3.305 \(\int \sec ^4(e+f x) (a+b \sec ^2(e+f x))^p \, dx\)

Optimal. Leaf size=129 \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac{(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{1}{2},-p,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]

[Out]

(Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(b*f*(3 + 2*p)) - ((a - 2*b*(1 + p))*Hypergeometric2F1[1/2,
-p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^p)/(b*f*(3 + 2*p)*(1 + (b*Tan[
e + f*x]^2)/(a + b))^p)

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Rubi [A]  time = 0.0992134, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {4146, 388, 246, 245} \[ \frac{\tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)}-\frac{(a-2 b (p+1)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right )}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

(Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^(1 + p))/(b*f*(3 + 2*p)) - ((a - 2*b*(1 + p))*Hypergeometric2F1[1/2,
-p, 3/2, -((b*Tan[e + f*x]^2)/(a + b))]*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2)^p)/(b*f*(3 + 2*p)*(1 + (b*Tan[
e + f*x]^2)/(a + b))^p)

Rule 4146

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/
2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[n/2]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^FracPart[p])/(1 + (b*x^n)/a)^Fr
acPart[p], Int[(1 + (b*x^n)/a)^p, x], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILt
Q[Simplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sec ^4(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \left (1+x^2\right ) \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{(a-2 b (1+p)) \operatorname{Subst}\left (\int \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{\left ((a-2 b (1+p)) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \left (1+\frac{b x^2}{a+b}\right )^p \, dx,x,\tan (e+f x)\right )}{b f (3+2 p)}\\ &=\frac{\tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}-\frac{(a-2 b (1+p)) \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac{b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b f (3+2 p)}\\ \end{align*}

Mathematica [A]  time = 2.26217, size = 126, normalized size = 0.98 \[ \frac{\tan (e+f x) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p \left ((2 b (p+1)-a) \text{Hypergeometric2F1}\left (\frac{1}{2},-p,\frac{3}{2},-\frac{b \tan ^2(e+f x)}{a+b}\right )+\left (a+b \tan ^2(e+f x)+b\right ) \left (\frac{b \tan ^2(e+f x)}{a+b}+1\right )^p\right )}{b f (2 p+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(a + b*Sec[e + f*x]^2)^p,x]

[Out]

((a + b*Sec[e + f*x]^2)^p*Tan[e + f*x]*((-a + 2*b*(1 + p))*Hypergeometric2F1[1/2, -p, 3/2, -((b*Tan[e + f*x]^2
)/(a + b))] + (a + b + b*Tan[e + f*x]^2)*(1 + (b*Tan[e + f*x]^2)/(a + b))^p))/(b*f*(3 + 2*p)*(1 + (b*Tan[e + f
*x]^2)/(a + b))^p)

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Maple [F]  time = 0.426, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( fx+e \right ) \right ) ^{4} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)

[Out]

int(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^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(sec(f*x+e)**4*(a+b*sec(f*x+e)**2)**p,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^p*sec(f*x + e)^4, x)