3.31 \(\int \cos (c+d x) (b \sec (c+d x))^n (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=117 \[ \frac{b^2 (C (1-n)-A n) \sin (c+d x) (b \sec (c+d x))^{n-2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2-n}{2},\frac{4-n}{2},\cos ^2(c+d x)\right )}{d (2-n) n \sqrt{\sin ^2(c+d x)}}+\frac{b C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n} \]

[Out]

(b^2*(C*(1 - n) - A*n)*Hypergeometric2F1[1/2, (2 - n)/2, (4 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-2 + n)*
Sin[c + d*x])/(d*(2 - n)*n*Sqrt[Sin[c + d*x]^2]) + (b*C*(b*Sec[c + d*x])^(-1 + n)*Tan[c + d*x])/(d*n)

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Rubi [A]  time = 0.11834, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {16, 4046, 3772, 2643} \[ \frac{b^2 (C (1-n)-A n) \sin (c+d x) (b \sec (c+d x))^{n-2} \, _2F_1\left (\frac{1}{2},\frac{2-n}{2};\frac{4-n}{2};\cos ^2(c+d x)\right )}{d (2-n) n \sqrt{\sin ^2(c+d x)}}+\frac{b C \tan (c+d x) (b \sec (c+d x))^{n-1}}{d n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*(C*(1 - n) - A*n)*Hypergeometric2F1[1/2, (2 - n)/2, (4 - n)/2, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(-2 + n)*
Sin[c + d*x])/(d*(2 - n)*n*Sqrt[Sin[c + d*x]^2]) + (b*C*(b*Sec[c + d*x])^(-1 + n)*Tan[c + d*x])/(d*n)

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 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 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 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]

Rubi steps

\begin{align*} \int \cos (c+d x) (b \sec (c+d x))^n \left (A+C \sec ^2(c+d x)\right ) \, dx &=b \int (b \sec (c+d x))^{-1+n} \left (A+C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n}+\frac{(b (C (-1+n)+A n)) \int (b \sec (c+d x))^{-1+n} \, dx}{n}\\ &=\frac{b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n}+\frac{\left (b (C (-1+n)+A 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 )^{1-n} \, dx}{n}\\ &=\frac{(C (1-n)-A n) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2-n}{2};\frac{4-n}{2};\cos ^2(c+d x)\right ) (b \sec (c+d x))^n \sin (c+d x)}{d (2-n) n \sqrt{\sin ^2(c+d x)}}+\frac{b C (b \sec (c+d x))^{-1+n} \tan (c+d x)}{d n}\\ \end{align*}

Mathematica [A]  time = 0.231657, size = 119, normalized size = 1.02 \[ \frac{\sqrt{-\tan ^2(c+d x)} (b \sec (c+d x))^n \left (A (n+1) \cos (c+d x) \cot (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n-1}{2},\frac{n+1}{2},\sec ^2(c+d x)\right )+C (n-1) \csc (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\sec ^2(c+d x)\right )\right )}{d (n-1) (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.752, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{n} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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*cos(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \sec \left (d x + c\right )^{2} + A\right )} \left (b \sec \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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*cos(d*x + c), x)