3.152 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=92 \[ \frac{a 2^{m+\frac{1}{2}} \tan (e+f x) (c-c \sec (e+f x))^2 (\sec (e+f x)+1)^{\frac{1}{2}-m} (a \sec (e+f x)+a)^{m-1} \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-m,\frac{7}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{5 f} \]

[Out]

(2^(1/2 + m)*a*Hypergeometric2F1[5/2, 1/2 - m, 7/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(1/2 - m)*(a + a*
Sec[e + f*x])^(-1 + m)*(c - c*Sec[e + f*x])^2*Tan[e + f*x])/(5*f)

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Rubi [A]  time = 0.108356, antiderivative size = 92, 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} \[ \frac{a 2^{m+\frac{1}{2}} \tan (e+f x) (c-c \sec (e+f x))^2 (\sec (e+f x)+1)^{\frac{1}{2}-m} (a \sec (e+f x)+a)^{m-1} \, _2F_1\left (\frac{5}{2},\frac{1}{2}-m;\frac{7}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + m)*a*Hypergeometric2F1[5/2, 1/2 - m, 7/2, (1 - Sec[e + f*x])/2]*(1 + Sec[e + f*x])^(1/2 - m)*(a + a*
Sec[e + f*x])^(-1 + m)*(c - c*Sec[e + f*x])^2*Tan[e + f*x])/(5*f)

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

Mathematica [A]  time = 0.24753, size = 89, normalized size = 0.97 \[ \frac{c^2 2^{m+\frac{1}{2}} \tan (e+f x) (\sec (e+f x)-1)^2 (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a (\sec (e+f x)+1))^m \text{Hypergeometric2F1}\left (\frac{5}{2},\frac{1}{2}-m,\frac{7}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + m)*c^2*Hypergeometric2F1[5/2, 1/2 - m, 7/2, (1 - Sec[e + f*x])/2]*(-1 + Sec[e + f*x])^2*(1 + Sec[e +
f*x])^(-1/2 - m)*(a*(1 + Sec[e + f*x]))^m*Tan[e + f*x])/(5*f)

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Maple [F]  time = 0.433, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{2}\, 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))^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c \sec \left (f x + e\right ) - c\right )}^{2}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c^{2} \sec \left (f x + e\right )^{3} - 2 \, c^{2} \sec \left (f x + e\right )^{2} + c^{2} \sec \left (f x + e\right )\right )}{\left (a \sec \left (f x + e\right ) + a\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))^2,x, algorithm="fricas")

[Out]

integral((c^2*sec(f*x + e)^3 - 2*c^2*sec(f*x + e)^2 + c^2*sec(f*x + e))*(a*sec(f*x + e) + a)^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))**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c \sec \left (f x + e\right ) - c\right )}^{2}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{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))^2,x, algorithm="giac")

[Out]

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