3.290 \(\int F^{c (a+b x)} \text {sech}^3(d+e x) \, dx\)

Optimal. Leaf size=124 \[ \frac {e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (1,\frac {e+b c \log (F)}{2 e};\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{e^2}+\frac {b c \log (F) \text {sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tanh (d+e x) \text {sech}(d+e x) F^{c (a+b x)}}{2 e} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 124, 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} \[ \frac {e^{d+e x} F^{c (a+b x)} (e-b c \log (F)) \, _2F_1\left (1,\frac {e+b c \log (F)}{2 e};\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )}{e^2}+\frac {b c \log (F) \text {sech}(d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tanh (d+e x) \text {sech}(d+e x) F^{c (a+b x)}}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sech[d + e*x]^3,x]

[Out]

(E^(d + e*x)*F^(c*(a + b*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]))/e^2 + (b*c*F^(c*(a + b*x))*Log[F]*Sech[d + e*x])/(2*e^2) + (F^(c*(a + b*x))*Sech[d + e
*x]*Tanh[d + e*x])/(2*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}^3(d+e x) \, dx &=\frac {b c F^{c (a+b x)} \log (F) \text {sech}(d+e x)}{2 e^2}+\frac {F^{c (a+b x)} \text {sech}(d+e x) \tanh (d+e x)}{2 e}+\frac {1}{2} \left (1-\frac {b^2 c^2 \log ^2(F)}{e^2}\right ) \int F^{c (a+b x)} \text {sech}(d+e x) \, dx\\ &=\frac {e^{d+e x} F^{c (a+b x)} \, _2F_1\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 ) (e-b c \log (F))}{e^2}+\frac {b c F^{c (a+b x)} \log (F) \text {sech}(d+e x)}{2 e^2}+\frac {F^{c (a+b x)} \text {sech}(d+e x) \tanh (d+e x)}{2 e}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 96, normalized size = 0.77 \[ \frac {F^{c (a+b x)} \left (2 e^{d+e x} (e-b c \log (F)) \, _2F_1\left (1,\frac {e+b c \log (F)}{2 e};\frac {1}{2} \left (\frac {b c \log (F)}{e}+3\right );-e^{2 (d+e x)}\right )+\text {sech}(d+e x) (b c \log (F)+e \tanh (d+e x))\right )}{2 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(c*(a + b*x))*Sech[d + e*x]^3,x]

[Out]

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

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (F^{b c x + a c} \operatorname {sech}\left (e x + d\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(F^(b*c*x + a*c)*sech(e*x + d)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{{\left (b x + a\right )} c} \operatorname {sech}\left (e x + d\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(F^((b*x + a)*c)*sech(e*x + d)^3, x)

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int F^{c \left (b x +a \right )} \mathrm {sech}\left (e x +d \right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(b*x+a))*sech(e*x+d)^3,x)

[Out]

int(F^(c*(b*x+a))*sech(e*x+d)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ 48 \, {\left (F^{a c} b c e e^{d} \log \relax (F) + F^{a c} e^{2} e^{d}\right )} \int \frac {e^{\left (b c x \log \relax (F) + e x\right )}}{b^{2} c^{2} \log \relax (F)^{2} - 8 \, b c e \log \relax (F) + 15 \, e^{2} + {\left (b^{2} c^{2} e^{\left (8 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (8 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (8 \, d\right )}\right )} e^{\left (8 \, e x\right )} + 4 \, {\left (b^{2} c^{2} e^{\left (6 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (6 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 6 \, {\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (4 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 4 \, {\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (2 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}}\,{d x} - \frac {8 \, {\left (6 \, F^{a c} e e^{\left (e x + d\right )} - {\left (F^{a c} b c e^{\left (3 \, d\right )} \log \relax (F) - 5 \, F^{a c} e e^{\left (3 \, d\right )}\right )} e^{\left (3 \, e x\right )}\right )} F^{b c x}}{b^{2} c^{2} \log \relax (F)^{2} - 8 \, b c e \log \relax (F) + 15 \, e^{2} + {\left (b^{2} c^{2} e^{\left (6 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (6 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (6 \, d\right )}\right )} e^{\left (6 \, e x\right )} + 3 \, {\left (b^{2} c^{2} e^{\left (4 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (4 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (4 \, d\right )}\right )} e^{\left (4 \, e x\right )} + 3 \, {\left (b^{2} c^{2} e^{\left (2 \, d\right )} \log \relax (F)^{2} - 8 \, b c e e^{\left (2 \, d\right )} \log \relax (F) + 15 \, e^{2} e^{\left (2 \, d\right )}\right )} e^{\left (2 \, e x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(c*(b*x+a))*sech(e*x+d)^3,x, algorithm="maxima")

[Out]

48*(F^(a*c)*b*c*e*e^d*log(F) + F^(a*c)*e^2*e^d)*integrate(e^(b*c*x*log(F) + e*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*l
og(F) + 15*e^2 + (b^2*c^2*e^(8*d)*log(F)^2 - 8*b*c*e*e^(8*d)*log(F) + 15*e^2*e^(8*d))*e^(8*e*x) + 4*(b^2*c^2*e
^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F) + 15*e^2*e^(6*d))*e^(6*e*x) + 6*(b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e
^(4*d)*log(F) + 15*e^2*e^(4*d))*e^(4*e*x) + 4*(b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d)*log(F) + 15*e^2*e^(2
*d))*e^(2*e*x)), x) - 8*(6*F^(a*c)*e*e^(e*x + d) - (F^(a*c)*b*c*e^(3*d)*log(F) - 5*F^(a*c)*e*e^(3*d))*e^(3*e*x
))*F^(b*c*x)/(b^2*c^2*log(F)^2 - 8*b*c*e*log(F) + 15*e^2 + (b^2*c^2*e^(6*d)*log(F)^2 - 8*b*c*e*e^(6*d)*log(F)
+ 15*e^2*e^(6*d))*e^(6*e*x) + 3*(b^2*c^2*e^(4*d)*log(F)^2 - 8*b*c*e*e^(4*d)*log(F) + 15*e^2*e^(4*d))*e^(4*e*x)
+ 3*(b^2*c^2*e^(2*d)*log(F)^2 - 8*b*c*e*e^(2*d)*log(F) + 15*e^2*e^(2*d))*e^(2*e*x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {F^{c\,\left (a+b\,x\right )}}{{\mathrm {cosh}\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/cosh(d + e*x)^3,x)

[Out]

int(F^(c*(a + b*x))/cosh(d + e*x)^3, 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)**3,x)

[Out]

Timed out

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