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

Optimal. Leaf size=208 \[ -\frac {\sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f \sqrt {\sin ^2(e+f x)}}+\frac {(1-n p) \sin (e+f x) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2-n p);\frac {1}{2} (4-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f (2-n p) \sqrt {\sin ^2(e+f x)}}+\frac {\sin (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (a \sec (e+f x)+a)} \]

[Out]

(c*(d*sec(f*x+e))^p)^n*sin(f*x+e)/f/(a+a*sec(f*x+e))-cos(f*x+e)*hypergeom([1/2, -1/2*n*p+1/2],[-1/2*n*p+3/2],c
os(f*x+e)^2)*(c*(d*sec(f*x+e))^p)^n*sin(f*x+e)/a/f/(sin(f*x+e)^2)^(1/2)+(-n*p+1)*cos(f*x+e)^2*hypergeom([1/2,
-1/2*n*p+1],[-1/2*n*p+2],cos(f*x+e)^2)*(c*(d*sec(f*x+e))^p)^n*sin(f*x+e)/a/f/(-n*p+2)/(sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3948, 3820, 3787, 3772, 2643} \[ -\frac {\sin (e+f x) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (1-n p);\frac {1}{2} (3-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f \sqrt {\sin ^2(e+f x)}}+\frac {(1-n p) \sin (e+f x) \cos ^2(e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (2-n p);\frac {1}{2} (4-n p);\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f (2-n p) \sqrt {\sin ^2(e+f x)}}+\frac {\sin (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (a \sec (e+f x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 3820

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

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]

Rubi steps

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

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Mathematica [F]  time = 1.18, size = 0, normalized size = 0.00 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a}\,{d x} \]

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \left (d \sec \left (f x +e \right )\right )^{p}\right )^{n}}{a +a \sec \left (f x +e \right )}\, dx \]

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)),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a}\,{d x} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^p\right )}^n}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(d/cos(e + f*x))^p)^n/(a + a/cos(e + f*x)),x)

[Out]

int((c*(d/cos(e + f*x))^p)^n/(a + a/cos(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

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)),x)

[Out]

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

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