\(\int \frac {(c (d \sec (e+f x))^p)^n}{a+a \sec (e+f x)} \, dx\) [234]

Optimal result

Integrand size = 27, antiderivative size = 208 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\frac {\left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{f (a+a \sec (e+f x))}-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{a f \sqrt {\sin ^2(e+f x)}}+\frac {(1-n p) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2-n p),\frac {1}{2} (4-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \sin (e+f x)}{a f (2-n p) \sqrt {\sin ^2(e+f x)}} \]

[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)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4033, 3905, 3872, 3857, 2722} \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=-\frac {\sin (e+f x) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (1-n p),\frac {1}{2} (3-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f \sqrt {\sin ^2(e+f x)}}+\frac {(1-n p) \sin (e+f x) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (2-n p),\frac {1}{2} (4-n p),\cos ^2(e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n}{a f (2-n p) \sqrt {\sin ^2(e+f x)}}+\frac {\sin (e+f x) \left (c (d \sec (e+f x))^p\right )^n}{f (a \sec (e+f x)+a)} \]

[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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 3872

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 3905

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 4033

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*} \text {integral}& = \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) \operatorname {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 \sin (e+f x)}{a f \sqrt {\sin ^2(e+f x)}}+\frac {(1-n p) \cos ^2(e+f x) \operatorname {Hypergeometric2F1}\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*}

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.76 \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\frac {\cot \left (\frac {1}{2} (e+f x)\right ) \left (c (d \sec (e+f x))^p\right )^n \left (-\left ((-1+n p) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n p}{2},1+\frac {n p}{2},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )+n p \left (1-\cos (e+f x)+\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+n p),\frac {1}{2} (1+n p),\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )\right )}{a f n p (1+\sec (e+f x))} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (c \left (d \sec \left (f x +e \right )\right )^{p}\right )^{n}}{a +a \sec \left (f x +e \right )}d x\]

[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)

Fricas [F]

\[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\int { \frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a} \,d x } \]

[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)

Sympy [F]

\[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\frac {\int \frac {\left (c \left (d \sec {\left (e + f x \right )}\right )^{p}\right )^{n}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \]

[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

Maxima [F]

\[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\int { \frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\int { \frac {\left (\left (d \sec \left (f x + e\right )\right )^{p} c\right )^{n}}{a \sec \left (f x + e\right ) + a} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c (d \sec (e+f x))^p\right )^n}{a+a \sec (e+f x)} \, dx=\int \frac {{\left (c\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^p\right )}^n}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \]

[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)