3.137 \(\int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{c 2^{n+\frac{1}{2}} \tan (e+f x) (1-\sec (e+f x))^{\frac{1}{2}-n} F_1\left (-\frac{3}{2};\frac{1}{2}-n,1;-\frac{1}{2};\frac{1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-(2^(1/2 + n)*c*AppellF1[-3/2, 1/2 - n, 1, -1/2, (1 + Sec[e + f*x])/2, 1 + Sec[e + f*x]]*(1 - Sec[e + f*x])^(1
/2 - n)*(c - c*Sec[e + f*x])^(-1 + n)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

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Rubi [A]  time = 0.10022, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3912, 137, 136} \[ -\frac{c 2^{n+\frac{1}{2}} \tan (e+f x) (1-\sec (e+f x))^{\frac{1}{2}-n} F_1\left (-\frac{3}{2};\frac{1}{2}-n,1;-\frac{1}{2};\frac{1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2^(1/2 + n)*c*AppellF1[-3/2, 1/2 - n, 1, -1/2, (1 + Sec[e + f*x])/2, 1 + Sec[e + f*x]]*(1 - Sec[e + f*x])^(1
/2 - n)*(c - c*Sec[e + f*x])^(-1 + n)*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2)

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 137

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^
FracPart[n]/((b/(b*c - a*d))^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*((b*c)/(b*c
- a*d) + (b*d*x)/(b*c - a*d))^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&
!IntegerQ[n] && IntegerQ[p] &&  !GtQ[b/(b*c - a*d), 0] &&  !SimplerQ[c + d*x, a + b*x]

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^2} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c-c x)^{-\frac{1}{2}+n}}{x (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{\left (2^{-\frac{1}{2}+n} a c (c-c \sec (e+f x))^{-1+n} \left (\frac{c-c \sec (e+f x)}{c}\right )^{\frac{1}{2}-n} \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}-\frac{x}{2}\right )^{-\frac{1}{2}+n}}{x (a+a x)^{5/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+n} c F_1\left (-\frac{3}{2};\frac{1}{2}-n,1;-\frac{1}{2};\frac{1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac{1}{2}-n} (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end{align*}

Mathematica [F]  time = 1.57377, size = 0, normalized size = 0. \[ \int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.287, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c-c\sec \left ( fx+e \right ) \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-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^2,x)

[Out]

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

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

[Out]

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

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

[Out]

integral((-c*sec(f*x + e) + 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 \sec{\left (e + f x \right )} + c\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-c*sec(f*x+e))**n/(a+a*sec(f*x+e))**2,x)

[Out]

Integral((-c*sec(e + f*x) + c)**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 (-c \sec \left (f x + e\right ) + 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-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

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