3.442 \(\int (d \cos (e+f x))^n (a+a \sec (e+f x))^2 \, dx\)

Optimal. Leaf size=179 \[ -\frac {2 a^2 \sin (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a^2 (1-2 n) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\cos ^2(e+f x)\right )}{f (1-n) (n+1) \sqrt {\sin ^2(e+f x)}}+\frac {a^2 \tan (e+f x) (d \cos (e+f x))^n}{f (1-n)} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4264, 3788, 3772, 2643, 4046} \[ -\frac {2 a^2 \sin (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}-\frac {a^2 (1-2 n) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac {1}{2},\frac {n+1}{2};\frac {n+3}{2};\cos ^2(e+f x)\right )}{f (1-n) (n+1) \sqrt {\sin ^2(e+f x)}}+\frac {a^2 \tan (e+f x) (d \cos (e+f x))^n}{f (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[e + f*x])^n*(a + a*Sec[e + f*x])^2,x]

[Out]

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

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 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

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

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Mathematica [C]  time = 1.29, size = 266, normalized size = 1.49 \[ \frac {i a^2 2^{-n-2} e^{-i (e+f x)} \left (e^{-i (e+f x)} \left (1+e^{2 i (e+f x)}\right )\right )^{n-1} (\cos (e+f x)+1)^2 \sec ^4\left (\frac {1}{2} (e+f x)\right ) \left (4 (n-1) n e^{2 i (e+f x)} \, _2F_1\left (1,\frac {n}{2};2-\frac {n}{2};-e^{2 i (e+f x)}\right )+(n-2) \left (1+e^{2 i (e+f x)}\right ) \left (4 n e^{i (e+f x)} \, _2F_1\left (1,\frac {n+1}{2};\frac {3-n}{2};-e^{2 i (e+f x)}\right )+(n-1) \left (1+e^{2 i (e+f x)}\right ) \, _2F_1\left (1,\frac {n+2}{2};1-\frac {n}{2};-e^{2 i (e+f x)}\right )\right )\right ) \cos ^{-n}(e+f x) (d \cos (e+f x))^n}{f (n-2) (n-1) n} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Cos[e + f*x])^n*(a + a*Sec[e + f*x])^2,x]

[Out]

(I*2^(-2 - n)*a^2*((1 + E^((2*I)*(e + f*x)))/E^(I*(e + f*x)))^(-1 + n)*(d*Cos[e + f*x])^n*(1 + Cos[e + f*x])^2
*(4*E^((2*I)*(e + f*x))*(-1 + n)*n*Hypergeometric2F1[1, n/2, 2 - n/2, -E^((2*I)*(e + f*x))] + (1 + E^((2*I)*(e
+ f*x)))*(-2 + n)*(4*E^(I*(e + f*x))*n*Hypergeometric2F1[1, (1 + n)/2, (3 - n)/2, -E^((2*I)*(e + f*x))] + (1
+ E^((2*I)*(e + f*x)))*(-1 + n)*Hypergeometric2F1[1, (2 + n)/2, 1 - n/2, -E^((2*I)*(e + f*x))]))*Sec[(e + f*x)
/2]^4)/(E^(I*(e + f*x))*f*(-2 + n)*(-1 + n)*n*Cos[e + f*x]^n)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \sec \left (f x + e\right ) + a^{2}\right )} \left (d \cos \left (f x + e\right )\right )^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2)*(d*cos(f*x + e))^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \left (d \cos \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((a*sec(f*x + e) + a)^2*(d*cos(f*x + e))^n, x)

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maple [F]  time = 5.46, size = 0, normalized size = 0.00 \[ \int \left (d \cos \left (f x +e \right )\right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(f*x+e))^n*(a+a*sec(f*x+e))^2,x)

[Out]

int((d*cos(f*x+e))^n*(a+a*sec(f*x+e))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \left (d \cos \left (f x + e\right )\right )^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((a*sec(f*x + e) + a)^2*(d*cos(f*x + e))^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\cos \left (e+f\,x\right )\right )}^n\,{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(e + f*x))^n*(a + a/cos(e + f*x))^2,x)

[Out]

int((d*cos(e + f*x))^n*(a + a/cos(e + f*x))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \left (d \cos {\left (e + f x \right )}\right )^{n}\, dx + \int 2 \left (d \cos {\left (e + f x \right )}\right )^{n} \sec {\left (e + f x \right )}\, dx + \int \left (d \cos {\left (e + f x \right )}\right )^{n} \sec ^{2}{\left (e + f x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))**n*(a+a*sec(f*x+e))**2,x)

[Out]

a**2*(Integral((d*cos(e + f*x))**n, x) + Integral(2*(d*cos(e + f*x))**n*sec(e + f*x), x) + Integral((d*cos(e +
f*x))**n*sec(e + f*x)**2, x))

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