3.17 \(\int F^{c (a+b x)} \sec ^4(d+e x) \, dx\)

Optimal. Leaf size=143 \[ -\frac{2 e^{2 i (d+e x)} F^{c (a+b x)} (-b c \log (F)+2 i e) \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{2 e},2-\frac{i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{3 e^2}-\frac{b c \log (F) \sec ^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac{\tan (d+e x) \sec ^2(d+e x) F^{c (a+b x)}}{3 e} \]

[Out]

(-2*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]))/(3*e^2) - (b*c*F^(c*(a + b*x))*Log[F]*Sec[d + e*x]^2)/(6*e^2)
+ (F^(c*(a + b*x))*Sec[d + e*x]^2*Tan[d + e*x])/(3*e)

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Rubi [A]  time = 0.0525938, antiderivative size = 143, normalized size of antiderivative = 1., 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 = {4448, 4451} \[ -\frac{2 e^{2 i (d+e x)} F^{c (a+b x)} (-b c \log (F)+2 i e) \, _2F_1\left (2,1-\frac{i b c \log (F)}{2 e};2-\frac{i b c \log (F)}{2 e};-e^{2 i (d+e x)}\right )}{3 e^2}-\frac{b c \log (F) \sec ^2(d+e x) F^{c (a+b x)}}{6 e^2}+\frac{\tan (d+e x) \sec ^2(d+e x) F^{c (a+b x)}}{3 e} \]

Antiderivative was successfully verified.

[In]

Int[F^(c*(a + b*x))*Sec[d + e*x]^4,x]

[Out]

(-2*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]))/(3*e^2) - (b*c*F^(c*(a + b*x))*Log[F]*Sec[d + e*x]^2)/(6*e^2)
+ (F^(c*(a + b*x))*Sec[d + e*x]^2*Tan[d + e*x])/(3*e)

Rule 4448

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

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 ^4(d+e x) \, dx &=-\frac{b c F^{c (a+b x)} \log (F) \sec ^2(d+e x)}{6 e^2}+\frac{F^{c (a+b x)} \sec ^2(d+e x) \tan (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)} \sec ^2(d+e x) \, dx\\ &=-\frac{2 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))}{3 e^2}-\frac{b c F^{c (a+b x)} \log (F) \sec ^2(d+e x)}{6 e^2}+\frac{F^{c (a+b x)} \sec ^2(d+e x) \tan (d+e x)}{3 e}\\ \end{align*}

Mathematica [A]  time = 0.198649, size = 111, normalized size = 0.78 \[ \frac{F^{c (a+b x)} \left (\sec ^2(d+e x) (2 e \tan (d+e x)-b c \log (F))+4 e^{2 i (d+e x)} (b c \log (F)-2 i e) \text{Hypergeometric2F1}\left (2,1-\frac{i b c \log (F)}{2 e},2-\frac{i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )\right )}{6 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(c*(a + b*x))*(4*E^((2*I)*(d + e*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]) + Sec[d + e*x]^2*(-(b*c*Log[F]) + 2*e*Tan[d + e*x])))/(6*e^2)

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Maple [F]  time = 0.235, size = 0, normalized size = 0. \begin{align*} \int{F}^{c \left ( bx+a \right ) } \left ( \sec \left ( ex+d \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(F^(c*(b*x+a))*sec(e*x+d)^4,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)^4,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 )^{4}, 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)^4,x, algorithm="fricas")

[Out]

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

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Sympy [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)**4,x)

[Out]

Timed out

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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 )^{4}\,{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)^4,x, algorithm="giac")

[Out]

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