3.2.52 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^2 \, dx\) [152]

Optimal. Leaf size=92 \[ \frac {2^{\frac {1}{2}+m} a \, _2F_1\left (\frac {5}{2},\frac {1}{2}-m;\frac {7}{2};\frac {1}{2} (1-\sec (e+f x))\right ) (1+\sec (e+f x))^{\frac {1}{2}-m} (a+a \sec (e+f x))^{-1+m} (c-c \sec (e+f x))^2 \tan (e+f x)}{5 f} \]

[Out]

1/5*2^(1/2+m)*a*hypergeom([5/2, 1/2-m],[7/2],1/2-1/2*sec(f*x+e))*(1+sec(f*x+e))^(1/2-m)*(a+a*sec(f*x+e))^(-1+m
)*(c-c*sec(f*x+e))^2*tan(f*x+e)/f

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 92, normalized size of antiderivative = 1.00, 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 = {4046, 72, 71} \begin {gather*} \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} \end {gather*}

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 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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]))

Rule 72

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 4046

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]

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)) \text {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 ) \text {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 = 0.25, size = 89, normalized size = 0.97 \begin {gather*} \frac {2^{\frac {1}{2}+m} c^2 \, _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))^2 (1+\sec (e+f x))^{-\frac {1}{2}-m} (a (1+\sec (e+f x)))^m \tan (e+f x)}{5 f} \end {gather*}

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)

________________________________________________________________________________________

Maple [F]
time = 0.23, size = 0, normalized size = 0.00 \[\int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{2}\, dx\]

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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{2} \left (\int \left (a \sec {\left (e + f x \right )} + a\right )^{m} \sec {\left (e + f x \right )}\, dx + \int \left (- 2 \left (a \sec {\left (e + f x \right )} + a\right )^{m} \sec ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (a \sec {\left (e + f x \right )} + a\right )^{m} \sec ^{3}{\left (e + f x \right )}\, dx\right ) \end {gather*}

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]

c**2*(Integral((a*sec(e + f*x) + a)**m*sec(e + f*x), x) + Integral(-2*(a*sec(e + f*x) + a)**m*sec(e + f*x)**2,
x) + Integral((a*sec(e + f*x) + a)**m*sec(e + f*x)**3, x))

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________