3.231 \(\int (c (d \sec (e+f x))^p)^n (a+a \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=275 \[ -\frac{a^3 (4 n p+1) \sin (e+f x) \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (1-n p),\frac{1}{2} (3-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{a^3 (4 n p+7) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n p}{2},\frac{1}{2} (2-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p (n p+2) \sqrt{\sin ^2(e+f x)}}+\frac{a^3 (2 n p+5) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1) (n p+2)}+\frac{\tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right ) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+2)} \]

[Out]

(a^3*(7 + 4*n*p)*Hypergeometric2F1[1/2, -(n*p)/2, (2 - n*p)/2, Cos[e + f*x]^2]*(c*(d*Sec[e + f*x])^p)^n*Sin[e
+ f*x])/(f*n*p*(2 + n*p)*Sqrt[Sin[e + f*x]^2]) - (a^3*(1 + 4*n*p)*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 - n*p
)/2, (3 - n*p)/2, Cos[e + f*x]^2]*(c*(d*Sec[e + f*x])^p)^n*Sin[e + f*x])/(f*(1 - n^2*p^2)*Sqrt[Sin[e + f*x]^2]
) + (a^3*(5 + 2*n*p)*(c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(f*(1 + n*p)*(2 + n*p)) + ((c*(d*Sec[e + f*x])^p)^
n*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(f*(2 + n*p))

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Rubi [A]  time = 0.436603, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3948, 3814, 3997, 3787, 3772, 2643} \[ -\frac{a^3 (4 n p+1) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{a^3 (4 n p+7) \sin (e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{f n p (n p+2) \sqrt{\sin ^2(e+f x)}}+\frac{a^3 (2 n p+5) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+1) (n p+2)}+\frac{\tan (e+f x) \left (a^3 \sec (e+f x)+a^3\right ) \left (c (d \sec (e+f x))^p\right )^n}{f (n p+2)} \]

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^3,x]

[Out]

(a^3*(7 + 4*n*p)*Hypergeometric2F1[1/2, -(n*p)/2, (2 - n*p)/2, Cos[e + f*x]^2]*(c*(d*Sec[e + f*x])^p)^n*Sin[e
+ f*x])/(f*n*p*(2 + n*p)*Sqrt[Sin[e + f*x]^2]) - (a^3*(1 + 4*n*p)*Cos[e + f*x]*Hypergeometric2F1[1/2, (1 - n*p
)/2, (3 - n*p)/2, Cos[e + f*x]^2]*(c*(d*Sec[e + f*x])^p)^n*Sin[e + f*x])/(f*(1 - n^2*p^2)*Sqrt[Sin[e + f*x]^2]
) + (a^3*(5 + 2*n*p)*(c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(f*(1 + n*p)*(2 + n*p)) + ((c*(d*Sec[e + f*x])^p)^
n*(a^3 + a^3*Sec[e + f*x])*Tan[e + f*x])/(f*(2 + n*p))

Rule 3948

Int[((c_.)*((d_.)*sec[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Sec[e + f*x])^p)^FracPart[n])/(d*Sec[e + f*x])^(p*FracPart[n]), Int[(a + b*Sec[e
+ f*x])^m*(d*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[n]

Rule 3814

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[b/(m + n - 1), Int[(a
+ b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n*(b*(m + 2*n - 1) + a*(3*m + 2*n - 4)*Csc[e + f*x]), x], x] /; Fr
eeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, 1] && NeQ[m + n - 1, 0] && IntegerQ[2*m]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^3 \, dx &=\left ((d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} (a+a \sec (e+f x))^3 \, dx\\ &=\frac{\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac{\left (a (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} (a+a \sec (e+f x)) (a (2+2 n p)+a (5+2 n p) \sec (e+f x)) \, dx}{2+n p}\\ &=\frac{a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac{\left (a (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} \left (a^2 (2+n p) (1+4 n p)+a^2 (1+n p) (7+4 n p) \sec (e+f x)\right ) \, dx}{(1+n p) (2+n p)}\\ &=\frac{a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac{\left (a^3 (1+4 n p) (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{n p} \, dx}{1+n p}+\frac{\left (a^3 (7+4 n p) (d \sec (e+f x))^{-n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int (d \sec (e+f x))^{1+n p} \, dx}{d (2+n p)}\\ &=\frac{a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}+\frac{\left (a^3 (1+4 n p) \left (\frac{\cos (e+f x)}{d}\right )^{n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-n p} \, dx}{1+n p}+\frac{\left (a^3 (7+4 n p) \left (\frac{\cos (e+f x)}{d}\right )^{n p} \left (c (d \sec (e+f x))^p\right )^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-1-n p} \, dx}{d (2+n p)}\\ &=\frac{a^3 (7+4 n p) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f n p (2+n p) \sqrt{\sin ^2(e+f x)}}-\frac{a^3 (1+4 n p) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f \left (1-n^2 p^2\right ) \sqrt{\sin ^2(e+f x)}}+\frac{a^3 (5+2 n p) \left (c (d \sec (e+f x))^p\right )^n \tan (e+f x)}{f (1+n p) (2+n p)}+\frac{\left (c (d \sec (e+f x))^p\right )^n \left (a^3+a^3 \sec (e+f x)\right ) \tan (e+f x)}{f (2+n p)}\\ \end{align*}

Mathematica [F]  time = 2.39777, size = 0, normalized size = 0. \[ \int \left (c (d \sec (e+f x))^p\right )^n (a+a \sec (e+f x))^3 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^3,x]

[Out]

Integrate[(c*(d*Sec[e + f*x])^p)^n*(a + a*Sec[e + f*x])^3, x]

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Maple [F]  time = 0.173, size = 0, normalized size = 0. \begin{align*} \int \left ( c \left ( d\sec \left ( fx+e \right ) \right ) ^{p} \right ) ^{n} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,x)

[Out]

int((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,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((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{3} \sec \left (f x + e\right )^{3} + 3 \, a^{3} \sec \left (f x + e\right )^{2} + 3 \, a^{3} \sec \left (f x + e\right ) + a^{3}\right )} \left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

integral((a^3*sec(f*x + e)^3 + 3*a^3*sec(f*x + e)^2 + 3*a^3*sec(f*x + e) + a^3)*((d*sec(f*x + e))^p*c)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n}\, dx + \int 3 \left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n} \sec{\left (e + f x \right )}\, dx + \int 3 \left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n} \sec ^{3}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sec(f*x+e))**p)**n*(a+a*sec(f*x+e))**3,x)

[Out]

a**3*(Integral((c*(d*sec(e + f*x))**p)**n, x) + Integral(3*(c*(d*sec(e + f*x))**p)**n*sec(e + f*x), x) + Integ
ral(3*(c*(d*sec(e + f*x))**p)**n*sec(e + f*x)**2, x) + Integral((c*(d*sec(e + f*x))**p)**n*sec(e + f*x)**3, x)
)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (f x + e\right ) + a\right )}^{3} \left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(d*sec(f*x+e))^p)^n*(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^3*((d*sec(f*x + e))^p*c)^n, x)