3.2.34 \(\int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^n \, dx\) [134]

Optimal. Leaf size=101 \[ \frac {2^{\frac {1}{2}+n} c F_1\left (\frac {5}{2};\frac {1}{2}-n,1;\frac {7}{2};\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac {1}{2}-n} (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{5 f} \]

[Out]

1/5*2^(1/2+n)*c*AppellF1(5/2,1,1/2-n,7/2,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(1-sec(f*x+e))^(1/2-n)*(a+a*sec(f*x+
e))^2*(c-c*sec(f*x+e))^(-1+n)*tan(f*x+e)/f

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Rubi [A]
time = 0.07, antiderivative size = 101, 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) (a \sec (e+f x)+a)^2 (1-\sec (e+f x))^{\frac {1}{2}-n} F_1\left (\frac {5}{2};\frac {1}{2}-n,1;\frac {7}{2};\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right ) (c-c \sec (e+f x))^{n-1}}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + n)*c*AppellF1[5/2, 1/2 - n, 1, 7/2, (1 + Sec[e + f*x])/2, 1 + Sec[e + f*x]]*(1 - Sec[e + f*x])^(1/2
- n)*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(-1 + n)*Tan[e + f*x])/(5*f)

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 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^n \, dx &=-\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \frac {(a+a x)^{3/2} (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 {\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} (a+a x)^{3/2}}{x} \, 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 {5}{2};\frac {1}{2}-n,1;\frac {7}{2};\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (1-\sec (e+f x))^{\frac {1}{2}-n} (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{-1+n} \tan (e+f x)}{5 f}\\ \end {align*}

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Mathematica [F]
time = 1.57, size = 0, normalized size = 0.00 \begin {gather*} \int (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^n \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \left (a +a \sec \left (f x +e \right )\right )^{2} \left (c -c \sec \left (f x +e \right )\right )^{n}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int 2 \left (- c \sec {\left (e + f x \right )} + c\right )^{n} \sec {\left (e + f x \right )}\, dx + \int \left (- c \sec {\left (e + f x \right )} + c\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx + \int \left (- c \sec {\left (e + f x \right )} + c\right )^{n}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*(-c*sec(e + f*x) + c)**n*sec(e + f*x), x) + Integral((-c*sec(e + f*x) + c)**n*sec(e + f*x)**2
, x) + Integral((-c*sec(e + f*x) + c)**n, x))

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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