3.291 \(\int F^{c (a+b x)} \text {sech}^4(d+e x) \, dx\)

Optimal. Leaf size=133 \[ \frac {2 e^{2 (d+e x)} F^{c (a+b x)} (2 e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{2 e}+1;\frac {b c \log (F)}{2 e}+2;-e^{2 (d+e x)}\right )}{3 e^2}+\frac {b c \log (F) \text {sech}^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tanh (d+e x) \text {sech}^2(d+e x) F^{c (a+b x)}}{3 e} \]

[Out]

2/3*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))*(2*e-b*
c*ln(F))/e^2+1/6*b*c*F^(c*(b*x+a))*ln(F)*sech(e*x+d)^2/e^2+1/3*F^(c*(b*x+a))*sech(e*x+d)^2*tanh(e*x+d)/e

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Rubi [A]  time = 0.06, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {5490, 5492} \[ \frac {2 e^{2 (d+e x)} F^{c (a+b x)} (2 e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{2 e}+1;\frac {b c \log (F)}{2 e}+2;-e^{2 (d+e x)}\right )}{3 e^2}+\frac {b c \log (F) \text {sech}^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac {\tanh (d+e x) \text {sech}^2(d+e x) F^{c (a+b x)}}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sech[d + e*x]^4,x]

[Out]

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

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

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Mathematica [A]  time = 0.20, size = 101, normalized size = 0.76 \[ \frac {F^{c (a+b x)} \left (4 e^{2 (d+e x)} (2 e-b c \log (F)) \, _2F_1\left (2,\frac {b c \log (F)}{2 e}+1;\frac {b c \log (F)}{2 e}+2;-e^{2 (d+e x)}\right )+\text {sech}^2(d+e x) (b c \log (F)+2 e \tanh (d+e x))\right )}{6 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Sech[d + e*x]^4,x]

[Out]

(F^(c*(a + b*x))*(4*E^(2*(d + e*x))*Hypergeometric2F1[2, 1 + (b*c*Log[F])/(2*e), 2 + (b*c*Log[F])/(2*e), -E^(2
*(d + e*x))]*(2*e - b*c*Log[F]) + Sech[d + e*x]^2*(b*c*Log[F] + 2*e*Tanh[d + e*x])))/(6*e^2)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (F^{b c x + a c} \operatorname {sech}\left (e x + d\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^4,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*sech(e*x + d)^4, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^4,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*sech(e*x + d)^4, x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int F^{c \left (b x +a \right )} \mathrm {sech}\left (e x +d \right )^{4}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*sech(e*x+d)^4,x)

[Out]

int(F^(c*(b*x+a))*sech(e*x+d)^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^4,x, algorithm="maxima")

[Out]

-128*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2*F^(a*c)*b*c*e^2*log(F))*integrate(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^(10*d)*log(F)^3 - 18*b^2*c^2*e*e^(10*d)*log(F)^2 + 10
4*b*c*e^2*e^(10*d)*log(F) - 192*e^3*e^(10*d))*e^(10*e*x) + 5*(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) + 10*(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) + 10*(b^3*c^3*e^(4*d)*log(F)^3 -
18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e^2*e^(4*d)*log(F) - 192*e^3*e^(4*d))*e^(4*e*x) + 5*(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)), x) + 16*(
8*F^(a*c)*b*c*e*log(F) + 16*F^(a*c)*e^2 + (F^(a*c)*b^2*c^2*e^(4*d)*log(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) - 8*(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 - 1
8*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)*lo
g(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) + 6*(b^3*c^3*
e^(4*d)*log(F)^3 - 18*b^2*c^2*e*e^(4*d)*log(F)^2 + 104*b*c*e^2*e^(4*d)*log(F) - 192*e^3*e^(4*d))*e^(4*e*x) + 4
*(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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/cosh(d + e*x)^4,x)

[Out]

int(F^(c*(a + b*x))/cosh(d + e*x)^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(c*(b*x+a))*sech(e*x+d)**4,x)

[Out]

Timed out

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