3.14 \(\int F^{c (a+b x)} \sec (d+e x) \, dx\)

Optimal. Leaf size=84 \[ \frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (1,\frac {e-i b c \log (F)}{2 e};\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right );-e^{2 i (d+e x)}\right )}{b c \log (F)+i e} \]

[Out]

2*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([1, 1/2*(e-I*b*c*ln(F))/e],[3/2-1/2*I*b*c*ln(F)/e],-exp(2*I*(e*x+d)))
/(b*c*ln(F)+I*e)

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Rubi [A]  time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4451} \[ \frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (1,\frac {e-i b c \log (F)}{2 e};\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right );-e^{2 i (d+e x)}\right )}{b c \log (F)+i e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sec[d + e*x],x]

[Out]

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

Rule 4451

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(I*n*(d + e*x))*
F^(c*(a + b*x))*Hypergeometric2F1[n, n/2 - (I*b*c*Log[F])/(2*e), 1 + n/2 - (I*b*c*Log[F])/(2*e), -E^(2*I*(d +
e*x))])/(I*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)} \sec (d+e x) \, dx &=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (1,\frac {e-i b c \log (F)}{2 e};\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right );-e^{2 i (d+e x)}\right )}{i e+b c \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 84, normalized size = 1.00 \[ \frac {2 e^{i (d+e x)} F^{c (a+b x)} \, _2F_1\left (1,\frac {1}{2}-\frac {i b c \log (F)}{2 e};\frac {3}{2}-\frac {i b c \log (F)}{2 e};-e^{2 i (d+e x)}\right )}{b c \log (F)+i e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (F^{b c x + a c} \sec \left (e x + d\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int F^{c \left (b x +a \right )} \sec \left (e x +d \right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [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))*sec(e*x+d),x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(c*(a + b*x))/cos(d + e*x),x)

[Out]

int(F^(c*(a + b*x))/cos(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{c \left (a + b x\right )} \sec {\left (d + e x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(F**(c*(a + b*x))*sec(d + e*x), x)

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