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

Optimal. Leaf size=69 \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^m \text{Hypergeometric2F1}\left (1,m+\frac{1}{2},m+\frac{3}{2},\frac{1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1) \sqrt{c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.129482, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3961, 68} \[ -\frac{\tan (e+f x) (a \sec (e+f x)+a)^m \, _2F_1\left (1,m+\frac{1}{2};m+\frac{3}{2};\frac{1}{2} (\sec (e+f x)+1)\right )}{f (2 m+1) \sqrt{c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [F]  time = 0.642719, size = 0, normalized size = 0. \[ \int \frac{\sec (e+f x) (a+a \sec (e+f x))^m}{\sqrt{c-c \sec (e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.305, size = 0, normalized size = 0. \begin{align*} \int{\sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m}{\frac{1}{\sqrt{c-c\sec \left ( fx+e \right ) }}}}\, 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))^(1/2),x)

[Out]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m/(c-c*sec(f*x+e))^(1/2),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 )}{\sqrt{-c \sec \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-c \sec \left (f x + e\right ) + c}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )}{c \sec \left (f x + e\right ) - c}, 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))^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a \left (\sec{\left (e + f x \right )} + 1\right )\right )^{m} \sec{\left (e + f x \right )}}{\sqrt{- c \left (\sec{\left (e + f x \right )} - 1\right )}}\, dx \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))**(1/2),x)

[Out]

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

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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 )}{\sqrt{-c \sec \left (f x + e\right ) + c}}\,{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))^(1/2),x, algorithm="giac")

[Out]

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