3.344 \(\int \sec (e+f x) (a+a \sec (e+f x))^m \, dx\)

Optimal. Leaf size=73 \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f} \]

[Out]

2^(1/2+m)*hypergeom([1/2, 1/2-m],[3/2],1/2-1/2*sec(f*x+e))*(1+sec(f*x+e))^(-1/2-m)*(a+a*sec(f*x+e))^m*tan(f*x+
e)/f

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3828, 3827, 69} \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a \sec (e+f x)+a)^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]))

Rule 3827

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0]

Rule 3828

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 73, normalized size = 1.00 \[ \frac {2^{m+\frac {1}{2}} \tan (e+f x) (\sec (e+f x)+1)^{-m-\frac {1}{2}} (a (\sec (e+f x)+1))^m \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {1}{2} (1-\sec (e+f x))\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(1/2 + m)*Hypergeometric2F1[1/2, 1/2 - m, 3/2, (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])/f

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.99, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^m}{\cos \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{m} \sec {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))**m,x)

[Out]

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

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