3.1.30 \(\int (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\) [30]

Optimal. Leaf size=113 \[ -\frac {b (A+A n+C n) \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{-1+n} \sin (c+d x)}{d (1-n) (1+n) \sqrt {\sin ^2(c+d x)}}+\frac {C (b \sec (c+d x))^n \tan (c+d x)}{d (1+n)} \]

[Out]

-b*(A*n+C*n+A)*hypergeom([1/2, 1/2-1/2*n],[3/2-1/2*n],cos(d*x+c)^2)*(b*sec(d*x+c))^(-1+n)*sin(d*x+c)/d/(-n^2+1
)/(sin(d*x+c)^2)^(1/2)+C*(b*sec(d*x+c))^n*tan(d*x+c)/d/(1+n)

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Rubi [A]
time = 0.06, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4131, 3857, 2722} \begin {gather*} \frac {C \tan (c+d x) (b \sec (c+d x))^n}{d (n+1)}-\frac {b (A n+A+C n) \sin (c+d x) (b \sec (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(c+d x)\right )}{d (1-n) (n+1) \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

-((b*(A + A*n + C*n)*Hypergeometric2F1[1/2, (1 - n)/2, (3 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-1 + n)*Si
n[c + d*x])/(d*(1 - n)*(1 + n)*Sqrt[Sin[c + d*x]^2])) + (C*(b*Sec[c + d*x])^n*Tan[c + d*x])/(d*(1 + n))

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3857

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 4131

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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 3.95, size = 262, normalized size = 2.32 \begin {gather*} -\frac {i 2^{1+n} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^n \left (1+e^{2 i (c+d x)}\right )^n \left (A \left (8+6 n+n^2\right ) \, _2F_1\left (\frac {n}{2},2+n;\frac {2+n}{2};-e^{2 i (c+d x)}\right )+2 (A+2 C) e^{2 i (c+d x)} n (4+n) \, _2F_1\left (\frac {2+n}{2},2+n;\frac {4+n}{2};-e^{2 i (c+d x)}\right )+A e^{4 i (c+d x)} n (2+n) \, _2F_1\left (2+n,\frac {4+n}{2};\frac {6+n}{2};-e^{2 i (c+d x)}\right )\right ) \sec ^{-2-n}(c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right )}{d n (2+n) (4+n) (A+2 C+A \cos (2 c+2 d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^n*(A + C*Sec[c + d*x]^2),x]

[Out]

((-I)*2^(1 + n)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^n*(1 + E^((2*I)*(c + d*x)))^n*(A*(8 + 6*n + n^2)*H
ypergeometric2F1[n/2, 2 + n, (2 + n)/2, -E^((2*I)*(c + d*x))] + 2*(A + 2*C)*E^((2*I)*(c + d*x))*n*(4 + n)*Hype
rgeometric2F1[(2 + n)/2, 2 + n, (4 + n)/2, -E^((2*I)*(c + d*x))] + A*E^((4*I)*(c + d*x))*n*(2 + n)*Hypergeomet
ric2F1[2 + n, (4 + n)/2, (6 + n)/2, -E^((2*I)*(c + d*x))])*Sec[c + d*x]^(-2 - n)*(b*Sec[c + d*x])^n*(A + C*Sec
[c + d*x]^2))/(d*n*(2 + n)*(4 + n)*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [F]
time = 0.20, size = 0, normalized size = 0.00 \[\int \left (b \sec \left (d x +c \right )\right )^{n} \left (A +C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

[Out]

int((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x)

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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((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n, x)

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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((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b \sec {\left (c + d x \right )}\right )^{n} \left (A + C \sec ^{2}{\left (c + d x \right )}\right )\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**n*(A+C*sec(d*x+c)**2),x)

[Out]

Integral((b*sec(c + d*x))**n*(A + C*sec(c + d*x)**2), x)

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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((b*sec(d*x+c))^n*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*(b*sec(d*x + c))^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n,x)

[Out]

int((A + C/cos(c + d*x)^2)*(b/cos(c + d*x))^n, x)

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