3.15 \(\int F^{c (a+b x)} \sec ^2(d+e x) \, dx\)

Optimal. Leaf size=80 \[ \frac{4 e^{2 i (d+e x)} F^{c (a+b x)} \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{2 e},2-\frac{i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)+2 i e} \]

[Out]

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

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Rubi [A]  time = 0.023731, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {4451} \[ \frac{4 e^{2 i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac{i b c \log (F)}{2 e};2-\frac{i b c \log (F)}{2 e};-e^{2 i (d+e x)}\right )}{b c \log (F)+2 i e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sec[d + e*x]^2,x]

[Out]

(4*E^((2*I)*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - ((I/2)*b*c*Log[F])/e, 2 - ((I/2)*b*c*Log[F])/e
, -E^((2*I)*(d + e*x))])/((2*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 ^2(d+e x) \, dx &=\frac{4 e^{2 i (d+e x)} F^{c (a+b x)} \, _2F_1\left (2,1-\frac{i b c \log (F)}{2 e};2-\frac{i b c \log (F)}{2 e};-e^{2 i (d+e x)}\right )}{2 i e+b c \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0155679, size = 80, normalized size = 1. \[ \frac{4 e^{2 i (d+e x)} F^{c (a+b x)} \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{2 e},2-\frac{i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)+2 i e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ( \sec \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))*sec(e*x+d)^2,x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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