3.165 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{-m} \, dx\)

Optimal. Leaf size=101 \[ -\frac{c 2^{\frac{1}{2}-m} \tan (e+f x) (1-\sec (e+f x))^{m+\frac{1}{2}} (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-1} \text{Hypergeometric2F1}\left (m+\frac{1}{2},m+\frac{1}{2},m+\frac{3}{2},\frac{1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1)} \]

[Out]

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

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Rubi [A]  time = 0.129243, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {3961, 70, 69} \[ -\frac{c 2^{\frac{1}{2}-m} \tan (e+f x) (1-\sec (e+f x))^{m+\frac{1}{2}} (a \sec (e+f x)+a)^m (c-c \sec (e+f x))^{-m-1} \, _2F_1\left (m+\frac{1}{2},m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3961

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

Rule 70

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

Rule 69

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

Rubi steps

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

Mathematica [C]  time = 1.44974, size = 257, normalized size = 2.54 \[ \frac{2^{m-1} \left (-i e^{-\frac{1}{2} i (e+f x)} \left (-1+e^{i (e+f x)}\right )\right )^{-2 m} \left (\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^{-m} \left (\frac{\left (1+e^{i (e+f x)}\right )^2}{1+e^{2 i (e+f x)}}\right )^m \sin ^{2 m}\left (\frac{1}{2} (e+f x)\right ) \left (\frac{\sec (e+f x)}{\sec (e+f x)+1}\right )^m \left (\text{Hypergeometric2F1}\left (1,-2 m,1-2 m,\frac{i \left (-1+e^{i (e+f x)}\right )}{1+e^{i (e+f x)}}\right )-\text{Hypergeometric2F1}\left (1,-2 m,1-2 m,-\frac{i \left (-1+e^{i (e+f x)}\right )}{1+e^{i (e+f x)}}\right )\right ) (a (\sec (e+f x)+1))^m (c-c \sec (e+f x))^{-m}}{f m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-1 + m)*((1 + E^(I*(e + f*x)))^2/(1 + E^((2*I)*(e + f*x))))^m*(-Hypergeometric2F1[1, -2*m, 1 - 2*m, ((-I)*
(-1 + E^(I*(e + f*x))))/(1 + E^(I*(e + f*x)))] + Hypergeometric2F1[1, -2*m, 1 - 2*m, (I*(-1 + E^(I*(e + f*x)))
)/(1 + E^(I*(e + f*x)))])*(Sec[e + f*x]/(1 + Sec[e + f*x]))^m*(a*(1 + Sec[e + f*x]))^m*Sin[(e + f*x)/2]^(2*m))
/((((-I)*(-1 + E^(I*(e + f*x))))/E^((I/2)*(e + f*x)))^(2*m)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*f*m*
(c - c*Sec[e + f*x])^m)

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Maple [F]  time = 0.52, size = 0, normalized size = 0. \begin{align*} \int{\frac{\sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m}}{ \left ( c-c\sec \left ( fx+e \right ) \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [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(sec(f*x+e)*(a+a*sec(f*x+e))**m/((c-c*sec(f*x+e))**m),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m/((c-c*sec(f*x+e))^m),x, algorithm="giac")

[Out]

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