### 3.900 $$\int \frac{F^{c (a+b x)}}{(f+f \cosh (d+e x))^2} \, dx$$

Optimal. Leaf size=151 $\frac{2 e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{d+e x}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) F^{c (a+b x)}}{6 e f^2}$

[Out]

(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^(d + e*x)]*(e -
b*c*Log[F]))/(3*e^2*f^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d/2 + (e*x)/2]^2)/(6*e^2*f^2) + (F^(c*(a + b*x))*
Sech[d/2 + (e*x)/2]^2*Tanh[d/2 + (e*x)/2])/(6*e*f^2)

________________________________________________________________________________________

Rubi [A]  time = 0.102074, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.136, Rules used = {5497, 5490, 5492} $\frac{2 e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{d+e x}\right )}{3 e^2 f^2}+\frac{b c \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) F^{c (a+b x)}}{6 e^2 f^2}+\frac{\tanh \left (\frac{d}{2}+\frac{e x}{2}\right ) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) F^{c (a+b x)}}{6 e f^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[F^(c*(a + b*x))/(f + f*Cosh[d + e*x])^2,x]

[Out]

(2*E^(d + e*x)*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^(d + e*x)]*(e -
b*c*Log[F]))/(3*e^2*f^2) + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d/2 + (e*x)/2]^2)/(6*e^2*f^2) + (F^(c*(a + b*x))*
Sech[d/2 + (e*x)/2]^2*Tanh[d/2 + (e*x)/2])/(6*e*f^2)

Rule 5497

Int[(Cosh[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[2^n*g^n
, Int[F^(c*(a + b*x))*Cosh[d/2 + (e*x)/2]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f - g, 0]
&& ILtQ[n, 0]

Rule 5490

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b
*x))*Sech[d + e*x]^(n - 2))/(e^2*(n - 1)*(n - 2)), x] + (Dist[(e^2*(n - 2)^2 - b^2*c^2*Log[F]^2)/(e^2*(n - 1)*
(n - 2)), Int[F^(c*(a + b*x))*Sech[d + e*x]^(n - 2), x], x] + Simp[(F^(c*(a + b*x))*Sech[d + e*x]^(n - 1)*Sinh
[d + e*x])/(e*(n - 1)), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*(n - 2)^2 - b^2*c^2*Log[F]^2, 0] && GtQ
[n, 1] && NeQ[n, 2]

Rule 5492

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sech[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(n*(d + e*x))*F
^(c*(a + b*x))*Hypergeometric2F1[n, n/2 + (b*c*Log[F])/(2*e), 1 + n/2 + (b*c*Log[F])/(2*e), -E^(2*(d + e*x))])
/(e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{F^{c (a+b x)}}{(f+f \cosh (d+e x))^2} \, dx &=\frac{\int F^{c (a+b x)} \text{sech}^4\left (\frac{d}{2}+\frac{e x}{2}\right ) \, dx}{4 f^2}\\ &=\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{e x}{2}\right )}{6 e f^2}+\frac{\left (1-\frac{b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) \, dx}{6 f^2}\\ &=\frac{2 e^{d+e x} F^{c (a+b x)} \, _2F_1\left (2,1+\frac{b c \log (F)}{e};2+\frac{b c \log (F)}{e};-e^{d+e x}\right ) (e-b c \log (F))}{3 e^2 f^2}+\frac{b c F^{c (a+b x)} \log (F) \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right )}{6 e^2 f^2}+\frac{F^{c (a+b x)} \text{sech}^2\left (\frac{d}{2}+\frac{e x}{2}\right ) \tanh \left (\frac{d}{2}+\frac{e x}{2}\right )}{6 e f^2}\\ \end{align*}

Mathematica [A]  time = 0.332149, size = 127, normalized size = 0.84 $\frac{2 \cosh \left (\frac{1}{2} (d+e x)\right ) F^{c (a+b x)} \left (4 e^{d+e x} \cosh ^3\left (\frac{1}{2} (d+e x)\right ) (e-b c \log (F)) \, _2F_1\left (2,\frac{b c \log (F)}{e}+1;\frac{b c \log (F)}{e}+2;-e^{d+e x}\right )+b c \log (F) \cosh \left (\frac{1}{2} (d+e x)\right )+e \sinh \left (\frac{1}{2} (d+e x)\right )\right )}{3 e^2 f^2 (\cosh (d+e x)+1)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[F^(c*(a + b*x))/(f + f*Cosh[d + e*x])^2,x]

[Out]

(2*F^(c*(a + b*x))*Cosh[(d + e*x)/2]*(b*c*Cosh[(d + e*x)/2]*Log[F] + 4*E^(d + e*x)*Cosh[(d + e*x)/2]^3*Hyperge
ometric2F1[2, 1 + (b*c*Log[F])/e, 2 + (b*c*Log[F])/e, -E^(d + e*x)]*(e - b*c*Log[F]) + e*Sinh[(d + e*x)/2]))/(
3*e^2*f^2*(1 + Cosh[d + e*x])^2)

________________________________________________________________________________________

Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ( f+f\cosh \left ( ex+d \right ) \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d))^2,x, algorithm="maxima")

[Out]

-16*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*b*c*e^2*log(F))*integrate(F^(b*c*x)/(b^3*c^3*f^2*log(F)^3 - 9*b^2*c^
2*e*f^2*log(F)^2 + 26*b*c*e^2*f^2*log(F) - 24*e^3*f^2 + (b^3*c^3*f^2*e^(5*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(5*d
)*log(F)^2 + 26*b*c*e^2*f^2*e^(5*d)*log(F) - 24*e^3*f^2*e^(5*d))*e^(5*e*x) + 5*(b^3*c^3*f^2*e^(4*d)*log(F)^3 -
9*b^2*c^2*e*f^2*e^(4*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(4*d)*log(F) - 24*e^3*f^2*e^(4*d))*e^(4*e*x) + 10*(b^3*c^
3*f^2*e^(3*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(3*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(3*d)*log(F) - 24*e^3*f^2*e^(3*d)
)*e^(3*e*x) + 10*(b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(2*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(2*d)*log
(F) - 24*e^3*f^2*e^(2*d))*e^(2*e*x) + 5*(b^3*c^3*f^2*e^d*log(F)^3 - 9*b^2*c^2*e*f^2*e^d*log(F)^2 + 26*b*c*e^2*
f^2*e^d*log(F) - 24*e^3*f^2*e^d)*e^(e*x)), x) + 4*(4*F^(a*c)*b*c*e*log(F) + 4*F^(a*c)*e^2 + (F^(a*c)*b^2*c^2*e
^(2*d)*log(F)^2 - 7*F^(a*c)*b*c*e*e^(2*d)*log(F) + 12*F^(a*c)*e^2*e^(2*d))*e^(2*e*x) - 4*(F^(a*c)*b*c*e*e^d*lo
g(F) - 4*F^(a*c)*e^2*e^d)*e^(e*x))*F^(b*c*x)/(b^3*c^3*f^2*log(F)^3 - 9*b^2*c^2*e*f^2*log(F)^2 + 26*b*c*e^2*f^2
*log(F) - 24*e^3*f^2 + (b^3*c^3*f^2*e^(4*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(4*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(4*
d)*log(F) - 24*e^3*f^2*e^(4*d))*e^(4*e*x) + 4*(b^3*c^3*f^2*e^(3*d)*log(F)^3 - 9*b^2*c^2*e*f^2*e^(3*d)*log(F)^2
+ 26*b*c*e^2*f^2*e^(3*d)*log(F) - 24*e^3*f^2*e^(3*d))*e^(3*e*x) + 6*(b^3*c^3*f^2*e^(2*d)*log(F)^3 - 9*b^2*c^2
*e*f^2*e^(2*d)*log(F)^2 + 26*b*c*e^2*f^2*e^(2*d)*log(F) - 24*e^3*f^2*e^(2*d))*e^(2*e*x) + 4*(b^3*c^3*f^2*e^d*l
og(F)^3 - 9*b^2*c^2*e*f^2*e^d*log(F)^2 + 26*b*c*e^2*f^2*e^d*log(F) - 24*e^3*f^2*e^d)*e^(e*x))

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{F^{b c x + a c}}{f^{2} \cosh \left (e x + d\right )^{2} + 2 \, f^{2} \cosh \left (e x + d\right ) + f^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{F^{a c} F^{b c x}}{\cosh ^{2}{\left (d + e x \right )} + 2 \cosh{\left (d + e x \right )} + 1}\, dx}{f^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))/(f+f*cosh(e*x+d))**2,x)

[Out]

Integral(F**(a*c)*F**(b*c*x)/(cosh(d + e*x)**2 + 2*cosh(d + e*x) + 1), x)/f**2

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (f \cosh \left (e x + d\right ) + f\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))/(f+f*cosh(e*x+d))^2,x, algorithm="giac")

[Out]

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