3.443 \(\int (d \cos (e+f x))^n (a+a \sec (e+f x)) \, dx\)

Optimal. Leaf size=132 \[ -\frac {a \sin (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a \sin (e+f x) (d \cos (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\cos ^2(e+f x)\right )}{d f (n+1) \sqrt {\sin ^2(e+f x)}} \]

[Out]

-a*(d*cos(f*x+e))^n*hypergeom([1/2, 1/2*n],[1+1/2*n],cos(f*x+e)^2)*sin(f*x+e)/f/n/(sin(f*x+e)^2)^(1/2)-a*(d*co
s(f*x+e))^(1+n)*hypergeom([1/2, 1/2+1/2*n],[3/2+1/2*n],cos(f*x+e)^2)*sin(f*x+e)/d/f/(1+n)/(sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4225, 16, 2748, 2643} \[ -\frac {a \sin (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a \sin (e+f x) (d \cos (e+f x))^{n+1} \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\cos ^2(e+f x)\right )}{d f (n+1) \sqrt {\sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[e + f*x])^n*(a + a*Sec[e + f*x]),x]

[Out]

-((a*(d*Cos[e + f*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*n*Sqrt[Sin[e +
f*x]^2])) - (a*(d*Cos[e + f*x])^(1 + n)*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2]*Sin[e +
f*x])/(d*f*(1 + n)*Sqrt[Sin[e + f*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 4225

Int[(csc[(a_.) + (b_.)*(x_)]*(B_.) + (A_))*(u_), x_Symbol] :> Int[(ActivateTrig[u]*(B + A*Sin[a + b*x]))/Sin[a
+ b*x], x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]

Rubi steps

\begin {align*} \int (d \cos (e+f x))^n (a+a \sec (e+f x)) \, dx &=\int (d \cos (e+f x))^n (a+a \cos (e+f x)) \sec (e+f x) \, dx\\ &=d \int (d \cos (e+f x))^{-1+n} (a+a \cos (e+f x)) \, dx\\ &=a \int (d \cos (e+f x))^n \, dx+(a d) \int (d \cos (e+f x))^{-1+n} \, dx\\ &=-\frac {a (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a (d \cos (e+f x))^{1+n} \, _2F_1\left (\frac {1}{2},\frac {1+n}{2};\frac {3+n}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{d f (1+n) \sqrt {\sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 105, normalized size = 0.80 \[ -\frac {a \sqrt {\sin ^2(e+f x)} (d \cos (e+f x))^n \left (n \cot (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\cos ^2(e+f x)\right )+(n+1) \csc (e+f x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(e+f x)\right )\right )}{f n (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[e + f*x])^n*(a + a*Sec[e + f*x]),x]

[Out]

-((a*(d*Cos[e + f*x])^n*((1 + n)*Csc[e + f*x]*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[e + f*x]^2] + n*Cot[e
+ f*x]*Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2])*Sqrt[Sin[e + f*x]^2])/(f*n*(1 + n)))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (f x + e\right ) + a\right )} \left (d \cos \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sec(f*x + e) + a)*(d*cos(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)*(d*cos(f*x + e))^n, x)

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maple [F]  time = 2.72, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (f x +e \right )\right )^{n} \left (a +a \sec \left (f x +e \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(f*x+e))^n*(a+a*sec(f*x+e)),x)

[Out]

int((d*cos(f*x+e))^n*(a+a*sec(f*x+e)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*sec(f*x + e) + a)*(d*cos(f*x + e))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))**n*(a+a*sec(f*x+e)),x)

[Out]

a*(Integral((d*cos(e + f*x))**n, x) + Integral((d*cos(e + f*x))**n*sec(e + f*x), x))

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