3.235 \(\int \frac{(c (d \sec (e+f x))^p)^n}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=248 \[ \frac{2 (2-n p) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},-\frac{n p}{2},\frac{1}{2} (2-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{(3-2 n p) \sin (e+f x) \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{1}{2} (1-n p),\frac{1}{2} (3-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{2 (2-n p) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f (\sec (e+f x)+1)}-\frac{\tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

(2*(2 - 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])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) - ((3 - 2*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])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) - (2*(2 - n*p)*(c*(d*S
ec[e + f*x])^p)^n*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - ((c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(3*f*(a
+ a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.451474, antiderivative size = 248, 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, 3817, 4020, 3787, 3772, 2643} \[ \frac{2 (2-n p) \sin (e+f x) \, _2F_1\left (\frac{1}{2},-\frac{n p}{2};\frac{1}{2} (2-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{(3-2 n p) \sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1}{2} (1-n p);\frac{1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f \sqrt{\sin ^2(e+f x)}}-\frac{2 (2-n p) \tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{3 a^2 f (\sec (e+f x)+1)}-\frac{\tan (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*(d*Sec[e + f*x])^p)^n/(a + a*Sec[e + f*x])^2,x]

[Out]

(2*(2 - 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])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) - ((3 - 2*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])/(3*a^2*f*Sqrt[Sin[e + f*x]^2]) - (2*(2 - n*p)*(c*(d*S
ec[e + f*x])^p)^n*Tan[e + f*x])/(3*a^2*f*(1 + Sec[e + f*x])) - ((c*(d*Sec[e + f*x])^p)^n*Tan[e + f*x])/(3*f*(a
+ a*Sec[e + f*x])^2)

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 3817

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

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

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

Mathematica [F]  time = 1.79699, size = 0, normalized size = 0. \[ \int \frac{\left (c (d \sec (e+f x))^p\right )^n}{(a+a \sec (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c*(d*Sec[e + f*x])^p)^n/(a + a*Sec[e + f*x])^2,x]

[Out]

Integrate[(c*(d*Sec[e + f*x])^p)^n/(a + a*Sec[e + f*x])^2, x]

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Maple [F]  time = 0.161, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c \left ( d\sec \left ( fx+e \right ) \right ) ^{p} \right ) ^{n}}{ \left ( a+a\sec \left ( fx+e \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int((c*(d*sec(f*x+e))^p)^n/(a+a*sec(f*x+e))^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((c*(d*sec(f*x+e))^p)^n/(a+a*sec(f*x+e))^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 (\frac{\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (c \left (d \sec{\left (e + f x \right )}\right )^{p}\right )^{n}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec{\left (e + f x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

Integral((c*(d*sec(e + f*x))**p)**n/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x)/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

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