3.141 \(\int \frac{(c-c \sec (e+f x))^n}{\sqrt{a+a \sec (e+f x)}} \, dx\)

Optimal. Leaf size=139 \[ \frac{2 \tan (e+f x) (c-c \sec (e+f x))^n \text{Hypergeometric2F1}\left (1,n+\frac{1}{2},n+\frac{3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) (c-c \sec (e+f x))^n \text{Hypergeometric2F1}\left (1,n+\frac{1}{2},n+\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.112058, antiderivative size = 139, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3912, 86, 65, 68} \[ \frac{2 \tan (e+f x) (c-c \sec (e+f x))^n \, _2F_1\left (1,n+\frac{1}{2};n+\frac{3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) (c-c \sec (e+f x))^n \, _2F_1\left (1,n+\frac{1}{2};n+\frac{3}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^(n - 1/2))/x, x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0]

Rule 86

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), In
t[(e + f*x)^p/(a + b*x), x], x] - Dist[d/(b*c - a*d), Int[(e + f*x)^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] &&  !IntegerQ[p]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [F]  time = 1.36948, size = 0, normalized size = 0. \[ \int \frac{(c-c \sec (e+f x))^n}{\sqrt{a+a \sec (e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.295, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c-c\sec \left ( fx+e \right ) \right ) ^{n}{\frac{1}{\sqrt{a+a\sec \left ( fx+e \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(1/2),x)

[Out]

int((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(1/2),x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((-c*sec(f*x + e) + c)^n/sqrt(a*sec(f*x + e) + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((-c*sec(f*x + e) + c)^n/sqrt(a*sec(f*x + e) + a), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))**n/(a+a*sec(f*x+e))**(1/2),x)

[Out]

Integral((-c*(sec(e + f*x) - 1))**n/sqrt(a*(sec(e + f*x) + 1)), x)

________________________________________________________________________________________

Giac [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-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out