3.2.36 \(\int \frac {(c-c \sec (e+f x))^n}{a+a \sec (e+f x)} \, dx\) [136]

Optimal. Leaf size=99 \[ -\frac {2^{\frac {1}{2}+n} c F_1\left (-\frac {1}{2};\frac {1}{2}-n,1;\frac {1}{2};\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac {1}{2}-n} (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{f (a+a \sec (e+f x))} \]

[Out]

-2^(1/2+n)*c*AppellF1(-1/2,1,1/2-n,1/2,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(1-sec(f*x+e))^(1/2-n)*(c-c*sec(f*x+e)
)^(-1+n)*tan(f*x+e)/f/(a+a*sec(f*x+e))

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Rubi [A]
time = 0.07, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 = {3997, 142, 141} \begin {gather*} -\frac {c 2^{n+\frac {1}{2}} \tan (e+f x) (1-\sec (e+f x))^{\frac {1}{2}-n} F_1\left (-\frac {1}{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}}{f (a \sec (e+f x)+a)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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])

Rule 142

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 3997

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]

Rubi steps

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

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Mathematica [F]
time = 0.99, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(c-c \sec (e+f x))^n}{a+a \sec (e+f x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {\left (c -c \sec \left (f x +e \right )\right )^{n}}{a +a \sec \left (f x +e \right )}\, dx\]

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)),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (- c \sec {\left (e + f x \right )} + c\right )^{n}}{\sec {\left (e + f x \right )} + 1}\, dx}{a} \end {gather*}

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)),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n}{a+\frac {a}{\cos \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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