\(\int F^{c (a+b x)} \text {sech}^2(d+e x) \tanh (d+e x) \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 120 \[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \tanh (d+e x) \, dx=-\frac {2 e^{2 d+2 e x} F^{c (a+b x)}}{e \left (1+e^{2 d+2 e x}\right )^2}+\frac {2 b c e^{2 d+2 e x} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2+\frac {b c \log (F)}{e}\right ),\frac {1}{2} \left (4+\frac {b c \log (F)}{e}\right ),-e^{2 d+2 e x}\right ) \log (F)}{e (2 e+b c \log (F))} \] Output:

-2*exp(2*e*x+2*d)*F^(c*(b*x+a))/e/(1+exp(2*e*x+2*d))^2+2*b*c*exp(2*e*x+2*d 
)*F^(c*(b*x+a))*hypergeom([2, 1+1/2*b*c*ln(F)/e],[2+1/2*b*c*ln(F)/e],-exp( 
2*e*x+2*d))*ln(F)/e/(2*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70 \[ \int F^{c (a+b x)} \text {sech}^2(d+e x) \tanh (d+e x) \, dx=\frac {F^{c (a+b x)} \left (2 b c \operatorname {Hypergeometric2F1}\left (1,\frac {b c \log (F)}{2 e},1+\frac {b c \log (F)}{2 e},-e^{2 (d+e x)}\right ) \log (F)-e \text {sech}^2(d+e x)+b c \log (F) (-1+\tanh (d+e x))\right )}{2 e^2} \] Input:

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

Output:

(F^(c*(a + b*x))*(2*b*c*Hypergeometric2F1[1, (b*c*Log[F])/(2*e), 1 + (b*c* 
Log[F])/(2*e), -E^(2*(d + e*x))]*Log[F] - e*Sech[d + e*x]^2 + b*c*Log[F]*( 
-1 + Tanh[d + e*x])))/(2*e^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.24, 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.

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

\(\Big \downarrow \) 6037

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {4 e^{4 d+4 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (\frac {b c \log (F)}{e}+4\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+6\right ),-e^{2 (d+e x)}\right )}{b c \log (F)+4 e}-\frac {4 e^{2 d+2 e x} F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (\frac {b c \log (F)}{e}+2\right ),\frac {1}{2} \left (\frac {b c \log (F)}{e}+4\right ),-e^{2 (d+e x)}\right )}{b c \log (F)+2 e}\)

Input:

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

Output:

(-4*E^(2*d + 2*e*x)*F^(a*c + b*c*x)*Hypergeometric2F1[3, (2 + (b*c*Log[F]) 
/e)/2, (4 + (b*c*Log[F])/e)/2, -E^(2*(d + e*x))])/(2*e + b*c*Log[F]) + (4* 
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])
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

-48*F^(a*c)*b^2*c^2*e*integrate(F^(b*c*x)/(b^3*c^3*log(F)^3 - 12*b^2*c^2*e 
*log(F)^2 + 44*b*c*e^2*log(F) - 48*e^3 + (b^3*c^3*e^(8*d)*log(F)^3 - 12*b^ 
2*c^2*e*e^(8*d)*log(F)^2 + 44*b*c*e^2*e^(8*d)*log(F) - 48*e^3*e^(8*d))*e^( 
8*e*x) + 4*(b^3*c^3*e^(6*d)*log(F)^3 - 12*b^2*c^2*e*e^(6*d)*log(F)^2 + 44* 
b*c*e^2*e^(6*d)*log(F) - 48*e^3*e^(6*d))*e^(6*e*x) + 6*(b^3*c^3*e^(4*d)*lo 
g(F)^3 - 12*b^2*c^2*e*e^(4*d)*log(F)^2 + 44*b*c*e^2*e^(4*d)*log(F) - 48*e^ 
3*e^(4*d))*e^(4*e*x) + 4*(b^3*c^3*e^(2*d)*log(F)^3 - 12*b^2*c^2*e*e^(2*d)* 
log(F)^2 + 44*b*c*e^2*e^(2*d)*log(F) - 48*e^3*e^(2*d))*e^(2*e*x)), x)*log( 
F)^2 + 4*(12*F^(a*c)*b*c*e*log(F) + (F^(a*c)*b^2*c^2*e^(4*d)*log(F)^2 - 10 
*F^(a*c)*b*c*e*e^(4*d)*log(F) + 24*F^(a*c)*e^2*e^(4*d))*e^(4*e*x) - (F^(a* 
c)*b^2*c^2*e^(2*d)*log(F)^2 - 2*F^(a*c)*b*c*e*e^(2*d)*log(F) - 24*F^(a*c)* 
e^2*e^(2*d))*e^(2*e*x))*F^(b*c*x)/(b^3*c^3*log(F)^3 - 12*b^2*c^2*e*log(F)^ 
2 + 44*b*c*e^2*log(F) - 48*e^3 + (b^3*c^3*e^(6*d)*log(F)^3 - 12*b^2*c^2*e* 
e^(6*d)*log(F)^2 + 44*b*c*e^2*e^(6*d)*log(F) - 48*e^3*e^(6*d))*e^(6*e*x) + 
 3*(b^3*c^3*e^(4*d)*log(F)^3 - 12*b^2*c^2*e*e^(4*d)*log(F)^2 + 44*b*c*e^2* 
e^(4*d)*log(F) - 48*e^3*e^(4*d))*e^(4*e*x) + 3*(b^3*c^3*e^(2*d)*log(F)^3 - 
 12*b^2*c^2*e*e^(2*d)*log(F)^2 + 44*b*c*e^2*e^(2*d)*log(F) - 48*e^3*e^(2*d 
))*e^(2*e*x))
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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