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

Optimal. Leaf size=83 \[ \frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}} \]

[Out]

AppellF1(1/2+m,1,1/2,3/2+m,1+sec(f*x+e),1/2+1/2*sec(f*x+e))*(a+a*sec(f*x+e))^m*2^(1/2)*tan(f*x+e)/f/(1+2*m)/(1
-sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3779, 3778, 136} \[ \frac {\sqrt {2} \tan (e+f x) (a \sec (e+f x)+a)^m F_1\left (m+\frac {1}{2};\frac {1}{2},1;m+\frac {3}{2};\frac {1}{2} (\sec (e+f x)+1),\sec (e+f x)+1\right )}{f (2 m+1) \sqrt {1-\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^m,x]

[Out]

(Sqrt[2]*AppellF1[1/2 + m, 1/2, 1, 3/2 + m, (1 + Sec[e + f*x])/2, 1 + Sec[e + f*x]]*(a + a*Sec[e + f*x])^m*Tan
[e + f*x])/(f*(1 + 2*m)*Sqrt[1 - Sec[e + f*x]])

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 3778

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^n*Cot[c + d*x])/(d*Sqrt[1 + Csc[c + d*x]
]*Sqrt[1 - Csc[c + d*x]]), Subst[Int[(1 + (b*x)/a)^(n - 1/2)/(x*Sqrt[1 - (b*x)/a]), x], x, Csc[c + d*x]], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] && GtQ[a, 0]

Rule 3779

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[(a^IntPart[n]*(a + b*Csc[c + d*x])^FracPart
[n])/(1 + (b*Csc[c + d*x])/a)^FracPart[n], Int[(1 + (b*Csc[c + d*x])/a)^n, x], x] /; FreeQ[{a, b, c, d, n}, x]
&& EqQ[a^2 - b^2, 0] &&  !IntegerQ[2*n] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int (a+a \sec (e+f x))^m \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int (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} x} \, dx,x,\sec (e+f x)\right )}{f \sqrt {1-\sec (e+f x)}}\\ &=\frac {\sqrt {2} F_1\left (\frac {1}{2}+m;\frac {1}{2},1;\frac {3}{2}+m;\frac {1}{2} (1+\sec (e+f x)),1+\sec (e+f x)\right ) (a+a \sec (e+f x))^m \tan (e+f x)}{f (1+2 m) \sqrt {1-\sec (e+f x)}}\\ \end {align*}

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Mathematica [B]  time = 6.80, size = 711, normalized size = 8.57 \[ \frac {30 \sin (e+f x) \cos ^2\left (\frac {1}{2} (e+f x)\right ) \cos (e+f x) (a (\sec (e+f x)+1))^m F_1\left (\frac {1}{2};m,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (3 F_1\left (\frac {1}{2};m,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (F_1\left (\frac {3}{2};m,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-m F_1\left (\frac {3}{2};m+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )}{f \left (45 \cos ^2\left (\frac {1}{2} (e+f x)\right ) (-2 m \cos (e+f x)+\cos (2 (e+f x))+2 m+1) F_1\left (\frac {1}{2};m,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ){}^2+40 \sin ^2\left (\frac {1}{2} (e+f x)\right ) \cos (e+f x) \tan ^2\left (\frac {1}{2} (e+f x)\right ) \left (F_1\left (\frac {3}{2};m,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-m F_1\left (\frac {3}{2};m+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ){}^2+6 \sin ^2\left (\frac {1}{2} (e+f x)\right ) F_1\left (\frac {1}{2};m,1;\frac {3}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right ) \left (-5 (-2 (m+2) \cos (e+f x)+\cos (2 (e+f x))+2 m+1) F_1\left (\frac {3}{2};m,2;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+5 m (-2 (m+2) \cos (e+f x)+\cos (2 (e+f x))+2 m+1) F_1\left (\frac {3}{2};m+1,1;\frac {5}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-48 \sin ^4\left (\frac {1}{2} (e+f x)\right ) \cot (e+f x) \csc (e+f x) \left (2 F_1\left (\frac {5}{2};m,3;\frac {7}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )-2 m F_1\left (\frac {5}{2};m+1,2;\frac {7}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+m (m+1) F_1\left (\frac {5}{2};m+2,1;\frac {7}{2};\tan ^2\left (\frac {1}{2} (e+f x)\right ),-\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[e + f*x])^m,x]

[Out]

(30*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Cos[(e + f*x)/2]^2*Cos[e + f*x]*(a*(1 +
Sec[e + f*x]))^m*Sin[e + f*x]*(3*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*(Appell
F1[3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x)/2]^
2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))/(f*(45*AppellF1[1/2, m, 1, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2]^2*Cos[(e + f*x)/2]^2*(1 + 2*m - 2*m*Cos[e + f*x] + Cos[2*(e + f*x)]) + 6*AppellF1[1/2, m, 1, 3/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2*(-5*AppellF1[3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan
[(e + f*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[e + f*x] + Cos[2*(e + f*x)]) + 5*m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[
(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*(1 + 2*m - 2*(2 + m)*Cos[e + f*x] + Cos[2*(e + f*x)]) - 48*(2*AppellF1[5/
2, m, 3, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] - 2*m*AppellF1[5/2, 1 + m, 2, 7/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2] + m*(1 + m)*AppellF1[5/2, 2 + m, 1, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Cot[e
+ f*x]*Csc[e + f*x]*Sin[(e + f*x)/2]^4) + 40*(AppellF1[3/2, m, 2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] - m*AppellF1[3/2, 1 + m, 1, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])^2*Cos[e + f*x]*Sin[(e + f*x)/2]^2
*Tan[(e + f*x)/2]^2))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )}^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m, x)

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (f x +e \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^m,x)

[Out]

int((a + a/cos(e + f*x))^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*sec(e + f*x) + a)**m, x)

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