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

Optimal. Leaf size=90 \[ -\frac{a 2^{m+\frac{1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac{1}{2}-m} (a \sec (e+f x)+a)^{m-1} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{2}-m,\frac{1}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{f (c-c \sec (e+f x))} \]

[Out]

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

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Rubi [A]  time = 0.111511, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {3961, 70, 69} \[ -\frac{a 2^{m+\frac{1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{\frac{1}{2}-m} (a \sec (e+f x)+a)^{m-1} \, _2F_1\left (-\frac{1}{2},\frac{1}{2}-m;\frac{1}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{f (c-c \sec (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(1/2 + m)*a*Hypergeometric2F1[-1/2, 1/2 - m, 1/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(1/2 - m)*(a +
a*Sec[e + f*x])^(-1 + m)*Tan[e + f*x])/(f*(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 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 \frac{\sec (e+f x) (a+a \sec (e+f x))^m}{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)^{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}+m} a c (a+a \sec (e+f x))^{-1+m} \left (\frac{a+a \sec (e+f x)}{a}\right )^{\frac{1}{2}-m} \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{1}{2}+m}}{(c-c x)^{3/2}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{2^{\frac{1}{2}+m} a \, _2F_1\left (-\frac{1}{2},\frac{1}{2}-m;\frac{1}{2};\frac{1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{\frac{1}{2}-m} (a+a \sec (e+f x))^{-1+m} \tan (e+f x)}{f (c-c \sec (e+f x))}\\ \end{align*}

Mathematica [F]  time = 0.555419, size = 0, normalized size = 0. \[ \int \frac{\sec (e+f x) (a+a \sec (e+f x))^m}{c-c \sec (e+f x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

[Out]

integral(-(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*} - \frac{\int \frac{\left (a \sec{\left (e + f x \right )} + a\right )^{m} \sec{\left (e + f x \right )}}{\sec{\left (e + f x \right )} - 1}\, dx}{c} \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)),x)

[Out]

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

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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 )}{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)),x, algorithm="giac")

[Out]

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