tanh[d + e*x])))/(2*e^2*(e + b*c*Log[F]))
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.27, 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.

\(\displaystyle \int \tanh ^2(d+e x) \text {sech}(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 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))*Sech[d + e*x]*Tanh[d + e*x]^2,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])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6037

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]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \operatorname {sech}\left (e x +d \right ) \tanh \left (e x +d \right )^{2}d x\]

Input:

int(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,x)
 

Output:

int(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,x)
 

Fricas [F]

\[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right ) \tanh \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*sech(e*x + d)*tanh(e*x + d)^2, x)
 

Sympy [F]

\[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\int F^{c \left (a + b x\right )} \tanh ^{2}{\left (d + e x \right )} \operatorname {sech}{\left (d + e x \right )}\, dx \] Input:

integrate(F**(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)**2,x)
 

Output:

Integral(F**(c*(a + b*x))*tanh(d + e*x)**2*sech(d + e*x), x)
 

Maxima [F]

\[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right ) \tanh \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,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* 
log(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)*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) + 6*(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) + 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)*l 
og(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)*l 
og(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)*lo 
g(F) - 15*e^3*e^(2*d))*e^(2*e*x))
 

Giac [F]

\[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right ) \tanh \left (e x + d\right )^{2} \,d x } \] Input:

integrate(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*sech(e*x + d)*tanh(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}\,{\mathrm {tanh}\left (d+e\,x\right )}^2}{\mathrm {cosh}\left (d+e\,x\right )} \,d x \] Input:

int((F^(c*(a + b*x))*tanh(d + e*x)^2)/cosh(d + e*x),x)
 

Output:

int((F^(c*(a + b*x))*tanh(d + e*x)^2)/cosh(d + e*x), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \text {sech}(d+e x) \tanh ^2(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \mathrm {sech}\left (e x +d \right ) \tanh \left (e x +d \right )^{2}d x \right ) \] Input:

int(F^(c*(b*x+a))*sech(e*x+d)*tanh(e*x+d)^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*sech(d + e*x)*tanh(d + e*x)**2,x)